KEY ANSWER TO MODEL QUESTION PAPER – 4 OF KSEEB

KEY ANSWER TO MODEL QUESTION PAPER – 4 OF KSEEB

Yakub S., GHS NADA, Belthangady Taluk, DK., Ph:9008983286 Email:[email protected]

1. d) a = (b x q) + r 2. a) C.V. = 휎

푋x100

3. c) 퐴푌퐴퐶

4. Confusing 5. a) 퐴퐵

퐵퐶

6. b) equilateral triangle 7. a) 6

36

8. c) 1

II.

9. n(AUB) = n(A) + n(B) 10. 200 = n2 + 4

n2 = 196 n = 14

11. n = 8 + 12 = 20 12. x2+ 5x -14

x2 +7x -2x -14 x(x+7)-2(x+7) (x+7)(x-2) Zeros of Polynomial are x= -7 and x = 2 .

13. d = 푥2 + 푦2 d = 122 + (−5)2 d = √144 + 25 d = √169 d = 13푢푛푖푡푠

14. A = 3휋푟2 A = 3x22

7x7

A = 66 sq.cm.

mailto:Email:[email protected]



III.

15. Let us assume that √3 +√2 is an irrational numbers Let √3 +√2 = 푝

푞 ,

(√3 +√2)2 = 푝2

푞2

(√3)2 + (√2)2 + 2x√3x√2 = 푝2

푞2

5 + 2√6 = 푝2

푞2

2 √6 = 푝2

푞2 – 5

√6 = 푝2−5푞2

2푞2

Which is a contradiction as the right hand side is a rational number while √6 is an irrational number.

∴ √3 +√2 is an irrational number

16. A – { Speek Kannada} B – { Speek english} n(AUB) = n(A) + n(B) – n(A∩B) n(AUB)= 100 +50 – 25 = 125 The people who speek kannada only: n(A) – n(A∩B) = 100 -25 =75

17. ಎ) Permutation

) Combination

) Permutation

) Combination

18. nC2 – n = 14 푛(푛−1)

2 - n = 14

푛2−푛−2푛2

= 14

푛2 − 3푛 = 28 푛(푛 − 3) = 28 7(7-3) =28 ∴ n = 7

19. P(A) = 푛(퐴)푛(푆)




S = Picking 2marbles from 20 marbles n(S) = 20C2 = 20x19

2 = 190

A – Both Marbles are red n(A) = 6C2 = 6x5

2 = 15

∴ P(A) = 15190

20. √24 x √33

21 4 x 3

13 = 2

312 x 3

412

√2312 x √3412 √8x8112 √64812

21. √5+ 33−√5

x3+ √53+ √5

(√5+ 3)(3+ 5)

9−5

3√5+ 9+5+3 5)4

14+6√5

4

2(7+3 5)4

(7+3 5)

2

x2 + 3x - 8

22. x+1 x3 + 4x2 - 5x + 6 x3 + x2

3x2 - 5x 3x2 + 3x

- 8x + 6 - 8x - 8 14

p(x) = x3 + 4x2 - 5x + 6 q(x) .g(x) + r = (x2 + 3x – 8). (x+1) + 14 q(x) .g(x) + r = x3 + 3x2 – 8x + x2 + 3x – 8 +14 = x3 + 4x2 – 5x + 6 ∴ p(x) = q(x) .g(x) + r Or f(3) = a(3)3 + 3(3)2 -13 f(3) = 2(3)3 – 4(3) + a = 27a + 27 – 13 = 54 – 12 + a = 27a + 14 = 42 + a 27a - 14 = 42 + a




26a = 56 a = 56

26 ⇒ 28

13

23. 2y2 +6y = 3

푦 =−6 ± 62 − 4.2. (−3)

2.2

푦 =−6 ± √36 + 24

4

푦 =−6 ± √60

4

푦 =−6 ± √4.15

4

푦 =−6 ± 2√15

4

푦 =2(−3 ± 15)

4

풚 = −ퟑ±√ퟏퟓퟐ

24. In ∆ PQR , ∠PQR = 900 QS⟘PR,PQ = 1.5cm,QR =2cm ∴ PR2 = PQ2 + QR2

PR2 = (1.5)2 + 22

PR2 = 2.25 + 4=6.25 ∴ PR 2.5cm PQ2 = PR.PS , (1.5)2 = PS(2.5) ⇒ PS = 2.25

2.5 = 0.9cm

∴ QS2 = PS.RS ⇒ 0.9 x 1.6= 1.44 ∴ QS = 1.2cm

25. 1+sin휃1−sin휃

= 1+sin휃

cos휃1−sin휃

cos휃 =

1cos휃+ sin휃

cos휃1

cos휃−sin휃cos휃

= sec 휃+ tan휃sec휃−tan 휃

sec휃+ tan휃sec휃−tan 휃

xsec휃+ tan 휃sec휃+ tan 휃

(sec휃+ tan 휃)2

sec2 휃− tan2 휃 but 1 + tan2 휃 = sec2 휃 ⇒ 1 = sec2 휃 − tan2 휃

∴ ퟏ+퐬퐢퐧휽ퟏ−퐬퐢퐧휽

= (퐬퐞퐜휽 + 퐭퐚퐧휽)ퟐ

26. y = mx + c




m = tan 600 = √3 c = -2 ∴ y = √ퟑx -2

27.

28. Curved Surface area A = 휋푟푙 = 22

7x 7x25

= 550 sq.m = Rs 50 per 1 sq.m 550 x 50 = 27500 OR 1. Curved Surface area of Cylinder = 2휋푟ℎ A = 2x 22

7x52.5 x 3

A = 990 sq.m 2. Curved surface area of Cone = 휋푟푙 A = 22

7x 52.5x 53

A = 8745 sq.m.




∴ Canvas required = 990 + 8745 = 9735 sq.m 29.

N = 3 R = 5 A = 6 N + R = 3 + 5 = 8 A + 2 = 6 + 2 = 8 ∴ N + R = A + 2

30.

IV.

31. S6 = 345 T6 – a = 55 ⇒ a + 5d – a = 55 ⇒ 5d = 55 ⇒ d = 11 ………(1) 345 = 6

2[2a + (6-1)11]

345 = 3[2a + 55] 345 = 6a + 165 6a = 345 – 165




a = 1806

= 30 ∴ Terms of AP- 30, 41, 52, 63, 74, 85

32.

휎 = ∑ 2

− ∑ 2

휎 = 110030

− 030

2

휎 = √ 36.67

흈 = ퟔ.ퟎퟓ

33. Let the Number of books be = x 48푥

- 48푥+4

= 1

48(x + 4) – 48x = x(x+4)

⇒ 48x + 192 – 48x = x2 + 4x

⇒ x2 + 4x – 192 = 0

⇒ x2 + 16x – 12x -192 = 0

⇒ x(x + 16) – 12(x + 16) = 0

⇒ (x + 16)(x- 12) = 0

⇒ x = 12

∴ Number of Books = 12

OR

Let the speed of the train be = x km./hour

CI f d=x-A fd d2 fd2

50 4 -10 -40 100 400 55 6 -5 -30 25 150 60 10 0 0 0 0 65 6 5 30 25 150 70 4 10 40 100 400

30 0 1100




196푥

+ 196푥+21

= 11

196(x + 21) + 196x = 11x(x+21)

⇒ 196x + 4116 + 196x = 11x2 + 231x

⇒ 11 x2 + 231x – 392x - 4116 = 0

⇒ 11 x2 - 161x - 4116 = 0

⇒ 푥 =−(−161)± (−161)2−4.11.4116

2.11

⇒ 푥 = 161±√25921+18110422

푥 = 161±45522

푥 = −231±45522

푥 = 56

∴ Speed of train = 56 Km/hour and speed of Car = 56+ 21 = 77km./hour

34. 흅 = 1800 , A = 흅ퟑ

, B = 휋6

LHS = tan(퐴 − 퐵) = tan(60 − 30) = tan 30 LHS = ퟏ

√ퟑ

RHS = tan퐴−tan퐵1+ tan퐴.tan퐵

RHS = tan 60−tan 301+ tan 60.tan 30

RHS = √3 − 1

√3

1+ √3. 1√3

RHS = (√3)2− 1

√31+ 1

RHS = 3 − 1

√32

RHS = 2

√32




RHS = 1√3

∴ LHS = RHS

∴ tan(퐴 − 퐵) = tan퐴−tan퐵1+ tan퐴.tan퐵

OR ( Question is wrong)

1 + cos 휃1 − cos 휃

– 1 − cos 휃1+ cos 휃

= (1 + cos 휃)2 − (1 − cos 휃)2

(1 − cos 휃)(1+ cos 휃)

= 1+2 cos 휃+cos2 휃 – 1+2 cos 휃− cos2 휃1− cos2 휃

= 4 cos 휃sin2 휃

= 4 cos 휃sin휃.sin휃

= 4 (cos 휃sin 휃

) ( 1sin 휃

)

= ퟒ 퐜퐨퐭 휽. 퐜퐨퐬퐞퐜 휽

35. Case-1). If two circles touch each other externally, thecentres and the point of contact are collinear.

Given:A and B are the centres of touching circles. P is the point of contact. To prove : A,P,and B are collinear. Construction: Draw the tangent XPY. Proof:In the figure ∠APX = 900……………..(1) ∵Radius drawn at the point of contact is ∠BPX = 900 ………… ..(2) perpendicular to the tangent ∠APX + ∠BPX = 900 +900 [ by adding (1) and (2) ∠APB = 1800 [ APB is a straight line ∴ APB is a straight line ∴ A, P andB are collinear.

Case-2 ). If two circles touch each other internally the centres and the point of contact are collinear. Given:A and B are centres of touching circles. P is point of contact. To prove : A,P,and B are collinear Construction: Draw the common tangent XPY . Join AP and BP. Proof:In the figure ∠APX = 900……………..(1) ∵Radius drawn at the point of contact




∠BPX = 900 ………… ..(2) is perpendicular to the tangent. ∠APX = ∠BPX = 900 [ From (1) and (2) AP and BP lie on the same line ∴ APB is a straight line ∴ A, P and B are collinear.

36. ದತ:In trapizium ABCD ಯ ,(Plz Note: Deleated Problem fo exam)

AD || BC, AE = EB, DF = FC To Prove: EF || AD or EF || BC Construction: Extend BA & CD to meet at P . Proof: In ∆PBC, AD||BC [ ∵ Given] ∴ 푃퐴

퐴퐵 = 푃퐷

퐷퐶 [∵ B.P.T.]

∴ 푃퐴2퐴퐸

= 푃퐷2퐷퐹

[ E and F are the midpoints of AB and DC]

∴ 푃퐴퐴퐸

= 푃퐷퐷퐹

∴ 퐸퐹||퐴퐷 [∵ Converse of BPT] ∴ EF || AD OR EF || BC OR In Trapzium ABCD , AB||CD, AO = 3x-9 CO = x-5 DO = 3 B0 = x-3 In ∆AOB and ∆COD , ∠OAB = ∠OCD [∵ AB||CD, Alternate angles] ∠OBA = ∠ODC [∵ AB||CD, Alternate angles] ∠AOB = ∠COD [ ∵ Vertically opposite angles ] ∆AOB ~ ∆COD 퐴푂퐶푂

= 퐵푂퐷푂

[ ∵ Theorem]

∴ 3푥−9푥−5

= 푥−33

∴ (3x -9)3 = (x-3)(x-5) 9x – 27 = x2 – 8x + 15 x2 – 17x + 42 = 0 x2 – 14x – 3x + 42 = 0




x(x – 14) – 3(x – 14)=0 (x-14) (x-3)=0 x = 14 ಮತು x = 3

V.

37. Theorem: Phythagoras Therem In a right angled triangle,the square of the hypotenuse is equal to the sum of the square of the other two sides. Given: ∆ABC In which ∠ABC = 900 To Prove : AB2 + BC2 = CA2 Construction: Draw BD ⟘ AC . Proof: In ∆ABC and ∆ADB , ∠ABC = ∠ADB = 900 [ ∵ Given and Construction ∠BAD =∠BAD [∵ Common angle ∴ ∆ABC ~ ∆ADB [∵ AA criteria

⇒ ABAD

= ACAB

⇒ AB2 = AC.AD……..(1) In ∆ABC and ∆BDC , ∠ABC = ∠BDC = 900 [ ∵ Given and construction ∠ACB = ∠ACB [∵ Common angle ∴ ∆ABC ~ ∆BDC [∵ AA criteria

⇒ BCDC

= ACBC

⇒ BC2 = AC.DC……..(2) (1) + (2) AB2+ BC2 = (AC.AD) + (AC.DC) AB2+ BC2 = AC.(AD + DC) AB2+ BC2 = AC.AC AB2+ BC2 = AC2 [ ∵AD + DC = AC]

38. y= x2 + x - 2




x -3 -2 -1 0 1 2 y 4 0 -2 -2 0 4

39.




Length of the tangents = 10.7 cm

40. a –d + a + a + d = 21 3a = 21 a = 7 (a –d)(a + d) = a + 6 (7 – d) ( 7 + d) = 7 + 6 72 – d2 = 13




49 – d2 = 13 d2 = 49 – 13 d2 = 36 d = 6 ∴ 3 terms : 1, 7 , 13 OR(Question wrong) corrected T3 = 7 ⇒ a + 2d = 7 ……….(1) T7 = 3T3 + 2 ⇒ a + 6d = 3x7 + 2 a + 6d = 21 + 2 a + 6d = 23 ………….(2)

(1) - (2) a + 2d = 7

a + 6d = 23 -4d = -16 d = 4 Apply d = 4 in (1) a + 2x4 = 7 a + 8 =7 a = -1 ∴ Sn = 푛

2[2a + (n-1)d]

S20 = 202

[2x-1 + (20-1)4] S20 = 10[-2 + 19x4] S20 = 10[-2 + 76] S20 = 10[74] S20 = 740


KEY ANSWER TO MODEL QUESTION PAPER – 4 OF KSEEB

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