APPLICATION OF DERIVATIVES

Tangent and Normal

Let y = f(x) be function with graph as shown in figure. Consider secant PQ. If Q tends to P along the curve

passing through the points Q1, Q

2, .....I.e. Q P, secant PQ will become tangent at P. A line through P

perpendicular to tangent is called normal at P.

Geometrical Meaning of dy

dx

As Q P, h 0 and slope of chord PQ tends to slope of tangent at P (see figure).

Slope of chord PQ = f(x h) f(x)

h

Q Plim

slope of chord PQ = h 0lim

f(x h) f(x)

h

slope of tangent at P = f(x) = dy

dx

Equation of tangent and normal

1 1(x , y )

dy

dx

= f(x1) denotes the slope of tangent at point (x

1, y

1) on the curve y = f(x). Hence the equation

of tangent at (x1, y

1) is given by

(y – y1) = f(x

1) (x – x

1) ; when, f(x

1) is real.

Also, since normal is a line perpendicular to tangent at (x1 , y

1) so its equation is given by

(y – y1) = –

1

1

f (x ) (x – x

1), when f(x

1) is nonzero real.

If f(x1) = 0, then tangent is the line y = y

1 and normal is the line x = x

1.

If h 0lim

1 1f(x h) f(x )

h

= or – , then x = x

1 is tangent (VERTICAL TANGENT) and y = y

1 is normal.

Example # 1 Find equation of tangent to y = ex at x = 0. Hence draw graph

Solution : At x = 0 y = e0 = 1

dy

dx = ex

x 0

dy

dx

= 1

Hence equation of tangent is 1 (x – 0) = (y – 1)

y = x + 1

Example # 2 Find the equation of all straight lines which are tangent to curve y = 1

x 1 and which are

parallel to the line x + y = 0.

Solution : Suppose the tangent is at (x1, y

1) and it has slope – 1.

1 1(x , y )

dy

dx = – 1.

– 2

1

1

(x 1) = – 1.

x1 = 0 or 2

y1 = – 1 or 1

Hence tangent at (0, – 1) and (2, 1) are the required lines (see figure) with equations

– 1(x – 0) = (y + 1) and – 1 (x – 2) = (y – 1)

x + y + 1 = 0 and y + x = 3

Example # 3 Find equation of normal to the curve y = |x2 – | x | | at x = – 2.

Solution : In the neighborhood of x = – 2, y = x2 + x.

Hence the point of contact is (– 2, 2)

dy

dx = 2x + 1

x 2

dy

dx

= – 3.

So the slope of normal at (– 2, 2) is .

Hence equation of normal is

1

3 (x + 2) = y – 2 3y = x + 8

Example # 4 Prove that sum of intercepts of the tangent at any point to the curve represented by x = 3cos4

& y = 3sin4 on the coordinate axis is constant.

Solution : Let P(3cos4 , 3sin4) be a variable point on the given curve.

m =dy

dx =

dy

ddx

d

= 3

3

3.4sin .cos

3.4cos sin

=–tan

equation of tangent at point P is

y – 3sin4 = –tan2 (x – 3cos4)

2 2

x y

3cos 3sin

= 1

sum of x-axis intercept and y-axis intercept = 3cos2 + 3sin2= 3 (which is constant)

Self Practice Problems :

(1) Find the slope of the normal to the curve x = 1 – a sin , y = b cos2 at =2

.

(2) Find the equation of the tangent and normal to the given curves at the given points.

(i) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)

(ii) y2 = 3x

4 x at (2, – 2).

(3) Prove that area of the triangle formed by any tangent to the curve xy = c2 and coordinate axes

is constant.

(4) A curve is given by the equations x = at2 & y = at3 . A variable pair of perpendicular lines

through the origin 'O' meet the curve at P & Q . Show that the locus of the point of intersection

of the tangents at P & Q is 4y2 = 3ax - a2 .

Ans. (1) – a

2b (2) (i) Tangent : y = 2x + 1, Normal :x + 2y = 7

(ii) Tangent : 2x + y = 2, Normal :x – 2y = 6

Tangent and Normal from an external point

Given a point P(a, b) which does not lie on the curve y = f(x), then the equation of possible tangents to

the curve y = f(x), passing through (a, b) can be found by solving for the point of contact Q.

f(h) = f(h) b

h a

And equation of tangent is y – b = f(h) b

h a

(x – a)

Example # 5 Tangent at P(2, 8) on the curve y = x3 meets the curve again at Q.

Find coordinates of Q.

Solution : Equation of tangent at (2, 8) is y = 12x – 16

Solving this with y = x3

x3 – 12x + 16 = 0

This cubic will give all points of intersection of line and

curve y = x3 i.e., point P and Q. (see figure)

But, since line is tangent at P so x = 2 will be a repeated root of equation x3 – 12x + 16 = 0 and

another root will be x = h. Using theory of equations :

sum of roots 2 + 2 + h = 0 h = – 4

Hence coordinates of Q are (– 4, – 64)

Self Practice Problems :

(5) How many tangents are possible from (1, 1) to the curve y – 1 = x3. Also find the equation of

these tangents.

(6) Find the equation of tangent to the hyperbola y = x 9

x 5

which passes through (0, 0) origin

Ans. (5) y = 1, 4y = 27x – 23 (6) x + y = 0; 25y + x = 0

Derivative as rate of change

In various fields of applied mathematics one has the quest to know the rate at which one variable is changing, with respect to other. The rate of change naturally refers to time. But we can have rate of change with respect to other variables also.

An economist may want to study how the investment changes with respect to variations in interest rates.

A physician may want to know, how small changes in dosage can affect the body's response to a drug.

A physicist may want to know the rate of charge of distance with respect to time.

All questions of the above type can be interpreted and represented using derivatives.

Definition : The average rate of change of a function f(x) with respect to x over an interval [a, a + h] is

defined as f(a h) f(a)

h

Definition : The instantaneous rate of change of f(x) with respect to x is defined as f(x) =h 0lim

f(a h) f(a)

h

,

provided the limit exists. Note : To use the word 'instantaneous', x may not be representing time. We usually use the word 'rate

of change' to mean 'instantaneous rate of change'.

Example # 6 How fast the area of a circle increases when its radius is 5cm; (i) with respect to radius (ii) with respect to diameter

Solution : (i) A = r2 , dA

dr = 2r

r 5

dA

dr

= 10 cm2/cm.

(ii) A = 4

D2 ,

dA

dD =

2

D

D 10

dA

dD

= 2

. 10 = 5 cm2/cm.

Example # 7 If area of circle increases at a rate of 2cm2/sec, then find the rate at which area of the inscribed

square increases.

Solution : Area of circle, A1 = r2. Area of square, A

2 = 2r2 (see figure)

1dA

dt = 2r

dr

dt , 2dA

dt = 4r .

dr

dt

2 = 2r . dr

dt r

dr

dt =

1

2dA

dt = 4 .

1

=

4

cm2/sec

Area of square increases at the rate 4

cm2/sec.

Example # 8 The volume of a cube is increasing at a rate of 7 cm3/sec. How fast is the surface area

increasing when the length of an edge is 4 cm?

Solution. Let at some time t, the length of edge is x cm.

v = x3 dv

dt = 3x2

dx

dt (but

dv

dt = 7)

dx

dt =

2

7

3x cm/sec.

Now S = 6x2

dS

dt = 12x

dx

dt

dS

dt = 12x.

2

7

3x =

28

x

when x = 4 cm, dS

dt = 7 cm2/sec.

Example # 9 Sand is pouring from pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground

in such a way that the height of the cone is always one - sixth of radius of base. How fast is the

height of the sand cone increasing when height is 4 cm?

Solution. V = 1

3 r2h

but h = r

6

V =1

3 (6h)2 h

V = 12 h3

dV

dt = 36 h2.

dh

dt

when, dV

dt = 12 cm3/s and h = 4 cm

dh

dt =

2

12

36 .(4) =

1

48 cm/sec.

Self Practice Problems :

(7) Radius of a circle is increasing at rate of 3 cm/sec. Find the rate at which the area of circle is

increasing at the instant when radius is 10 cm.

(8) A ladder of length 5 m is leaning against a wall. The bottom of ladder is being pulled along the

ground away from wall at rate of 2cm/sec. How fast is the top part of ladder sliding on the wall

when foot of ladder is 4 m away from wall.

(9) Water is dripping out of a conical funnel of semi-vertical angle 45° at rate of 2cm3/s. Find the

rate at which slant height of water is decreasing when the height of water is 2 cm.

(10) A hot air balloon rising straight up from a level field is tracked by a range finder 500 ft from the

lift-off point. At the moment the range finder's elevation angle is /4, the angle is increasing at

the rate of 0.14 rad/min. How fast is the balloon rising at that moment.

Ans. (7) 60 cm2/sec (8) 8

3 cm/sec (9)

1

2 cm/sec. (10) 140 ft/min.

Error and Approximation :

Let y = f(x) be a function. If these is an error x in x then corresponding error in y is y = f(x + x) – f(x).

We have x 0lim

f(x x) f(x)

x

=

dy

dx = f(x)

We define the differential of y, at point x, corresponding to the increment x as f(x) x and denote it by dy.

i.e. dy = f(x) x.

Let P(x, f(x)), Q((x + x), f(x + x)) (as shown in figure)

y = QS,

x = PS, dy = RS

In many practical situations, it is easier to evaluate dy but not y.

Example # 10. Find the approximate value of 251/3. Sol. Let y = x1/3

Let x = 27 and x = –2

Now y = (x + x)1/3 – x1/3 = (25)1/3 – 3

dy

dx x = 251/3 – 3

At x = 27, 251/3 = 3 – 0.074 = 2.926

Monotonicity of a function :

Let f be a real valued function having domain D(DR) and S be a subset of D. f is said to be

monotonically increasing (non decreasing) (increasing) in S if for every x1, x

2 S, x

1 < x

2 f(x

1) f(x

2). f

is said to be monotonically decreasing (non increasing) (decreasing) in S if for every x1, x

2 S, x

1 < x

2

f(x1) f(x

2)

f is said to be strictly increasing in S if for x1, x

2 S, x

1 < x

2 f(x

1) < f(x

2). Similarly, f is said to be strictly

decreasing in S if for x1, x

2 S, x

1 < x

2 f(x

1) > f(x

2) .

Notes : (i) f is strictly increasing f is monotonically increasing (non decreasing). But converse need not

be true.

(ii) f is strictly decreasing f is monotonically decreasing (non increasing). Again, converse need

not be true.

(iii) If f(x) = constant in S, then f is increasing as well as decreasing in S

(iv) A function f is said to be an increasing function if it is increasing in the domain. Similarly, if f is

decreasing in the domain, we say that f is monotonically decreasing

(v) f is said to be a monotonic function if either it is monotonically increasing or monotonically

decreasing

(vi) If f is increasing in a subset of S and decreasing in another subset of S, then f is non monotonic

in S.

Application of differentiation for detecting monotonicity :

Let be an interval (open or closed or semi open and semi closed)

(i) If f(x) > 0 x , then f is strictly increasing in

(ii) If f(x) < 0 x , then f is strictly decreasing in

Note : Let I be an interval (or ray) which is a subset of domain of f. If f (x) > 0, x except for countably

many points where f (x) = 0, then f(x) is strictly increasing in .

{f (x) = 0 at countably many points f (x) = 0 does not occur on an interval which is a subset of }

Let us consider another function whose graph is shown below for x (a, b).

Here also f(x) 0 for all x (a, b). But, note that in this case, f(x) = 0 holds for all x (c, d) and (e,b).

Thus the given function is increasing (monotonically increasing) in (a, b), but not strictly increasing.

Example # 11 : Let f(x) = x3. Find the intervals of monotonicity.

Solution : f(x) = 3x2

f(x) > 0 everywhere except at x = 0. Hence f(x) will be strictly increasing function for

x R {see figure}

Example # 12 : Let f(x) = x – sinx. Find the intervals of monotonicity.

Solution : f(x) = 1 – cosx

Now, f(x) > 0 every where, except at x = 0, ± 2, ± 4 etc. But all these points are discrete

(countable) and do not form an interval. Hence we can conclude that f(x) is strictly increasing in

R. In fact we can also see it graphically.

Example # 13 : Find the intervals in which f(x) = x3 – 2x2 – 4x + 7 is increasing.

Solution : f(x) = x3 – 2x2 – 4x + 7

f(x) = 3x2 – 4x – 4

f(x) = (x – 2) (3x + 2)

for M.. f(x) 0 x 2

,3

[2, )

Example # 14 : Find the intervals of monotonicity of the following functions.

(i) f(x) = x2 (x – 2)2 (ii) f(x) = x n x

Solution : (i) f(x) = x2 (x – 2)2 f(x) = 4x (x – 1) (x – 2)

observing the sign change of f(x)

Hence increasing in [0, 1] and in [2, )

and decreasing for x (– , 0] and [1, 2]

(ii) f(x) = x n x

f(x) = 1 + n x

f(x) 0 n x – 1 x 1

e

increasing for x 1

,e

and decreasing for x

10 ,

e

.

Note : If a function f(x) is increasing in (a, b) and f(x) is continuous in [a, b], then f(x) is increasing on

[a, b]

Example # 15 : f(x) = [x] is a step up function. Is it a strictly increasing function for x R.

Solution : No, f(x) = [x] is increasing (monotonically increasing) (non-decreasing), but not strictly

increasing function as illustrated by its graph.

Example # 16 : If f(x) = sin4x + cos4x + bx + c, then find possible values of b and c such that f(x) is monotonic

for all x R

Solution : f(x) = sin4x + cos4x + bx + c

f(x) = 4 sin3x cosx – 4cos3x sinx + b = – sin4x + b.

Case - (i) : for M.I. f(x) 0 for all x R

b sin4x for all x R b 1

Case - (ii) : for M.D. f(x) 0 for all x R

b sin4x for all x R b – 1

Hence for f(x) to be monotonic b (– , – 1] [1, ) and c R.

Example # 17 : Find possible values of 'a' such that f(x) = e2x – 2(a2 – 21) ex + 8x + 5 is monotonically

increasing for x R

Solution : f(x) = e2x – 2(a2 – 21) ex + 8x + 5

f(x) = 2e2x – 2(a2 – 21) ex + 8 0 ; c x R

ex + x

4

e a2 – 21

4 a2 – 21 x

x

4e 4

e

a [–5, 5]

Self Practice Problems :

(11) Find the intervals of monotonicity of the following functions.

(i) f(x) = – x3 + 6x2 – 9x – 2

(ii) f(x) = x + 1

x 1

(iii) f(x) = x . 2x xe

(iv) f(x) = x – cosx

(12) Let f(x) = x – tan–1x. Prove that f(x) is monotonically increasing for x R.

(13) If f(x) = 2ex – ae–x + (2a + 1) x – 3 monotonically increases for x R, then find range of

values of a

(14) Let f(x) = e2x – aex + 1. Prove that f(x) cannot be monotonically decreasing for x R for any

value of 'a'.

(15) The values of 'a' for which function f(x) = (a + 2) x3 – ax2 + 9ax – 1 monotonically decreasing

for x R.

Ans. (11) (i) in [1, 3] ; D in (– , 1] (3, )

(ii) in (– , – 2] [0, ) ; D in [– 2, – 1) (–1, 0]

(iii) in 1

, 12

; D in 1

,2

[1, )

(iv) I for x R

(13) a 0 (15) – < a – 3

Monotonicity of function about a point :

1. A function f(x) is called as a strictly increasing function about a point (or at a point) a Df if it is strictly

increasing in an open interval containing a (as shown in figure).

2. A function f(x) is called a strictly decreasing function about a point x = a, if it is strictly decreasing in an

open interval containing a (as shown in figure).

Note : If x = a is a boundary point then use the appropriate one sided inequality to test monotonicity of f(x).

e.g. : Which of the following functions (as shown in figure) is increasing, decreasing or neither increasing nor

decreasing at x = a.

(i) (ii)

(iii) (iv)

Test for increasing and decreasing functions about a point

Let f(x) be differentiable.

(1) If f(a) > 0 then f(x) is increasing at x = a.

(2) If f(a) < 0 then f(x) is decreasing at x = a.

(3) If f(a) = 0 then examine the sign of f(x) on the left neighbourhood and the right neighbourhood

of a.

(i) If f(x) is positive on both the neighbourhoods, then f is increasing at x = a.

(ii) If f(x) is negative on both the neighbourhoods, then f is decreasing at x = a.

(iii) If f(x) have opposite signs on these neighbourhoods, then f is non-monotonic at x = a.

Example # 18 : Let f(x) = x3 – 3x + 2. Examine the monotonicity of function at points x = 0, 1, 2.

Solution : f(x) = x3 – 3x + 2

f(x) = 3(x2 – 1)

(i) f(0) = – 3 decreasing at x = 0

(ii) f(1) = 0

also, f(x) is positive on left neighbourhood and f(x) is negative in right neighbourhood.

neither increasing nor decreasing at x = 1.

(iii) f(2) = 9 increasing at x = 2

Note : Above method is applicable only for functions those are continuous at x = a.

Self Practice Problems :

(16) For each of the following graph comment on monotonicity of f(x) at x = a.

(i) (ii) (iii)

(17) Let f(x) = x3 – 3x2 + 3x + 4, comment on the monotonic behaviour of f(x) at (i) x = 0 (ii) x = 1.

(18) Draw the graph of function f(x) = x 0 x 1

[x] 1 x 2

. Graphically comment on the monotonic

behaviour of f(x) at x = 1. Is f(x) M.. for x [0, 2] ?

Ans. (16) (i) neither M.. nor M.D. (ii) M.D. (iii) M.

(17) M.. both at x = 0 and x = 1.

(18) M.. at x = 1; f(x) is M.. for x [0, 2].

Global Maximum :

A function f(x) is said to have global maximum on a set E if there exists at least one c E such that

f(x) f(c) for all x E.

We say global maximum occurs at x = c and global maximum (or global maximum value) is f(c).

Local Maxima :

A function f(x) is said to have a local maximum at x = c if f(c) is the greatest value of the function in a

small neighbourhood (c – h, c + h), h > 0 of c.

i.e. for all x (c – h, c + h), x c, we have f(x) f(c).

Global Minimum :

A function f(x) is said to have a global minimum on a set E if there exists at least one c E such that

f(x) f(c) for all x E.

Local Minima :

A function f(x) is said to have a local minimum at x = c if f(c) is the least value of the function in a small

neighbourhood (c – h, c + h), h > 0 of c.

i.e. for all x (c –h, c + h), x c, we have f(x) f(c).

Extrema :

A maxima or a minima is called an extrema.

Explanation : Consider graph of y = f(x), x [a, b]

x = c2, x = c

4 are points of local maxima, with maximum values f(c

2), f(c

4) respectively.

x = c1, x = c

3 are points of local minima, with minimum values f(c

1), f(c

3) respectively

x = c2 is a point of global maximum

x = c3 is a point of global minimum

Consider the graph of y = h(x), x [a, b)

a c1c1 c2 c3 c4 b

h(a)

h(c )2

h(c )1

h(c )4

h(c )3

x = c1, x = c

4 are points of local maxima, with maximum values h(c

1), h(c

4) respectively.

x = c2 are points of local minima, with minimum values h(c

2) respectively.

x = c3 is neither a point of maxima nor a minima.

Global maximum is h(c4)

Global minimum is h(a)

Example # 19 : Let f(x) =| x | 0 | x | 2

1 x 0

. Examine the behaviour of f(x) at x = 0.

Solution : f(x) has local maxima at x = 0 (see figure).

Example # 20 : Let f(x) =

3 23

2

(b b b 1)–x 0 x 1

(b 3b 2)

2x 3 1 x 3

Find all possible values of b such that f(x) has the smallest value at x = 1.

Solution. Such problems can easily be solved by graphical approach (as in figure).

Hence the limiting value of f(x) from left of x = 1 should be either greater or equal to the value of

function at x = 1.

x 1lim

f(x) f(1)

– 1 +3 2

2

(b b b 1)

(b 3b 2)

– 1

2(b 1)(b 1)

(b 1) (b 2)

0

b (– 2, –1) [1, + ) Self Practice Problems :

(19) In each of following graphs identify if x = a is point of local maxima, minima or neither

(i) (ii) (iii)

(20) Examine the graph of following functions in each case identify the points of global maximum/minimum and local maximum / minimum.

(i) (ii) (iii)

Ans. (19) (i) Maxima (ii) Neither maxima nor minima (iii) Minima

(20) (i) Local maxima at x = 2, Local minima at x = 3, Global maximum at x = 2. No global minimum

(ii) Local minima at x = – 1, No point of Global minimum, no point of local or Global maxima

(iii) Local & Global maximum at x = 1, Local & Global minimum at x = 0.

Maxima, Minima for differentiable functions :

Mere definition of maxima, minima becomes tedious in solving problems. We use derivative as a tool to

overcome this difficulty.

A necessary condition for an extrema :

Let f(x) be differentiable at x = c.

Theorem : A necessary condition for f(c) to be an extremum of f(x) is that f(c) = 0.

i.e. f(c) is extremum f(c) = 0

Note : f(c) = 0 is only a necessary condition but not sufficient

i.e. f(c) = 0 f(c) is extremum.

Consider f(x) = x3

f(0) = 0

but f(0) is not an extremum (see figure).

Sufficient condition for an extrema :

Let f(x) be a differentiable function.

Theorem : A sufficient condition for f(c) to be an extremum of f(x) is that f(x) changes sign as x passes through c.

i.e. f(c) is an extrema (see figure) f(x) changes sign as x passes through c.

x = c is a point of maxima. f(x) changes sign from positive to negative.

x = c is a point of local minima (see figure), f(x) changes sign from negative to positive.

Stationary points :

The points on graph of function f(x) where f(x) = 0 are called stationary points.

Rate of change of f(x) is zero at a stationary point.

Example # 21 : Find stationary points of the function f(x) = 4x3 – 6x2 – 24x + 9.

Solution : f(x) = 12x2 – 12x – 24

f(x) = 0 x = – 1, 2

f(– 1) = 23, f(2) = – 31

(– 1, 23), (2, – 31) are stationary points

Example # 22 : If f(x) = x3 + ax2 + bx + c has extreme values at x = – 1 and x = 3. Find a, b, c.

Solution. Extreme values basically mean maximum or minimum values, since f(x) is differentiable

function so

f(– 1) = 0 = f(3)

f(x) = 3x2 + 2ax + b

f(3) = 27 + 6a + b = 0

f(– 1) = 3 – 2a + b = 0

a = – 3, b = – 9, c R

First Derivative Test :

Let f(x) be continuous and differentiable function.

Step - Find f(x)

Step - . Solve f(x) = 0, let x = c be a solution. (i.e. Find stationary points)

Step - . Observe change of sign

(i) If f(x) changes sign from negative to positive as x crosses c from left to right then x = c is a

point of local minima

(ii) If f(x) changes sign from positive to negative as x crosses c from left to right then x = c is a

point of local maxima.

(iii) If f(x) does not changes sign as x crosses c then x = c is neither a point of maxima nor minima.

Example # 23 : Find the points of maxima or minima of f(x) = x2 (x – 2)2.

Solution. f(x) = x2 (x – 2)2

f(x) = 4x (x – 1) (x – 2)

f(x) = 0 x = 0, 1, 2

examining the sign change of f(x)

Hence x = 1 is point of maxima, x = 0, 2 are points of minima.

Note : In case of continuous functions points of maxima and minima are alternate.

Example # 24 : Find the points of maxima, minima of f(x) = x3 – 12x. Also draw the graph of this functions.

Solution. f(x) = x3 – 12x

f(x) = 3(x2 – 4) = 3(x – 2) (x + 2)

f(x) = 0 x = ± 2

For tracing the graph let us find maximum and minimum values of f(x).

x f(x)

2 16

2 16

Example # 25 : Show that f(x) = (x3 – 6x2 + 12x – 8) does not have any point of local maxima or minima. Hence

draw graph

Solution. f(x) = x3 – 6x2 + 12x – 8

f(x) = 3(x2 – 4x + 4)

f(x) = 3(x – 2)2

f(x) = 0 x = 2

but clearly f(x) does not change sign about x = 2. f(2+) > 0 and f(2–) > 0. So f(x) has no point

of maxima or minima. In fact f(x) is a monotonically increasing function for x R.

Example # 26 : Let f(x) = x3 + 3(a – 7)x2 + 3(a2 – 9) x – 1. If f(x) has positive point of maxima, then find possible

values of 'a'.

Solution. f(x) = 3 [x2 + 2(a – 7)x + (a2 – 9)]

Let , be roots of f(x) = 0 and let be the smaller root. Examining sign change of f(x).

Maxima occurs at smaller root which has to be positive. This basically implies that both roots

of f(x) = 0 must be positive and distinct.

(i) D > 0 a < 29

7

(ii) – b

2a > 0 a < 7

(iii) f(0) > 0 a (– , – 3) (3, )

from (i), (ii) and (iii) a (– , – 3) 29

3,7

Self Practice Problems :

(21) Find the points of local maxima or minima of following functions (i) f(x) = (x – 1)3 (x + 2)2 (ii) f(x) = x3 + x2 + x + 1.

Ans. (i) Maxima at x = – 2, Minima at x = – 4

5

(ii) No point of local maxima or minima.

Maxima, Minima for continuous functions : Let f(x) be a continuous function. Critical points :

The points where f(x) = 0 or f(x) is not differentiable are called critical points.

Stationary points Critical points.

Example # 27 : Find critical points of f(x) = max (sinx, cosx) , x (0, 2).

Solution :

From the figure it is clear that f(x) has three critical points x = 4

,

2

,

5

4

.

Important Note :

For f(x) defined on a subset of R, points of extrema (if exists) occur at critical points

Example # 28 : Find the possible points of Maxima/Minima for f(x) = |x2 – 2x| (x R)

Solution. f(x) =

2

2

2

x 2x x 2

2x x 0 x 2

x 2x x 0

f(x) =

2(x 1) x 2

2(1 x) 0 x 2

2(x 1) x 0

f(x) = 0 at x = 1 and f(x) does not exist at x = 0, 2. Thus these are critical points.

Example # 29 : Let f(x) =3 2x x 10x x 0

3sinx x 0

. Examine the behaviour of f(x) at x = 0.

Solution : f(x) is continuous at x = 0.

f(x) = 23x 2x 10 x 0

3cosx x 0

f(0+) = 3 and f(0–) = – 10 thus f(x) is non-differentiable at x = 0 x = 0 is a critical point.

Also derivative changes sign from negative to positive, so x = 0 is a point of local minima.

Example # 30 : Find the critical points of the function f(x) = 4x3 – 6x2 – 24x + 9 if (i) x [0, 3] (ii) x [–3, 3]

(iii) x [– 1, 2].

Solution : f(x) = 12(x2 – x – 2)

= 12(x – 2) (x + 1)

f(x) = 0 x = – 1 or 2

(i) if x [0, 3] , x = 2 is critical point.

(ii) if x [– 3, 3], then we have two critical points x = – 1, 2.

(iii) If x [– 1, 2], then no critical point as both x = –1 and x = 2 become boundary points.

Note : Critical points are always interior points of an interval.

Global extrema for continuous functions :

(i) Function defined on closed interval

Let f(x), x [a, b] be a continuous function

Step - I : Find critical points. Let it be c1, c

2 ......., c

n

Step - II: Find f(a), f(c1).........., f(c

n), f(b)

Let M = max· { f(a), f(c1),..........., f(c

n), f(b)}

m = min · {f(a), f(c1), ........f(c

n), f(b)}

Step - M is global maximum.

m is global minimum.

(ii) Function defined on open interval.

Let f(x), x (a, b) be continuous function.

Step - I Find critical points . Let it be c1, c

2, .......c

n

Step - II Find f(c

1), f(c

2), ........., f(c

n)

Let M = max · {f(c1), .......f(c

n)}

m = min· {f(c1),............,f(c

n)}

Step - III x aLim

f(x) = 1 (say),

–x bLim

f(x) = 2 (say).

Let = min. {1,

2}, L = max. {

1,

2}

Step - IV

(i) If m then m is global minimum

(ii) If m > then f(x) has no global minimum

(iii) If M L then M is global maximum

(iv) If M < L , then f(x) has no global maximum

Example # 31 : Find the greatest and least values of f(x) = x3 – 12x x [– 1, 3]

Solution : The possible points of maxima/minima are critical points and the boundary points.

for x [– 1, 3] and f(x) = x3 – 12x

x = 2 is the only critical point.

Examining the value of f(x) at points x = –1, 2, 3. We can find greatest and least values.

x f(x)

1 11

2 16

3 9

Minimum f(x) = – 16 & Maximum f(x) = 11.

Self Practice Problems :

(22) Let f(x) = x3 + x2 – x – 4

(i) Find the possible points of Maxima/Minima of f(x) for x R.

(ii) Find the number of critical points of f(x) for x [1, 3].

(iii) Discuss absolute (global) maxima/minima value of f(x) for x [–2, 2]

(iv) Prove that for x (1, 3), the function does not has a Global maximum.

Ans. (i) x = –1, 1

3 (ii) zero

(iii) f(–2) = –6 is global maximum, f(2) = 6 is global maximum

Example # 32 : Let f(x) =

2

1/ 2

;x x 1 x 0

; x 0

1 3log x ; 0 x

2 2

Discuss global maxima, minima for = 0 and = 1. For what values of does f(x) has global

maxima

Solution : Graph of y = f(x) for = 0

3/2–1/2

–1

–1

1

No global maxima, minima

Graph of y = f(x) for = 1

3/2–1/2

–1

–1

1

Global maxima is 1, which occurs at x = 0

Global minima does not exists

x 0Lim

f(x) = 0,

x 0Lim

f(x) = 1, f(0) =

For global maxima to exists

f(0) 1 1.

Example # 33 : Find extrema of f(x) = 2x 4

2x 3

. Draw graph of g(x) =

1

f(x) and comment on its local and

global extrema.

Solution : f(x) = 2

2

2(x 3x 4)

(2x 3)

=

2

2(x 4)(x 1)

(2x 3)

= 0

–4 3

2

3

4

1

1

4

1

local minima occurs at x = –4

local maxima occurs at x = 1

g(x) = 2

1 2x 3

f(x) x 4

g'(x) = 2 2

2(x 4)(x 1)

(x 4)

local maxima at x = –4

local minima at x = 1

global maxima & minima do not exists

–4

1 –3/2 –4

1

Self Practice Problems :

(23) Let f(x) = x +1

x . Find local maximum and local minimum value of f(x). Can you explain this

discrepancy of locally minimum value being greater than locally maximum value.

(24) If f(x) = 2(x ) x 0

cosx x 0

, find possible values of such that f(x) has local maxima at x = 0.

Answers : (23) Local maxima at x = –1, f(–1) = – 2 ; Local minima at x = 1, f(1) = 2.

(24) [ –1, 1)

Maxima, Minima by higher order derivatives :

Second derivative test :

Let f(x) have derivatives up to second order

Step - I. Find f(x)

Step - II. Solve f(x) = 0. Let x = c be a solution

Step - III. Find f(c) Step - IV.

(i) If f(c) = 0 then further investigation is required

(ii) If f(c) > 0 then x = c is a point of minima.

(iii) If f(c) < 0 then x = c is a point of maxima.

For maxima f(x) changes from positive to negative (as shown in figure).

f(x) is decreasing hence f(c) < 0

Example # 34 : Find the points of local maxima or minima for f(x) = sin2x – x, x (0, ).

Solution : f(x) = sin2x – x

f(x) = 2cos2x – 1

f(x) = 0 cos 2x = 1

2 x =

6

,

5

6

f(x) = – 4 sin 2x

f 6

< 0 Maxima at x = 6

f 5

6

> 0 Minima at x = 5

6

Self Practice Problems :

(25) Let f(x) = sinx (1 + cosx) ; x (0, 2). Find the number of critical points of f(x). Also identify which of these critical points are points of Maxima/Minima.

Ans. Three

x = 3

is point of maxima.

x = is not a point of extrema.

x = 5

3

is point of minima.

nth Derivative test :

Let f(x) have derivatives up to nth order

If f(c) = f(c) = ..........= fn–1(c) = 0 and

fn(c) 0 then we have following possibilities

(i) n is even, f(n)(c) < 0 x = c is point of maxima

(ii) n is even, f(n)(c) > 0 x = c is point of minima.

(iii) n is odd, f(n)(c) < 0 f(x) is decreasing about x = c

(iv) n is odd, f(n) > 0 f(x) is increasing about x = c.

Example # 35 : Find points of local maxima or minima of f(x) = x5 – 5x4 + 5x3 – 1

Solution. f(x) = x5 – 5x4 + 5x3 – 1

f(x) = 5x2 (x – 1) (x – 3)

f(x) = 0 x = 0, 1, 3

f(x) = 10x (2x2 – 6x + 3)

Now, f(1) < 0 Maxima at x = 1

f(3) > 0 Minima at x = 3

and, f(0) = 0 nd derivative test fails

so, f(x) = 30 (2x2 – 4x + 1)

f(0) = 30

Neither maxima nor minima at x = 0.

Note : It was very convenient to check maxima/minima at first step by examining the sign change of

f(x) no sign change of f(x) at x = 0

f(x) = 5x2 (x – 1) (x – 3)

Application of Maxima, Minima :

For a given problem, an objective function can be constructed in terms of one parameter and then

extremum value can be evaluated by equating the differential to zero. As discussed in n th derivative test

maxima/minima can be identified.

Useful Formulae of Mensuration to Remember :

Area of a circular sector = 1

2 r2 , when is in radians.

volume of cube = 3 , Total Surface area of cube = 6 2

volume of cuboid = bh, Total Surface area of cube = 2( b + bh + h)

3-D Figures Volume Total Surface area Curved/lateral Surface area

Cone 1

3r2h

r + r2 Curved Surface area = r

Cylinder r2h 2rh + 2r2 Curved Surface area =2rh

Sphere 4

3r3

4r2

Prism (area of base) × (height) lateral surface area

+ 2 (area of base)

lateral Surface area =

(perimeter of base) × (height)

Right

Pyramid

1

3×(area of base) × (height)

Curved surface area

+ (area of base)

Curved Surface area =

2

×(perimeter of base) ×(slant height)

(Note that lateral surfaces of a prism are all rectangle).

(Note that slant surfaces of a pyramid are triangles).

Example # 36: If the equation x3 + px + q = 0 has three real roots, then show that 4p3 + 27q2 < 0.

Solution: f(x) = x3 + px + q, f(x) = 3x2 + p

f(x) must have one maximum > 0 and one minimum < 0. f(x) = 0

x = ± p

, p 03

f is maximum at x = – p

3

and minimum at x =

p

3

fp p

– f 03 3

2p p 2p p

q q 03 3 3 3

q2 + 34p

27 < 0, 4p3 + 27q2 < 0.

Example # 37 : Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.

Solution : x + y = 60

x = 60 – y xy3 = (60 – y)y3

Let f(y) = (60 – y) y3 ; y (0, 60)

for maximizing f(y) let us find critical points

f(y) = 3y2 (60 – y) – y3 = 0

f(y) = y2 (180 – 4y) = 0

y = 45

f(45+) < 0 and f(45–) > 0. Hence local maxima at y = 45.

So x = 15 and y = 45.

Example # 38 : Rectangles are inscribed inside a semicircle of radius r. Find the rectangle with maximum area. Solution : Let sides of rectangle be x and y (as shown in figure).

A = xy. Here x and y are not independent variables and are related by Pythogorus theorem with r.

2x

4 + y2 = r2 y =

22 x

r4

A(x) = x 2

2 xr

4

A(x) = 4

2 2 xx r

4

Let f(x) = r2x2 – 4x

4 ; x (0, r)

A(x) is maximum when f(x) is maximum

Hence f(x) = x(2r2 – x2) = 0 x = r 2

also f(r 2 ) < 0 and f(r 2 ) > 0

confirming at f(x) is maximum when x = r 2 & y = r

2.

Aliter Let us choose coordinate system with origin as centre of circle (as shown in figure).

A = xy

A = 2 (rcos) (rsin) A = r2 sin2 0,2

Clearly A is maximum when = 4

x = r 2 and y =

r

2 .

Example # 39. Show that the least perimeter of an isosceles triangle circumscribed about a circle of radius ‘r’

is 6 3 r .

Solution : AQ = r cot = AP

AO = r cosec

Q

x

Bx N x C

x

P

(

A

r rO

x

AO ON = tan

x = (r cosec + r) tan

x = r(sec + tan)

Perimeter = p = 4x + 2AQ

p = 4r(sec + tan) + 2rcot

p = r(4sec + 4tan + 2cot)

dp

d = r[4sec tan + 4sec2 – 2cosec2]

for max or min dp

d = 0 2sin3 + 3sin2 – 1 = 0

(sin + 1) (2sin2 + sin – 1) = 0

(sin + 1)2 (2sin – 1) = 0 sin = 1/2 = 30° = /6

pleast

= r 4.2 4

2 33 3

= r

8 4 6

3

= 6 3 3

3 r = 6 3 r

Example # 40 : Let A(1, 2) and B(– 2, – 4) be two fixed points. A variable point P is chosen on the straight line

y = x such that perimeter of PAB is minimum. Find coordinates of P.

Solution. Since distance AB is fixed so for minimizing the perimeter of PAB, we basically have to minimize (PA + PB)

Let A be the mirror image of A in the line y = x (see figure). F(P) = PA + PB

F(P) = PA + PB

But for PAB

PA + PB AB and equality hold when P, A and B becomes collinear. Thus for minimum path

length point P is that special point for which PA and PB become incident and reflected rays with respect to the mirror y = x.

Equation of line joining A and B is y = 2x intersection of this line with y = x is the point P.

Hence P (0, 0).

Note : Above concept is very useful because such problems become very lengthy by making

perimeter as a function of position of P and then minimizing it.

Self Practice Problems :

(26) Find the two positive numbers x and y whose sum is 35 and the product x2 y5 maximum.

(27) A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the slops to form a box. What should be the side of the square to be cut off such that volume of the box is maximum possible.

(28) Prove that a right circular cylinder of given surface area and maximum volume is such that the height is equal to the diameter of the base.

(29) A normal is drawn to the ellipse 2x

25 +

2y

16 = 1. Find the maximum distance of this normal from

the centre.

(30) A line is drawn passing through point P(1, 2) to cut positive coordinate axes at A and B. Find

minimum area of PAB.

(31) Two towns A and B are situated on the same side of a straight road at distances a and b respectively perpendiculars drawn from A and B meet the road at point C and D respectively. The distance between C and D is c. A hospital is to be built at a point P on the road such that the distance APB is minimum. Find position of P.

Ans. (26) x = 25, y = 10. (27) 3 cm (29) 1 unit

(30) 4 units (31) P is at distance of ac

a b from C.

Use of monotonicity for proving inequalities

Comparison of two functions f(x) and g(x) can be done by analysing the monotonic behaviour of

h(x) = f(x) – g(x)

Example # 41 : For x 0,2

prove that sin2 x < x2 < tan2 x

Solution : Let f(x) = x2 – sin2x

= (x + sinx) (x – sinx) > 0

for x 0,2

sin2 x < x2 .....(i)

Let g(x) = x2 – tan2x

= (x + tanx) (x + tanx) < 0

for x 0,2

x2 < tan2 x .....(ii)

From (i) & (ii) sin2 x < x2 < tan2 x

Example # 42 : For x (0, 1) prove that x – 3x

3 < tan–1 x < x –

3x

6 hence or otherwise find

x 0lim

1tan x

x

Solution : Let f(x) = x –3x

3 – tan–1x f(x) = 1 – x2 –

2

1

1 x f(x) = –

4

2

x

1 x

f(x) < 0 for x (0, 1) f(x) is M.D.

f(x) < f(0) x – 3x

3 – tan–1x < 0

x –3x

3 < tan–1x ...........(i)

Similarly g(x) = x – 3x

6 – tan–1x ,g(x) = 1 –

2x

2 –

2

1

1 x g(x) =

2 2

2

x (1 x )

2(1 x )

g(x) > 0 for x (0, 1) g(x) is M.I.

g(x) > g(0) x – 3x

6– tan–1x > 0

x – 3x

6 > tan–1x ........(ii)

from (i) and (ii), we get x – 3x

3 < tan–1x < x –

3x

6 Hence Proved

Also, 1 – 2x

3 <

1tan x

x

< 1 – 2x

6, for x > 0

Hence by sandwich theorem we can prove thatx 0lim

1tan x

x

= 1 but it must also be noted that as x 0,

value of 1tan x

x

1 from left hand side i.e. 1tan x

x

< 1

x 0lim

1tan x

x

= 0

NOTE : In proving inequalities, we must always check when does the equality takes place because the point of equality is very important in this method. Normally point of equality occur at end point of the interval or will be easily predicted by hit and trial.

Example # 43 : For x 0,2

, prove that sin x > x – 3x

6

Solution : Let f(x) = sin x – x + 3x

6

f(x) = cos x – 1 + 2x

2

we cannot decide at this point whether f(x) is positive or negative, hence let us check for

monotonic nature of f(x)

f(x) = x – sinx

Since f(x) > 0 f(x) is M.I. for x 0,2

f(x) > f(0) f(x) > 0

f(x) is M.. f(x) > f(0)

sin x – x +3x

6 > 0 sin x > x –

3x

6 . Hence proved

Example # 44 : Examine which is greater : sin x tan x or x2. Hence evaluate x 0lim 2

sinx tanx

x

, where x 0,2

Solution : Let f(x) = sinx tanx – x2

f(x) = cos x . tan x + sin x . sec2x – 2x

f(x) = sin x + sin x sec2x – 2x

f(x) = cos x + cos x sec2x + 2sec2x sin x tan x – 2

f(x) = (cos x + sec x – 2) + 2 sec2x sin x tan x

Now cos x + sec x – 2 = 2

cosx sec x and 2 sec2x tan x . sin x > 0 because x 0,2

f(x) > 0 f(x) is M.I.

Hence f(x) > f(0)

f(x) > 0 f(x) is M.I. f(x) > 0

sin x tan x – x2 > 0

Hence sin x tan x > x2 2

sinx tanx

x> 1

x 0lim 2

sinx tanx

x

= 1.

Example # 45 : Prove that f(x) =

x1

1x

is monotonically increasing in its domain. Hence or otherwise draw

graph of f(x) and find its range

Solution : f(x) =

x1

1x

, for Domain of f(x), 1 +

1

x > 0

x 1

x

> 0 (–, –1) (0, )

Consider f(x) =

x1

1x

2

1 x 1n 1

1x x1

x

f(x) =

x1

1x

1 1n 1

x x 1

Now

x1

1x

is always positive, hence the sign of f(x) depends on sign of n

11

x

–

1

1 x

i.e. we have to compare n 1

1x

and

So lets assume g(x) = n1

1x

–

1

x 1

g(x) = 1

11

x

2

1

x

+

2

1

(x 1) g(x) =

2

1

x(x 1)

(i) for x (0, ), g(x) < 0 g(x) is M.D. for x (0, )

g(x) > xlim

g(x)

g(x) > 0. and since g(x) > 0 f(x) > 0

(ii) for x (– , – 1), g(x) > 0 g(x) is M.I. for x (– , –1)

g(x) > xlim

g(x) g(x) > 0 f(x) > 0

Hence from (i) and (ii) we get f(x) > 0 for all x (– , –1) (0, )

f(x) is M.I. in its Domain

For drawing the graph of f(x), its important to find the value of f(x) at boundary points

i.e. ± , 0, –1

xlim

x1

1x

= e

x 0lim

x1

1x

= 1 and

x 1lim

x1

1x

=

so the graph of f(x) is

Range is y (1, ) – {e}

Example # 46 : Compare which of the two is greater (100)1/100 or (101)1/101.

Solution : Assume f(x) = x1/x and let us examine monotonic nature of f(x)

f(x) = x1/x . 2

1 nx

x

f(x) > 0 x (0,e)

and f(x) < 0 x (e,)

Hence f(x) is M.D. for x e

and since 100 < 101

f(100) > f(101)

(100)1/100 > (101)1/101

Self Practice Problems :

(32) Prove the following inequalities

(i) x > tan–1(x) for x (0, )

(ii) ex > x + 1 for x (0, )

(iii) x

1 x n (1 + x) x for x (0, )

Rolle’s Theorem :

If a function f defined on [a, b] is

(i) continuous on [a, b]

(ii) derivable on (a, b) and

(iii) f(a) = f(b),

then there exists at least one real number c between a and b (a < c < b) such that f(c) = 0

Geometrical Explanation of Rolle’s Theorem :

Let the curve y = f(x), which is continuous on [a, b] and derivable on (a, b), be drawn (as shown in

figure).

A(a, f(a)), B(b, f(b)), f(a) = f(b), C(c, f(c)), f(c) = 0.

C

1 (c

1, f(c

1)), f(c

1) = 0

C2 (c

2, f(c

2)), f(c

2) = 0

C3 (c

3, f(c

3)), f(c

3) = 0

The theorem simply states that between two points with equal ordinates on the graph of f(x), there

exists at least one point where the tangent is parallel to x-axis.

Algebraic Interpretation of Rolle’s Theorem :

Between two zeros a and b of f(x) (i.e. between two roots a and b of f(x) = 0) there exists at least one

zero of f(x)

Example # 47 : If 2a + 3b + 6c = 0 then prove that the equation ax2 + bx + c = 0 has at least one real root

between 0 and 1.

Solution : Let f(x) = 3ax

3 +

2bx

2 + cx

f(0) = 0 and f(1) = a

3 +

b

2 + c = 2a + 3b + 6c = 0

If f(0) = f(1) then f(x) = 0 for some value of x (0, 1)

ax2 + bx + c = 0 for at least one x (0, 1)

Self Practice Problems :

(33) If f(x) satisfies condition in Rolle’s theorem then show that between two consecutive zeros of

f(x) there lies at most one zero of f(x).

(34) Show that for any real numbers , the polynomial P(x) = x7 + x3 + , has exactly one real root.

Lagrange’s Mean Value Theorem (LMVT) :

If a function f defined on [a, b] is

(i) continuous on [a, b] and

(ii) derivable on (a, b)

then there exists at least one real numbers between a and b (a < c < b) such that f(b) f(a)

b a

= f(c)

Proof : Let us consider a function g(x) = f(x) + x, x [a, b]

where is a constant to b determined such that g(a) = g(b).

= – f(b) f(a)

b a

Now the function g(x), being the sum of two continuous and derivable functions it self

(i) continuous on [a, b]

(ii) derivable on (a, b) and

(iii) g(a) = g(b).

Therefore, by Rolle’s theorem there exists a real number c (a, b) such that g(c) = 0

But g(x) = f(x) +

0 = g(c) = f(c) +

f(c) = – = f(b) f(a)

b a

Geometrical Interpretation of LMVT :

The theorem simply states that between two points A and B of the graph of f(x) there exists at least one

point where tangent is parallel to chord AB.

C(c, f(c)), f(c) = slope of AB.

Alternative Statement : If in the statement of LMVT, b is replaced by a + h, then number c between a

and b may be written as a + h, where 0 < < 1. Thus

f(a h) f(a)

h

= f(a + h) or f(a + h) = f(a) + hf (a + h), 0 < < 1

Example # 48 : Verify LMVT for f(x) = – x2 + 4x – 5 and x [–1, 1]

Solution : f(1) = –2 ; f(–1) = –10

f(c) = f(1) f( 1)

1 ( 1)

–2c + 4 = 4 c = 0

Example # 49 : Using Lagrange’s mean value theorem, prove that if b > a > 0,

then 2

b a

1 b

< tan–1 b – tan–1 a <

2

b a

1 a

Solution : Let f(x) = tan–1 x ; x [a, b] applying LMVT

f(c) = 1 1tan b tan a

b a

for a < c < b and f(x) =

21 x

,

Now f(x) is a monotonically decreasing function

Hence if a < c < b

f(b) < f(c) < f(a)

2

1

1 b <

1 1tan b tan a

b a

<

2

1

1 a Hence proved

Example # 50 : Let f : R R be a twice differentiable function such that f 4

= 0, f 5

4

= 0 & f(3) = 4 then

show that there exists a c (0, 2) such that f(c) + sinc – cosc < 0.

Solution : Consider g(x) = f(x) – sinx + cosx

g'(x) = f'(x) – cosx – sinx

g"(x) = f"(x) + sinx – cosx

By LMVT

g(3) g4

34

= g(c1) ,

4

< c

1 < 3 and

5g g(3)

4

53

4

= g(c2), 3 < c

2 <

4

g(c1) > 0 , g(c

2) < 0

By LMVT

2 1

2 1

g (c ) g (c )

c c

= g(c), c

1 < c < c

2 g(c) < 0 f(c) + sinc – cosc < 0

for some c (c1 , c

2), c (0, 2)

Self Practice Problems :

(35) Using LMVT, prove that if two functions have equal derivatives at all points of (a, b), then they

differ by a constant

(36) If a function f is

(i) continuous on [a, b],

(ii) derivable on (a, b) and

(iii) f(x) > 0, x (a, b), then show that f(x) is strictly increasing on [a, b].

Marked questions are recommended for Revision.

PART - I : SUBJECTIVE QUESTIONS

Section (A) : Equation of Tangent / Normal and Common Tangents / Normals

A-1. (i) Find the equation of tangent to curve y = 3x2 + 4x + 5 at (0, 5).

(ii) Find the equation of tangent and normal to the curve x2 + 3xy + y2 = 5 at point (1, 1) on it.

(iii) Find the equation of tangent and normal to the curve x = 2

2

2at

1 t, y =

3

2

2at

1 t at the point for

which t = 1

2

(iv) Find the equation of tangent to the curve y = 2x sin1/ x x 0

0 x 0

at (0,0)

A-2 (i) Find equations of tangents drawn to the curve y2 – 2x2 – 4y+ 8 = 0 from the point (1, 2).

(ii) Find the equation of all possible normals to the curve x2 = 4y drawn from the point (1,2)

A-3. (i) Find the point on the curve 9y2 = x3 where normal to the curve has non zero x-intercept and

both the x intercept and y-intercept are equal.

(ii) If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, then find coordinates of P

(iii) The normal to the curve 5x5 – 10x3 + x + 2y + 6 = 0 at the point P(0, –3) is tangent to the curve

at some other point(s). Find those point(s)?

A-4.(i) Find common tangent between curves y = x3 and 112x2 + y2 = 112

(ii) Find common normals of the curves y = 2

1

x and x2 + y2 – y = 0

A-5. (i) If the tangent to the curve xy + ax + by = 0 at (1, 1) is inclined at an angle tan–1 2 with positive

x-axis in anticlockwise, then find a and b ?

(ii) The curve y = ax3 + bx2 + 3x + 5 touches y = (x + 2)2 at (–2, 0) then a

b2 is

Section (B) : Angle between curves, Orthogonal curves, Shortest/Maximum distance

between two curves

B-1. Find the cosine of angle of intersection of curves y = 2x nx and y = x2x–1 at (1, 0).

B-2. Find the angle between the curves y = lnx and y = (lnx)2 at their point of intersections.

B-3. Find the angle between the curves y2 = 4x + 4 and y2 = 36 (9 – x).

B-4. Show that if the curves ax2 + by2 = 1 and Ax2 + By2 = 1 are orthogonal then ab(A – B) = AB(a – b).

B-5. Find the shortest distance between line y = x – 2 and y = x2 + 3x + 2

B-6. Find shortest distance between y2 = 4x and (x – 6)2 + y2 = 1

Section (C) : Rate of change and approximation

C-1. The length x of rectangle is decreasing at a rate of 3 cm/min and width y is increasing at a rate of 2

cm/min. When x = 10 cm and y = 6 cm, find the rate of change of (i) the perimeter, (ii) the area of

rectangle.

C-2. x and y are the sides of two squares such that y = x – x2 . Find the rate of change of the area of the

second square with respect to the first square.

C-3. A man 1.5 m tall walks away from a lamp post 4.5 m high at a rate of 4 km/hr.

(i) How fast is his shadow lengthening?

(ii) How fast is the farther end of shadow moving on the pavement?

C-4. Find the approximate change in volume V of a cube of side 5m caused by increasing its side length by

2%.

Section (D) : Monotonicity on an interval, about a point and inequalities, local

maxima/minima

D-1. Show that f(x) = x

1 x – n (1 + x) is an increasing function for x > – 1.

D-2. Find the intervals of monotonicity for the following functions.

(i) 4 3x x

4 3 – 3x2 + 5 (ii) 2

3log x + log3x

D-3. If g(x) is monotonically increasing and f(x) is monotonically decreasing for x R and if (gof) (x) is

defined for x R, then prove that (gof)(x) will be monotonically decreasing function. Hence prove that

(gof) (x + 1) (gof) (x – 1).

D-4. Let f(x) =

2x ; x 0

ax ; x 0

. Find real values of 'a' such that f(x) is strictly monotonically increasing at x = 0.

D-5. Check monotonocity at following points for

(i) f(x) = x3 – 3x + 1 at x = –1, 2

(ii) f(x) = | x – 1 | + 2 | x – 3 | – | x + 2 | at x = – 2, 0, 3, 5

(iii) f(x) = x1/3 at x = 0

(iv) f(x) = x2 + 2

1

x at x = 1, 2

(v) f(x) = 3 2x 2x 5x , x 0

3sinx , x 0

at x = 0

D-6. Prove that

1 1sin sin

10 9

1 1

10 9

.

D-7. Let f and g be differentiable on R and suppose f(0) = g(0) and f(x) g(x) for all x 0. Then show that

f(x) g(x) for all x 0.

D-8.Let f(x) =2

3 x 0 x 1

x nb x 1

. Find the set of values of b such that f(x) has a local minima at x = 1.

D-9. Find the points of local maxima/minima of following functions

(i) f(x) = 2x3 – 21x2 + 36x – 20 (ii) f(x) = – (x – 1)3 (x + 1)2

(iii) f(x) = x nx

D-10. Find points of local maxima / minima of

(i) f(x) = (2x – 1)(2x – 2)2 (ii) f(x) = x2e–x

(iii) f(x) = 3cos4x + 10cos3x + 6cos2x – 3, x [0, ] (iv) f(x) = 2x + 3x2/3

(v) f(x) = 2

2

x – 2

x – 1

D-11. Draw graph of f(x) = x|x – 2| and, hence find points of local maxima/minima.

Section (E) : Global maxima, Global minima, Application of Maxima and Minima

E-1. Find the absolute maximum/minimum value of following functions (i) f(x) = x3 ; x [–2, 2]

(ii) f(x) = sinx + cosx ; x [0, ]

(iii) f(x) = 4x – 2x

2 ; x

92,

2

(iv) f(x) = 3x4 – 8x3 + 12x2 – 48x + 25 ; x [0, 3]

(v) f (x) = sin x + 1

2 cos 2 x ; x 0 ,

2

E-2. Let f(x) = x2 ; x (– 1, 2). Then show that f(x) has exactly one point of local minima but global maximum

is not defined.

E-3. John has 'x' children by his first wife and Anglina has 'x + 1' children by her first husband. They both

marry and have their own children. The whole family has 24 children. It is given that the children of the

same parents don't fight. Then find then maximum number of fights that can take place in the family. E-4. If the sum of the lengths of the hypotenuse and another side of a right angled triangle is given, show

that the area of the triangle is a maximum when the angle between these sides is /3.

E-5. Find the volume of the largest cylinder that can be inscribed in a sphere of radius ' r ' cm.

E-6. Show that the semi vertical angle of a right circular cone of maximum volume, of a given slant height is

tan1 2 .

E-7. A running track of 440 m. is to be laid out enclosing a football field, the shape of which is a rectangle

with semi circle at each end . If the area of the rectangular portion is to be maximum, find the length of

its sides.

E-8. Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve

y = 12 x² .

E-9. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as

possible when revolved around one of its side .

E-10. The combined resistance R of two resistors R1 & R

2 (R

1, R

2 > 0) is given by,

1

R =

1

1

R +

2

1

R.

If R1 + R

2 = constant. Prove that the maximum resistance R is obtained by choosing R

1 = R

2.

Section (F) : Rolle's Theorem, LMVT

F-1. Let f : [1, 2] [1, 4] and g : [1, 2] [2, 7] be two continuous bijective functions such thaf

f(1) = 4 & g(2) = 7. The number of solutions of the equation f(x) = g(x) in (1, 2), is :

F-2. Verify Rolle’s theorem for the function, f(x) = loge

2x ab

x(a b)

+ p, for [a, b] where 0 < a < b.

F-3. Using Rolle’s theorem prove that the equation 3x2 + px – 1 = 0 has at least one real root in the interval

(– 1, 1).

F-4. Using Rolle's theorem show that the derivative of the function f(x) = xsin for x 0

x

0 for x 0

vanishes at an

infinite set of points of the interval (0, 1).

F-5. Let f(x) be differentiable function and g(x) be twice differentiable function. Zeros of f(x), g(x) be a, b

respectively (a < b). Show that there exists at least one root of equation f(x) g(x) + f(x) g(x) = 0 on

(a, b).

F-6. If f(x) = tanx, x 0,5

then show that 2

f5 5 5

F-7. If f(x) and g(x) are differentiable functions for 0 x 23 such that f(0) = 2, g(0) = 0, f(23) = 22, g(23) =

10, then show that f'(x) = 2g'(x) for at least one x in the interval (0, 23).

F-8. If f(x) =

3 3 3

x a b

2 2 2

sin x sin a sin b

xe ae be

x a b

1 x 1 a 1 b

where 0 < a < b < 2, then show that the equation f (x) = 0 has atleast one root in the interval (a, b)

F-9. A function y = f(x) is defined on [0, 6] as f(x) = 3

–8x ; 0 x 1

(x – 3) ; 1 x 4

2 ; 4 x 6

Show that for the function y = f(x), all the three conditions of Rolle's theorem are violated on [0, 6] but

still f'(x) vanishes at a point in (0, 6)

F-10. f is continuous in [a, b] and differentiable in (a, b) (where a > 0 ) such that f(a)

a=

f(b)

b. Prove that there

exist x0 (a, b) such that f(x

0) = 0

0

f(x )

x.

PART - II : ONLY ONE OPTION CORRECT TYPE

Section (A) : Equation of Tangent / Normal and Common Tangents / Normals

A-1. Equation of normal drawn to the graph of the function defined as f(x) = 2sinx

x, x 0 and f(0) = 0 at the

origin is

(A) x + y = 0 (B) x y = 0 (C) y = 0 (D) x = 0

A-2. Equation of the normal to the curve y = x + 2 at the point (1, 1)

(A) 2x y 1 = 0 (B) 2x y + 1 = 0 (C) 2x + y 3 = 0 (D) none of these

A-3. The angle between x-axis and tangent of the curve y = (x+1) (x–3) at the point (3, 0)

(A) tan–1 8

15

(B) tan–1 15

8

(C) tan–1 4 (D) none of these

A-4. The numbers of tangent to the curve y – 2 = x5 which are drawn from point (2,2) is / are

(A) 3 (B) 1 (C) 2 (D) 5

A-5. The equation of tangent drawn to the curve xy = 4 from point (0, 1) is

(A) 1 1

y– = – (x + 8)2 16

(B) 1 1

y– = – (x – 8)2 16

(C) 1 1

y+ = – (x – 8)2 16

(D) 1 1

y–8= – x – 16 2

A-6. The curve y exy + x = 0 has a vertical tangent at point

(A) (1, 1) (B) (0, 1) (C) (1, 0) (D) no point

A-7. If the tangent to the curve x = a ( + sin ), y = a (1 + cos ) at = 3

makes an angle (0 < ) with

x-axis, then =

(A) 3

(B)

2

3

(C)

6

(D)

5

6

A-8. If the normal at the point (3t, 4/t) of the curve xy = 12 cuts the curve again at (3t1, 4/t1) then find 't1' in

terms of 't'

(A) 3

9

16t

(B)

3

16

9t

(C)

3

9

16t (D)

3

16

9t

A-9. The common tangent of the curves y = x2 + 1

x and y2 = 4x is

(A) y = x + 1 (B) y = x –1 (C) y = – x + 1 (D) y = – x –1

A-10_. The area of triangle formed by tangent at (1,1) on y = x2 + bx + c with coordinate axis is equal to 2 then

the integral value of b is

(A) –3 (B) 3 (C) 2 (D) –2

Section (B) : Angle between curves, Orthogonal curves, Shortest/Maximum distance

between two curves

B-1. The angle of intersection of y = ax and y = bx is given by

(A) tan = log(ab)

1 log(ab) (B)

log(a /b)

1 logalogb (C)

log(a /b)

1 log(a /b) (D) None of these

B-2. The angle between curves x2 + 4y2 = 32 and x2 – y2 = 12 is

(A) 3

(B)

4

(C)

6

(D)

2

B-3. Find the angle at which two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect

(A) 0 (B) 6

(C)

3

(D)

2

B-4. The value of a2 if the curves 2 2

2

x y

4a = 1 and y3 = 16x cut orthogonally is

(A) 3/4 (B) 1 (C) 4/3 (D) 4

B-5. The shortest distance between curves y2 = 8x and y2 = 4 (x–3) is

(A) 2 (B) 2 2 (C) 3 2 (D) 4 2

B-6. The shortest distance between curves 2 2x y

32 18 = 1 and

227

x y 14

(A) 15 (B) 11

2 (C)

15

4 (D)

11

4

Section (C) : Rate of change and approximation

C-1. Water is poured into an inverted conical vessel of which the radius of the base is 2 m and height 4 m, at

the rate of 77 litre/minute. The rate at which the water level is rising at the instant when the depth is 70

cm is (use = 22/7)

(A) 10 cm/min (B) 20 cm/min (C) 40 cm/min (D) 30 cm/min

C-2 On the curve x3 = 12y. The interval in which abscissa changes at a faster rate then its ordinate

(A) (–3, 0) (B) (–, –2) (2, ) (C) (–2, 2) (D) (–3, 3)

C-3. A kite is 300 m high and there are 500 m of cord out. If the wind moves the kite horizontally at the rate

of 5 km/hr. directly away from the person who is flying it, find the rate at which the cord is being paid?

(A) 4 (B) 8 (C) 3 (D) cannot be determined

C-4. The approximate value of tan 46° is (take = 22/7) :

(A) 3 (B) 1.035 (C) 1.033 (D) 1.135

C-5. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate

of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases ,

is-

(A) 5

6cm/min (B)

1

54cm/min (C)

1

18 cm/min (D)

1

36 cm/min

Section (D) : Monotonicity on an interval, about a point and inequalities, local

maxima/minima

D-1. The complete set of values of ‘a’ for which the function f(x) = (a + 2) x3 – 3ax2 + 9ax – 1 decreases for

all real values of x is.

(A) (– , – 3] (B) (– , 0] (C) [– 3, 0] (D) [– 3, )

D-2. Let f(x) = x3 + ax2 + bx + 5 sin2 x be an increasing function in the set of real numbers R. Then a & b

satisfy the condition :

(A) a2 3b 15 > 0 (B) a2 3b + 15 0 (C) a2 3b 15 < 0 (D) a > 0 & b > 0

D-3. The function 2

x 1|

x

is monotonically decreasing at the point

(A) x = 3 (B) x = 1 (C) x = 2 (D) none of these

D-4. If f(x) = 1 + 2 x2 + 4 x4 + 6 x6 +...... + 100 x100 is a polynomial in a real variable x, then f(x) has:

(A) neither a maximum nor a minimum (B) only one maximum

(C) only one minimum (D) one maximum and one minimum

D-5_. Which of the following statement is/are true ?

(1) f(x) = sin x is increasing in interval ,2 2

(2) f(x) = sin x is increasing in interval ,2 2

3 5

,2 2

(3) f(x) = sin x is increasing at all point of the interval ,2 2

3 5

,2 2

(4) f(x) = sin x is increasing in intervals ,2 2

3 5

,2 2

(A) all are correct (B) all are false

(C) (2) and (4) are correct (D) (1), (3) & (4) are correct

D-6_. Let f(x) =

x x [1,2]

5 x x (2,4)

2 x 4

7 x x (4,6]

then which of the following statement is / are correct about f(x) ?

(A) Function is strictly increasing at point x = 2

(B) Function is strictly increasing at point x = 4

(C) Function is not increasing at point x = 2 and x = 4

(D) None of these

Section (E) : Global maxima, Global minima, Application of Maxima and Minima

E-1. The greatest, the least values of the function, f(x) = 2 21 2x x , x [21] are respectively

(A) 2, 1 (B) 2, 1 (C) 2, 0 (D) –2, 3

E-2. Let f(x) = (1 + b2)x2 + 2bx + 1 and let m(b) be the minimum value of f(x). As b varies, the range of m(b)

is

(A) [0, 1] (B) 1

0,2

(C) 1

, 12

(D) (0, 1]

E-3. The radius of a right circular cylinder of greatest curved surface which can be inscribed in a given right

circular cone is

(A) one third that of the cone (B) 1/ 2 times that of the cone

(C) 2/3 that of the cone (D) 1/2 that of the cone

E-4. The dimensions of the rectangle of maximum area that can be inscribed in the ellipse

(x/4)2 + (y/3)2 = 1 are

(A) 8, 2 (B) 4, 3 (C) 2 8, 3 2 (D) 2 , 6

E-5. The largest area of a rectangle which has one side on the xaxis and the two vertices on the curve

y = 2xe is

(A) 2 e 1/2 (B) 2 e 1/2 (C) e 1/2 (D) none of these

E-6. Let f(x) =

3 2

2

2

x x 10x 5 , x 1

2x log b 2 , x 1

the set of values of b for which f(x) has greatest value at x =

1 is given by :

(A) 1 b 2 (B) b = {1, 2}

(C) b (, 1) (D) 130, 2 U 2 , 130

E-7. Four points A, B, C, D lie in that order on the parabola y = ax2 + bx + c. The coordinates of A, B & D are

known as A( 2, 3); B( 1, 1) and D(2, 7). The coordinates of C for which the area of the quadrilateral

ABCD is greatest, is

(A) (1/2, 7/4) (B) (1/2, 7/4) (C) ( 1/2, 7/4) (D) ( 1/2, –7/4)

Section (F) : Rolle's Theorem, LMVT

F-1. The function f(x) = x3 – 6x2 + ax + b satisfy the conditions of Rolle's theorem on [1, 3]. Which of these

are correct ?

(A) a =11, b R (B) a = 11, b = – 6 (C) a = –11, b = 6 (D) a = – 11, b R

F-2. The function f(x) = x(x + 3)e–x/2 satisfies all the conditions of Rolle's theorem on [–3, 0]. The value of c

which verifies Rolle's theorem, is

(A) 0 (B) – 1 (C) – 2 (D) 3

F-3. If f(x) satisfies the requirements of Lagrange’s mean value theorem on [0, 2] and if f(0) = 0 and f (x) 1

2

x [0, 2], then

(A) | f(x) | 2 (B) f(x) 1

(C) f(x) = 2x (D) f(x) = 3 for at least one x in [0, 2]

F-4_. If ab > 0 and 3a + 5b + 15 c = 0 then which of the following statement is “INCORRECT”?

(A) there exist exactly one root of equation ax4 + bx2 + c = 0 in (–1,0)

(B) there exist exactly one root of equation ax4 + bx2 + c = 0 in (0,1)

(C) there exist exactly two root of equation ax4 + bx2 + c = 0 in (–1,1)

(D) number of roots of equation ax4 + bx2 + c = 0 can be two in (–1,0)

F-5. Consider the function for x [–2, 3]

f(x) = 3 2

–6 ; x 1

x – 2x – 5x 6; x 1

x – 1

. The value of c obtained by applying Rolle's theorem for which

f(c) = 0 is

(A) 0 (B) 1 (C) 1/2 (D) 'c' does not exist

PART - III : MATCH THE COLUMN

1. Column – I Column – II

(A) If curves y2 = 4ax and y = x

2ae

are orthogonal then ‘a’ (p) 3

can take value

(B) If is angle between the curves y = [| sin x | + | cos x|], (q) 1

([] denote GIF) and x2 + y2 = 5 then cosec2 is

(C)_ If curves y2 = 4a (x + a) and y2 = 4b (x + b) intersects

each other orthogonally then a

bcan be equal to____ (r) 5/4

(D)^ If y = x2 + 3x + c and x =y2 + 3y + c has only

one common point (h, k) then |h + k + c| can be equal to…… (s) 2

2. Column-I Column-II

(A) The number of point (s) of maxima of f(x) = x2 +2

1

x is (p) 0

(B) (sin– 1 x)3 + (cos– 1 x)3 is maximum at x = (q) 2

(C) If [a, b], (b < 1) is largest interval in which (r) 8

3

f(x) = 3x4 + 8x3 – 6x2 – 24x + 19 is strictly increasing

then a

b is

(D) If a + b = 8, a, b > 0 then minimum value of3 3a b

48

is (s) –1

3. Column – Column –

(A) f(x) = x

sinx

e, x [0,] (p) Conditions in Rolle's theorem are satisfied.

(B) f(x) = sgn ((ex – 1)nx), x 1 3

,2 2

(q) Conditions in LMVT are satisfied.

(C) f(x) = (x–1)2/5, x [0,3] (r) At least one condition in Rolle's theorem is not

satisfied.

(D) f(x) =

0x,0

}0{–]1,1[–x,

1e

1–ex

x

1

x

1

(s) At least one condition in LMVT is not satisfied.

4. Column – I Column – II

(A) A rectangle is inscribed in an equilateral triangle of side 4cm. (p) 65

Square of maximum area of such a rectangle is

(B) The volume of a rectangular closed box is 72 and the base (q) 45

sides are in the ratio 1 : 2. The least total surface area is

(C) If x and y are two positive numbers such that x + y = 60 and x3y is (r) 12

maximum then value of x is

(D) The sides of a rectangle of greatest perimeter which is inscribed (s) 108

in a semicircle of radius 5 are a and b. Then a3 + b3 =

Marked questions are recommended for Revision.

PART - I : ONLY ONE OPTION CORRECT TYPE

1. The line x y

1a b touches the curve y = be–x/a at the point

(A) (– a, 2b) (B) a b

,2 2

(C) b

a,e

(D) (0, b)

2.The equation of normal to the curve x3 + y3 = 8xy at point where it is meet by the curve y2 = 4x, other

than origin is

(A) y = x (B) y = –x + 4 (C) y = 2x (D) y = –2x

3. The length of segment of all tangents to curve x2/3 + y2/3 = a2/3 intercepted between coordinate axes is

(A) 2|a| (B) |a| (C) | a |

2 (D)

3 | a |

2

4. If tangents are drawn from the origin to the curve y = sin x, then their points of contact lie on the curve

(A) x – y = xy (B) x + y = xy (C) x2 – y2 = x2y2 (D) x2 + y2 = x2y2

5. Number of tangents drawn from the point (–1/2, 0) to the curve y = e{x}. (Here { } denotes fractional part

function).

(A) 2 (B) 1 (C) 3 (D) 4

6. Let f(x) = 2

2

x , x 0

x 8 , x 0

Equation of tangent line touching both branches of y = f(x) is

(A) y = 4x + 1 (B) y = 4x + 4 (C) y = x + 4 (D) y = x + 1

7. Minimum distance between the curves f(x) = ex & g(x) = n x is

(A) 1 (B) 2 (C) 2 (D) e

8. The point(s) on the parabola y2 = 4x which are closest to the circle, x2 + y2 24y + 128 = 0 is/are:

(A) (0, 0) (B) 2 , 2 2 (C) (4, 4) (D) none

9. If f(x) = xa sgnx

a ; g(x) =

xa sgnxa

for a > 1, a 1and x R, where { } & [ ] denote the fractional part

and integral part functions respectively, then which of the following statements holds good for the

function h(x), where (n a) h(x) = (n f(x) + n g(x)).

(A) ‘h’ is even and increasing (B) ‘h’ is odd and decreasing

(C) ‘h’ is even and decreasing (D) ‘h’ is odd and increasing

10. If f : [1, 10] [1, 10] is a non-decreasing function and g : [1, 10] [1, 10] is a non-increasing function.

Let h(x) = f(g(x)) with h(1) = 1, then h(2)

(A) lies in (1, 2) (B) is more than 2 (C) is equal to 1 (D) is not defined

11. If f(x) = |ax – b| + c|x| is stricly increasing at atleast one point of non differentiability of the function

where a > 0, b > 0, c > 0 then

(A) c > a (B) a > c (C) b > a + c (D) a = b

12. If g(x) is a curve which is obtained by the reflection of f(x) = x xe e

2

by the line y = x then

(A) g(x) has more than one tangent parallel to x-axis

(B) g(x) has more than one tangent parallel to y-axis

(C) y = –x is a tangent to g(x) at (0, 0)

(D) g(x) has no extremum

13. The set of values of p for which all the points of extremum of the function

f(x) = x3 3 px2 + 3 (p2 1) x + 1 lie in the interval ( 2, 4), is:

(A) ( 3, 5) (B) ( 3, 3) (C) ( 1, 3) (D) ( 1, 4)

14. Which of the following statement is correct about e and e.

(A) e > e (B) e < e (C) e = e (D) None of these

15^. The complete set of values of the parameter ‘a’ for which the point of minimum of the function

f(x) = 1 + a2 x – x3 satisfies the inequality 2

2

x x 2

x 5x 6

< 0 is

(A) ( 3 3, 2 3) (2 3, 3 3) (B) ( 3 3, 2 3)

(C) ( 3 3, 2 3) (D) ( 3 2, 2 3)

16. If f(x) = sin3x + sin2 x ; –/2 < x < /2, then the interval in which should lie in order that f(x) has

exactly one minima and one maxima

(A) (–3/2, 3/2) (B) (–2/3, 2/3) – {0} (C) R (D) none of these

17. In a regular triangular prism the distance from the centre of one base to one of the vertices of the other

base is . The altitude of the prism for which the volume is greatest, is :

(A) 2

(B) 3

(C) 3

(D) 4

18. The maximum area of the rectangle whose sides pass through the angular points of a given rectangle

of sides a and b is

(A) 2 (ab) (B) 1

2 (a + b)2 (C)

1

2 (a2 + b2) (D)

3a

b

19. Let ABC is given triangle having respective sides a,b,c. D,E,F are points of the sides BC,CA,AB

respectively so that AFDE is a parallelogram. The maximum area of the parallelogram is

(A) 1

4bcsinA (B)

1

2 bcsinA (C) bcsinA (D)

1

8bcsinA

20. Square roots of 2 consecutive natural number greater than N2 is differ by

(A) > 1

2N (B)

1

2N (C) <

1

2N (D) >

1

N

21. If Rolle's theorem is applicable to the function f(x) =nx

x , (x > 0) over the interval [a, b] where a , b

, then the value of a2 + b2 can be

(A) 20 (B) 25 (C) 45 (D) 10

22. If f(x) be a twice differentiable function such that f(x) = x2 for x = 1, 2, 3, then

(A) f (x) = 2 x [1, 3] (B) f(x) = 2 for some x (1, 3)

(C) f(x) = 2 x (1, 3) (D) f (x) = 2x x (1, 3)

PART-II: NUMERICAL VALUE QUESTIONS

INSTRUCTION :

The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto two digit.

If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal

placed.

1. The number of distinct line(s) which is/are tangent at a point on curve 4x3 = 27 y2 and normal at other

point, is :

2. The sum of the ordinates of point of contacts of the common tangent to the parabolas y = x2 + 4x + 8

and y = x2 + 8x + 4, is

3. Value of p for which equationn x px = 0 has exactly two solution is

4. A light shines from the top of a pole 50 ft. high. A ball is dropped from the same height from a point 30

ft. away from the light. If the shadow of the ball moving at the rate of 200 ft/sec along the ground 1/2

sec. later [ Assume the ball falls a distance s = 16 t2 ft. in ' t ' sec.], then || is :

5. A variable ABC in the xy plane has its orthocentre at vertex 'B' , a fixed vertex 'A' at the origin and the

third vertex 'C' restricted to lie on the parabola y = 1 +27 x

36 . The point B starts at the point (0, 1) at time

t = 0 and moves upward along the y axis at a constant velocity of 2 cm/sec . If the area of the triangle

increasing at the rate of 'p' cm2/sec when t = 7

2 sec, then p is.

6. Function defined by f(x) =

2 2

2 2

x x

x x

e e

e e

is injective in [ – 2, ), then least value of is

7. The value of x 0lim

3 3

5

tan x–sin x

x3x2

2sinx tan

is (where [ . ] denotes the GIF)

[Hint: For x 0,2

identify which is greater (2sinx + tanx) or (3x)}

8. If f(x) = 2ex – ae–x + (2a – 3) x – 3 monotonically increases for x R, then minimum value of a is

9. If the set of all values of the parameter 'a' for which the function

f(x) = sin2x – 8(a + 1) sin x + (4a2 + 8a – 14) x increases for all x R and has no critical points for all x

R, is (–, –m – n ) ( n , ) then (m2 + n2) is (where m, n are prime numbers) :

10. If n2< log2 2 3 <n3, then number of roots of the equation 4cos (ex) = 2x + 2–x, is

11. For –1 p 1, the equation 4x3 – 3x – p = 0 has ‘n’ distinct real roots in the interval 1

, 12

and one of

its root is cos(kcos–1p), then the value of 10n + 1

2k is :

12. Least value of the function, f(x) = 2x2 1 +

2x

200

2 1 is

13. Real root of the equation

(x – 1)2013 + (x – 2)2013 + (x – 3)2013 +...............+(x – 2013)2013 = 0 is a four digit number. Then the sum of

square of the digits is :

14. The exhaustive set of values of 'a' for which the function f(x) = a

3 x3 + (a + 2) x2 + (3a 10)

x + 2 possess a negative point of minimum is (p, q) then value of p is :

15. If f(x) is a polynomial of degree 6, which satisfiesx 0Lim

1/ x

3

f (x)1

x

= e2 and has local maximum at x = 1

and local minimum at x = 0 and x = 2, then the value of

45 18

f9 5

is :

16. Maximum value of 2

2 2–3 4x – x 4 (x – 5) (where 1 x 3) is

17. The three sides of a trapezium are equal each being 6 cms long. Let cm2 be the maximum area of the

trapezium. The value of3

2

is :

18. A sheet of poster has its area 18 m². The margin at the top & bottom are 75 cms. and at the sides 50

cms. Let , n are the dimensions of the poster in meters when the area of the printed space is

maximum. The value of 2 2 100

7

n is :

19. The fuel charges for running a train are proportional to the square of the speed generated in m.p.h. and

costs Rs. 48/- per hour at 16 mph. What is the most economical speed if the fixed charges i.e. salaries

etc. amount to Rs. 300/- per hour.

20. Let f(x) = Max. {x2, (1 – x)2, 2x(1 – x)} where x [0, 1] If Rolle's theorem is applicable for f(x) on largest

possible interval [a, b] then the value of (4a + 6b + 11c) when c (a, b) such that f'(c) = 0, is

21. For every twice differentiable function f(x) the value of |f(x)| 3 x R and for some

f() + (f'())2 = 80. Number of integral values that (f'(x))2 can take between (0, 77) are equal to ___

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

1. If tangent to curve 2y3 = ax2 + x3 at point (a, a) cuts off intercepts , on co-ordinate axes, where 2 +

2 = 61, then the value of 'a' is equal to

(A) 20 (B) 25 (C) 30 (D) 30

2. For the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5, at point (2, – 1)

(A) length of subtangent is 7/6. (B) slope of tangent = 6/7

(C) length of tangent = (85) / 6 (D) none of these

3. Which of the following statements is/are correct ?

(A) x + sinx is increasing function

(B) sec x is neither increasing nor decreasing function

(C) x + sinx is decreasing function

(D) sec x is an increasing function

4. If f(x) = 2x + cot–1 x + n 21 x x , then f(x) :

(A) increases in [0, ) (B) decreases in [0, )

(C) neither increases nor decreases in [0, ) (D) increases in (– , )

5. Let g(x) = 2f(x/2) + f(1 – x) and f(x) < 0 in 0 x 1 then g(x)

(A) decreases in 2

0,3

(B) decreases 2

, 13

(C) increases in 2

0,3

(D) increases in 2

, 13

6. Let f(x) = xm/n for x R where m and n are integers, m even and n odd and 0 < m < n. Then

(A) f(x) decreases on (– , 0] (B) f(x) increases on [0, )

(C) f(x) increases on (– , 0] (D) f(x) decreases on [0, )

7. Let f and g be two differentiable functions defined on an interval such that f(x) 0 and g(x) 0 for all

x and f is strictly decreasing on while g is strictly increasing on then

(A) the product function fg is strictly increasing on

(B) the product function fg is strictly decreasing on I

(C) fog(x) is monotonically increasing on

(D) fog (x) is monotonically decreasing on

8. Let (x) = (f(x))3 – 3(f(x))2 + 4f(x) + 5x + 3 sin x + 4 cos x x R, where f(x) is a differentiable function

xR, then

(A) is increasing whenever f is increasing (B) is increasing whenever f is decreasing

(C) is decreasing whenever f is decreasing (D) is decreasing if f(x) = – 11

9. If p, q, r be real, then the intervals in which, f(x) =

2

2

2

x p pq pr

pq x q qr

pr qr x r

,

(A) increase is x < – 2

3 (p2 + q2 + r2), x > 0 (B) decrease is (–

2

3 (p2 + q2 + r2), 0)

(C) decrease is x < –2

3 (p2 + q2 + r2), x > 0 (D) increase is (–

2

3 (p2 + q2 + r2), 0)

10. If f(x) = 2x

2 2cos x ; g(x) =

2x

6x 6sinx where 0 < x < 1, then

(A) 'f' is increasing function (B) 'g' is decreasing function

(C) f(x)

g(x) is increasing function (D) g(f(x)) is decreasing function

11. Which of the following statement is / are correct ?

(A) The function y = 2

4

2x – 1

x is neither increasing nor decreasing.

(B) If f(x) is strictly increasing real function defined on R and c is a real constant, then number of

Solutions of f(x) = c is always equal to one.

(C) Let f(x) = x ; x (0, 1). f(x) does not has any point of local maxima/minima

(D) f(x) = {x} has maximum at x = 6 (here {.} denotes fractional part function).

State, in order, whether S1, S

2, S

3, S

4 are true or false

12. Let f(x) = x

sinx & x 0,

2

Then the interval in which at least one root of equaiton lie 2

x f12

+ 3

x f4

+ 4

5x f

12

= 0

(A) f , f12 4

(B) 0, f12

(C)

5f ,

12

(D)

5f , f

4 12

13. Let f(x) = (x2 1)n (x2 + x + 1). f(x) has local extremum at x = 1 if

(A) n = 2 (B) n = 3 (C) n = 4 (D) n = 6

14. If f(x) = x

1 x tanx, x 0,

2

, then

(A) f(x) has exactly one point of minima (B) f(x) has exactly one point of maxima

(C) f(x) is increasing in 0,2

(D) maxima occurs at x0 where x

0 = cosx

0

15. If f(x) = a n |x| + bx2 + x has its extremum values at x = – 1 and x = 2, then

(A) a = 2 (B) b = – 1/2 (C) a = – 2 (D) b = 1/2

16. If f(x) = 21 x , 0 x 1

x , x 1

, then

(A) Maximum of f(x) exist at x = 1 (B) Maximum of f (x) doesn't exists

(C) Minimum of f–1(x) exist at x = – 1 (D) Minimum of f–1(x) exist at x = 1

17. If f(x) = tan–1x – (1/2) n x. Then

(A) the greatest value of f(x) on 1/ 3, 3

is /6 + (1/4) n 3

(B) the least value of f(x) on 1/ 3, 3

is /3 – (1/4) n 3

(C) f(x) decreases on (0, )

(D) f(x) increases on (– , 0)

18. Let f(x) = 4 3 2

40

3x 8x 18x 60 . Which of the following statement(s) about f(x) is (are) correct ?

(A) f(x) has local minima at x = 0.

(B) f(x) has local maxima at x = 0.

(C) Absolute maximum value of f(x) is not defined.

(D) f(x) is local maxima at x = – 3, x = 1.

19. A function f(x) = 2

2

x 3x 2

x 2x 3

is -

(A) 1 is not in its domain (B) minimum at x = – 3 and maximum at x = 1

(C) no point of maxima and minima (D) increasing in its domain

20. Let 4x2 + 12xy + 10y2 – 4y + 3 = 0.

(A) Maximum value of y is 3. (B) Minimum value of y is 1.

(C) Maximum value of x is 3. (D) Minimum value of x is 1.

21. For the function f(x) = x cot–1x, x 0

(A) there is atleast one x (0, 1) for which cot–1x = 2

x

1 x

(B) for atleast one x in the interval (0, ), 2

f x

–f(x) < 1

(C) number of solution of the equation f(x) = sec x is 1

(D) f’(x) is strictly decreasing in the interval (0, )

22^. If f(x) = (x – 4) (x – 5) (x – 6) (x – 7) then,

(A) f"(x) = 0 has no real roots.

(B) three roots of f(x) = 0 lie in (4, 5) (5, 6) (6, 7).

(C) one root of the equation f(x) = 0 is 2

11

(D) graph of y = f"(x) is symmectric about x = 2

11

23. Which of the following statements are true :

(A) If f(x) is differentiable function such that f(a) f(b) then there exist no c(a, b) such that f’(c) = 0

(B) The function x100 + sinx – 1 is strictly increasing in [0, 1]

(C) If a, b, c are in A.P, then at least one root of the equation 3ax2 – 4bx + c = 0 is positive

(D) The number of solution(s) of equation 3 tanx + x3 = 2 in (0, /4) is 2

24. Let f(x) be a differentiable function and f() = f() = 0 ( < ), then in the interval (, )

(A) f(x) + f '(x) = 0 has at least one root (B) f(x) – f '(x) = 0 has at least one real root

(C) f(x) . f '(x) = 0 has at least one real root (D) none of these

25. Which of the following inequalities are valid –

(A) |tan–1 x – tan–1y| |x – y| x, y R (B) |tan–1 x – tan–1y| |x – y|

(C) |sin x – sin y| |x – y| (D) |sin x – sin y| |x – y|

26. For all x in [1, 2]

Let f"(x) of a non-constant function f(x) exist and satisfy |f(x)| 2. If f(1) = f(2), then

(A) There exist some a (1, 2) such that f(a) = 0

(B) f(x) is strictly increasing in (1, 2)

(C) There exist atleast one c (1, 2) such that f(c) > 0

(D) |f(x)| < 2 x [1, 2]

27_. For every thrice differentiable function f : R [–3, 3] with (f(0))2 + (f(0))2 = 85,

Which of the following statement(s) is (are) TRUE?

(A) There exist x0 (a, a + 6) such that (f'(x0))2 1

(B) (f'(x))2 can take all real value that lies between (0, 76]

(C) There exist (–6, 0) such that (f())2 + (f'())2 = 10

(D) There exists (–6, 6) such that |f()| = |f()|

(E) There exist atleast one c (–3, 3) such that f'(c) f'"(c) < 0.

PART - IV : COMPREHENSION Comprehension # 1 Lengths of tangent, normal, subtangent and subnormal : Let P (h, k) be any point on curve y = f(x). Let tangent drawn at point P meets x-axis at T & normal at

point P meets x-axis at N (as shown in figure) and m = (h, k )

dy

dx

= slope of tangent.

(i) Length of Tangent = PT = 2

1| k | 1

m

(ii) Length of Normal = PN = 2| k | 1 m

(iii) Length of subtangent = Projection of segment PT on x-axis = TM = k

m

(iv) Length of subnormal = projection of line segment PN on x axis =MN = |km|

1^. Find the product of length of tangent and length of normal for the curve y = x3 + 3x2 + 4x – 1 at point x = 0.

(A) 17

4 (B)

15

4 (C) 17 (D)

4

17

2. Determine 'p' such that the length of the subtangent and subnormal is equal for the curve y = epx + px at the point (0, 1).

(A) ±1 (B) ±2 (C) ± 1

2 (D) ±

1

4

3. Find length of subnormal to x = 2 cos t, y = – 3sin t at t = 4

.

(A) 2

9 (B) 1 (C)

7

2 (D)

9

2

Comprehension # 2

Consider a function f defined by f(x) = sin–1 sinx sin x

2

, x [0, ], which satisfies

f(x) + f(2 – x) = , x [, 2] and f(x) = f(4 – x) for all x [2, 4], then

4. If is the length of the largest interval on which f(x) is increasing, then =

(A) 2

(B) (C) 2 (D) 4

5. If f(x) is symmetric about x = , then =

(A) 2

(B) (C)

4

(D) 2

6. Maximum value of f(x) on [0, 4] is :

(A) 2

(B) (C)

4

(D) 2

Comprehension # 3

Concavity and convexity :

If f(x) > 0 x (a, b), then the curve y = f(x) is concave up (or convex down) in (a,b) and

If f(x) < 0 x (a, b) then the curve y = f(x) is concave down (or convex up) in (a, b).

a

b

concave upward or convex downward

a

b

concave downward or convex upward

Inflection point :

The point where concavity of the curve changes is known as point of inflection (at inflection

point f(x) is equal to 0 or undefined).

c

inflection point

y>0 y< 0

7. Number of point of inflection for f(x) = (x – 1)3 (x– 2)2 , is

(A) 1 (B) 2 (C) 3 (D) 4

8. Exhaustive set of values of ‘a’ for which the function f(x) = x4 + ax3 + 23x

2 + 1 will be concave upward

along the entire real line, is :

(A) [–1,1] (B) [–2,2] (C) [0,2] (D) [0,4]

Comprehension # 4

For a double differentiable function f(x) if f(x) 0 then f(x) is concave upward and if f(x) 0 then f(x) is

concave downward

Here M 1 2

1 2

k k, 0

k k

If f(x) is a concave upward in [a, b] and , [a, b] then 1 2 1 2

1 2 1 2

k f( ) k f( ) k kf

k k k k

,

where k1, k2 R+

If f(x) is a concave downward in [a, b] and , [a, b] then 1 2 1 2

1 2 1 2

k f( ) k f( ) k kf

k k k k

,

where k1, k2 R+

then answer the following

9. Which of the following is true

(A) sin sin

sin2 2

; , (0, ) (B)

sin sinsin

2 2

; , (, 2)

(C)sin sin

sin2 2

; , (0, ) (D) None of these

10. Which of the following is true

(A) 12 2

3

2

32

(B) 2 n n

3

2n

3

(C) 1 1tan tan

2

1tan

2

) a, b R– (D) e 2e

3

2

3e

11. Let , and are three distinct real numbers and f(x) < 0. Also f(x) is increasing function and let

A = –1 –1 –1f ( ) f ( ) f ( )

3

and B = –1f

3

, then order relation between A and B is ?

(A) A > B (B) A < B (C) A = B (D) none of these

Marked questions are recommended for Revision.

* Marked Questions may have more than one correct option.

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

1. Let f be a function defined on R (the set of all real numbers) such that

f(x) = 2010 (x – 2009) (x – 2010)2 (x – 2011)3 (x – 2012)4, for all x R.

If g is a function defined on R with values in the interval (0, ) such that f(x) = n (g(x)), for all x R,

then the number of points in R at which g has a local maximum is [IIT-JEE 2010, Paper-2, (3, 0)/ 79]

2. Let f, g and h be real-valued functions defined on the interval [0, 1] by f(x) = 2xe +

2–xe ,

g(x) = 2xxe +

2–xe and h(x) = 22 xx e +

2–xe . If a, b and c denote, respectively, the absolute maximum

of f, g and h on [0, 1], then [IIT-JEE 2010, Paper-1, (3, –1)/ 84]

(A) a = b and c b (B) a = c and a b (C) a b and c b (D) a = b = c

3. Match the statements given in Column-I with the intervals/union of intervals given in Column-II

[IIT-JEE 2011, Paper-2, (8, 0), 80]

Column-I Column-II

(A) The set 2

2izRe :

1 z

(p) (–, –1) (1, )

z is a complex number, | z | 1, z 1 is

(B) The domain of the function f(x) = sin–1x 2

2(x 1)

8(3)

1 3

is (q) (–, 0) (0, )

(C) If f() =

1 tan 1

tan 1 tan

1 tan 1

, (r) [2, )

then the set f( ) : 02

is

(D) If f(x) = x3/2 (3x – 10), x 0, then f(x) is increasing in (s) (–, –1] [1, )

(t) (–, 0] [2, )

4. The number of distinct real roots of x4 – 4x3 + 12x2 + x – 1 = 0 is [IIT-JEE 2011, Paper-2, (4, 0), 80]

5. Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum

at x = 3. If p(1) = 6 p(3) = 2 , then p(0) is [IIT-JEE 2012, Paper-1, (4, 0), 70]

6. Let f : R R be defined as f(x) = |x| + |x2 – 1|. The total number of points at which f attains either a

local maximum or a local minimum is [IIT-JEE 2012, Paper-1, (4, 0), 70]

7. The number of points in (– ,), for which x2 – x sinx – cosx = 0, is

[JEE (Advanced) 2013, Paper-1, (2, 0)/60]

(A) 6 (B) 4 (C) 2 (D) 0

8*. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into

an open rectangular box by folding after removing squares of equal area from all four corners. If the

total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides

of the rectangular sheet are [JEE (Advanced) 2013, Paper-1, (4, – 1)/60]

(A) 24 (B) 32 (C) 45 (D) 60

9. A vertical line passing through the point (h, 0) intersects the ellipse 2 2x y

4 3 = 1 at the points P and Q.

Let the tangents to the ellipse at P and Q meet at the point R. If (h) = area of the triangle PQR, 1 =

1/ 2 h 1max

(h) and

2 =

1/ 2 h 1min

(h), then 8

5

1 – 8

2 = [JEE (Advanced) 2013, Paper-1, (4, – 1)/60]

10*. The function f(x) = 2|x| + |x + 2| – ||x + 2| – 2|x|| has a local minimum or a local maximum at x =

[JEE (Advanced) 2013, Paper-2, (3, –1)/60]

(A) – 2 (B) –2

3 (C) 2 (D)

2

3

Paragraph for Question Nos. 11 to 12

Let f : [0, 1] R (the set of all real numbers) be a function. Suppose the function f is twice

differentiable, f(0) = f(1) = 0 and satisfies f(x) – 2f(x) + f(x) ex, x [0, 1].

11. Which of the following is true for 0 < x < 1 ? [JEE (Advanced) 2013, Paper-2, (3, –1)/60]

(A) 0 < f(x) < (B) – 1

2 < f(x) <

1

2 (C) –

1

4 < f(x) < 1 (D) – < f(x) < 0

12. If the function e–x f(x) assumes its minimum in the interval [0, 1] at x = 1

4, which of the following is true ?

[JEE (Advanced) 2013, Paper-2, (3, –1)/60]

(A) f (x) < f(x), (B) f (x) > f(x), 0 < x < 1

4

(C) f (x) < f(x), 0 < x < 1

4 (D) f (x) < f(x),

3

4 < x < 1

13. A line L : y = mx + 3 meets y - axis at E(0, 3) and the arc of the parabola y2 = 16x, 0 y 6 at the point

F(x0 , y

0). The tangent to the parabola at F(x

0, y

0) intersects the y-axis at G(0, y

1). The slope m of the line

L is chosen such that the area of the triangle EFG has a local maximum

Match List I with List II and select the correct answer using the code given below the lists :

[JEE (Advanced) 2013, Paper-2, (3, –1)/60]

List - I List - II

P. m = 1. 1

2

Q. Maximum area of EFG is 2. 4

R. y0 = 3. 2

S. y1 = 4. 1

Codes :

P Q R S

(A) 4 1 2 3

(B) 3 4 1 2

(C) 1 3 2 4

(D) 1 3 4 2

14*. Let a R and let f : R R be given by f(x) = x5 – 5x + a. Then

[JEE (Advanced) 2014, Paper-1, (3, 0)/60]

(A) f(x) has three real roots if a > 4 (B) f(x) has only one real root if a > 4

(C) f(x) has three real roots if a < – 4 (D) f(x) has three real roots if – 4 < a < 4

15. The slope of the tangent to the curve (y – x5)2 = x(1 + x2)2 at the point (1, 3) is [JEE (Advanced) 2014, Paper-1, (3, 0)/60]

16. A cylindrical container is to be made from certain solid material with the following constraints: It has

fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at the top. The bottom of the

container is solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the

container.

If the volume of the material used to make the container is minimum when the inner radius of the

container is 10 mm, then the value of V

250is [JEE (Advanced) 2015, P-1 (4, 0) /88]

17*. Let f, g : [–1, 2] R be continuous function which are twice differentiable on the interval (–1, 2). Let the

values of f and g at the points –1, 0 and 2 be as given in the following table :

1–10)x(g

063)x(f

2x0x1–x

In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct

statement(s) is (are) [JEE (Advanced) 2015, P-2 (4, –2)/ 80]

(A) f'(x) – 3g'(x) = 0 has exactly three solutions in (–1, 0) (0, 2)

(B) f'(x) – 3g'(x) = 0 has exactly one solution in (–1, 0)

(C) f'(x) – 3g'(x) = 0 has exactly one solution in (0, 2)

(D) f'(x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)

18. Let f : R (0, ) and g : R R be twice differentiable functions such that f " and g" are continuous

functions on R. Suppose f '(2) = g(2) = 0, f "(2) 0 and g'(2) 0, If 2x

lim )x('g)x('f

)x(g)x(f= 1, then

[JEE (Advanced) 2016, Paper-2, (4, –2)/62]

(A) f has a local minimum at x = 2 (B) f has a local maximum at x = 2

(C) f "(2) > f(2) (D) f(x) – f "(x) = 0 for at least one x R

Answer Q.19, Q.20 and Q.21 by appropriately matching the information given in the three

columns of the following table.

Let f(x) = x + logex – xlogex, x (0, )

Column1 contains information about zeros of f(x), f(x) and f(x).

Column2 contains information about the limiting behavior of f(x), f(x) and f(x) at infinity.

Column3 contains information about increasing/decreasing nature of f(x) and f(x).

Column-1 Column-2 Column-3

() f(x) = 0 for some x (1, e2) (i) limxf(x) = 0 (P) f is increasing in (0, 1)

() f(x) = 0 for some x (1, e) (ii) limxf(x) = – (Q) f is decreasing in (e, e2)

() f(x) = 0 for some x (0, 1) (iii) limxf(x) = – (R) f is increasing in (0, 1)

(V) f(x) = 0 for some x (1, e) (iv) limxf(x) = 0 (S) f is decreasing in (e, e2)

19. Which of the following options is the only INCORRECT combination ?

[JEE(Advanced) 2017, Paper-1,(3, –1)/61]

(A) () (iii) (P) (B) () (iv) (Q) (C) () (iii) (P) (D) () (i) (R)

20. Which of the following options is the only CORRECT combination?

[JEE(Advanced) 2017, Paper-1,(3, –1)/61]

(A) () (ii) (R) (B) () (iv) (P) (C) () (iii) (S) (D) (V) (i) (S)

21. Which of the following options is the only CORRECT combination?

[JEE(Advanced) 2017, Paper-1,(3, –1)/61]

(A) () (iii) (R) (B) (V) (iv) (S) (C) () (ii) (Q) (D) () (i) (P)

22. If f : R R is a twice differentiable function such that f (x) > 0 for all x R, and f

2

1 =

2

1, f(1) = 1,

then [JEE(Advanced) 2017, Paper-2,(3, –1)/61]

(A) f (1) 0 (B) f (1) > 1 (C) 0 < f (1) 2

1 (D)

2

1 < f (1) 1

23. If f : R R is a differentiable function such that f(x) > 2f(x) for all x R, and f(0) = 1, then

[JEE(Advanced) 2017, Paper-2,(4, –2)/61]

(A) f(x) > e2x in (0, ) (B) f(x) < e2x in (0, )

(C) f(x) is increasing in (0, ) (D) f(x) is decreasing in (0, )

24. If f(x) =

cos(2x) cos(2x) sin(2x)

cos x cos x sinx

sinx sinx cos x

, then [JEE(Advanced) 2017, Paper-2,(4, –2)/61]

(A) f (x) attains its minimum at x = 0 (B) f (x) attains its maximum at x = 0

(C) '(x)f = 0 at more than three points in (–, ) (D) '(x)f = 0 at exactly three points in (–, )

25. For every twice differentiable function f : R [–2, 2] with (f(0))2 + (f(0))2 = 85, which of the following

statement(s) is (are) TRUE? [JEE(Advanced) 2018, Paper-1,(4, –2)/60]

(A) There exist r, s R, where r < s, such that f is one-one on the open interval (r, s)

(B) There exists x0 (–4, 0) such that |f(x0)| 1

(C) xlim f(x) 1

(D) There exists (–4, 4) such that f() + f() = 0 and f() 0

26. Let f : R R be given by f(x) =

3x

3x1

1x0

0x

)3/10(x–)2–x(n)2–x(

)3/8(–x7x4–x)3/2(

1x–x

,1x3x10x10x5x

23

2

2345

Then which of the following options is/are Correct ? [JEE(Advanced) 2019, Paper-1,(4, –1)/62]

(A) f is NOT differentiable at x = 1 (B) f is increasing on (–,0)

(C) f is onto (D) f has a local maximum at x = 1

27. Let f(x) = 2x

xsin, x > 0.

Let x1 < x2 < x3 .... < xn < ..... be all points of local maximum of f

and y1 < y2 < y3 < ...... < yn < ....... be all the points of local minimum of f

Then which of the following options is/are correct ?

(A) |xn – yn| > 1 for every n (B) x1 < y1

(C) xn +1 – xn > 2 for every n (D) xn

2

1n2,n2 for every n

PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)

1. Let f : R R be defined by [AIEEE 2010 (8, –2), 144]

f(x) = k – 2x , if x –1

2x 3 , if x –1

If f has a local minimum at x = – 1, then a possible value of k is

(1) 0 (2) 1

–2

(3) –1 (4) 1

2. Let f : R R be a continuous function defined by f(x) = x –x

1

e 2e [AIEEE 2010(8, –2), 144]

Statement -1 : f(c) = 1

3 , for some c R.

Statement -2 : 0 < f(x) 1

2 2, for all x R.

(1) Statement -1 is true, Statement-2 is true ; Statement -2 is not a correct explanation for Statement -1.

(2) Statement-1 is true, Statement-2 is false.

(3) Statement -1 is false, Statement -2 is true.

(4) Statement -1 is true, Statement -2 is true; Statement-2 is a correct explanation for Statement-1.

3. The equation of the tangent to the curve y = x + 2

4

x, that is parallel to the x-axis, is

[AIEEE 2010 (4, –1), 144]

(1) y = 1 (2) y = 2 (3) y = 3 (4) y = 0

4. Let f be a function defined by - [AIEEE 2011 II(4, –1), 120]

f(x) =

tanx, x 0

x

, x 01

Statement - 1 : x = 0 is point of minima of f

Statement - 2 : f '(0) = 0.

(1) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for statement-1.

(2) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1

(3) Statement-1 is true, statement-2 is false.

(4) Statement-1 is false, statement-2 is true.

5. The shortest distance between line y – x = 1 and curve x = y2 is : [AIEEE 2011 (4, –1), 120]

(1) 3

4 (2)

3 2

8 (3)

8

3 2 (4)

4

3

6. A spherical balloon is filled with 4500 cubic meters of helium gas. If a leak in the balloon causes the

gas to escape at the rate of 72 cubic meters per minute, then the rate (in meters per minute) at which

the radius of the balloon decreases 49 minutes after the leakage began is :[AIEEE 2012(4, –1), 120]

(1) 9

7 (2)

7

9 (3)

2

9 (4)

9

2

7. Let a, b R be such that the function f given by f(x) = n |x| + bx2 + ax, x 0 has extreme values at x =

– 1 and x = 2.

Statement-1 : f has local maximum at x = – 1 and at x = 2. [AIEEE 2012 (4, –1), 120]

Statement-2 : a = 1

2 and b =

1

4

.

(1) Statement-1 is false, Statement-2 is true.

(2) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

(3) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

(4) Statement-1 is true, statement-2 is false.

8. The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1]

[AIEEE - 2013, (4, –1),120]

(1) lies between 1 and 2 (2) lies between 2 and 3

(3) lies between –1 and 0 (4) does not exist.

9. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for

some c]0, 1[ : [JEE(Main) 2014, (4, – 1), 120]

(1) f'(c) = g'(c) (2) f'(c) = 2g'(c) (3) 2f'(c) = g'(c) (4) 2f'(c) = 3g'(c)

10. If x = –1 and x = 2 are extreme points of f(x) = log|x| + x2 + x then :

[JEE(Main) 2014, (4, –1), 120]

(1) = 2, =1

2 (2) = 2, =

1

2 (3) = –6, =

1

2 (4) = –6, = –

1

2

11. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side

= x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is

minimum, then [JEE(Main) 2016, (4, – 1), 120]

(1) (4 – ) x = r (2) x = 2r (3) 2x = r (4) 2x = ( + 4) r

12. Consider f(x) = tan–1 1 sinx

1– sinx

, x 0,2

. A normal to y = f(x) at x = 6

also passes through the

point : [JEE(Main) 2016, (4, – 1), 120]

(1) 2

0,3

(2) , 06

(3) , 04

(4) (0, 0)

13. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the

maximum area (in sq. m) of the flower-bed, is : [JEE(Main) 2017, (4, – 1), 120]

(1) 12.5 (2) 10 (3) 25 (4) 30

14. The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes

through the point : [JEE(Main) 2017, (4, – 1), 120]

(1) 1 1

– ,–2 2

(2) 1 1

,2 2

(3) 1 1

,–2 3

(4) 1 1

,2 3

15. The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is

[JEE(Main) 2017, (4, – 1), 120]

(1) 2 2 1 (2) 2 2 – 1 (3) 4 2 – 1 (4) 4 2 1

16. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angels , then the value of b is :

[JEE(Main) 2018, (4, – 1), 120]

(1) 4 (2) 9

2 (3) 6 (4)

7

2

17. Let f(x) = x2 + 2

1

xand g(x) = x –

1

x, x R – {–1, 0, 1}. If h(x) =

f(x)

g(x), then the local minimum value of

h(x) is : [JEE(Main) 2018, (4, – 1), 120]

(1) – 2 2 (2) 2 2 (3) 3 (4) – 3

18. Let A(4, –4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the

parabola, where O is the origin, such that the area of ACB is maximum. Then, the area (in sq. units) of

ACB, is : [JEE(Main) 2019, Online (09-01-19),P-2 (4, – 1), 120]

(1) 2

130 (2)

4

331 (3)

4

131 (4) 32

19. A helicopter is flying along the curve given by y – x3/2 = 7, (x 0). A soldier positioned at the point

1,7

2

wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

[JEE(Main) 2019, Online (10-01-19),P-2 (4, – 1), 120]

(1) 1

2 (2)

1 7

3 3 (3)

1 7

6 3 (4)

5

6

20. Let f(x) = 2 2

x

a x –

2 2

(d x)

b (d x)

, xR, where a, b and d are non-zero real constant. Then :

[JEE(Main) 2019, Online (11-01-19),P-2 (4, – 1), 120]

(1) f is neither increasing nor decreasing function of x

(2) f is an increasing function of x

(3) fis not a continuous function of x

(4) f is a decreasing function of x

21. Let f : [0,2] R be a twice differentiable function such that f" (x) > 0, for all x(0,2).

If (x) = f(x) + f(2–x), then is : [JEE(Main) 2019, Online (08-04-19),P-1 (4, – 1), 120]

(1) decreasing on (0,1) and increasing on (1,2).

(2) increasing on (0,2)

(3) increasing on (0,1) and decreasing on (1,2)

(4) decreasing on (0.2)

22. The shortest distance between the line y = x and the curve y= x –2 is :

[JEE(Main) 2019, Online (08-04-19),P-1 (4, – 1), 120]

(1) 24

7 (2) 2 (3)

8

7 (4)

24

11

23. If m is the minimum value of k for which the function f(x) = x2x–kx is increasing in the interval [0,3]

and M is the maximum value of f in [0,3] when k = m, then the ordered pair (m,M) is equal to :

[JEE(Main) 2019, Online (12-04-19),P-1 (4, – 1), 120]

(1) 23,4 (2) 33,4 (3) 33,3 (4) 63,5

24. A Water tank has the shape of an inverted right circular cone, whose semi–vertical angle is tan–1

2

1.

Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/ min), at

which the level of water is rising at the instant when the depth of water in the tank is 10 m; is :

[JEE(Main) 2019, Online (09-04-19),P-2 (4, – 1), 120]

(1) 1/5 (2) 2/ (3) 1/15 (4) 1/10

25. Let f(x) be a polynomial of degree 5 such that x = ±1 are its critical points. If 3x 0

f(x)lim 2

x

= 4, then

which one of the following is not true? [JEE(Main) 2020, Online (07-01-20),P-2 (4, –1), 120]

(1) x = 1 is a point of minima and x = –1 is a point of maxima of f.

(2) f(1) –4f(–1) = 4.

(3) x = 1 is a point of maxima and x = –1 is a point of minimum of f.

(4) f is an odd function.

26. Let f(x) = xcos–1 (–sin|x|), x

2,

2– , then which of the following is true?

(1) f'(0) = – 2

[JEE(Main) 2020, Online (08-01-20),P-1 (4, –1), 120]

(2) f' is decreasing in

0,

2– and increasing in

2,0

(3) f is not differentiable at x = 0

(4) f' is increasing in

0,

2– and decreasing in

2,0

27. If c is a point at which Rolle's theorem holds for the function, f(x) = loge

x7

x2

in the interval [3, 4],

where R, then f"(c) is equal to : [JEE(Main) 2020, Online (08-01-20),P-1 (4, –1), 120]

(1) 7

3 (2) –

24

1 (3) –

12

1 (4)

12

1

28. Let f be any function continuous on [a, b] and twice differentiable on (a, b). If for all x (a, b), f'(x) > 0

and f"(x) < 0, then for any c (a, b), )c(f)b(f

)a(f)c(f

is greater than

[JEE(Main) 2020, Online (09-01-20),P-1 (4, –1), 120]

(1) 1 (2) ac

cb

(3)

cb

ac

(4)

ab

ab

EXERCISE - 1

PART -I Section (A) : A-1. (i) y = 4x + 5 (ii) y + x = 2 (tangent) , y = x (normal)

(iii) 16x + 13y = 9a (iv) y = 0

A-2 (i) 2x + y = 4, y = 2x (ii) x + y = 3

A-3. (i) 8

4,3

(ii) (9/4, 3/8) (iii) (1, –1) , (–1, –5)

A-4. (i) y = 12x – 16 or y = 12x + 16 (ii) x – 2y + 1 = 0 or 2y + x – 1 = 0

A-5. (i) a = 1, b = –2 (ii) 1

Section (B) :

B-1. 1 B-2. 45° at (1, 0) and tan–1 2

e

e 2

at (e, 1)

B-3. 90° B-5. 3

2 B-6. 20 1

Section (C) : C-1. (i) – 2 cm/min (ii) 2 cm2/min C-2. 2x2 – 3x + 1 C-3. (i) 2 km/hr (ii) 6 km/h

C-4. 7.5 m3

Section (D) :

D-2. (i) M.D. in (– , –3] M.I. in [– 3 , 0] M.D. in [0 , 2]

M.I. in [2 , )

(ii) M.D. in 1

0,3

M.I. in 1

,3

D-4. a R+

D-5. Ans. (i) Neither increasing nor decreasing at x = –1 and increasing at x = 2 (ii) at x = – 2 decreasing at x = 0 decreasing at x = 3 neither increasing nor decreasing at x = 5 increasing

(iii) Strictly increasing at x = 0 (iv) Strictly increasing at x = 2, neither I nor D at x = 1 (v) Strictly increasing at x = 0

D-8.b (0, e]

D-9. (i) local max at x = 1, local min at x = 6

(ii) local max. at x = –1

5 , local min. at x = – 1

(iii) local mini at x =1

e , No local maxima

D-10. (i) local maxima at x = log2

4

3 and local minima at x = 1

(ii) local min at 0, local max at 2

(iii) local max at x = 0, 2

3

, local min at x =

2

,

(iv) local maxima at –1 and local minma at 0

(v) local minima at x = ± 2 , 0

(v)

(0,2)

local minima at x = ±, 2 0

D-11. local max at x = 1, local min at x = 2.

Section (E) :

E-1. (i) max = 8, min. = – 8

(ii) max = 2 , min = – 1

(iii) max. = 8, min. = – 10 (iv) max. = 25, min = – 39

(v) max. at x = /6, max. value = 3/4; min. at x = 0 and /2, min. value = 1/2

E-3. F = 191 E-5. 34 r

3 3

E-7. 110 m ,

220

m E-8. 32 sq. units

E-9. 12cm, 6 cm

Section (F) :

F-1. 1

PART -II Section (A) : A-1. (A) A-2. (A) A-3. (C) A-4. (C) A-5. (B) A-6. (C) A-7. (D)

A-8. (B) A-9. (A) A-10. (A)

Section (B) :

B-1. (B) B-2. (D) B-3. (D) B-4. (C) B-5. (B) B-6. (D)

Section (C) : C-1. (B) C-2 (C) C-3. (A) C-4. (B) C-5. (C)

Section (D)

D-1. (A) D-2. (B) D-3. (A) D-4. (C) D-5. (D) D-6. (C)

Section (E) : E-1. (C) E-2. (D) E-3. (D) E-4. (C) E-5. (A) E-6. (D) E-7. (A)

Section (F) : F-1. (A) F-2. (C) F-3. (B) F-4. (D) F-5. (C)

PART -III 1. (A p, q, r,s); (B r) ; (C p,q,r,s); (D q)

2. (A) (p), (B) (s), (C) (q), (D) (r)

3. (A) (p,q), (B) (r,s), (C) (r,s), (D) (r,s) 4. (A) (r), (B) (s), (C) (q),(D) (p)

EXERCISE - 2

PART -I

1. (D) 2.(A) 3. (B) 4. (C) 5. (B) 6. (B) 7. (B)

8. (C) 9. (D) 10. (C) 11. (A) 12. (D) 13. (C) 14. (A)

15. (A) 16. (D) 17. (B) 18. (B) 19. (A) 20. (C) 21. (A)

22. (B)

PART -II

1. 02.00 2. 24.00 3. 00.36 or 00.37 4. 07.50

5. 09.42 or 09.43 6. 02.00 7. 02.82 or 02.83 8. 00.50

9. 29.00 10. 04.00 11. 11.50 12. 26.28

13. 50.00 14. 03.33 15. 32.00 16. 36.00

17. 40.50 18. 19.85 or 19.86 19. 40.00 20. 10.83

21. 76.00

PART -III

1. (CD) 2. (ABC) 3. (AB) 4. (AD) 5. (BC) 6. (AB) 7. (AD)

8. (AD) 9. (AB) 10. (ABC) 11. (AC) 12. (AD) 13. (ACD) 14. (BD)

15. (AB) 16. (AC) 17. (ABC) 18. (ACD) 19. (AC) 20. (AB) 21. (BD)

22. (BCD) 23. (BC) 24. (ABC) 25. (AC) 26. (ACD) 27. (ABCDE)

PART -IV

1. (A) 2. (C) 3. (D) 4. (C) 5. (B) 6. (A) 7. (C)

8. (B) 9. (C) 10. (D) 11. (A)

EXERCISE - 3

PART -I

1. 1 2. (D) 3. (A) (s), (B) (t), (C) (r),(D) (r)

4. 2 5. (9) 6. (5) 7. (C) 8. (AC) 9. 9 10. (AB)

11. (D) 12. (C) 13. (A) 14. (BD) 15. (8) 16. 4 17. (B,C)

18. (A,D) 19. (D) 20. (C) 21. (C) 22. (B) 23. (A,C) 24. (BC)

25. (ABD) 26. (A,C,D) 27. (ACD)

PART -II

1. (3) 2. (4) 3. (3) 4. (2) 5. (2) 6. (3) 7. (2)

8. (4) 9. (2) 10. (1) 11. (2) 12. (1) 13. (3) 14. (2)

15. (3) 16. (2) 17. (2) 18. (3) 19. (3) 20. (2) 21. (1)

22. (1) 23. (2) 24. (1) 25. (1) 26. (2) 27. (4) 28. (3)

TOPIC : APPLICATION OF DERIVATIVES EXERCISE # 1

PART–1 Section (A)

A-1. (i) dy

dx= 6x + 4

(0,5)

dy

dx

= 4

Equation of tangent is y 5

x 0

= 4 y = 4x + 5

(ii) dy

dxat (1,1) = –

(1,1)

2x 3y

3x 2y

= – 1

equation of tangent is (y–1) = – (x–1) and equation of normal is (y–1) = (x–1)

(iii) At t = 1

2, the value of x =

2a

5 and y =

a

5

Also dx

dt =

2 2

4at

(1 t ) and

dy

dt=

2 2

2 2

2at (3 t )

(1 t )

dy

dx=

t

2(3+ t2)

1t

2

dy

dx

= 13

16

equation of tangent is a

y5

=

13

16

2ax

5

13x – 16y = 2a

equation of normal is a

y5

13

16 + x –

2a

5= 0 16x + 13y = 9a

(iv) dy

dx at (0, 0) =

h 0lim

f(0 h) f(0)

h

=

h 0lim

2h sin1/h 0

h

= h 0lim

(h sin (1/h)) = 0 equation of tangent is y 0

x 0

= 0 y = 0

A-2 y2 – 2x2 – 4y+ 8 = 0

dy dy

2y 4x 4 0dx dx

dy 4x

dx 2y 4

(h, k)

dy 4h

dx 2k 4

Equation of tangent is (y – k) =

4h

2k 4 (x – h).

It passes through (1, 2) (2 – k)(2k – 4) = 4h (1 – h) or, 4k – 2k2 – 8 + 4k = 4h – 4h2 or, 4k – k2 – 4 = 2h – 2h2 or, –2h2 + k2 + 2h – 4k + 4 = 0

k2 – 2h2 – 4k + 8 = 0

2h – 4 = 0 or h = 2 k = 0 or 4

Equation of tangent at (2,0); y = 8

( 4) (x – 2)

or y = –2x + 4 or y + 2x = 4 and equation of tangent at (2, 4) is y = 2x

(ii) Let point Q be 2h

h,4

and point P be the point of contact on the curve.

Now, mPQ = slope of the normal at Q.

Differentiating x2 = 4y w.r.t we get 2x = 4dy

dx or

dy

dx=

x

2

or Slope of the normals at Q =

x h

dy 2

dy h

or

2h2

24

h 1 h

[From (1)]

or 3h

4 – 2h = – 2h + 2 or h3 = 8 or h = 2

Hence, the coordinates of point Q are (2, 1) and so the equation of the required normal becomes x + y = 3

A-3. (i) Slope of normal equal to –1.

dy

dx =

2x

6y= 1 x2 = 6y and 9y2 = x3 x = 0, 4 point is

84,

3

(ii) Differentiating equation of curve w.r.t. x,dy

2ydx (1, 1)

dy

dx=

1 ( 2)

2

= –

1

2

Equation of tangent is (y – 1) = 1

2 (x – 1) or 2y + x = 3.

Solving the equations of tangent and curve:

y2 = (–2y + 3) (2 – 3 + 2y)2 or y2 = (3 – 2y) (2y – 1)2

or y2 = (3 – 2y) (4y2 + 1 – 4y) or y2 = 12y2 + 3 – 12y – 8y3 – 2y + 8y2

or 8y3 – 19y2 + 14y – 3 = 0 or (y – 1) (8y2 – 11y + 3) = 0

or (y – 1) (8y – 3)(y – 1) = 0 or y = 1, 3/8

P(9/4, 3/8)

(iii) Differentiating, 25x4 – 30x2 + 1 + 2y = 0

At P(0, – 3), y = – 1

2

The normal at P is y + 3 = 2x

Eliminating y with the given equation x(x2 – 1)2 = 0 x = 0, 1, – 1

The line is tangent at (1, – 1) and (– 1, – 5).

A-4.(i) Slope of y = x3 at (h, k) is 3h2

Slope of y = x3 at (h, k) is 3h2

equation of tangent is =y k

x h

= 3h2

y = 3h2x – 3h3 + k y = 3h2x – 2h3 (–2h3)2 = 1(3h2)2 + 112

4h6 – 9h4 –112 = 0

(h2 – 4) (4h4 + 7h2 + 28) = 0 h = 2, –2

Tangent are y = 12x – 16 or y = 12x + 16

=(2 – x)2 + 2x(2 – x)(–1)

(ii) dy

dx for y =

2

1

x is

3

2

x

(h,x)

dy

dx=

3

2

h

equation of normal is y k

x h

=

3h

2

2y – 2

2

h = h3x – h4

it passes from 1

0,2

1– 2

2

h = – h4 h6 + h2 – 2 = 0

(h2 –1) (h4 + h2 + 2) = 0 h = ± 1

common normals are x – 2y + 1 = 0 or 2y + x – 1 = 0

A-5. (i) xy + ax + by = 0

As (1,1) lies on curve, so

1 + a + b = 0 & = tan–1 2 or dy

2dx

Differentiating equation of curve w.r.t. x, dy

xdx

+ y + a + b dy

dx = 0,

put x = 1, y = 1 dy

dx = –

1 a

1 b

= 2

or 1 + a + 2 + 2b = 0 or 3 + a + 2b = 0 b = – 2 & a = 1

(ii) (–2, 0) lies on both curves –8a + 4b – 6 + 5 = 0 .....(i)

(dy

dx at (–2, 0) for second curve) = (

dy

dx at (–2, 0) for first curve) = 0

12a – 4b + 3 = 0 ...(ii)

Solving (i) and (ii) we get a =1

2, b =

3

4

Section (B)

B-1. For C1 , x

x

(1,0) (1,0)

dy 2nx.2 . n2 2

dx x

For C2, 2x 2x 1

(1,0)(1,0)

dyx . nx 2 2x(x) 2

dx

= 0 cos = 1

B-2. For C1, x 1 x 1

dy 11

dx x

For C1, x e x e

dy 1 1

dx x e

For C2, x 1 x 1

dy 2lnx0

dx x

For C2, x e x e

dy 2lnx 2

dx x e

angle between curves at (1,0) is /4 and angle between curves at (e,1) is tan–1 2

e

e 2

B-3. Given curves are y2 = 4x + 4 and y2 = 36(9 – x) ......(i)

On solving, we get the point (8, 6) and (8,–6)

On differentiating equation (i), we get

2y dy

dx = 4 and 2y

dy

dx = – 36

dy

dx =

2

y and

dy

dx =

–18

y

At point (8, 6), m1 =

dy

dx =

1

3and m

2 =

dy

dx= – 3

m1m

2 = – 1

B-4. ax2 + by2 = 1 dy

dx =

ax

by

Ax2 + By2 = 1 dy

dx =

Ax

By

Product of slopes = – 1

aAx2 + bBy2 = 0

Eliminating x2, y2

a b 1

A B 1

aA bB 0

= 0

(AbB – aAB) – (abB – abA) = 0

AB(b – a) – ab (B – A) = 0

ab(A – B) = AB(a – b)

B-5. Let C1 is y = x – 2 and C2 is y = x2 + 3x + 2

Slope of common normal is – 1 (if possible)

Now, for C2, dy

2x 3 1dx

x = – 1

Point on C2 where normal has slope equal to –1 is (–1, 0)

Shortest distance between C1 and C2 is distance of (–1, 0) from y = x – 2 which is3

2

B-6. Equation of normal to y2 = 4x is y = –tx + 2t + t3 at point (t2, 2t). If it is common normal to

(x – 6)2 + y2 = 1 then (6, 0) satisfies the above equation of normal

t3 – 4t = 0 t = 0, 2, –2

feet of these common normal are (0,0), (4,4), (4,–4)

Distance of these feet from (6, 0) are 6, 20, 20 respectively

shortest distance between y2 = 4x and (x – 6)2 + y2 = 1 is 20 1

Section (C) C-1. (i) Let P be perimeter

P = 2x + 2y

dP

dt =

dx dy2 2

dt dt

dA

dt =

dx dyy x

dt dt – 6 + 4 = – 2

(ii) Let A be area A = xy

dA

dt =

dx dyy x

dt dt = – 18 + 20 = 2

C-2. We have to obtain 2

2

d(y )

d(x ) =

dy2y

dx

2x =

y

x .

dy

dx

y = x – x2 dy

dx = 1 – 2x

2

2

d(y )

d(x ) =

y

x (1– 2x) =

2(x x )(1 2x)

x

= 2x2 – 3x + 1

C-3. Let AC be pole, DE be man and B be farther end of shadow as shown in figure From triangles ABC and DBE

4.5 1.5

x y y

3y = 1.5 x

dy

dt = 2,

d

dt (x + y) =

dx dy

dt dt

C-4. V = a3 V =dV

da a = 3a2a = 3a2 ×

2a

100 =

36 5

100

= 7.5 m3

Section (D)

D-1.

3 / 2

2 x – 2 1 xf (x)

2 x 1

=

2

3 / 2

x 1 –1

2(x 1)

0, x > –1 f(x) is increasing.

D-2. (i) Let f(x) = 4 3x x

4 3 – 3x2 + 5

f(x) = x(x + 3) (x – 2)

M.D. in (– , –3]

M.I. in [– 3, 0]

M.D. in [0, 2]

M.I. in [2, )

(ii) Let f(x) = (log3x)2 + log

3x

=

2

2

nx nx

n3n3 f(x) =

2

2 nx n3

x( n3)

M.D. in 1

0,3

M.I. in 1

,3

D-3. g(x) is monotonically increasing

g(x) 0 & f(x) is M.D. f(x) 0

d

dx (fog) (x) =

d

dx (f(g(x) = f (g (x) g (x) 0

as f(x) 0 & g(x) 0

(fog) (x) is monotonically decreasing f(x) 0

Also x + 1 > x – 1

f(x + 1) < f(x – 1) as f(x) is M.D.

g(f(x + 1) < g(f(x – 1)) as g(x) is M..

D-4. a ; x 0 ,

f (x)2x ; x 0.

f(x) > 0 a > 0

D-5. (i) f'(x) = 3x2 – 3

= 3(x – 1) (x + 1)

at x = 1 point of minima

Neither increasing nor decreasing at x = 2 increasing (ii) at x = – 2 decreasing

at x = 0 decreasing

at x = 3 neither increasing nor decreasing

at x = 5 increasing

(iii) f(x) = 2 / 3

1

3x f(x) > 0 x R – {0}

strictly increasing at x = 0

(iv) f(x) = 2

3

2(x 1)(x – 1)(x 1)

x

strictly increasing at x = 2, neither I nor D at x = 1

(v) f(x) = 23x 4x 5 , x 0

3cosx , x 0

f(0–) = +5

and f(0+) = 3 strictly increasing at x = 0

D-6. Let f(x) = sinx

x f'(x) =

2

xcosx sinx

x

=

2

cosx(x tanx)

x

< 0 x 0,

2

f(x) is decreasing in 0,2

f1

10

> f1

9

1 1sin sin

10 9

1 1

10 9

D-7. Let h(x) = f(x) – g(x)

h(0) = f(0) – g(0) h(0) = 0

h(x) = f(x) – g(x) 0 for x 0

h(x) is decreasing for x 0

x 0

h(x) h(0)

h(x) 0

f(x) – g(x) 0

f(x) g(x)

D-8.f(x) = –1 , 0 x 1

2x x 1

f(x) changes sign from negative to positive.

f(1–) = 2,f (1+) = 1 + nb and f(1) = 1 + nb.

f(1–) f(1) 2 1 + nb

nb 1 0 < b e

D-9. (i) f(x) = 6(x – 1) (x – 6)

Local maxima at x = 1

Local minima at x = 6

(ii) f(x) = –(x – 1)2 (x + 1) (5x + 1)

Local minima at x = – 1

Local maxima at x = –1

5

Neither local minima nor local maxima at x = 1.

(iii) f (x) = nx + 1

Local minima at x = 1

e

No local maxima

D-10. (i) g(t) = (t – 1) (t – 2)2

g(t) = (3t – 4) (t – 2)

4/3 2

– ++

local maxima at x = log2

4

3 and local minima at x = 1

(ii) f(x) = xe–x(2 – x)

local min at 0, local max at 2

0 2

+ ––

(iii) f(x) = – 3sin2x (2cos x + 1) (cos x + 2)

local max at x = 0, 2

3

local min at x = 2

,

(iv) f(x) =1/ 3

1/ 3

2(1 x )

x

– ++

–1 0

local maxima at –1 and local minma at 0

(v)

(0,2)

local minima at x = ±, 2 0

D-11. f(x) = 2

2

–x 2x : x 2

x – 2x : x 2

Section (E) E-1. (i) f(x) = 3x3

f(x) = 0 x = 0

x = – 2, f(–2) = – 8

x = 0, f(0) = 0

x = 2, f(2) = 8

Minimum = – 8, maximum = 8

(ii) f(x) = cos x – sin x

f(x) = 0 x = 4

x = 0, f(0) = 1

x =4

, f

4

= 2

x = , f() = – 1

Minimum = –1, Maximum = 2

(iii) f(x) = 4 – x

f(x) = 0 x = 4

x = – 2, f(–2) = – 10

x = 4, f(4) = 8

x = 9

2, f

9

2

=63

8

Minimum = –10, Maximum = 8

(iv) f(x) = cos x – sin 2x

f(x) = 0 cos x = 0, sin x =1

2 x =

2

, x =

6

x = 0, f(0) = 1

2

x = 6

, f

6

= 3

4

x = 2

, f

2

= 1

2

Minimum =1

2, Maximum =

3

4

E-2.

x = 0 is local minima and global maximum is not defined

E-3. Let No. of children of john & anglina = y

x + (x + 1) + y = 24

y = 23 – 2x

Number of fights

F = x(x + 1) + x(23 – 2x) + (x + 1) (23 – 2x)

F = – 3x2 + 45x + 23

df

dx = 0 – 6x + 45 = 0 x = 7.5

But 'x' wil be integral.

check x = 6 or x = 7

F = 191

E-4. x + y = k constant x + y = k

Area of triangle be A.

A = 1

2x 2 2y – x A =

1

2 x k k – 2x

dA k (k – 3x)

dx 2 k – 2x Area is maximum when x =

k

3.

y = 2k

3 cos =

k

32k

3

= 1

2 =

3

.

E-5. R2 + r2 = h2 R2 = h2 – r2

volume of cylinder,

V = R2 (2h) = (2h) ( 2 2r h )2

dV

dh = 2 (r2 – h2) + 2h(–2h) = 0

r2 = 3h2 h = r

3

2

2

d V

dh < 0 at h =

r

3 V

max = 2

r

3

22 r

r3

=

34 r

3 3

E-6. h = cos

r =sin

V =1

3 r2 h

V = 1

3 3 sin2 cos

dV

d =

1

3 3 (2sincos2 – sin3)

dV

d =

1

3 3 sin (2 – 3 sin2) = 0 at

sin = 2

3

tan = 2

E-7. 2 + 2r = 440

A = 2r = – 2r2 + 440r

dA

dr = – 4r + 440 = 0

at r = 110

E-8. Let A be area of rectangle.

A = (2x) (y) = 2x (12 – x2)

dA

dx = 6(2 + x) (2 – x)

At x = 2, A has largest value. Largest A = 32 E-9. Let x, y be dimensions of rectangle.

2x + 2y = 36. Let V be volume sweeped

V = x2y

V = x2(18 – x)

dV

dx = x.3.(12 – x)

At x = 12, V has maximum value. y = 6

E-10. 1

R =

1

1

R +

1

1

C R

R = R1 (C – R

1) /C

1

1

C 2RdR

dR C

1

dR

dR = 0 at R

1 =

C

2

R2 =

C

2

So R1 = R

2

Section (F) F-1. f(x) , g(x)

let h(x) = f(x) – g(x)

h(1) =2, h(2) = – 6

h(x) = 0 has atleast one root in (1, 2).

but h(x) = f(x) – g(x) < 0

h(x) is always decreasing so only one root

F-2. f(x) is continuous on [a, b], f(x) is

differentiable on (a, b), f(a) = p = f(b).

Conditions in Rolle's theorem are satisfied.

f(x) = 2

x a b

x ab

2

ab1–

x

1

a b

f(x) = 0 x = ab (a , b)

i.e f(c) = 0, c = ab

conclusion of Rolle's theorem also valid.

F-3. Let f(x) = 3x2 + px –1 f(x) = x3 + 2px

2– x + c

f(–1) = p

2+ c = f(1)

f(x) satisfies conditions in Rolle's theorem f(c) = 0 for atleast one c (–1,1)

3x2 + px –1 = 0 has atleast one root in (–1, 1).

F-4. f(x) = 0 sin x

= 0

x

= n

x = 1

n , n N

x = ......., 1

n,

1

n 1 , .......,

1

3 ,

1

2 1.

Consider interval 1 1

,n 1 n

1f

n 1

= 0 =

1f

n

By Rolle’s theorem f(x) vanishes at least once in 1 1

,n 1 n

Infinite number of such intervals are there. Hence f(x) vanishes at infinite number of points in (0, 1)

F-5. Let h(x) = f(x) g(x) h(a) = 0 = h(b)

By Rolle’s theorem on [a, b] h(x) = 0, for at least one c (a, b). f (c) g(c) + f(c) g(c) = 0

F-6. f’(c) =

f – f(0)5

– 05

; f’(c) = sec2c which is strictly increasing f’(c) =

f – f(0)5

– 05

; f’(c) = sec2c

f’(0) < f’(c) < f’5

sec20 < 2 2

tan5 sec sec

5 4

5

2tan

5 5 5

F-7. Let (x) = f(x) – 2g(x) ; x [0, 23] (x) = f'(x) – 2g(x)

Also (0) = f(0) – 2g(0) = 2 – 0 = 2

(23) = f(23) – 2g(23) = 22 – 20 = 2

Since f(x) and g(x) are differentiable in [0, 23] hence (x) is also continous in [0, 23] and differentiable in

[0, 23], so all the conditions of Rolle's theorem are satisfied. Hence there exist a number c, 0 < c < 23

for which '(x) = 0

F-8. f(a) = f(b) = 0 and we know sin3x, xex, 2

x

1 x are continuous and differentiable for x R therefore f(x) is

also continuous and differentiable in [a, b] hence by Rolle's theorem there exist a real number c [a, b]

such that f (c) = 0.

F-9. f(0) = 0 and f(6) = 2

so f(0) f(6)

f(x) is discontinuous at x = 4 and nondifferentiable at x = 1, 4 but f'(3) = 0

F-10. Let g(x) =f(x)

x , x [a, b]. By Rolle’s theorem g’(x

0) = 0 0 0 0

2

0

f (x )x 1.f(x )

x

= 0 f(x

0) = 0

0

f(x )

x

PART - II

Section (A)

A-1. f(0) = x 0im

2sinx– 0

x

x – 0 = 1 (slope of tangent)

slope of normal is – 1

Equation of normal is y – 0 = – (x – 0)

A-2. dy 1

–dx 2 x

(1, 1)

dy

dx = –

1

2

Slope of normal = 2

Equation of normal is 2x – y =1

A-3. (3,0)

dy

dx

is 4 = tan–1 4

A-4. dy

dx= 5x4

Equation of tangent at (h,k) is y k

x h

= 5 h4 Which passes through (2, 2)

(2 – k) = 5h4 (2 – h) –h5 = 10h4 – 5h5 4h5 = 10h4 h = 0, 5

2

Equation of tangents are y – 2 = 0 and 16(y – 2) = 55 (x–2)

A-5. Equation of tangent is

y – 4/h = – 4/h2 (x – h)

It passes through (0, 1) 1 – 4/h2 = 4h/h2

h – 4 = 4 h = 8 tangent is y – 1/2 = – 1/16 (x – 8).

A-6. y exy + x = 0

Differentiating w.r.t. to y

1 – exy dx dx

. y x 0dy dy

dx

dy = 0

1 – xexy = 0

xexy = 1 x = 1 , y = 0

Point is (1, 0)

A-7.

dydy d

dxdx

d

= a (– sin )

a (1 cos )

3

dy

dx

= – 3

3 = –

1

3

tan = – 1

3 = –

6

=

5

6

A-8. (3t,4 / t)

dy

dx

= 2

(3t,4 / t)

12

x

= 2

4

3t

equation of normal y 4 / t

x 3t

=

23t

4which passes through (3t1, 4/t1)

1

4

t–

4

t =

23t

4(3t1– 3t)

2

1

4 9t

tt 4

t1 =

3

16

9t

A-9. dy

dx for y = x2 +

1

x is 2x –

2

1

x

dy

dx at (h, k) is 2h –

2

1

h

equation of tangent at (h, k) is 2y (h 1/h)

x h

= 2h –

2

1

h

y = 2

12h

h

x – 2h2 +

1

h+ h2 +

1

h

y = 2

12h

h

x +

2

h– h2

2

h– h2 =

2

1

2h – 1/h

32 h

h

=

2

3

h

2h 1 h6 – 2h3 + 1 = 0

(h3 –1)2 = 0 (h –1)2 (h2 + h +1)2 = 0 h =1 equation of tangent is y = x + 1

A-10_. 1 = 1 + b + c c = – b

dy

dx = 2x + b = 2 + b at (1, 1)

equation of tangent is y –1 = (2 + b) (x –1) x intercept = b 1

b 2

y intercept = – (b + 1) 1 b 1

(b 1)2 b 2

= 2

b2 + 2b + 1 = 4b + 8 on –4b – 8

b2 – 2b – 7 = 0 on b2 + 6b + 9 = 0 integral b = – 3

Section (B)

B-1. For C1, x

(0,1)(0,1)

dya na na

dx

For C2, x

(0,1)(0,1)

dyb nb nb

dx angle between curve is

loga logb

1 logalogb

=

log(a /b)

1 logalogb

B-2. Both curves are confocal, where focus of both curves is ( 24,0) and ( 24,0)

angle between curves x2 + 4y2 = 32 and x2 – y2 = 12 is 2

B-3.

1C

dy

dx

=2 2x y

2xy

,

2C

dy

dx

= 2 2

2xy

x y

1C

dy

dx

×

2C

dy

dx

= – 1

B-4. For C1, 2

dy 4x

dx a y

and For C2,

2

dy 16

dx 3y

2 2

4x 161

a y 3y

2 3

4 16x1

3a y

a2 = 4/3

B-5. Equation of normal to the curve y2 = 8x and y2 = 4 (x–3) are y = mx –4m – 2m3 and y = m(x–3) – 2m – m3

respectively.

– 4m – 2m3 = –3m – 2m – m3 m3 – m = 0 m = 0, 1,–1

feet of common normal with slope equal to –1 on the curves y2 = 8x and y2 = 4 (x–3) are (4,2) and (2,4) respectively

Now, distance between points (4, 2) and (2, 4) is 2 2

B-6. Equation of normal at 2 2x y

32 18 = 1 at (h, k) is

2x 18y

h k

=14 which passes through

7,0

4

56

0 14h

h = 4 k = 3

Now, distance between (4, 3) and 7

,04

is

29

94

=

15

4

shortest distance between curves 2 2x y

32 18 = 1 and

227

x y 14

is

11

4

Section (C)

C-1. V = 21r h

3

r 2 1

h 4 2

V = 31 h

3 4 = 3h

12

77 × 103 =22 1

7 4 × 70 × 70 ×

dh

dt ( 1 litre = 103 c.c.)

dh

dt = 20 cm/min.

C-2 x3 = 12y

3x2 dx

dt = 12

dy

dt

dx

dt >

dy

dt

12dy

dt .

2

1

3x >

dy

dt

x2 < 4 x (–2, 2)

C-3. From figure z2 = x2 + y2

z dz

dt= x

dx

dt

If z = 500 then x = 400 500 dz

dt = 400(5)

dz

dt = 4

C-4. Let y = tan xy = tan (x + x) – tan x dy

dxx = tan (x + x) – tan x

(sec2 x)x = tan (x + x) – tan x

put x = 45°, x = 1° 2

180

= tan46° – 1 tan46°= 1 +

90

C-5. V = 34 (10 r)3

, 0 r 15

dV

dt = – 50.

4 (10 + r)2 dr

dt = – 50

dr

dt =

–1

18 (where r = 5)

Section (D)

D-1. f(x) = 3(a + 2)x2 – 6ax + 9a 0 x R

a + 2 < 0 and D 0

a < –2 and a (––3] [0, )

a (––3]

D-2. f(x) = 3x2 + 2ax + b + 5 sin 2x 0 x R

sin 2x – 1

f(x) 3x2 + 2ax + b – 5 x R 3x2 + 2ax + b – 5 0 x R

4a2 – 4. 3. (b – 5) 0

a2 – 3b + 15 0

D-3. f(x) = 2

2

1– xx 1, x 0

x

x – 1, x 1,

x

The given function is not differentiable at x = 1

f(x) = 2 3

3 2

1 2– , x 1, x 0

x x

2 1– , x 1

x x

Now f(x) < 03

3

x – 20 given x 1

x

2 – x0 when x 1

x

f(x) decreasing x (0, 1) (2, ) and f(x) increases x (–, 0) (1, 2)

here f(x) is decreasing at all points in x (, 1) (2, ) so will also be decreasing at x = 3 at x = 1 minima and at x = 2 maxima

D-4. f (x) = (2 2 + 42x2 + 62 x4 +...... + 1002 x98) x

Minimum at x = 0

D-6. (4,3) (2,3)

(6,1) (4,2)

(4,1)

(2,2)

(1,1)

1 2 4 6

Section (E) E-1. f(x) = 2 – |x + 1|

From figure it is clear that greatest, least values are respectively 2, 0

E-2. Since coefficient of x2 is (+ve)

m(b) = D

4a m(b) = –

2 2

2

(4b 4(1 b )

4(1 b )

m(b) =

2

1

1 b

2b 0 1 + b2 1 2

10 1

1 b

m(b) (0, 1]

E-3. H H – h

R r

S = 2rh

= 2H 2r

r –R

dS

dr = 2H

2r1–

R

Maximum at r = R

2

E-4. x = 4 cos, y = 3 sin

Let A be area. A = 4 (4 cos ) (3 sin ) = 24 sin 2

A is maximum when 2 =2

Dimensions are.

2 . 4 2.3,

2 2

E-5. Let A be area A = (2x)2–x(e ) , x > 0

dA

dx = – 2

1x

2

1x –

2

2–xe

At x =1

2, A is maximum. Largest area is 2

1

2 e–1/2

E-6. f(1–) f(1) and f(1+) f(1) – 2 + log2 (b2 – 2) 5

0 < b2 – 2 128 2 < b2 130

E-7. A(– 2, 3) 3 = 4a – 2b + c

B(– 1, 1) 1 = a – b + c

D(2, 7) 7 = 4a + 2b + c

y = x2 + x + 1

C (h, h2 + h + 1), – 1 < h < 2

Area = 3

2 (– h2 + h + 6)

Maximum at h = 1

2 C

1 7,

2 4

Section (F) F-1. f(x) = x3 – 6x2 + ax + b

f(x) satisfies condition in Rolle's theorem on [1, 3]

f(1) = f(3) 1 – 6 + a + b = 27 – 54 + 3a + b

2a = 22

a = 11 and b R.

F-2. f (x) = 0 x = – 2, 3

x = – 2 (–3, 0)

c = – 2

F-3. For x(0, 2)

f'(c) =f(x) – f(0)

x – 0

(Here c(0, x))

f(x) = 2.f '(x)

f(x) 1

F-4_. Let f(x) = 5 3ax bx

5 3 + cx then f (1) = 0, f(–1) = 0, f(0) = 0

there exist atleast one root of equation ax4 + bx2 + c = 0 in (–1, 0) and there exist atleast one root of

equation f’(x) = 0 in (0, 1)

there exist atleast one root of equation ax4 + bx2 + c = 0 in (–1, 0) and there exist atleast one root of

equation f’(x) = 0 in (0, 1)

But b

a

equal to negative so equation ax4 + bx2 + c = 0 has two real and two non-real roots. Hence

there exist exactly one root of equation ax4 + bx2 + c = 0 in (–1, 0) and there exist exactly one root of

equation f’(x) = 0 in (0, 1)

F-5. 3 2 2x 2x 5x 6 (x 1)(x x 6)

x 1 (x 1)

= x2 – x – 6 f(x) =

2x x 6 ; x 1

6 ; x 1

x 1lim

f(x) = –6 ; x 1lim

f(x) = – 6 LHL = RHL = f(1)

f(x) is continuous

LHD at x = 1 is 1

RHD at x = 1 is 1

f(x) is differentiable at x = 1

f(–2) = 0 ; f(3) = 0 all the conditions of Rolle’s are holding

f'(x) = 2x – 1 = 0

x = 1

2

1

2 [–2, 3]

PART - III

1. (A) y2 = 4ax dy

dx =

2a

y

y = x

2ae

dy

dx =

1

2a

x

2ae

= 1

2a

y

Product of slopes = 2a

y

y

2a

= – 1

(B) 1 | sin x | + | cos x | 2

y = [| sin x | + | cos x|] y = 1

P(2, 1)

x2 + y2 = 5

dy

dx =

x

y

tan =

2 0

1 0

cosec2 =

5

4

(C) Let y2 = 4a (x + a) ........(1) and y2 = 4b (x + b) .........(2) intersect each other at (h, k) then h = – (b + a) (h, k) h = – (b + a)

Now (h,k)

dy

dx

for curve (1) is 4a

2k and

(h,k)

dy

dx

for curve (2) is 4b

2k

4a

2k ×

4b

2k = –1

2

4ab

k= –1

4ab

4a(h a)= –1

b

h a= –1

b

b= –1

Which is always true a

b can take any value from interval] R– {0}

(D) Inverse curves touches each other at line y = x

y = x is tangent to both curves equation x = x2 + 3x + c has both equal roots

c = 1 and x = –1 = h and y = –1 = k |h + k + c| = 1

2. (A) f(x) = 2x – 3

2

x =

2 2

3

2(x – 1)(x 1)

x

No point of local maxima

(B) Given expression = (sin– 1 x)3 + (cos– 1 x)3

=

3

2

– 3sin– 1 x –1– sin x .2 2

=3

2

(sin–1x)2 –

23

4

sin–1x +

3

8

This is quadratic in sin–1x. Therefore it will give maximum value when sin–1x = –2

x = –1

(C) f’(x) = 12 (x + 2)(x + 1) (x – 1)

a = – 2, b = – 1

(D) 3 3a b

48

=

3 3a (8 a)

48

=

8

48(3a2 – 24a + 64)

Minimum 3 3a b

48

is

8

48

2(4.3.64 24 )

4.3

=

8

3

3. (A) f(x) is continuous and differentiable f(0) = f() Hence condition in Rolle’s theorem and LMVT are satisfied.

(B) f(1–) = –1, f(1) = 0, f(1+) = 1

f(x) is not continuous at x = 1, belonging to 1 3

,2 2

Hence, atleast one condition in LMVT and Rolle’s theorem is not satisfied

(C) f’(x) = 2

5 (x – 1)–3/5, x 1

At x = 1, f(x) is not differentiable. Hence at least one condition in LMVT and Rolle’s theorem is not satisfied.

(D) At x = 0

L.H.D. =–

1

x

1

x

x 0

e 1x 0

e 1limt

x 0

=

0 1

0 1

= –1

R.H.D. = 1 At x = 0, f(x) is not differentiable Hence at least one condition in LMVT and Rolle’s theorem is not satisfied.

4. (A) Let PQ = x

Then BP = 4 x

2

PS = 4 x

2

tan60º =

3 (4 x)

2

area A of rectangle = 3

2 (4 – x) x

dA

dx =

3

2 (4 – 2x) = 0 x = 2

2

2

d A

dx = – 3 < 0

A is maximum, when x = 2.

Maximum area = 3

2 2.2 = 2 3 .

Square of maximum area = 12 (B) Dimensions be x, 2x, h

72 = x. 2x. h

36 = x2h ....(1)

S = 4x2 + 6xh

S = 4x2 + 636

x

dS

dx = 8x –

2

216

x=

3 3

2

8(x 3 )

x

For least S, x = 3 and least S is 108.

(C) f(x) = x3y = x3(60 – x)

f(x) = 4x2 (45 – x)

f(x) is maximum at x = 45

(D) x2 + y2 = 5

a

2 = 5 cos, b = 5 sin

Let f() be perimeter f() = 2a + 2b = 2 5 (2cos + sin)

f()= 2 5 (– 2sin + cos)

f()= 2 5 (–2cos – sin)

f() = 0 tan 1

2 = and f () < 0 f() is greatest

a = 4, b = 1

a3 + b3 = 65

EXERCISE # 2

1. y

b= 1 –

x

a

y

b = e–x/a e–x/a = 1 –

x

a put t = –

x

a et = 1 + t

Draw graph of y = et, y = 1 + t

From graph it is clear that t = 0 is the only Solution

x = 0 y = b (0, b)

2.x3 + y3 = 8xy, y2 = 4x x3 + 8x3/2 = 8x 2x1/2

x3 – 8x3/2 = 0 x6 – 64x3 = 0

x3 = 0 or x3 = 64

x = 4, y = 4.Point of intersection (4, 4) 3x2 + 3y2 y = 8y + 8xy

y = –1

slop of normal is 1

Equation of normal is y – 4 = 1 (x – 4) x = y

3. Here x2/3 + y2/3 = a2/3

Differentiating w.r.t. to x 1/ 3 1/ 32 2 dyx y 0

3 3 dx

1/ 3dy k

(h, k)dx h

Equation (y – k) = –

1/ 3k

h

(x – h)

P(0, k1/3 (h2/3 + k2/3)) or P(0, a2/3 k1/3)

And, Q (h1/3 a2/3, 0)

PQ = 2/3 4 /3 4 /3 2/3h a a k = |a| = constant.

4. The tangent at (x1, sin x

1) is y – sin x

1 = cosx

1 (x – x

1)

It passes through the origin sin x1 = x

1 cos x

1 = x

1 2

11 sin x

y1

2 = sin2 x1 = x

12(1 – y

12) (x

1y

1) lies on the curve

y2 = x2(1 – y2) x2 – y2 = x2y2

5. y = e{x} = ex–a in x [a, a + 1) dy

dx = ex–a = e{x}

equation of tangent (Y – y) =dy

dx (X – x)

passing through (–1/2, 0) (0 – y) = e{x} (–1/2 – x)

–1 = 1

x2

x = 1

2

point 1/ 21, e

2

Number of tangent = 1

6. Let y = mx + c be tangent touching both branches. f(x) = –x2, y = mx + c , x < 0

x2 + mx + c = 0 , m > 0 ( x < 0 ) (negative roots)

D = 0 m2 = 4c f(x) = x2 + 8, y = mx + c, x > 0 x2 – mx + 8 – c = 0 , m > 0 (positive roots)

D = 0 m2 = 32 – 4c c = 4, m2 = 16 c = 4, m = 4

7. f(x) = ex & g(x) = nx are image of each other in line mirror y = x hence minimum distance between these will be equal to distance between parallel tangents of f(x) & g(x) which are parallel y = x.

ex = 1 & = 1

O

Q

P

y y=ex

y=x

y = logx

x = 0 & x = 1

P (0, 1) ; Q = (1, 0); PQ = 2

8.

shortest distance always lie along the common normal

Equation of normal at (t2, 2t) to the parabola is y + xt = 2t + t3 ..... (i) above equation passes through the center of the circle c(0, 12)

12 = 2t + t3 t3 + 2t – 12 = 0 t = 2

point is (4, 4)

9. f(x) = |x|{a sgnx}a ; g(x) =

|x|[a sgnx]a

for a > 1, a 1 and x R

n a h(x) = n f(x) + ng(x)

(n a) h(x) = |x|a sgnx na + |x|[ a sgnx] n a h(x) = |x|a sgnx + |x|[ a sgnx]

h(x) = a|x| sgn x

Now h(–x) = a|–x| sgn (–x) = –h(x) h(x) is an odd function Also graph of h(x) is

It is clear from the graph that h(x) is an increasing function

10. x > 1 f(x) f(1)

x > 1 g(x) g(1) f(g(x)) f (g(1))

h(x) 1 .... (i)

Range of h(x) is subset of [1, 10] h(x) 1 .... (ii)

By (i), (ii) we have h(x) =1 h(2) = 1

11. f(x) =

b – (a c)x , x 0

b (c – a)x , 0 x b /a

(a c)x – b , x b /a

(i) c > a0 b/a

(ii) c = a0 b/a

(iii) c < a0 b/a

12. f(x) = x xe e

2

> 0

f(x) increasing hence g(x) is also increasing function

13. f(x) = 3x2 – 3p2x + 3p2 – 3 = 3((x – p)2 – 1) = 3(x –(p + 1)) (x – (p – 1))

p – 1 > – 2 and p + 1 < 4 p > – 1 and p < 3 – 1 < p < 3

14. f(x) = 1/ x

2

x

x (1 – nx) f(x) 0, when x e

f(x) is decreasing function, when x e

> e f() < f(e)

1/ < e1/e e > e

Statement-1 is True, Statement-2 is False

15^. 2

2

x x 2

x 5x 6

< 0 x (–3, – 2)

For maximum or minimum of the function, put f’(x) = 0 a2 – 3x2 = 0 x = –a

3,

a

3

If a > 0, then point of minima is x = a

3 –3 <

a

3 < – 2 or 2 3 < a < 3 3

if a < 0, then point of minima is x = a

3 –3 <

a

3 < – 2 –3 3 < a < –2 3

16. f(x) = sinx cos x (3 sin x + 2)

f(x) = 0

sin x = 0 or cos x = 0 or sin x =–2

3

x = 0 or sin x = –2

3

(as cos x = 0 is not possible).

If = 0 then f(x) 0

no extrema,

hence 0

– 1 < –2

3

< 0 or 0 <

–2

3

< 1

0 < < 3

2 or –

3

2 < < 0

17. 2 = h2 + x2

Area of base (triangle) is 23a

4

3x = 3

2 a Volume V =

3

4 ha2 = h

3

4. 4. 3. x2 = 3 3 h (2 – h2)

dV

dh= 3 3 (2 – 3h2)

V is maximum when h =3

.

18.

Area = ab + 2 21 1a sin cos b sin cos . 2

2 2

= ab +

2 2(a b )

2

sin 2

Maximum area is ab + 2 2(a b )

2

19.

From similar triangles ABC, FBD c x

y

=

c

b

Area of AFDE = xy sin A = b

c (c – x) sin A

It is maximum when x = c

2

Maximum area = bc

4 sinA

statement -1 is true statement-2 is obvious.

20. Let f(x) = x If x 7

f(x 1) – f(x)

f '(c)x 1– x

x 1– x = 1

2 c, N2 < x < c < x + 1 ; c > N2

1 1

x 1 – x2N2 c

21. f(x) =logx

x is differentiable and continous for every x > 0. Now for RMV to be applicable f(a) = f(b)

na nb

a b ab = ba a = 2, b = 4

hence a2 + b2 = 20.

22. Let g(x) = f(x) – x2 g(x) has atleast 3 real roots which are x = 1,2,3

By langrange mean value theorem (LMVT)

g(x) has atleast 2 real roots in x (1, 3) g(x) has atleast 1 real roots in x (1, 3)

f(x) – 2 = 0 for atleast 1 real roots in x (1, 3) f(x) = 2 for atleast one x (1, 3)

PART-II

1. Parametric form of curve is x = 3t2, y = 2t3 dy

dx = t

Let P 2 3

1 13t , 2t , Q 2 3

2 23t , 2t

Conditions are

(i) P

dy

dx

Q

dy

dx

= –1

(ii) P

dy

dx = Slope of line segment PQ

t1 t

2 = –1 ...(i);

t1 =

2

3

2 2

2 1 2 1

2 1

t t t t

t t

...(ii)

3(–1 + t1

2) = 2 2 2

2 1t 1 t

t1

2 = 2

1

2

t + 1

t1

2 = 2, –1

t1

2 = 2 t1 = ± 2 t

2 =

1

2

If t1 = 2 , t

2 = –

1

2 P(6, 4 2 ), Q

3 1,

2 2

Required line is y = 2 x – 2 2

If t1 = – 2 , t

2 =

1

2 P(6, –4 2 ), Q

3 1,

2 2

Required line is y = – 2 x + 2 2

2. Any point on y = x2 + 4x + 8 is P(h,h2 + 4h + 8). p

dy

dx= 2h + 4

Tangent at P is (2h + 4) x – y = h2 – 8.

It is sufficient to have only one solution for equations y = x2 + 8x + 4,

y = (2h + 4) x + 8 – h2

x2 + (4 – 2h) x + h2 – 4 = 0 D = 0

(4 – 2h)2 – 4(h2 – 4) = 0 h = 2

8x – y + 4 = 0

coordinates of point of contact are (2, 20) and (0, 4)

3. |nx| = px

It is sufficient to find values of p for which y = |nx| and y = px has three points in common.

If y = px is passing through points O, A, B then we obtain three roots. Let us consider line y = px. When it is passing through points O, P. (tangent)

Let P (, p) (> 1) p= |n|

p= n .....(1)

dy 1

dx x

p =1

.....(2)

(1) , (2) = e

Slope of tangent (OP) is 1

e .

For three roots condition is 0 < p < 1

e.

4. y = 50 – 16t2 So, dy

32tdt

tan = y 50

x 30 x

y =

50

30 x

. x

dy

dt=

dy dx

dx dt =

2

1500 dx

dt(30 x)

dx

dt = – 16

2(375)

1500= – 1500 ft/sec.= 100 ft/sec = –15 =15

5. at t = 0 ; x = 0, y = 1

B

(0,k)

(0,1)

A

C

36

h71,h

2

dy

dt= 2cm/sec

A = 21 7h

h 12 36

2dA 1 7h h 14h dh

1dt 2 36 2 36 dt

; dA 1 14 6 6

8 3dt 2 36 (7 / 2)

(At, t = 7/2 sec, change in y -co-ordinate = 7 hence, pt. C has

y-co-ordinate = 8 and x- co-ordinate = 6 at t = 7/2 sec.) = (4 + 7) × 6

7 ×2 =

132

7 cm2/sec

6. f(x) =

2 2 2 2 2 2 2 2

2 2

x x x x x x x x

2x x

2x e e e e e e 2xe 2xe

e e

=

2 2 2 2

2 2

2 2x x x x

2x x

2x e e e e

e e

=

2 2 2

x x

8x

e e

0 x [0, ) least value of is 2.

7. Let f(x) = 2 sinx + tanx – 3x

f (x) = 2 cosx + sec2x – 3 =

2

2

cosx – 1 2cosx 1

cos x

> 0

f(x) is M.I.

x > 0

f(x) > f(0)

2 sinx + tanx > 3x

3x < 2 sinx + tan x 3x

2sinx tanx < 1 for x 0 ,

2

and 3xLt

3sinx tanxx 0 = 1

3xLt

3sinx tanxx 0

= 0

and x 0lim

3 3

5

tan – sinx x

x=

x 0lim

3 3

5

tan (1– cos )x x

x=

x 0lim

3

3

tan x

x 2

(1– cos )x

x(cos2x + cosx + 1) =

3

2

Hence x 0lim

3 3

5

tan x–sin x

x3x2

2sinx tan

= (2 + 0)3/2 = 2 2 = 2.828

8. f(x) = 2ex – ae–x + (2a – 3) x – 3

f(x) = 2ex + ae–x + (2a – 3) 0 x R a (3 – 2 )

(1 )

x x

x

e e

e

Let y = (3 – 2 )

(1 )

x x

x

e e

e Let ex = t

y = t 3 – 2

1 2

t

t

t (0, )

dy

dt =

2

(2 3)(2 – 1)–

(1 2 )

t t

t

hence maximum value of y will be at t = 1

2 ymax. =

1

2 hence minimum value of a is

1

2

9. f(x) = 4 2 2cos x 2(a 1)cosx a 2a 4

Let (t) = t2 – 2(a + 1)t + a2 + 2a – 4, –1 t 1

It is sufficient to find values of t when (t) = 0 has no root in [–1, 1]

D = 20

Case- : (–1) > 0 , a + 1 < –1, D = 20 > 0

a , 2 5 2, 5, and a < –2 a , 2 5

Case - : (1) > 0, a + 1 > 1, D = 20 > 0 a , 5 5, and a > 0 a 5,

10. Using graph of expressions on both the sides, we get only two roots.

11. – 1 p 1 Consider f(x) = 4x3 – 3x – p = 0

1

f2

· 0

f(1) 0

f(x) has at least one root between 1

, 12

Also f(x) = 12x3 – 3 > 0 1

, 12

f is increasing on 1

, 12

f(x) has only one real root between 1

, 12

To find root put x = cos

cos 3 = p = 1

3 cos–1 p

Root is cos –11cos p

3

12. Let y = 2x2 1 +

2x

200

2 1

let 2x2 = t y = t –1 +

200

1t t [1,)

dy

dt =

2

2

( 1) – 200

( 1)

t

t

=

2

( 1 10 2)( 1– 10 2)

( 1)

t t

t

hence function has minimum value at t = – 1 + 10 2 ymin. = 20 2 – 2 = 26.28

13. f(x) = (x – 1)2013 + (x – 2)2013 + ...........+ (x – 2013)2013]

f '(x) = 2013 [(x – 1)2012 + (x – 2)2012 + ...........+ (x – 2013)2012]

f '(x) >0 f(x)

f(x) (–, ) f(x) can have only one real root

Due to symmetric nature, real root is 1 2013

2

=

2014

2 = 1007

14. f(x) = a

3 x3 + (a + 2) x2 + (3a 10)

f(x) =g(x) = ax2 + 2 (a + 2) x + (a – 1) = a(x – ) (x – ) ( > )

Case - I : If a = 0 f'(x) = 4x – 10 x = 10

4 is point of minimum (which is not negative)

Case- : If a > 0 then will be point of minima which is negative hence both root of g(x) = 0

must be negative.

(0) 0

0

– / 2 0

g

D

b a

a 10 7 57

,3 2

Case-III : If a < 0 then will be point of minima which is negative hence can be negative, zero or

positive but g(0) < 0 is hence can be negative only hence again g(x) = 0 must have both negative

roots

(0) 0

0

– / 2 0

g

D

b a

a hence exhaustive set of values of a is 10 7 57

,3 2

15. f(x) = a0 + a

1x + a

2x2 + a

3x3 + a

4x4 + a

5x5 + a

6x6

x 0Lim 3

f (x)1 1

x

a

0 = a

1 = a

2 = a

3 = 0

x 0Lim

1/ x

2

3

f (x)1 e

x

x 0Lim

24 5 6(a a x a x )

e

= e2 a4 = 2

f(x) = 2x4 + a5x5 + a

6x6

f(x) = x3 (8 + 5a5x + 6a

6x2) f(1) = 0, f(2) = 0

5

12a

5 , 6

2a

3 f(x) = 2x4 5 612 2

x x5 3

16. Let y = 23 4x x

x2 + y2 – 4x + 3 = 0 (x – 2)2 + y2 = 1, center C = (2, 0)

Consider point P(5, – 4)

CP = 9 16 = 5

Maximum value of 2

2 2–3 4x – x 4 (x – 5) is (5 + 1)2 = 36.

17. Area = 36 (1 + cos) sin = 36.4. cos3 (/2) sin(/2)

Maximum occurs when tan 2

= 1

3 =

3

18. xy = 18

Area of printed space = 3

x – 1 y –2

= 3 3x 18

18 –2 2 x

Maximum when 3x 18

2 x x = 2 3 y = 3 3

19. Fuel charges per hour = k2 48 = k. 162

Fuel charges per hour =3

16 2

Charges per hour =3

16 2 + 300

Expenses of journey = (3

162 + 300)

s

where = speed s = distance

Maximum occures when 3 300

16

(ax +b

x , a, b, > 0, x > 0, has minimum when ax =

b

x)

2 = 16.100

= 40 20.

From figure we can see Rolle’stheorem is applicable for x1 2

,3 3

and f'(c)= 0 = 2 – 4c

c = 1

2a + b + c =

1 2 1 3

3 3 2 2

21. There exist atleast one c( + d) such that f'(c) = f( d) f( )

d

f'(c)

6 6,

d d

.

If d is very large then f'(c) will approach to zero. Also (f'())2 [77, 83] Because f'(x) is continuous

function so (f'(x))2 can take all values from (0, 77] (f'(x))2 can take 76 integral values in (0, 77)

PART - III

1. 2y3 = ax2 + x3 6y2 dy

dx = 2ax + 3x2

(a, a)

dy

dx =

2

2

5a

6a =

5

6

Tangent at (a, a) is 5x – 6y = – a =–a

5, =

a

6

2 + 2 = 61 2 2a a

25 36 = 61

a2 = 25.36 a = ± 30

2. x = 2 t2 + 3t – 10 = 0 t = 2, – 5

y = – 1 t2 – t – 2 = 0 t = 2, – 1

t = 2 (common value)

dy 4t – 2

dx 2t 3

.

t 2

dy 6

dx 7

= –1

6

7

= – 7

6=

–1

6

7

361

49 = –

85

6

3. Let f(x) = x + sinx f(x) = 1 + cos x

As f(x) 0 x R, f(x) is increasing Let g(x) = sec x

g (x) = sec x tan x

g (x) changes sign. g(x) is neither increasing nor decreasing.

4. f(x) = 2 – 2

1

1 x –

2

1

x 1 = 1 –

2

1

1 x + 1–

2

1

x 1 =

2

2 2

x 11– 0

1 x x 1

5. g(x) =x

2f f (1– x)2

and g(x) = f(x/2) – f(1 – x)

Now g(x) is increasing if g(x) 0

fx

2

f(1 – x)

[ f(x) < 0 i.e. f(x) is decreasing]

x

2 1 – x x 2 – 2x 3x 2 x 2/3 0 x

2

3 g(x) increases in 0 x 2/3

and g(x) 0 for decreasing x

f2

f(1 – x) x

2 1 – x x 2/3 2/3 x 1

6. f(x) =

m–n

nmx

n

m – n is odd.

f (x) < 0 x (–, 0)

f (x) > 0 x (0, )

7. Let h(x) = f(x) g(x)

h (x) = f(x) g(x) + g(x) f(x)

As f(x) < 0, g(x) 0 f(x) g(x) 0 and g(x) > 0, f(x) 0 f(x) g(x) 0

h(x) 0 h(x) is increasing.

Let x1, x

2

x1 < x

2

g(x1) < g(x

2)

f(g(x1)) > f(g(x

2))

fog(x1) > fog(x

2)

fog(x) is monotonically decreasing.

8. (x) = (3(f(x))2 – 6(f(x)) + 4)f(x) + 5 + 3 cos x – 4 sin x

5 – 9 16 5 + 3 cosx – 4 sin x 5 + 9 16

adding (3(f(x))2 – 6(f(x)) + 4)f(x)

(3(f(x))2 – 6(f(x)) + 4)f(x) (x) (3(f(x))2 – 6(f(x)) + 4)f(x) + 10

3(f(x))2 – 6f(x) + 4 = 3 (f(x) – 1)2 + 1 > 0

(3(f(x))2 – 6(f(x)) + 4)f(x) 0 when ever f(x) is increasing.

(x) 0 (x) is increasing, when ever f(x) is increasing.

If f(x) = – 11 then

(3(f(x))2 – 6f(x) + 4) f(x) + 10 = – 33 (f(x) – 1)2 – 1 < 0

(x) < 0 (x) is decreasing.

9. f(x) =

2

2

2

x p pq pr

pq x q qr

pr qr x r

= x3 + (p2 + r2 + q2) x2

f(x) = 3x2 + 2x(p2 + q2 + r2) = x {3x + 2(p2 + q2 + r2)}

0

– ++

–2

3(p + q + r )

2 2 2

Here f(x) is increasing if x < –2

3 (p2 + q2 + r2) and x > 0

decreasing is if –(p2 + q2 + r2) <x < 0

10. f(x) =2

2x xcosec

4 2

f(x) = 2x x xcosec cot

2 2 2

x xtan –

2 2

For 0 < x < 1, tan x x

2 2 f (x) > 0 f(x) is increasing.

11. (A) y = 2

4

2X 1

X

is even funciton.

Even function is nonmonotonic. [True]

(B) If range of f(x) is not R and c does not belong to range of f(x) then it is not necessary to have

one [False]

(C) f(x) = 1 > 0

f(x) is increasing

f(0), f(1) is not defined. Hence no local maxima/minima. [True]

(D) f(6) = 0 ; f(6 – h) > f(6)

& f(6 + h) > f(6)

Minima at x = 6 [False]

12. f '(x) = 2

sinx xcosx

sin x

Let g(x) = sin x – x cos x

g'(x) = cos x + x sin x – cos x

g'(x) > 0 x (0, /2)

g(x)

g(0) < g(x)

0 < sin x – x cos x

f '(x) > 0

f(x) Let = /12, = /4, r = 5/12

f() < f() < f(r)

Let H(x) = 2 (x –f()) (x – f(r)) + 3 (x – f()) (x – f(r)) + 4 (x – f()) (x – f())

H (f(r)) > 0, H(f()) < 0, H(f(r)) > 0

one root in (f(r), f()) (f(r), f())

and other in (f(), f(r)) (f(), f(r))

13. f(x) = (x – 1)n – 1 (x + 1)n – 1 [2(n + 1)x3 + (2n + 1)x2 + 2(n – 1)x – 1]

At x = 1 2(n + 1)x3 + (2n + 1)x2 + 2 (n – 1)x – 1 0

for n N

n – 1 must be odd

n is even

14. f(x) = 2

2

sec x(cosx x) (cosx – x)

(1 x tanx)

The only factor in f(x) which changes sign is cosx – x.

Let us consider graph of y = cos x and y = x

It is clear from figure that for x (0, x0), cos x – x > 0 and for x 0x ,

2

cos x – x < 0, f(x) has maxima at x0

15. f(x) = a

x + 2bx + 1

f(– 1) = 0

– a – 2b + 1 = 0

a + 2b = 1

f(2) = 0

a

x+ 4b + 1 = 0 a + 8b + 2 = 0

– 6b = 3 b = –1

2 , a = 2

16. From graph f–1 (x) =2

–x ; x –1

1– x ; –1 x 0

Maximum of f(x) exist at x = 1 Minimum of f–1(x) exist at x = – 1

17. f(x) =2

1 1 1–

2 x1 x , x > 0 =

2

2

–(x – 1)0

2x(1 x )

x > 0.

f(x) is decreasing x > 0.

On 1

, 33

, greatest value is f1

3

= 1

–6 2

n

1

3

and least value is f( 3 ) =3

–

1

2n. 3

18. f(x) =4 3 2 2

– 40.12x(x 3)(x – 1)

(3x 8x – 18x 60)

f(x) = 0

at x = 0, x = – 3, x = 1

so at x = 0, f(x) has local minima.

and at x = –3, x = 1; f(x) has local maxima

f(1) =40

53, f(– 3) =

– 40

75. f(–3) < 0, f(1) > 0 and f(x) 0

f(x) is undefined at point(s) in (–3, 1). Hence f(x) has no absolute maxima.

19. f(x) = x 2

x 3

; x 1 , –3 f’(x) =

2

5

(x 3)

20. (2x)2 + 2.2x . 3y + (3y)2 + (y –1) (y –3) = 0 (2x + 3y)2 + (y –1) (y –3) = 0

x R

So D 0

144y2 – 16 (10y2 – 4y + 3) 0

16 [–y2 + 4y – 3] 0

y2 – 4y + 3 0

(y – 1) (y – 3) 0

1 y 3

So ymax

= 3 and ymin

= 1

21. f(0) = 0 f(1)

there will be no x (0, ) ( Rolle’s theorem is not applicable)

for which f’(x) = 0 i.e, cot–12

xx

1 x

f”(x) =

2 22

2 2 2 22 2

1 x – 2x–1 –1 x – 1–

1 x 1 x1 x x 1

f’’(x) = 2 2

–20

(x 1)

f’(x) is strictly decreasing xlim

f’(x) = –1

2x

–xlim cot x 0

1 x

f(0+) = –1

2x 0

xlim cot x –

21 x

2f x – f(x)

f '(c)2 /

c 0,2

(LMVT is applicable)

f '(c)2

2 2

f x – f(x)2

2

f x – f(x) 1

f’(x) 0; f(x) is increasing

f(x)[f(0), f())

f(0) = 0

–1

–1

x x x 0

cot xlim f(x) lim xcot x lim

1/ x = 2

2x

–1lim (–x ) 1

1 x

f(x) [0, 1) f(x) = sec x will have no solution

22^. f(4) = f(5) = f(6) = f(7) = 0

By Rolle's theorem on interval

[4, 5], [5, 6], [6, 7] we have

f(x) = 0 for at least once in each intervals (4, 5), (5, 6), (6, 7).

23. (A) let f(x) = tan–1x

f’(x) = 2

1

1 x

|f(c)| = 2

1

1 c < 1

1 1tan x tan x

x y

< 1

(B) Let f(x) = x100 + sin x – 1

f(x) = 100x99 + cosx > 0, x [0, 1]

f(x) is increasing.

(C) Suppose f(x) = ax3 – 2bx2 + cx, then clearly f(0) = 0

and f(1) = a – 2b + c = 0,

f(0) = f(1)

By Rolle’s theorem f(x) = 3ax2 – 4bx + c = 0

for atleast one x in (0, 1) which is positive

(D) f(x) = 3 tanx + x3 – 2

f(x) = 3sec2x + 3x2 > 0

f(x) is always increasing

24. Apply Roll's theorem on f(x), g(x) = ex f(x) and h(x) = e–x f(x)

25. (A) Let x x + h and y x

|tan–1x – tan–1y| |x – y|

|tan–1(x + h) – tan–1x| |h|

1d(tan x)

dx

1

2

1

1 x 1 hence true

(C) |sin x – sin y| |x – y|

x x + h y x

sin(x h) sinx

h

1

|cos x| 1 hence true

Alternative solutions

For x = y, this is true

Let x, y R and x y

consider f(t) = tan–1t, t [x, y]

Using LMVT, 1 1tan y tan x

y x

=

2

1

1 c, c (x, y)

tan–1 y – tan–1x = 2

y x

1 c

y – x ........(i)

similarly x > y, tan–1 x – tan–1y x – y ........(ii)

From (i) and (ii) we get 1 1tan x tan y |x – y|

Similarly considering g(t) = sin t in [x, y] we get siny sinx

y x

= cos c

sin y – sin x = (cos c) (y – x) y – x ........(iii)

and sin x – sin y x – y ........(iv)

(iii), (iv) |sin x – siny| |x – y|

26. |f(c)| 2 |f(x) – f(a)| 2|x – a| |f(x) – f(a)| < 2

a is that point on the interval when f(a) = f(2) – f(1)

2 – 1 = 0 as f(1) = f(2)

27_. (A) Apply L. M. V. T. in (a, a + 6) we get f' (c) = f(a 6) – f(a)

6

Now f(a + 6) – f(a) [–6, 6] f'(c) [–1, 1] (f'(c))2 1

(B) Apply L. M. V. T. in (–d, 0) where d is positive then f'(c) = f(0) f( d)

d

f' (c)

6 6,

d d

If d is approaching to infinity then f'(c) approaches to zero. And (f'(0))2 [76, 85], Also f' (x) is

continuous function. So (f'(x))2 can take all real value from (0, 76]

(C) By using option (A) we get there exist a value x0 (–6, 0) such that (f'(x))2 1

also f(x0)2 9 (f'(x0))2 + (f(x0))2 10

But f(x))2 + (f'(x)2 is continuous function and (f(0)2 + (f'(0)2 = 85 hence there exists [x0, 0]

such that (f'())2 + (f())2 = 10 there exist (– 6, 0) such that (f'())2 + (f())2 = 10.

(D) Similarly by using option (C) we say that there exists (0, 6) such that (f())2 + (f'())2 = 10

Assume H(x) = (f(x))2 + (f'(x))2 where H(0) = 85

Let p is largest negative number lies between (–6, 0) such that H(p) = 10

Similarly let q is smallest positive number lies between (0, 6) such that H(q) = 10

Apply Rolle's in H (x) in interval (p, q) we get there exists r in (p, q) such that H'(r) = 0

2f'(r) (f(r) + f" (r)) = 0

Because in (p, q), (f(x))2 9 & H(x) 10

(f'(x))2 1 f'(r) 0 f(r) + f"(r) = 0 |f(r)| = |f"(r)|

(E) There exist one c (–3, 3) such that f'(c) = )3(–f–)3(f

[–1,1] and f'(0)2 [76, 85]

let f'(0) 76 then f'(x) is mix up of increasing as well as decreasing function

f'"(c) < 0 for some c (–3, 3) f'(c) f'"(c) < 0

PART - IV

1^. Here, m = x 0

dy

dx

dy

dx = 3x2 + 6x + 4 m = 4

and, k = y(0) k = – 1

= 2

1| k | 1

m =

1| ( 1) | 1

16 =

17

4

2. |yy'| = |y/y'| at (0, 1) (y') at (0, 1) equal ± 1 (pepx + p)(0,1) = ± 1 2p = ± 1 p = ± 1/2

3. Length of subnormal = dy

ydx

=

3cos4

3sin4

2 sin4

= 3

2

3

2 =

9

2

(4 to 6)

Let g(x) = x sinx

2

, x [0, ]. g(x) is increasing function of x.

range of g(x) is 0 ,2

f(x) = x sinx

2

, x [0, ]

Now let t 2, then f(t) + f(2 – t) =

i.e f(t) +2 t sin(2 t)

2

=

i.e f(t) + – t

2 –

sin t

2 =

i.e f(t) = t sin t

2

f(x) = x sinx

2

for x 2

Thus f(x) = x sinx

2

for 0 x 2

Also f(x) = f(4 – x) for all x [2, 4] f(x) is symmetric about x = 2

from graph of f(x)

= 2 – 0 = 2

=

Maximum value is f(2) = = 2

7. y = (x – 1)3 (x – 2)2

dy

dx = 3(x – 1)2 (x – 2)2 + 2(x – 2) (x – 1)3 = (x – 1)2 (x – 2) [3(x – 2) + 2(x – 1)] = (x – 1)2 (x – 2) (5x – 8)

(x2 – 2x + 1) (5x2 – 18x + 16)

2

2

d y

dx = (2x – 2) (5x2 – 18x + 16) + (10x – 18) (x2 – 2x + 1) = 0 = 20x3 – 42x2 + 11x – 50 = 0

= 10x3 – 42x2 + 57x – 25 = 0

(x – 1) (10x2 – 32x + 25) = 0

x = 1 or x = 32 24

20

no. of points of inflections = 3

8. f(x) = x4 + ax3 + 23x

12

f(x) = 4x3 + 3ax2 + 3x

f(x) = 12x2 + 6ax + 3

Now, f(x) will be concave upward along the entire real line iff

f(x) 0 x R

12x2 + 6ax + 3 > 0 D 0

36a2 – 144 0

a2 – 4 0 a [– 2, 2]

9. sin x is concave downward in (0, ) and sin x is concave upward in (, 2)

10. ex, 2x, tan–1x (If x R–) is concave upward and nx is concave downward

11. f(x) concave downward (f(x) < 0)

f(x) increasing (f(x) 0)

Let g(x) = f–1(x) = x

f(g(x)) = x

f(g(x)). gx = 1

g(x) =1

f (g(x)) > 0

g(x) = – 2

1

(f (g(x)) × f(x).g(x)

g(x) > 0

g(x) = f–1 (x) concave upward

EXERCISE # 3

PART - I

1. f(x) = 2010 (x – 2009) (x – 2010)2 (x – 2011)3 (x – 2012)4

f(x) = n (g(x))

g(x) = ef(x)

g(x) = ef(x) . f(x) only point of maxima [Applying first derivative test]

2. Clearly f(x) = 2xe +

2–xe

f(x) = 2x (2xe –

2–xe ) 0 increasing fmax

= f(1) = e + 1

e

g(x) = x2xe +

2–xe g'(x) = 2xe + 2x2

2xe – 2x 2–xe > 0 increasing

gmax

= g(1) = e + 1

e

h(x) = x2 2xe +

2–xe h(x) = 2x2xe + 2x3

2xe – 2x2–xe = 2x

2 2 2x 2 x xe x e e > 0

hmax

= h(1) = e +1

e, so a = b = c

3. (A) Re 2 2

2i (x iy)

1 (x y 2xyi)

= Re

2 2

2y 2ix

1 x y 2xyi)

= Re

2y 2ix

2y (y ix)

= Re (–1/y) =

1

y

= –1 y 1 = 1

y

1 or

1

y

– 1

Alternate

Re i

2i

2ie

1 e

= Re

2i(cos isin )

1 (cos2 isin2 )

= Re2

2i(cos isin )

2sin 2isin cos )

= Re

i (cos isin )

sin (sin icos )

= Re

(cos isin )

sin (cos isin )

= Re 1

sin

as –1 sin 1 (–, 0) (0, )

(B) –1 x 2

2x 2

8.3

1 3

1 –1

2

8t

9 t 1

–1 2

8t

9 t 1 0

2

2

9 t 8t

9 t

2

8t

9 t – 1 0

0 2

2

t 8t 9

t 9

2

2

8t 9 t

9 t

0 0

(t 9) (t 1)

(t 3) (t 3)

(t 9) (t 1)

(t 3) (t 3)

0

t (– , –9] [–1 , 1] [9, ) x (– , 0) [2 , )

t (– , –9] [–1 , 1] [9, ) x (– , 0) [2 , )

(C) f() = 2 sec2 f() 2 f() [2, )

(D) f(x) = x3/2 (3x – 10) f ’(x) = x3/2 3 +3

2 x1/2 (3x –10)

asf ’(x) 0 1/ 2 3x 3x (3x 10)

2

0 3x +

9x

2 – 15 0

15x

2 – 15 0 x 2 x [2, )

4. f(x) = x4 – 4x3 + 12x2 + x – 1

f(x) = 4x3 – 12x2 + 24x + 1

f(x) = 12x2 – 24x + 24 = 12 (x2 – 2x + 2) > 0 x R

f(x) is S.I. function

Let is a real root of the eqution f(x) = 0

f(x) is MD for x (– , ) and M.I. for x (, ) where < 0

f(0) = – 1 and < 0 f() is also negative

f(x) = 0 has two real & distinct roots.

5. p = (x – 1) (x – 3) = (x2 – 4x + 3)

p(x) = (x3/3 –2x2 + 3x) +

p(1) = 6

6 = (1/3 – 2 + 3) +

6 = (1/3 + 1) +

18 = 4 + 3 ...(i)

p(3) = 2

2 = (27/3 – 2 × 9 + 9) +

2 =

= 2 = 3

p(x) = 3(x – 1) (x – 3)

p(0) = 3(–1)(–3) = 9

6. f(x) = |x| + |x2 – 1|

f(x) =

2

2

2

2

–x x – 1 x –1

–x – x 1 –1 x 0

x – x 1 0 x 1

x x – 1 x 1

f(x) =

2

2

2

2

x – x – 1 x –1

–x – x 1 –1 x 0

–x x 1 0 x 1

x x – 1 x 1

7. x2 = x sinx + cos x

f(x) = x2

g(x) = x sin x + cos x

g(x) = sin x + x cos x – sin x

g(x) = x cos x

Only two solution.

8*.

a

a

aa

aa

a

a a

a a

Let = 8x, b = 15 x

Volume = (8x – 2a) (15x – 2a) (a) = 4a3 – 46a2x + 120 ax2

dV

da = 6a2 – 46ax + 60 x2

at x 5

dV

da

= 0

x = 3 and 5

6

2

2

d V

da = 6a – 23x

2

2

at a 5 & x 3

d V

da

< 0,

So, at x = 3 gives maxima 2

25

at a 5 & x6

d V

da

> 0

So, at x = 5

6 gives minima.

dV

da = 0 when a = 5 given

( 4a2 = 100 given for maximum volume)

at a = 5

bydV

da= 0

6x2 – 23x + 15 = 0

x = 3 or 5/6

So by x = 3 (for max volume)

8x = 24, 15x = 45

9. Point of intersection of tangents at P and Q is R(2 sec, 0)

Area of PQR = 1

2 · 2 3 sin · (2 sec – 2 cos )

= 2 3 · 3sin

cos

; where cos

1 1,

4 2

Nowd

d

2 3

2

2 3 cos .3sin cos sin ( sin )

cos

> 0

As increases, increases when cos decreases, increases

min.

occurs at cos = 1/2, Therefore 2 = 2 3 .

3 / 2

1 1/ 4

1/ 2

= 4 3 .

3 3

8=

36

8

max.

occurs at cos = 1/4, Therefore 1 = 2 3 .

3 / 2

1 1/16

1/ 4

15. 15

4.4.4=

2 3.15. 3 5

16

1 =

45

85

Now 8

5

1 – 8

2 = 45 – 36 = 9

10*. f(x) = 2|x| + |x + 2| – | |x + 2| – 2 |x| |

=

–2x – 4 x –2

2x 4 –2 x –2 / 3

–4x –2 / 3 x 0

4x 0 x 2

2x 4 x 2

Graph of y = f(x) is minima at x = –2, 0; maxima at x = –2/3

11. f(x) – 2f(x) + f(x) ex

f(x)·e–x – f(x)e–x – f(x)e–x + f(x)e–x 1

d

dx (f(x)e–x) –

d

dx (f(x) · e–x) 1

d

dx (f(x) e–x – f(x) e–x) 1

2

2

d

dx (e–xf(x)) 1 x [0, 1]

Let (x) = e–x f(x)

(x) is concave upward f(0) = f(1) = 0

= 8 3 .

(0) = 0 = (1) (x) < 0 f(x) < 0

12. (x) < 0, x (0, 1/4) and (x) > 0, x (1/4, 1) e–x f(x) – e–x f(x) < 0, x (0, 1/4)

f(x) < f(x), 0 < x < 1/4

13. tangent at Fyt = x + 4t2

a : x = 0 y = 4t (0, 4t)

(4t2, 8t) satisfies the line

8t = 4mt2 + 3

4mt2 – 8t + 3 = 0

Area =

2

0 3 11

0 4t 12

4t 8t 1

= 1

2 (4t2 (3 – 4t)) = 2t2 (3 – 4t)

A = 2[3t2 – 4t3] dA

dt = 2[6t – 12t2] = 24 t(1 – 2t)

t = 1/2 maxima

G(0, 4t) G(0, 2)

y1 = 2

(x0, y

0) = (4t2, 8t) = (1, 4)

y0 = 4

Area = 3 1

24 2

=

3 2 12

4 2

14*. f(x) = x5 – 5x + a = 0

x5 – 5x = –a

x(x4 – 5) = –a

x(x2 – 5 )(x2 + 5 ) = –a

x(x – 51/4)(x + 51/4)(x2 + 5 ) = –a ..(1)

f'(x) = 5x4 – 5 = 0

(x2 –1)(x2 + 1) = 0

(x – 1)(x +1)(x2 + 1) = 0

–1 1

+–+

(0, 4)

51/410

–1

–51/4

(0, –4)

15. (y – x5)2 = x(1 + x2)2

2(y – x5) 4dy5x

dx

= (1 + x2)2 + 2x (1 + x2) 2x

at point (1, 3)

2(3 – 1) dy

5dx

= 4 + 8

dy

5dx

= 12

4 = 3

dy

dx = 8

16. Volume of material V = r2h

V1 = (r + 2)22 + (r + 2)2h – r2h V

1 = 2(r + 2)2 + h(4 + 4r)

V1 = 2(r + 2)2 + 4h(r + 1) V

1 = 2 2

2

2(r 1)V(r 2)

r

1

2 3

dV 2v 1 22 2(r 2) 0

dr r r

3

2V 2 1024 0

10

3

24v24

10

v = 103

v4

250

17*. Let h(x) = f(x) – 3g(x)

h( 1) 3

h(0) 3

h'(x) = 0 has atleast one root in (–1, 0) and atleast one root in (0, 2)

h(2) = 3

But since h"(x) = 0 has no root in (–1, 0) & (0, 2) therefore h'(x) = 0 has exactly 1 root in (–1, 0) &

exactly 1 root in (0, 2)

18. x 2

f(x) g(x)lim 1

f '(x) g'(x)

x 2

f(x) g(x)lim

f '(x) g'(x)

0

0

Indeterminant form as f'(2) = 0, g(2) = 0

Using L.H.

x 2

f '(x) g(x) g'(x)f(x)lim

f "(x) g'(x) g"(x)f '(x)

=

f '(2) g(2) g'(2)f(2)

f "(2) g'(2) g"(2)f '(2)

=

g'(2)f(2)1

f "(2)g'(2) f"(2) = f(2)

and f'(2) = 0 & range of f(x) (0, ) so f"(2) = f(2) = +ve

so f(x) has point of minima at x = 2

and f(2) = f"(2) so f(x) = f"(x) have atleast one solution in x R

(19 to 21)

f(x) = x + nx – xnx

f '(x) = x

11 nx – x

x

1=

x

1 –nx

f "(x) = 0x

1

x

12

x (0, )

f'(x) is strictly decreasing function for x (0, )

)x('fLim

)x('fLim

0x

x f'(x) = 0 has only one real root in (0, )

f '(1) = 1 > 0

f '(e) = 01e

1 f '(x) = 0 has one root in (1, e)

Let f'() = 0, where (1, e)

+ + – –

1 e

f(x) is increasing in () and decreasing in (, )

f(1) = 1 and f(e2) = e2 + 2 – 2e2 = 2 – e2 < 0

f(x) = 0 has one root in (1, e2)

From column 1 : I and II are correct.

From column 2 : ii, iii, and iv are correct.

From column 3 : P, Q, S are correct

22. f(x) > 0 for all x R, f(1/2) = 1/2, f(1) = 1

f(x) increases

Let g(x) = f(x) – x , x [1/2,1]

Then g(x) = 0 has atleast one real root in (1/2, 1)

f(x) = 1 has atleast one real root in (1/2, 1)

Hence f(x) increases f(1) > 1

23. f'(x) – 2f(x) > 0

2xdf(x).e 0

dx

g(x) = f(x).e–2x is an increasing function. for x > 0 , g(x) > g(0)

f(x).e–2x > 1

f(x) > e2x

Now f'(x) > 2f(x) > 2.e2x

f(x) is an increasing function

24. f(x) =

xcosxsinxsin

xsinxcosxcos

x2sinx2cosx2cos

= cos2x – cos2x (–cos2x + sin2x) + sin2x (–2sinxcosx)

f(x)= cos4x + cos2x

f(x) = 2cos22x + cos2x –1

Let cos2x = t

f(x) = 2t2 + t –1 and t [–1, 1]

f(x) attains its minima at t = 4

1 [–1, 1]

f(x), t = 4

1 [–1, 1]

min

)x(f = 8

91

4

1

16

2

max

)x(f = 2 + 1 –1 = 2……..(when cos2x = 1)

f ' (x) = –4sin4x –2sin2x

f ' (x) = 0 4sin4x + 2sin2x = 0

8sin 2x cos 2x + 2sin 2x = 0 2sin 2x (4cos 2x + 1) = 0

sin 2x = 0 or cos 2x = 4

1

–

–

y= cos2x y= sin2x

Hence option (B), (C)

25. f2(0) + (f(0))2 = 85 f : R [–2, 2]

(A) This is true of every continuous function

(B) f(c) = f( 4) f(0)

4 0

|f(c)| = f( 4) f(0)

4

–2 f(–4) 2

–2 f(0) 2

___________

–4 f(-4) – f(0) 4

This |f(c)| 1

(C) xlim

f(x) = 1

Note f(x) should have a bound which can be concluded by considering

f(x) = 2 sin 85 x

2

f(x) = 85 cos 85 x

2

f2(0) + (f(0)2) = 85

and xlim f(x)

does not exist

(D) Consider H(x) = f2(x) + (f(x)2

H(0) = 85

By (B) choice there exists some x0 such that (f(x0))2 1 for some x0 in (-4, 0)

hence H(x0) = f2(x0) + (f(x0))2 4 + 1

H(x0) 5

Hence let p (-4, 0) for which H(p) = 5

(note that we have considered p as largest such negative number)

similarly let q be smallest positive number (0, 4) such that H(q) = 5

Hence By Rolle's theorem is (p, q)

H(c) = 0 for some c (–4, 4) and since H(x) is greater than 5 as we move from x = p

to x = q and f2(x) 4 (f(x))2 1 in (p, q) Thus H(c) = 0 ff + ff = 0

so f + f = 0 and f 0

26. f(x) =

3x3

10x)2xln()2x(

3x13

8x7x4x

3

2

1x01xx

0x1x3x10x10x5x

23

2

2345

f'(x) =

3x)2xln(

3x17x8x2

1x01x2

0x2)1x(5

2

4

x5 + 5x4 + 10x3 + 10x2 + 3x + 1 takes value between – to 1

Also (x – 2) ln(x – 2) – x + 3

10 takes value between

3

1 to

So, range of f(x) is R. So option (A) is correct f"(1–) = 2 and f"(1+) = – 4

so f'(x) is non-diff at x = 1 so option (B) is correct f'(x) has local maxima at x = 1 so option (C) is correct

27. f'(x) = 4x

xtan2

xxcosx2

2

1 1

2

3 2 5/2 0 y1 x1

For f'(x)

y1 x1 y2 x2

– + – + –

–

Min Max Min Max

PART - II

1. x 1lim

f(x) = 1

f(–1) = k + 2

x ( 1)lim

f(x) = k + 2

f has a local minimum at x = – 1

f(–1+) f(–1) f(–1–)

1 k + 2 k + 2

k – 1

possible value of k is – 1

Hence correct option is (3)

2. ex + 2e–x 2 2 (AM GM)

x x

1

e 2e

1

2 2

1

2 2 f(x) > 0 so statement- 2 is correct

As f(x) is continuous and 1

3 belongs to range

10,

2 2

of f(x),

f(c) =1

3 for some C.

Hence correction option is (4).

3. y = x + 2

4

x

y = 1 – 3

8

x = 0 x3 = 8 x = 2

y = 2 + 2

4

2 = 3

(2, 3) is point of contact Thus y = 3 is tangent Hence correct option is (3)

4. f(x) =

tanxx 0

x

1 x 0

In right neighbourhood of ‘0’ tan x > x

tanx

1x

In left neighbourhood of ‘0’ tan x < x

tanx

1x

as (x < 0)

at x = 0, f(x) = 1

x = 0 is point of minima so statement 1 is true. statement 2 obvious

5. y – x = 1 y2 = x

2y dy

dx = 1

dy

dx =

1

2y = 1

y = 1

2

x = 1

4

tangent at 1 1

,4 2

1

2 y =

1

2

1x

4

y = x +

1

4 y – x =

1

4

distance =

11

4

2

= 3

4 2 =

3 2

8

6. V = 4

3 r3 4500 =

34 r

3

dV

dt = 4r2

dr

dt

45 × 25 × 3 = r3

r = 15 m

after 49 min = (4500 – 49.72)= 972 m3

972 = r3

r3 = 3 × 243 = 3× 35 r = 9

72 = 4 × 9 × 9 dr

dt

dr

dt=

2

9

r3 = 3×243 = 3× 35 r = 9

72 = 4 × 9 × 9 dr

dt

dr

dt =

2

9

7. f '(x) =1

x + 2bx + a

at x = – 1 –1 – 2b + a = 0 a – 2b = 1 ...(i)

at x = 2 1

2 + 4b + a = 0

a + 4b = 1

–2

...(ii)

On solving (i) and (ii) a = 1

2, b =

1–

4

f '(x) = 1 x 1

–x 2 2

= 22 – x x

2x

=

–(x 1)(x – 2)

2x

So maxima at x = – 1, 2

8. f(x) = 2x3 + 3x + k

f(x) = 6x2 + 3 > 0 x R

f(x) is strictly increasing function

f(x) = 0 has only one real root, so two roots are not possible

9. Consider f(x) – 2g(x) = h(x)

Then, h(x) is continuous and differentiable in [0, 1]

Also h(0) = 2 & h(1) = 2

Hence h(x) satisfies conditions of Rolles Theorem in (0, 1)

Thus, There exist a 'c' such that h'(c) = 0 where c (0, 1)

f'(c) = 2g'(c)

10. f(x) = 2n x x x

(1). f ‘(x) = 2 x 1x

=

22 x x

x

Since x = –1, 2 are extreme points f ‘(x) = 0 at these points.

Hence 2 – 1 + = 0

8 + 2 + = 0

– 6 – 3 = 0 = – 1

2 & = 2.

11. 4x + 2r = 2 ….(i)

x2 + r2 = minimum So f(r) = 22

r2

r1

0r222

r

dr

df 2

r = 4

1

using equation (i) x = 2

)r1( x = 2r

12. at x = 6

y =

3

f(x) = tan–1

x xcos sin

2 2x x

cos sin2 2

x 0,2

= tan–1 x

tan4 2

f(x) = x

4 2

f'(x) =

1

2

slope of normal = –2

equation of normal y – 3

= –2 x

6

y = –2x + 2

3

13. 2r + = 20 2r + r = 20 = 20 – 2r

r A =

2r

360

=

2r

2.20 – 2r

r = r (10–r)

A = 10r – r2 dA

dr = 10 – 2r = 0 r = 5

= 10

5 = 2

Maximum area = 1

2 × 25 × 2 = 25 sq. m.

14. y(x –2) (x–3) = x + 6

Intersection with y-axis; Put x = 0 y = 1 Point of Intersection is (0, 1)

Now, y =6x5x

6x2

y' =

22

2

6x5x

5x2)(6x6x5x

y' =

36

306 = 1 at (0, 1)

Equation of normal is given by (y –1) = –1 (x – 0) x + y –1 = 0

15.

r

r 2

let radius of circle be r, its center lies on y-axis as y-axis bisects the 2 rays of y = |x|

Now 4 – r2r 12412

4r

NOTE : The correct solution should be

C

P(t,4–t2)

r

y=4–x2

due to symmetry center of the circle must be on y-axis let center be (0, k)

Length of perpendicular from (0, k) to y = x, i.e. r = 2

k

Equation of circle : x2 + (y – k)2 = 2

k2

solving circle and parabola, 4 – y + y2 – 2ky + 2

k2

= 0

y2 – (2k + 1) y +

4

2

k2

= 0

Because circle touches the parabola D = 0

(2k + 1)2 = 4

4

2

k2

4k2 + 4k + 1 = 2k2 + 16

On solving we get k = 4

1364

Therefore radius = 2/k 1.3546

However among the given choices the following method will yield one of the choice.

16. y2 = 6x and 9x2 + by2 = 16

2y dy

dx = 6

dy

dx =

3

y

18x + 2by dy

dx = 0

9x + by dy

dx = 0

by

x9

dx

dy

3 –9x

1y by

(b) 6x = 27x

b = 27

6 b =

9

2

17. f(x) x2 + ,x

12

g(x) = x = x

1 h (x) =

x

1x

2x

1x

x

1x

x

1x

)x(g

)x(f

2

2

2

x – x

1= t

h(t) = t

2t

t

2t2

|t| 2 AM GM . t

2.t

2

t

2t

t + 22t

2

18.

A(4, –4)

C(t2 , 2t)

B(9 , 6)

ACB = 1

2

2

4 4 1

9 6 1

t 2t 1

= 30 + 5t – 5t2

d

dt

= 0 5 – 10t = 0 t =

1

2

2

2

d

dt

= – 10 < 0

C1

,14

so = 30 + 5 5

2 4 =

4

131

19. y = 7 + x3/2

Let the point on curve be 3 / 2

1 1P x ,7 x and given point be A 1

,72

For nearest point normal at P passes through A

So slope of line AP = Slope of normal at P

1 1

3 / 2

1

x ,y 11

x dx 2

1 dy 3 xx2

2

1 13x 1 2x 2

1 13x 2x 1 0 (x1 + 1) (3x1 – 1) = 0

x1 = 1

3 (x1 = – 1 is not possible as x1 > 0) Hence point P is

1 1,7

3 3 3

So AP = 1 1 1 7

36 27 6 3

20. f(x) = )xa(

xa

xxa

22

22

222

–22

22

222

)xd(b

)xd(b

)xd()xd(b

=

2/322

2

2/322

2

)xd(b

b

)xa(

a

Hence f(x) is increasing.

21. f"(x) > 0, y = f(x) ; x (0, 2)

(x) = f(x) + f(2 – x)

'(x) = f'(x) – f'(2 – x)

for (x) to be increasing

'(x) > 0 f'(x) > f'(2 – x)

x > 2 – x (f'(x) is increasing in (0, 2))

x > 1

x (1, 2)

For (x) to be decreasing

'(x) < 0 f'(x) < f'(2 – x)

x (0, 1)

22.

Q

P(2 + 2, )

(2,0)

Q

Shortest distance between y2 = x – 2 and y = x

dx

dy at point P will be 1. Differentiating the curve

2yy' = 1 y' = 12

1

y2

1

P

2

1,

4

9

minimum distance = PQ = 24

7

2

2

1

4

9

23. f(x) = x 2kx – x

f'(x) = 2

2

k – 2x xkx – x

2 kx – x =

2

22

x–kx2

x2–kx)x–kx(2 =

2

2

3kx – 4x

2 kx – x =

2

x 3k – 4x

2 kx – x

for increasing function

for f'(x) 0 x 0,3

kx – x2 0, x 0,3 and x (3k – 4x) 0, x 0,3

x(x – k) 0, x 0,3 and x (4x – 3k) 0, x 0,3

k 3 and k 4 k 4 m = 4

maximum(f(x) when k = 4 is 324 3 – 3 =3 3 = M

(m, M) = (4, 3 3 )

24. 3dv5m / min

dt

21v r h

3

h

r

r 1

tanh 2

2r = h 3 31 h h

v3 4 12

2dv dhh

dt 4 dt

5 = 2 dh

104 dt

dh 20 1

dt 100 5

m/min

25. f(x) = ax5 +bx4 + cx3 4

x

cxbxax2lim

3

345

0x

2 + c = 4 c = 2

f'(x) = 5ax4 + 4bx3 + 6x2 = x2 6bx4ax5 2

f'(1) = 0 5a + 4b + 6 = 0

f'(–1) = 0 5a – 4b + 6 = 0

b = 0

a = – 5

6 f(x) = 35 x2x

5

6

f'(x) = –6x4 + 6x2 = 6x2 (–x2 +1) = – 6x2 (x+1) (x–1)

–1 + 1–

1– 1

Minimal at x = –1 Maxima at x = 1

26. f(x) = x ( – cos–1 (sin|x|)) =

|x|sinsin

2x 1

= x

|x|

2

f(x) =

0xx2

x

0xx2

x

f(x) =

0xx22

0xx22

f(x) is increasing in

2,0 and decreasing in

0,

2

27. f(3) = f(4) = 12 f'(x) = 12xx

12–x2

2

f'(c) = 0

c = 12

f"(c) = 12

1

28. Lets use LMVT for x [a,c]a–c

)a(f–)c(f = f'() , (a,c)

also use LMVT for x [c,b] c–b

)c(f–)b(f = f'() , (c,b)

f ''(x) < 0 f '(x) is decreasing f '() > f '() a–c

)a(f–)c(f >

c–b

)c(f–)b(f

)c(f–)b(f

)a(f–)c(f >

c–b

a–c

( f(x) is increasing)

HIGH LEVEL PROBLEMS (HLP) 1. Given x = f'(t) sint + f"(t) cost

y = f'(t) cost - f"(t) sint

from given equation dx

dt = f t f t cost &

dy

dt = – f t f t sin t

Velocity =

2 2dx dy

dt dt

=

2 22 2f t f t cos t f t f t sin t = f(t) + f(t)

2. f '(x) = 24ax – x + 2

x(4a – 2x)

2 4ax – x =

2

2

6ax 2x

4ax x

< 0, x (4a, 3a)

so f(x) is decreasing in [4a, 3a]

3. Here f is a differentiable function then f is continuous function. So by L.M.V. theorem for any a (0, 4)

f(a) =f(4) f(0)

4 0

....... (1)

Again from mean value for any b (0, 4)

f(b) = f(4) f(0)

2

.......(2)

Now multiplying (1) and (2), we get 2 2f (4) f (0)

8

= f(a) · f(b) f2(4) – f2(0) = 8f (a) · f(b)

4. f’(x) = 0 x =1

a,

2

3a

since, we have a cubic polynomial with coefficient of x3 +ve , minima will occur after maxima. Case - 1 : If a > 0

then 1

a=

1

3 a = 3 also f

1

3

> 0 b < – 1

2

Case - 2 : If a < 0

then – 2

3a

=

1

3 a = – 2

also f 1

3

> 0 2

2

( 2)

3

–

( 2)

2

.

2

1

3– 2

1

3

– b > 0 4 1 2

b 027 9 3

b < 11

27

5. f(k) = 3

f(k+h) = a2 – 2 + sin h

h

h 0lim

f(k+h) = a2 – 1

h 0lim

f(k – h) = h 0lim

(3 + |k–h–k|) = h 0lim

(3 + |–h|) = 3 a2 – 1 > 3

a2 > 4 | a | > 2 6. f(x) = x3 – 3x + k, k = [a]

f(x) = 3(x – 1) (x + 1) – 1 is maxima is 1 is minima

for three roots f(–1) f(1) < 0 (k + 2) (k – 2) < 0

k (–2, 2) – 2 < [a] < 2 – 1 a < 2

7. f(x) = sin{x}

a + cos

{x}

a a > 0

it attains max. if {x}

a =

4

{x}

a

10,

a

1

a >

4

, for f to have is maxima 0 < a <

4

8. Let f(x) = x4 + 4x3 – 8x2 + k

f’(x) = 4x3 + 12x2 – 16 x = 4x (x2 + 3x – 4) = 4x (x + 4) (x – 1) f’(x) = 0 x = – 4 , 0, 1

f’’ (x) = 12x2 + 24x – 16 = 4(3x2 + 6x – 4)

f’’ (–4) = 20 > 0

f’’(0) = – 16 < 0

f’’(1) = 20 > 0 x = – 4 and x = 1 are points of local minima whereas

x = 0 is point of local maxima

for f(x) = 0 to have 4 real roots

f(–4) < 0 k < 128

f(0) > 0 k > 0

f(1) < 0 k < 3 k (0, 3)

(09 to 11) Graph of y = f(x)

9.

Three points of intersection. Three solutions

10.

11.

12. (i) f( 1) f(0)

1 0

= f(), where – 1 < < 0

|f())| = |f(0) – f(1)| |f(0)| + |f(–1)| |f()| 1 + 1 = 2

similarly for 0 < < 1

(ii)

2

2

| f(x) | 1 (f(x)) 1F(x) 5

| f (x) | 2 f (x)) 4

(iii) Obvious from (i) and (ii) that there exits atleast one max.

(iv) Also from (i) and (ii) option iv is quite obvious.

13. As (a, b) lies on y = x2 + 1 b = a2 + 1

(a, b)

dy

dx= 2a

Tangent y – a2 – 1 = 2a (x – a)

x = 0 y = 1 – a2

x = 1 y = – a2 + 2a + 1

Area = 1

2 (1) (1 – a2 – a2 + 2a + 1) = – a2 + a + 1

It is greatest when a =1

2 b = 1 +

1

4=

5

4.

14. (2, – 1) – 1 = 2a b

( 1) (–2)

2a + b = 2

y= 2 2

a(x – 1)(x – 4) – (ax b)(2x – 5)

(x – 1) (x – 4)

y = 0 at x = 2 b = 0 a = 1

y = x

(x – 1) (x – 4)

y = 2 2

(2 x)(2 – x)

(x – 1) (x – 4)

At x = 2, ychanges sign from positive to negative x = 2 is point of maxima.

15. Since = s s a s b s c = 12

s s a s b s c

Taking logarithm of both sides, we get ln = 1

lns ln s a ln s b ln s c2

1 d

dc

=

d s a d s b d s c1 1 ds 1 1 1. . .

2 s dc s a dc s b dc s c dc

.....................(1)

But s = 1

a b c2

ds

dc=

1

2

d s a ds da 1 10 ,

dc dc dc 2 2

and d s b ds db 1 1

0dc dc dc 2 2

and

d s c ds 1 11 1

dc dc 2 2

Now from (1), 1 d

.dc

=

1 1 1 1 1 1 1 1 1

. . . .2 s 2 s a 2 s b 2 s c 2

=

1 1 1 1 1

4 s s a s b s c

Hence d = 1 1 1 1

dc4 s s a s b s c

16. 3 – x2 > | x – a |

Case (i) a < 0 and y = x – a is tangent of y = 3 – x2 (see figure)

(0, 3)

P

a3 3

– 2x = 1x = – 1

2 P

1 11,

2 4

Since y = x – a passes through 1 11

,2 4

a = x – y = – 11 1

4 2

= –

13

4(minimum value of a)

(0, 3)

3 3 a

Case (ii) a > 0 and y = – x + a passes through (0, 3), then a = 3 (maximum value of a) (see figure)

a 13

, 34

17. Let f(x) = xm + a1 xm – 1 + a

2 xm – 1 + ...... + a

0 if possible, let f(x) = 0 has 'm' real roots, then by Roll's

thearem, f '(x) = 0 must have "(m – 1)" real roots, f ''(x) = 0 must have "(m – 2)" real roots and so on,

fm – 2(x) = 0 must have 2 real roots, m!

2 x2 + a

1 (m – 1)! x + a

2 (m – 2)! = 0 must have 2 real roots

or m (m – 1)

2 x2 + a

1 (m – 1) + a

2 = 0 must have 2 real roots

D = a1

2 (m – 1)2 – 2m (m – 1) a2 = (m – 1) [(m – 1) a1

2 – 2a2]

which is –ve, so our allumption is wrong. Hence proved.

18.

Slope of OQ > slope of OP –1x f (x)

f(x) x f(x). f–1 (x) < x2

19. g'(x) = 2f '2x

2

. 2x

2+ f ' 227

x2

(–2x) = x

22x 27

f ' f ' x2 2

g'(x) = 0

x = 0 or 2x 27

2 2 – x2

x = –3, 0, 3

for g'(x) – + – +

–3 0 3

so g(x) is increasing in x (– , –3] and in [0, 3] and g(x) is decreasing in [–3, 0] and in [3,)

20. Let f(x) = nx f "(x) = 2

1–

x

So 1 2 n 1 2 nf(x ) f(x ) ...... f(x ) (x x .... x )f

n n

for x1, x

2, .......... , x

n R+

1 2 nn (x ) n (x ) .... n (x )

n

1 2 nx x ...... xn

n

1

n1 2 n(x x ........x ) 1 2 nx x ..... x

n

G.M. A.M.

Again 1 2 n 1 2 n

1 1 1 1 1 1f f ..... f .....

x x x x x xf

n n

1

n

1 2 n

1

x x .x

1 2 n

1 1 1.....

x x x

n

1 2 n

n

1 1 1....

x x x

(x1 x

2 ...... x

n)1/n G.M. A.M.

21. (i) 1 + x2 > (x sin x + cosx)

Let f(x) = 1 + x2 – x sinx – cosx, x [0, )

f(x) = 2x – sinx – x cos x + sin x = x(2 – cosx)

f(x) > 0 for x (0, )

f(x) is an increasing function

x > 0

f(x) > f(0)

1 + x2 > x sinx + cosx

(ii) f(x) = sin x – sin 2x – 2x

f(x) = cos x – 2 cos 2x – 2 = cos x – 2(2 cos2 x – 1) – 2

= cos x – 4 cos2x = cosx (1 – 4 cosx), x 0,3

cos x 1

2cos x(1 – 4 cos x) < 0

f(x) < 0 x 0,3

f(x) f(0) sinx – sin2x – 2x 0

sinx – sin2x 2x

(iii) f(x) = 2x

2 + 2x + 3 – 3ex + xex

f(x) = x + 2 – 3ex + ex + xex = x + 2 – 2ex + xex

f(x) = 1 – 2ex + ex + xex = 1 – ex + xex

f(x) = – ex + ex + xex = xex

f(x) 0 x 0

f(x) f(0) f(x) 0

f(x) f(0) f(x) 0

f(x) f(0) f(x) > 0

2x

2 + 2x + 3 3ex – xex

(iv) f(x) = x sin x – 2sin x

2 f(x) = x cos x + sin x – sin x cosx = x cos x + sin x (1 – cos x)

f(x) > 0 for x 0,2

f(x) > f(0) or x sin x –2sin x

2 > 0

and f(x) < f 2

, x sin x – 2sin x

2 <

2

–

1

2

x sinx – 2sin x

2 <

1

2 (– 1)

22. f(x) = 221 4b b

1b 1

x3 + 5x + 6

f(x) = 221 4b b

3 1b 1

x2 + 5

f(x) is increasing f(x) 0 x R

1 – 221 4b b

b 1

0

(b 7) (3 b)

b 1

1

Case- If b + 1 > 0, then 2

(b 7) (3 b)

(b 1)

1 and – 7 b 3

b – 5 and b 2 b [2, 3]

Case- If b + 1 < 0 b [–7, –1)

23. y = x n x – 2x

2+

1

2

y = 1 + n x – x

y = 1

x – 1 =

1 x

x

y > 0 x (0, 1)

y(x) < y(1)

y(x) < 0

y(x) > y(1)

x log x – 2x

2 +

1

2 > 0

x log x >2x

2 +

1

2

24. f’(x) = 0 x = a

3b

f a

3b

= 2a

3

a

3b

f a

3b

= 2a

3

a

3b

f(–1) = b – a f(1) = a – b

Given that 2a a

3 3b=

2a a

3 3b = | b – a | = | a – b | = 1

34a

27b= 1 b =

34a

27 a – b = 1 a –

34a

27 = 1 4a3 – 27a + 27 = 0

a = –3, 3

2 a = – 3 – 1 : b =

31

2 = –

1

2= – 4

Also, b – a = 1 34a

27 – a 4a3 – 27a – 27 = 0

(a – 3) (2a + 3)2 = 0

a = 3 b = 4

Rejecting – ve values, therefore a = 3, b = 4

25. Let f(x) = sinx

x

f’(x) = 2

xcosx sinx

x

=

2

cosx(x tanx)

x

< 0 x 0,

2

; ( tanx > x)

f’’(x) = 2

3

x sinx 2xcosx 2sinx

x

Let g(x) = – x2sinx – 2xcosx + 2sin x

g’(x) = – x2cosx < 0 x (0, /2)

for x > 0 , we have g(x) < g(0) i.e. g (x) < 0

f’(x) < 0 and f’’(x) < 0 x 0,2

f A B C

3

> f(A) f(B) f(C)

3

A B Csin

3

A B C

3

>

sinA sinB sinC

A B C

3

sin A sin B sin C

A B C <

9 3

2

26. Area (ABCD) = Area of ADB + Area of BDC

A = 1

2 ps sin +

1

2 qr sin

dA

dB =

1

2 ps(+cos) +

1

2 qrcos

d

d

= 0

d

d

=

ps

qr

cos

cos

BD2 = p2 + s2 – 2pscos = q2 + r2 – 2qr cos

Differentiating we get – 2ps (–sin) = – 2qr (– sin) d

d

d

d

=

ps

qr

sin

sin

– ps

qr

cos

cos

=

ps

qr

sin

sin

sin cos + cos sin = 0 sin( + ) = 0 + =

Also , dA

d =

1

2 ps

sin( )

sin

= 0 + =

If + < then dA

d > 0

If + > , then dA

d < 0

By 1st derivative test A has maxima when + = A, B, C, D are concyclic

27. Let u = 2x + 2–x

u3 = 8x + 8–x + 3(2x) (2–x) (2x + 2–x) u3 – 3u = 8x + 8–x

also, 4x + 4–x = u2 – 2 f(x) = u3 – 3u – 4 (u2 – 2) = u3 – 4u2 – 3u + 8

Let g(u) = u3 – 4u2 – 3u + 8 ; u > 2

g’(u) = 3u2 – 8u – 3 = (3u + 1) (u – 3)

putting g’(u) = 0 ; we get u = 3

g’’ (u) = 6u – 8 g’’(3) = 1 > 0

u = 3 is point of minima g(3) = 27 – 36 – 9 + 8 = –10 minimum f(x) = – 10

28. Let f(x) = e

e

log (2x 1)

log x

for x > 1

Now f’(x) = e e

2

e

(2x 1) log (2x 1) 2xlog x

x(2x 1) {log (2x 1)}

Let g(x) = (2x – 1) loge (2x – 1) – 2xlog

ex

g’(x) = 2loge (2x – 1) – 2log

ex + 2 – 2 = 2 log

e

12

x

> 0 for x > 1

for x > 1 , we have g(x) > g(1) g(x) > 0 f’(x) > 0 for x > 1 f(x) is increasing for x > 1

4 > 3 > 2 f(4) > f(3) > f(2) e

e

log 7

log 4> e

e

log 5

log 3> e

e

log 3

log 2.

29. = 1

2 bc sin =

1

2 c2 sin ( b = c)

In ABE, using cosine rule, 2 = c2 + 2b

4 – bc cos =

25c

4– c2 cos

c2 = 24

5 4cos

= 22. sin

5 4cos

d

d

=

2

2

2 .((5 4cos )cos sin (4sin ))

(5 4sin )

=

2

2

2 .(5cos 4)

(5 4sin )

For to be maximum, cos = 0.8

30. y = 1 – x2

Consider point P (x0, 1 – 2

0x ) 0 < x0 1

equation of tangent at P is y – (1 – 2

0x ) = – 2x0 (x – x

0)

intersection with x-axis at A 2

00

0

(1 x )x , 0

2x

intersection with y-axis at B(0, 2 2

0x + (1 – 2

0x ) area of OAB =1

2

2 2

0

0

(x 1)

2x

=

1

4

2 2

0

0

(x 1)

x

0

dA

dx =

2

0

2

0

(x 1)

4x

[3 2

0x – 1] = 0 x0 =

1

3

0

dA

dx changes sign from +ve to – ve

at x0 =

1

3 So point of minimum A

min =

4 3

9

31. 3 = h2 + r2

r2 = 3 – h2

V =1

3 r2 h =

1

3 (3 – h2) h

dV

dh =

1

3 (3 – 3h2)

dV

dh = 0 at h = 1

2

2

d V

dh < 0 at h = 1

Vmax

= 2

3

32. V = k a

a

V will be minimum when a

+

a

will be minimum

A.M. G.M.

a

a

2

a

a

a

a

2

minimum of a

a

= 2

Vmin

= K 2 which is independent of a.

33. P (R cos, Rsin) a(R cos)2 + 2b(Rcos) (Rsin) + a(R sin)2 = c

R2 [a cos2 + b sin2 + a sin2] = c R2 = c

a bsin2

for minimum distance sin2 = 1

R2 = c

a b R =

c

a b

34. Let (x) = x + 21 x

(x) = 1 + 2

x

1 x

If x < 0, –|x| = x

(x) = 2

2

1 x | x |

1 x

=

2 2

2

1 x x

1 x

> 0

If x > 0, (x) > 0

Hence (x) is increasing

As we know ex x + 1 (ex) (x + 1)

ex + 2x1 e x + 1 + 21 (x 1)

35. Let x > –1

Consider f(x) = (1 + x)n(1 + x) – tan–1 x

f(x) = n(1 + x) + 1 – 2

1

1 x

f(x) = 2 2 4

2 2

(1 x) 3x x

(1 x) (1 x )

> 0 f(x) is increasing

For x < 0, f(x) < f(0) f(x) < 0 f(x) is decreasing f(x) > f(0) f(x) > 0

(1 + x)n(1 + x) – tan–1 x > 0

n(1 + x) > 1tan x

x 1

For x > 0, f(x) > f(0) f(x) > 0 f(x) is increasing f(x) > f(0) f(x) > 0 n(1 + x) > 1tan x

x 1

Hence larger of these is n(1 + x).

36. cos x (0, 1) f(cos x) < 0, f(cos x) > 0

g(x) = sin x cos x f (sinx) f (cosx)

sinx cosx

.

Consider (t) = f (t)

t

, t (0, 1)

(t) = 2

f (t)t f (t)

t

f(sin x) > 0 f(t) > 0

f(sin x) < 0 f(t) < 0 (t) > 0 t (0, 1)

(t) is increasing. For x 0,4

cos x > sin x

(cos x) > (sin x) f (cosx)

cosx

>

f (sinx)

sinx

g(x) < 0 g(x) is decreasing. Similarly, for x ,4 2

g(x) is increasing

37. Let P(x) = x2013 – x2012 – 1007x2 + 1007x + k

as P(x) is a polynomial function in x hence it is everywhere continous and differentiable

also P(0) = 0 + k = k ; P(10071/2011) = k

hence by Rolle's theorem P'(x) = f(x) = 0

for atleast one real value of 'x' in given interval.

39. Clearly g(x) satisfies condition in LMVT g(5) g(0)

5 0

= g(c), c (0, 5)

f(5) f(0)

6 1

5

= g(c) – 5

6 = g(c)

39. Let two consecutive zero of f(x) be a and b f(a) = 0 = f(b).

If possible, suppose g(x) has no zero. Define (x) = f(x)

g(x)

(x) satisfies conditions in Rolle’s theorem, (c) = 0 for at least one c (a, b)

f(c) g(c) – f(c) g(c) = 0

Which is a contradiction to given condition f(x) g(x) f(x) g(x)

Hence our supposition that g(x) has no zero is wrong g(x) has at least one zero.

40. Let f(x) = (x + a) – (x) f(x) = (x + a) – (x) + k

f(0) = (a) – (0) + k

f(2a) = (3a) – (2a) + k f(0) = f(2a)

By Rolle's theorem on [0, 2a], f(c) = 0 for at least one c (0, 2a)

(x + a) = (x) has at least one root in (0, 2a)

41. f(x) = 8ax – a sin 6x – 7x – sin 5x

f(x) = 8a – 6a cos 6x – 7 – 5cos 5x= 8a – 7 – 6a cos 6x – 5cos 5x

f(x) is an increasing function

f(x) 0

8a – 7 6a + 5

2a 12

a 6

a (6, )

42. g'(x) = h'(x)

h(x) g"(x) =

2

2

h(x).h"(x) (h'(x))

(h(x))

g"(x) < 0

g(x) is concave down on J.

43. f(x) = 2ax + 2(a + 2)

Point of inflection is x = – a 2

a

– a 2

a

< 0 a (–, –2) (0, )

44. Period of y = 2sin5

6

x is

2

56

=

12

5 & y = (x – 1)(x – 2) + 1 which is a quadratic. So for given

information (1, 1) is the common point to two curves and the possible graph would the

2

1

0 1

5

65

12

can not be nagative must be positive for these twp graphs to touch each other at (1, 1)

(Which can be the only possible point of contact) dy

dx of both curves must be same.

y = 2sin5

6

x

dy

dx =

52cos x

6

.5

6

at x 1

dy

dx

= 5 5

2cos .6 6

= 5

6

.cos

6

= – 5 3

.3 2

...(1)

y = x2 – 3x + 2 + 1 dy

dx = 2x – 3

at x 1

dy

dx

= 2 –3 = – ....(2)

for (1) & (2) = 5

2 3

3

5

=

1

2

45. Let d be distance between (k, 0) and any point (x, y) on curve.

d = 2 2(k – x) y

d = 2 2–x 2(1– k)x k ( y2 = 2x – 2x2).

Maximum d = 2 2

4(–1)k – 4(1– k)

4(–1) Maximum d = 22k – 2k 1

46_. f(x) = (x + 1)(px2 + (q – p)x + p)

g(x) = px2 + (q – p)x + p

g(2) = 3p + 2q < 0

g(3) = 7p + 3q > 0

one root is –1, one root lies between (2, 3), one root lies between

2

1,

3

1

47_.

c1 c2 x=0 x=1 x=2

1. A particle moving on a curve has the position at time t given by x = f'(t) sin t + f''(t) cos t, y = f'(t) cos t -

f''(t) sin t, where f is a thrice differentiable function. Then prove that the velocity of the particle at time t

is f'(t) + f'''(t).

2. Find the interval in which f(x) = x 24ax x (a < 0) is decreasing

3. f : [0, 4] R is a differentiable function. Then prove that for some a, b (0, 4) , f2(4) – f2(0) = 8f(a) .

f(b)

4. If all the extreme value of function f(x) = a2x3 – a

2x2 – 2x – b are positive and the minimum is at the

point x0 =

1

3 then show that when a = –2 b <

11

27

and when a = 3 b < –

1

2

5. If f(x) = 2

3 | x k | , x k

sin (x k)a 2 , x k

x k

has minimum at x = k , then show that |a| > 2

6. The equation x3 – 3x + [a] = 0, where [.] denotes the greatest integer function, will have three real and

distinct roots then find the set of all posible values of a.

7. Let f(x) = sin {x}

a + cos

{x}

a where a > 0 and { . } denotes the fractional part function. Then find the set

of values of 'a' for which f can attain its maximum values.

8. Find the values of the parameter ‘k’ for which the equation x4 + 4x3 – 8x2 + k = 0 has all roots real.

Comprehension (Q. No. 9 to 11)

A function f(x) having the following properties;

(i) f(x) is continuous except at x = 3

(ii) f(x) is differentiable except at x = – 2 and x = 3

(iii) f(0) = 0, x 3lim

f(x) – , xlim

f(x) = 3, xlim

f(x) = 0

(iv) f (x) > 0 x (–, – 2) (3, ) and f (x) 0 x (– 2, 3)

(v) f (x) > 0 x (– , – 2) (– 2, 0) and f (x) 0 x (0, 3) (3, )

then answer the following questions

9. Find the Maximum possible number of solutions of f(x) = | x |

10. Show that graph of function y = f (– | x |) is continuous but not differentiable at two points, if f (0) = 0

11. Show that f(x) + 3x = 0 has five solutions if f (0) > – 3 and f(– 2) > 6

12. Let F(x) = (f(x))2 + (f(x))2, F(0) = 7, where f(x) is thrice differentiable function such that

|f(x)| 1 x [–1, 1], then prove the followings.

(i) there is atleast one point in each of the intervals (–1, 0) and (0, 1) where |f(x)| 2

(ii) there is atleast one point in each of the intervals (–1, 0) and (0, 1) where F(x) 5

(iii) there exits atleast one maxima of F(x) in (–1, 1)

(iv) for some c (–1, 1), F(c) 7, F(c) = 0 and F(c) 0

13. A figure is bounded by the curves, y = x2 + 1, y = 0, x = 0 and x = 1. At what point (a, b), a tangent

should be drawn to the curve y = x2 + 1 for it to cut off a trapezium of the greatest area from the figure.

14. If y = ax b

(x 1) (x 4)

has a turning value at (2, 1) find a and b, show that the turning value is a

maximum.

15. With the usual meaning for a, b, c and s, if be the area of a triangle, prove that the error in resulting

from a small error in the measurement of c, is given by

1 1 1 1

d dc4 s s a s b s c

16. Find the possible values of 'a' such that the inequality 3 – x2 > |x – a| has atleast one negative solution

17. If (m – 1) a1

2 – 2m a2 < 0, then prove that xm + a

1 xm – 1 + a

2 xm – 2 + ..... + a

m–1 x + a

0 = 0 has at least one

non real root (a1, a

2, ....., a

m R)

18. If f '(x) > 0, f''(x) > 0 x (0, 1) and f(0) = 0, f(1) = 1, then prove that f(x) f–1 (x) < x2 x (0, 1)

19. Find the interval of increasing and decreasing for the function g(x) = 2f2x

2

+ f 227x

2

, where

f "(x) < 0 for all x R.

20. Using calculus prove that H.M G.M. A.M for positive real numbers.

21. Prove the following inequalities

(i) 1 + x2 > (x sinx + cosx) for x [0, ).

(ii) sin x – sin 2x 2x for all x 0,3

(iii) 2x

2 + 2x + 3 (3 – x)ex for all x 0

(iv) 0 < x sinx – 2sin x

2 <

1

2 ( – 1) for 0 < x <

2

22. Find the interval to which b may belong so that the function f(x) = 221 4b b

1b 1

x3 + 5x + 6 is

increasing at every point of its domain.

23. If 0 < x < 1 prove that y = x n x –2x

2 +

1

2 is a function such that

2

2

d y

dx> 0. Deduce that

x n x >2x

2 –

1

2 .

24. Find positive real numbers ‘a’ and ‘b’ such that f(x) = ax – bx3 has four extrema on [–1 , 1] at each of

which | f(x) | = 1

25. For any acute angled ABC , find the maximum value of sinA

A +

sinB

B +

sin C

C

26. Suppose p,q,r,s are fixed real numbers such that a quadrilateral can be formed with sides p,q,r,s in

clockwise order. Prove that the vertices of the quadrilateral of maximum area lie on a circle .

27. Find the minimum value of f(x) = 8x + 8– x – 4(4x + 4– x), x R

28. Using calculus , prove that log23 > log

35 > log

47.

29. Find the cosine of the angle at the vertex of an isosceles triangle having the greatest area for the given

constant length of the median drawn to its lateral side .

30. A tangent to the curve y = 1 x2 is drawn so that the abscissa x

0 of the point of tangency belongs to the

interval (0, 1]. The tangent at x0 meets the xaxis and yaxis at A & B respectively. Then find the

minimum area of the triangle OAB, where O is the origin

31. A cone is made from a circular sheet of radius 3 by cutting out a sector and keeping the cut edges of

the remaining piece together. Then find the maximum volume attainable for the cone

32. Suppose velocity of waves of wave length in the Atlantic ocean is k a

a

, where k and a

are constants. Show that minimum velocity attained by the waves is independent of the constant a.

33. Find the minimum distance of origin from the curve ax2 + 2bxy + ay2 = c where a > b > c > 0

34. Prove that ex + 2x1 e (1 + x) + 22 2x x x R

35. Find which of the two is larger n (1 + x) or 1tan x

1 x

.

36. Let f (sinx) < 0 and f (sin x) > 0, x 0,2

and g(x) = f(sin x) + f(cos x), then find the intervals of

monotonicity of g(x).

37. If f(x) = (2013)x2012 – (2012)x2011 – 2014x + 1007, then show that for x [0, 10071/2011], f(x) = 0 has at

least one real root.

39. A function f is differentiable in the interval 0 x 5 such that f(0) = 4 & f(5) = – 1. If g(x) =f(x)

x 1 , then

prove that there exists some c (0, 5) such that g(c) = –5

6.

39. Let f(x) and g(x) be differentiable functions having no common zeros so that f(x) g(x) f(x) g(x). Prove

that between any two zeros of f(x), there exist atleast one zero of g(x).

40. If (x) is a differentiable function x R and a R+ such that (0) = (2a), (a) = (3a) and (0) (a)

then show that there is at least one root of equation (x + a) = (x) in (0, 2a)

41. Find the set of values of the parameter 'a' for which the function ;

f(x) = 8ax – a sin 6x – 7x – sin 5x increases & has no critical points for all x R, is

42. Let h be a twice differentiable positive function on an open interval J. Let

g(x) = n (h(x)) x J

Suppose (h'(x))2 > h''(x) h(x) for each x J. Then prove that g is concave downward on J.

43. If the complete set of value(s) of 'a' for which the function f (x) = 3a x

3 + (a + 2) x2 + (a 1) x + 2

possess a negative point of inflection is (, ) (, ), then || + || is :

44. If two curves y = 2sin5

6

x and y = x2 – 3x + 2 + 1 touch each other at some point then the value

of 3

5

is

180 x

5

45. The maximum distance of the point (k, 0) from the curve 2x2 + y2 – 2x = 0 is equal to

46_. Let f(x) = px3 + qx2 + qx + p; where 3p + 2q < 0, 7p + 3q > 0 then prove that

(i) Equation 3px2 + 2qx + q = 0 has at least one root lie between (–1, 1/2)

(ii) Equation 3px2 + 2qx + q = 0 has at least one root lie between (1/3, 3)

(iii) Equation 3px4 + 4qx3 + 6qx2 + 12px = 0 has two root lie in (–1, 1)

(iv) Equation 3px4 + 4qx3 + 6qx2 + 12px = 0 has one root less then –1 and one root greater than 1

47_. Let f, g : [0, 1] R are differentiable functions with f(0) g(0). f and g has exactly one point of local

minimum in (0, 1) and have same minimum value but different point of minimum. h(x) = f(x) – g(x).

f({x}) and g({x}) are differentiable function x R {where {.} denotes G..F.}. Prove that

(i) (f({x})) = 0 has atleast 3 solution in (0, 2)

(ii) (h({x})) = 0 has atleast 3 solution in (0, 2)

2. [4a, 3a] 6. a [–1, 2) 7. 4

0,

8. k [0,3] 9. 3 13. 1 5

,2 4

14. a = 1, b = 0 16. a 13

, 34

18. If f '(x) > 0, f''(x) > 0 x (0, 1) and f(0) = 0, f(1) = 1, then prove that f(x) f–1 (x) < x2 x (0, 1)

19. g(x) is increasing if x (–, 3] [0, 3]

g(x) is decreasing if x [–3, 0] [3, )

22. [– 7, – 1) [2, 3] 24. a = 3 , b = 4 25. 9 3

2

27. – 10 29. cos A = 0.8 30. 4 3

9

31. 2/3 33. c

a b 35. n (1 + x)

36. Increasing when x ,4 2

, decreasing when x 0,4

.

41. a (6, ) 43. 2 44. 1/2

45.

1kk

0kk1

)1,0[k1k2k2 2