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CBSENCERTSolutionsforClass12mathematicsChapter5

Miscellaneousexerciseonchapter5

Q.1. Differentiatew.r.t.xthefunction3x2-9x+59.

Solution: Givenfunctionis3x2-9x+59

Lety=3x2-9x+59,

Differentiatingbothsideswithrespecttoxbyapplyingchainruleweget, dydx=93x2-9x+58·ddx3x2-9x+5∵ddxf(g(x))=f'(g(x))·g'(x)=93x2-9x+58·6x-9∵ddxxn=nxn-1=273x2-9x+58·2x-3Therefore,thederivativeof3x2-9x+59is27(2x-3)(3x2-9x+5)8.

Q.2. Differentiatew.r.t.xthefunctionxx+xa+ax+aa,forsomefixeda>0andx>0.

Solution: Givenfunctionisxx+xa+ax+aa

Letu=xxandy=u+xa+ax+aa

Differentiatingbothsideswithrespecttoxweget,dydx=dudx+ddxxa+ddxax+ddxaa⇒dydx=dudx+axa-1+axloga+0…(i)Here,u=xx.Takinglogonbothsides,logu=logxx=xlogx

Differentiatingbothsideswithrespecttoxandapplyingproductruleweget,

1ududx=x·ddxlogx+logx·ddxx=x·1x+logx·1∵(uv)'=u'v+uv'

⇒dudx=u1+logx=xx1+logxPuttingthevalueofdudxinequation(i),wegetdydx=xx1+logx+axa-1+axlogaTherefore,thederivativeofxx+xa+ax+aawithrespecttoxisxx1+logx+axa-1+axloga.

Q.3. Differentiatew.r.t.xthefunctionxx2-3+x-3x2,forx>3.

Solution: Givenfunctionisxx2-3+x-3x2Letu=xx2-3andv=x-3x2andy=u+v

Differentiatingbothsideswithrespecttoxweget,

dydx=dudx+dvdx…(i)

Here,u=xx2-3.

Takinglogonbothsides,

logu=x2-3logx∵logam=mlogaDifferentiatingbothsideswithrespecttoxandapplyingproductruleweget,1ududx=x2-3·ddxlogx+logx·ddxx2-3∵(uv)'=u'v+uv'=x2-3·1x+logx·2x∵ddxxn=nxn-1⇒dudx=ux2-3+2x2logxx⇒dudx=xx2-3x2-3+2x2logxx=xx2-4x2-3+2x2logx…(ii)

Again,v=x-3x2.


logv=x2logx-3Differentiatingbothsideswithrespecttoxandapplyingproductruleweget,1vdvdx=x2·ddxlogx-3+logx-3·ddxx2=x2·1x-3+logx-3·2x=x2x-3+2x·logx-3∵ddxlogx=1x⇒dvdx=vx2x-3+2x·logx-3=x-3x2x2x-3+2x·logx-3…(iii)Puttingthevalueofdudxfrom(ii)andvalueofdvdxfrom(iii)inequation(i),wehave,dydx=xx2-4x2-3+2x2logx+x-3x2x2x-3+2x·logx-3

Q.4. Finddydx,ify=121-cost,x=10t-sin t,-π2<t<π2.

Solution: Giventhaty=121-cost,x=10t-sint,-π2<t<π2.

Forx=10t-sin t,Differentiatingbothsideswithrespecttotweget,

dxdt=ddt10(t-sint)=101-costSimilarlyfory=12(1-cost),Differentiatingbothsideswithrespecttotweget,dydt=ddt12(1-cost)=120+sint=12sintNow,dydx=dydtdxdt=12sint101-cost=62sint2cost252sin2t2∵sin2A=2sinAcosAandcos2A=2cos2A-1=65cott2Therefore,thevalueofdydxintermsoftis65cott2.

Q.5. Finddydx,ify=sin-1 x+sin-1 1-x2,0<x<1.

Solution: Giventhaty=sin-1x+sin-11-x2,0<x<1.

Here,y=sin-1 x+sin-1 1-x2.

Differentiatingbothsideswithrespecttoxweget,dydx=ddxsin-1x+ddxsin-11-x2∵ddx(f(x)+g(x))=ddxf(x)+ddxg(x)⇒dydx=11-x2+11-1-x22ddx1-x2∵Chainrule:ddxf(g(x))=f'(g(x))·g'(x)⇒dydx=11-x2+1x·121-x2ddx1-x2⇒dydx=11-x2+1x·121-x2-2x∵ddxxn=nxn-1⇒dydx=11-x2-11-x2=0Therefore,thevalueofdydxis0.

Q.6. Ifx1+y+y1+x=0,for-1<x<1.Provethatdydx=-11+x2.

Solution: Giventhatx1+y+y1+x=0⇒x1+y=-y1+x

Squaringbothsides,

x21+y=y21+x⇒x2+x2y=y2+y2x⇒x2-y2+x2y-y2x=0⇒x+yx-y+xyx-y=0⇒x-yx+y+xy=0⇒(x+y+xy)=0[∵x≠y⇒x-y≠0]⇒y1+x=-x⇒y=-x1+xDifferentiatingbothsideswithrespecttoxandapplyingquotientruleweget,dydx=-1+xddxx-xddx1+x1+x2∵uv'=u'v-uv'v2=-1+x-x1+x2=-11+x2(Henceproved)

Q.7. Ifx-a2+y-b2=c2forsomec>0,

provethat1+dydx232d2ydx2isaconstantindependentofaandb.

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Solution: Giventhatx-a2+y-b2=c2

Differentiatingwithrespecttox,wehave

ddxx-a2+ddxy-b2=ddxc2⇒2x-a+2y-bdydx=0⇒dydx=-x-ay-b

d2ydx2=-y-bddxx-a-x-addxy-by-b2=-y-b1-x-adydxy-b2

⇒d2ydx2=-y-b1-x-a-x-ay-by-b2=-y-b2+x-a2y-b3=-c2y-b3

Puttingthevaluesin1+dydx232d2ydx2,wehave

1+-x-ay-b232-c2y-b3=1+x-a2y-b232-c2y-b3

=y-b2+x-a2y-b232-c2y-b3=c2y-b232-c2y-b3=c3y-b3-c2y-b3=-c3c2=-c,whichisaconstantindependentofaandb.

Q.8. Ifcosy=xcosa+y,withcosa≠±1,provethatdydx=cos2 a+ysin a.

Solution: Giventhatcosy=xcosa+y⇒x=cosycosa+y,therefore,Differentiatingwithrespecttoy,wehave

dxdy=cosa+yddycosy-cosyddycosa+ycos2 a+y

⇒dxdy=cosa+y-siny-cosy-sina+ycos2a+y⇒dxdy=-sinycosa+y+cosysina+ycos2a+y=sina+y-ycos2a+y=sinacos2a+y⇒dydx=cos2a+ysinα

Q.9. Ifx=acost+tsin tandy=asin t-tcost,findd2ydx2.

Solution: Giventhatx=acost+tsintandy=asint-tcost

Here,x=acost+tsin t,y=asin t-tcost

Therefore,dxdt=a-sint+tcost+sint=atcost

and

dydt=acost--tsin t+cost=atsin t

dydx=dydtdxdt=atsintatcost=tant

⇒d2ydx2=sec2 t·dtdx=sec2 t·1atcost=sec3 tat

Q.10. Iffx=x3,showthatf''xexistsforallrealxandfindit.

Solution: Givenfunctionisfx=x3

Rewritingthefunctionfx=x3inthefollowingform:

fx=x3ifx>0x3ifx=0(-x)3ifx<0Hereweneedtofindthedifferentiabilityoff''(x)at,x>0,x=0andx<0Ifx>0,fx=x3⇒f'x=3x2⇒f''x=6xIfx<0,fx=-x3⇒f'x=-3x2⇒f''x=-6x

Nowifx=0,

f(x)=x3=x3=0

LHD=limh→0-f(0+h)-f(0)-h=limh→0--h3-h=0AndRHD=limh→0+f(0+h)-f(0)h=limh→0-h3h=0Here,LHD=RHDTherefore,f'(x)existsatx=0.Herethefunctionf'(x)canbere-writtenas:f'x=3x2ifx>03x2ifx=0-3x2ifx<0Similarlyitcanbeshownthatf'(x)isdifferentiableatx=0.Hence,f''xexistsforallrealxanditcanberepresentedasfollows:f''x=6xifx≥0-6xifx<0

Q.11. Usingmathematicalinductionprovethatddxxn=nxn-1forallpositiveintegersn.

Solution: LetPn:ddxxn=nxn-1

Puttingn=1,wehaveLHS=ddxx1=1andRHS=1x1-1=x0=1

Hence,Pnistrueforn=1.

LetPk:ddxxk=kxk-1istrue.

Toprove:Pk+1:ddxxk+1=k+1xkisalsotrue.LHS=ddxxk+1=ddxxk·x=xkddxx+xddxxk=xk·1+x·kxk-1=1+kxk=RHSHence,Pnistrueforn=k+1.Therefore,bytheprincipleofmathematicalinductionPnistrueforallnaturalnumbersn.Hence,ddxxn=nxn-1forallpositiveintegersn.

Q.12. Differentiatew.r.t.xthefunctionsin3 x+cos6 x.

Solution: Givenfunctionissin3x+cos6x

Lety=sin3 x+cos6 x,

Differentiatingbothsideswithrespecttoxandapplyingchainruleforeachtermweget,dydx=ddxsin3x+ddxcos6x∵ddxf(x)±g(x)=ddxf(x)±ddxg(x)dydx=3sin2x·ddxsinx+6cos5x·ddxcosx∵ddxf(g(x))=f'(g(x))·g'(x)=3sin2x·cosx+6cos5x·-sinx=3sinxcosxsinx-2cos4xTherefore,thederivativeofsin3x+cos6xis3sinxcosxsinx-2cos4x.

Q.13. Usingthefactthatsin A+B=sin AcosB+cosAsin Bandthedifferentiation,obtainthesumformulaforcosines.

Solution: GiventhatsinA+B=sinAcosB+cosAsinB,


ddxsinA+B=sinAddxcosB+cosBddxsinA+cosAddxsinB+sinBddxcosA⇒cosA+B·dAdx+dBdx=sinA-sinBdBdx+cosBcosAdAdx+cosAcosBdBdx+sinB-sinAdAdx⇒cosA+B·dAdx+dBdx=cosAcosB-sinAsinBdBdx+cosAcosB-sinAsinBdAdx⇒cosA+B·dAdx+dBdx=cosAcosB-sinAsinBdAdx+dBdx⇒cosA+B=cosAcosB-sinAsinB

Q.14. Doesthereexistafunctionwhichiscontinuouseverywherebutnotdifferentiableatexactlyonepoint?Justifyyouranswer.




Solution: Functionfx=x-1iscontinuousforallrealpointsbutnotdifferentiableatx=1.

Herethefunctioniscontinuousatx=1as,

limx→1-f(x)=0andlimx→1-f(x)=0also,f(1)=0Therefore,thefunctioniscontinuousatx=1.

Now,weknowthat:

f'(x)=limh→0f(x+h)-f(x)hSo,f'(1)=limh→0f(1+h)-f(1)h

Here,LHD=limh→0-f(1+h)-f(1)h=limh→0-1+h-1-1-1h=-1andRHD=limh→0+f(1+h)-f(1)h=limh→0+1+h-1-1-1h=1LHD≠RHDSo,f(x)isnotdifferentiableatx=1.

Q.15. Ify=fxgxhxlmnabcprovethatdydx=f'xg'xh'xlmnabc

Solution: Giventhaty=fxgxhxlmnabc,therefore,

dydx=f'xg'xh'xlmnabc+fxgxhxdldxdmdxdndxabc+fxgxhxlmndadxdbdxdcdx

⇒dydx=f'xg'xh'xlmnabc+fxgxhx000abc+fxgxhxlmn000⇒dydx=f'xg'xh'xlmnabc+0+0⇒dydx=f'xg'xh'xlmnabc

Q.16. Ify=eacos-1 x,-1≤x≤1,showthat1-x2d2ydx2-xdydx-a2y=0.

Solution: Giventhat:y=eacos-1x


dydx=ddxeacos-1x=eacos-1xddxacos-1x⇒dydx=eacos-1xa·-11-x2=-ay1-x2Squaringboththesides,wehavedydx2=a2y21-x2⇒1-x2dydx2=a2y2

Differentiatingagainwithrespecttox,wehave

1-x2·2dydx·d2ydx2+dydx2d1-x2dx=a22ydydx

⇒dydx21-x2d2ydx2+dydx-2x=2a2ydydx⇒2dydx1-x2d2ydx2-xdydx=2a2ydydx⇒1-x2d2ydx2-xdydx=a2y⇒1-x2d2ydx2-xdydx-a2y=0

Q.17. Differentiatew.r.t.xthefunction5x3cos2x.

Solution: Givenfunctionis5x3cos2x

Lety=5x3cos2x,takinglogonbothsides,

logy=log5x3cos2x=3cos2x·log5x∵logam=mloga


1ydydx=3cos2x·ddxlog5x+log5x·ddx3cos2x∵(uv)'=u'v+uv'

⇒dydx=y3cos2x·15x·5+log5x·3-sin2x·2∵ddxlogx=1x⇒dydx=35x3cos2xcos2xx-2sin2x·log5xTherefore,thederivativeof5x3cos2xwithrespecttoxis35x3cos2xcos2xx-2sin2x·log5x.

Q.18. Differentiatew.r.t.xthefunctionsin-1 xx,0≤x≤1.

Solution: Givenfunctionissin-1xx,0≤x≤1

Lety=sin-1 xx,

Differentiatingbothsideswithrespecttoxandapplyingproductruleweget,dydx=11-xx2·ddxxx∵ddxf(g(x))=f'(g(x))·g'(x)=11-x3·xddxx+x·ddxx∵Productrule:(uv)'=u'v+uv'=11-x3·x12x+x·1∵ddxxn=nxn-1=11-x3·x+2x2x=3x2x1-x3=32x1-x3Therefore,thederivativeofsin-1xxwithrespecttoxis32x1-x3.

Q.19. Differentiatew.r.t.xthefunctioncos-1 x22x+7,-2<x<2.

Solution: Givenfunctioniscos-1x22x+7,-2<x<2

Lety=cos-1 x22x+7,

Differentiatingbothsideswithrespecttoxandapplyingquotientruleruleweget,dydx=cos-1x2·ddx2x+7-2x+7ddxcos-1x22x+72∵uv'=uv'-vu'v2=cos-1x2·122x+7·2-2x+7-11-x22·122x+7=cos-1x2·12x+7+2x+714-x22x+7=cos-1x2·4-x2+2x+72x+72x+74-x2Therefore,thederivativeofcos-1x22x+7withrespecttoxiscos-1x2·4-x2+2x+72x+72x+74-x2.

Q.20. Differentiatew.r.t.xthefunctioncot-1 1+sin x+1-sin x1+sin x-1-sin x,0<x<π2

Solution: Givenfunctioniscot-11+sinx+1-sinx1+sinx-1-sinx,0<x<π2

Lety=cot-1 1+sin x+1-sin x1+sin x-1-sin x,

therefore,y=cot-1cos2x2+sin2x2+2sinx2cosx2+cos2x2+sin2x2-2sinx2cosx2cos2x2+sin2x2+2sinx2cosx2-cos2x2+sin2x2-2sinx2cosx2∵sin2A+cos2A=1andsin2A=2sinAcosA=cot-1cosx2+sinx22+cosx2-sinx22cosx2+sinx22-cosx2-sinx22∵a2+b2±2ab=(a±b)2cot-1cosx2+sinx2+cosx2-sinx2cosx2+sinx2-cosx2+sinx2=cot-12cosx22sinx2=cot-1cotx2=x2Differentiatingbothsideswithrespecttoxweget,dydx=12∵ddxxn=nxn-1Therefore,thederivativeofcot-11+sinx+1-sinx1+sinx+1-sinxwithrespecttoxis12.

Q.21. Differentiatew.r.t.xthefunctionlogxlogx,x>1.

Solution: Givenfunctionislogxlogx,x>1

Lety=logxlogx,takinglogonbothsides,

logy=loglogxlogx=logx·loglogx∵logam=mlogaDifferentiatingbothsideswithrespecttoxandapplyingproductruleweget,1ydydx=logx·ddxloglogx+loglogx·ddxlogx∵(uv)'=u'v+uv'1ydydx=logx·1logx·ddxlogx+loglogx·ddxlogx∵ddxf(g(x))=f'(g(x))·g'(x)⇒dydx=ylogx·1logx·1x+loglogx·1x⇒dydx=logxlogx1+loglogxxTherefore,thederivativeoflogxlogxwithrespecttoxislogxlogx1+loglogxx.

Q.22. Differentiatew.r.t.xthefunctioncosacosx+bsin x,forsomeconstantsaandb.




Solution: Givenfunctioniscosacosx+bsinx

Lety=cosacosx+bsin x.

Differentiatingbothsideswithrespecttoxandapplyingchainruleweget,dydx=-sinacosx+bsinx·ddxacosx+bsinx∵ddxf(g(x))=f'(g(x))·g'(x)=-sinacosx+bsinx-asinx+bcosx=sinacosx+bsinxasinx-bcosxTherefore,thederivativeofcosacosx+bsinxwithrespecttoxissinacosx+bsinxasinx-bcosx.

Q.23. Differentiatew.r.t.xthefunctionsin x-cosxsin x-cosx,π4<x<3π4

Solution: Givenfunctionissinx-cosxsinx-cosx,π4<x<3π4

Lety=sin x-cosxsin x-cosx.

Takinglogonbothsides,logy=logsinx-cosxsinx-cosx=sinx-cosx·logsinx-cosx∵logxm=mlogx


1ydydx=sin x-cosx·ddxlogsin x-cosx+logsin x-cosx·ddxsin x-cosx∵(uv)'=u'v+uv'

1ydydx=sinx-cosx·1sinx-cosxddxsinx-cosx+logsinx-cosx·ddxsinx-cosx∵ddxf(g(x))=f'(g(x))·g'(x)andddxlogx=1x⇒dydx=ysinx-cosx·cosx+sinxsinx-cosx+logsinx-cosxcosx+sinx⇒dydx=sinx-cosxsinx-cosxcosx+sinx1+logcosx-sinxTherefore,thederivativeofsinx-cosxsinx-cosxwithrespecttoxissinx-cosxsinx-cosxcosx+sinx1+logcosx-sinx.




Exercise5.1

Q.1. Provethatthefunctionfx=5x-3iscontinuousatx=0,atx=-3andatx=5.

Solution: Givenfunction:f(x)=5x-3.

Toprove:Thefunctioniscontinuousatx=0,x=-3andx=5.

Afunctionf(x)iscontinuousatapointk,ifTheleft-handlimitissameastheright-handlimitandalsothelimitingvalueissameasthevalueofthefunctionatthatpoint.i.e.limx→k+f(x)=limx→k-f(x)=f(k).Atx=0,f0=50-3=-3LHL=limx→0-fx=limx→0-5x-3=5(0)-3=-3RHL=limx→0+fx=limx→0+5x-3=5(0)-3=-3Here,atx=0,LHL=RHL=f0=-3Hence,thefunctionfiscontinuousatx=0.

Nowatx=-3,f-3=5-3-3=-18

LHL=limx→-3-fx=limx→-3-5x-3=5(-3)-3=-18

RHL=limx→-3+fx=limx→-3+5x-3=5(-3)-3=-18Here,atx=-3,LHL=RHL=f-3=-18Hence,thefunctionfiscontinuousatx=-3.

Atx=5,f5=55-3=22

LHL=limx→5-fx=limx→5-5x-3=5(5)-3=22

RHL=limx→5+fx=limx→5+5x-3=5(5)-3=22Here,atx=5,LHL=RHL=f5=22Hence,thefunctionfiscontinuousatx=5.Therefore,thegivenfunctioniscontinuousatx=0,x=-3andx=5

Q.2. .Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=x+1,Ifx≥1x2+1,Ifx<1

Solution: Givenfunctionisdefinedbyfx=x+1,Ifx≥1x2+1,Ifx<1Letkbeanyrealnumber.

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).

Accordingtoquestion,k<1ork=1ork>1

Firstcase:Ifk<1,fk=k2+1andlimx→k fx=limx→k x2+1=k2+1.Here,limx→k fx=fkHence,thefunctionfiscontinuousforallrealnumberssmallerthan1.

Secondcase:Ifk=1,f1=1+1=2LHL=limx→1-fx=limx→1-x2+1=1+1=2RHL=limx→1+ fx=limx→1+ x+1=1+1=2,Here,atx=1,LHL=RHL=f1.Hence,thefunctionfiscontinuousatx=1.

Thirdcase:Ifk>1,fk=k+1andlimx→kfx=limx→kx+1=k+1.Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumbersgreaterthan1.

Therefore,thefunctionfiscontinuousforallrealnumbers.

Q.3. Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=x3-3,Ifx≤2x2+1,Ifx>2.

Solution: Givenfunctionisdefinedbyfx=x3-3,Ifx≤2x2+1,Ifx>2Letkbeanyrealnumber.



Firstcase:Ifk<2,

fk=k3-3andlimx→k fx=limx→k x3-3=k3-3.Here,limx→k fx=fkHence,thefunctionfiscontinuousforallrealnumberslessthan2.

Secondcase:Ifk=2,f2=23-3=5LHL=limx→2-fx=limx→2-x3-3=23-3=5RHL=limx→2+ fx=limx→2+ x2+1=22+1=5Hereatx=2,LHL=RHL=f2Hence,thefunctionfiscontinuousatx=2.

Thirdcase:Ifk>2,fk=k2+1andlimx→kfx=limx→kx2+1=k2+1.

Here,limx→k fx=fkHence,thefunctionfiscontinuousforrealnumbersgreaterthan2.

Hence,thefunctionfiscontinuousforallrealnumbers.

Q.4. Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=x10-1,Ifx≤1x2,Ifx>1.

Solution: Givenfunctionisdefinedbyfx=x10-1,Ifx≤1x2,Ifx>1Letkbeanyrealnumber.



Firstcase:Ifk<1,fk=k10-1andlimx→k fx=limx→k x10-1=k10-1.Here,limx→k fx=fkHence,thefunctionfiscontinuousforallrealnumberslessthan1.

Secondcase:Ifk=1,f1=110-1=0LHL=limx→1-fx=limx→1-x10-1=0RHL=limx→1+ fx=limx→1+ x2=1Hereatx=1,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=1.

Thirdcase:Ifk>1,fk=k2andlimx→kfx=limx→kx2=k2.Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealvaluesgreaterthan1.

Hence,thefunctionfisdiscontinuousonlyatx=1.




Q.5. Isthefunctiondefinedbyfx=x+5,Ifx≤1x-5,Ifx>1acontinuousfunction?

Solution: Givenfunctionisdefinedbyfx=x+5,Ifx≤1x-5,Ifx>1Let,kbeanyrealnumber.



Firstcase:Ifk<1,fk=k+5andlimx→k fx=limx→k x+5=k+5.Here,limx→k fx=fkHence,thefunctionfiscontinuousforallrealnumberslessthan1.

Secondcase:Ifk=1,f1=1+5=6LHL=limx→1-fx=limx→1-x+5=6RHL=limx→1+ fx=limx→1+ x-5=-4,Hereatx=1,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=1.

Thirdcase:Ifk>1,fk=k-5andlimx→kfx=limx→kx-5=k-5Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumbersgreaterthan1.

Therefore,thefunctionfisdiscontinuousonlyatx=1.

Q.6. Discussthecontinuityofthefunctionf,wherefisdefinedbyfx=3,If0≤x≤14,If1<x<35,If3≤x≤10

Solution: Givenfunctionisdefinedbyfx=3,If0≤x≤14,If1<x<35,If3≤x≤10

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).Letkbeanyrealnumber.Accordingtoquestion,kcanbe

0≤k<1ork=1or1<k<3ork=3or3<k≤10

Firstcase:If0≤k<1,fk=3andlimx→k fx=limx→k 3=3.Here,limx→k fx=fkHence,thefunctionfiscontinuousfor0≤x≤1.

Secondcase:Ifk=1,f1=3LHL=limx→1-fx=limx→1-3=3RHL=limx→1+ fx=limx→1+ 4=4,Hereatx=1,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=1.

Thirdcase:If1<k<3,fk=4andlimx→k fx=limx→k 4=4.Herelimx→k fx=fkHence,thefunctionfiscontinuousfor1<x<3.

Fourthcase:Ifk=3,LHL=limx→3-fx=limx→3-4=4andRHL=limx→3+ fx=limx→3+ 5=5,Hereatx=3,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=3.

Fifthcase:If3<k≤10,fk=5andlimx→kfx=limx→k5=5.Here,limx→kfx=fkHence,thefunctionfiscontinuousfor3<x≤10.

Hence,thefunctionfisdiscontinuousonlyatx=1andx=3.

Q.7. Discussthecontinuityofthefunctionf,wherefisdefinedbyfx=2x,Ifx<00,If0≤x≤14x,Ifx>1

Solution: Givenfunctionisdefinedbyfx=2x,Ifx<00,If0≤x≤14x,Ifx>1Letkbeanyrealnumber.Accordingtoquestion,Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).k<0ork=0or0≤k≤1ork=1ork>1

Firstcase:Ifk<0,fk=2kandlimx→k fx=limx→k 2x=2k.Here,limx→k fx=fkHence,thefunctionfiscontinuousforallrealnumberslessthan0.

Secondcase:Ifk=0,f0=0LHL=limx→0-fx=limx→0-2x=0RHL=limx→0+ fx=limx→0+ 0=0.Here,limx→k fx=fkHence,thefunctionfiscontinuousatx=0.

Thirdcase:If0≤k≤1,fk=0andlimx→k fx=limx→k 0=0.Here,limx→k fx=fkHence,thefunctionfiscontinuousat0≤x≤1.

Fourthcase:Ifk=1,LHL=limx→1-fx=limx→1-0=0RHL=limx→1+ fx=limx→1+ 4x=4,Here,atx=1,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=1.

Fifthcase:Ifk>1,fk=4kandlimx→kfx=limx→k4x=4k.Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumbersgreaterthan1.

Hence,thefunctionfisdiscontinuousonlyatx=1.

Q.8. Discussthecontinuityofthefunctionf,wherefisdefinedbyfx=-2,Ifx≤-12x,If-1<x≤12,Ifx>1.




Solution: Givenfunctionisdefinedbyfx=-2,Ifx≤-12x,If-1<x≤12,Ifx>1.

Letkbeanyrealnumber.

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).Accordingtothequestion,k<-1ork=-1or-1<x≤1ork=1ork>1.

Firstcase:Ifk<-1,

fk=-2andlimx→k fx=limx→k -2=-2.Here,limx→k fx=fk

Hence,thefunctionfiscontinuousforallrealnumberslessthan-1.

Secondcase:Ifk=-1,f-1=-2

LHS=limx→-1-fx=limx→-1--2=-2

RHL=limx→-1+fx=limx→-1+2x=-2.Here,limx→kfx=fkHence,thefunctionfiscontinuousatx=-1.

Thirdcase:If-1<x≤1,

fk=2kandlimx→k fx=limx→k 2x=2k.

Here,limx→kfx=fkHence,thefunctionfiscontinuousat-1<x≤1.

Fourthcase:Ifk=1,

LHL=limx→1-fx=limx→1-2x=2

RHL=limx→1+fx=limx→1+2=2.Here,limx→kfx=fkHence,thefunctionfiscontinuousatx=1.

Fifthcase:Ifk>1,

fk=2andlimx→k fx=limx→k 2=2.

Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumbersgreaterthan1.Therefore,thefunctionfiscontinuousforallrealnumbers.

Q.9. Findtherelationshipbetweenaandbsothatthefunctionfdefinedbyfx=ax+1,Ifx≤3bx+3,Ifx>3iscontinuousatx=3.

Solution: Givenfunctionsisdefinedbyfx=ax+1,Ifx≤3bx+3,Ifx>3

Giventhatthefunctioniscontinuousatx=3.

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).Therefore,LHL=RHL=f3EquatingtheLHLandRHLweget,⇒limx→3-fx=limx→3+fx⇒limx→3-ax+1=limx→3+bx+3⇒3a+1=3b+3⇒3a=3b+2⇒a=b+23Hence,therelationshipbetweenaandbisa=b+23

Q.10. Forwhatvalueofλifthefunctiondefinedbyfx=λx2-2x,Ifx≤04x+1,Ifx>0isContinuousatx=0?

Whataboutcontinuityatx=1?

Solution: Givenfunctionisdefinedasfx=λx2-2x,Ifx≤04x+1,Ifx>0

Giventhatthefunctioniscontinuousatx=0.

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k+f(x)=f(k).Therefore,LHL=RHL=f0⇒limx→0-fx=limx→0+fx=f0⇒limx→0-λx2-2x=limx→0+4x+1=λ02-20⇒λ02-20=40+1=λ0⇒0·λ=1⇒λ=10Hence,thereisnorealvalueofλforwhichthegivenfunctionbecontinuous.

Ifx=1,

f1=41+1=5andlimx→1 fx=limx→1 41+1=5.Here,limx→1 fx=f1

Therefore,thefunctionfiscontinuousforallrealvaluesofλ.

Q.11. Showthatthefunctiondefinedbygx=x-xisdiscontinuousatallintegralpoints.Herexdenotesthegreatestintegerlessthanorequaltox.

Solution: Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).

Givenfunctionisdefinedbygx=x-xLetkbeanyinteger

LHL=limx→k-fx=limx→k-x-x=k-k-1=1RHL=limx→k+fx=limx→k+x-x=k-k=0,Here,atx=k,LHL≠RHL.Therefore,thefunctionfisdiscontinuousforallintegers.

Q.12. Examinethecontinuityofthefunctionfx=2x2-1atx=3.

Solution: Givenfunctionisfx=2x2-1.

Toexamine:continuityofthegivenfunctionatx=3.

Afunctionf(x)iscontinuousatx=k,if,limx→k+f(x)=limx→k-f(x)=f(k).Atx=3,f3=232-1=17LHL=limx→3-fx=limx→3-2x2-1=2(3)2-1=17RHL=limx→3+fx=limx→3+2x2-1=2(3)2-1=17Here,atx=3,LHL=RHL=f3=17Therefore,thefunctionfiscontinuousatx=3.

Q.13. Isthefunctiondefinedbyfx=x2-sin x+5continuousatx=π?

Solution: Givenfunctionisfx=x2-sinx+5.

Herewehavetoexaminethecontinuityofthefunctionatx=π.

Now,LHL=limx→π-fx=limh→0fπ-h=limh→0π-h2-sinπ-h+5=limh→0π-h2-sinh+5∵sinπ-x=sinx=π-02-sin0+5=π2-0+5=π2+5

RHL=limx→π+fx=limh→0fπ+h=limh→0π+h2-sinπ+h+5

=limh→0π+h2--sinh+5=limh→0π+h2+sinh+5∵sinπ+x=-sinx

=π2-sin0+5=π2-0+5=π2+5Also,fπ=π2-sinπ+5=π2-0+5=π2+5∵sinπ=0LHL=RHL=f(π)Thus,fxiscontinuousatx=π.

Q.14. Discussthecontinuityofthefollowingfunction:fx=sin x+cosx




Solution: Afunctionf(x)iscontinuousatapointk,iflimx→k-f(x)=limx→k+f(x)=f(k).

Letgx=sin x

Letkbeanyrealnumber,atx=k,gk=sinkLHL=limx→k-gx=limx→k-sinx=limh→0sink-h=limh→0sinkcosh-cosksinh=sinkRHL=limx→k+gx=limx→k+sinx=limh→0sink+h=limh→0sinkcosh+cosksinh=sinkHere,atx=k,LHL=RHL=gk.Hence,thefunctiongiscontinuousforallrealnumbers.

Lethx=cosx

Letkbeanyrealnumber.x=k,hk=cosk

LHL=limx→k-hx=limx→k-cosx=limh→0cosk-h=limh→0coskcosh+sinksinh=coskRHL=limx→k+hx=limx→k+cosx=limh→0cosk+h=limh→0coskcosh-sinksinh=coskHere,atx=k,LHL=RHL=hk.Hence,thefunctionhiscontinuousforallrealnumbers.Weknowthatifgandharetwocontinuousfunctions,thenthefunctions g+h,g-handgharealsoacontinuousfunctions.Therefore,fx=sinx+cosxisacontinuousfunction.

Q.15. Discussthecontinuityofthefollowingfunctions:fx=sin x-cosx


Also,f(x)±g(x)arecontinuousatkifbothf(x)andg(x)arecontinuousatthesamepoint

Letgx=sinxLetkbeanyrealnumber.Atx=k,gk=sinkLHL=limx→k-gx=limx→k-sinx=limh→0sink-h=limh→0sinkcosh-cosksinh=sinkRHL=limx→k+gx=limx→k+sinx=limh→0sink+h=limh→0sinkcosh+cosksinh=sinkHere,atx=k,LHL=RHL=gk.Hence,thefunctiongiscontinuousforallrealnumbers.

Lethx=cosx


Atx=k,hk=coskLHL=limx→k-hx=limx→k-cosx=limh→0cosk-h=limh→0coskcosh+sinksinh=coskRHL=limx→k+hx=limx→k+cosx=limh→0cosk+h=limh→0coskcosh-sinksinh=coskHere,atx=k,LHL=RHL=hk.Hence,thefunctionhiscontinuousforallrealnumbers.Weknowthatifgandharetwocontinuousfunctions,thenthefunctions g+h,g-handghalsobecontinuousfunctions.Therefore,fx=sin x-cosxisacontinuousfunction.

Q.16. Discussthecontinuityofthefollowingfunction:fx=sin x·cosx.


Also,f(x)·g(x)arecontinuousatkifbothf(x)andg(x)arecontinuousatthesamepoint

Letgx=sinxLetkbeanyrealnumber.Atx=k,gk=sinkLHL=limx→k-gx=limx→k-sinx=limh→0sink-h=limh→0sinkcosh-cosksinh=sinkRHL=limx→k+gx=limx→k+sinx=limh→0sink+h=limh→0sinkcosh+cosksinh=sinkHere,atx=k,LHL=RHL=gk.Hence,thefunctiongiscontinuousforallrealnumbers.

Lethx=cosx

Letkbeanyrealnumber.x=k,hk=coskLHL=limx→k-hx=limx→k-cosx=limh→0 cosk-h=limh→0 coskcosh+sin ksin h=coskRHL=limx→k+ hx=limx→k+ cosx=limh→0 cosk+h=limh→0 coskcosh-sin ksin h=coskHere,atx=k,LHL=RHL=hk.Hence,thefunctionhiscontinuousforallrealnumbers.

Weknowthatifgandharetwocontinuousfunctions,thenthefunctions g+h,g-handghalsobeacontinuousfunction.Therefore,fx=sinx.cosxisacontinuousfunction.

Q.17. Discussthecontinuityofthecosine,cosecant,secantandcotangentfunctions.


Letgx=sin xLetkbeanyrealnumber.Atx=k,gk=sin k

LHL=limx→k-gx=limx→k-sinx=limh→0sink-h=limh→0sinkcosh-cosksinh=sinkRHL=limx→k+gx=limx→k+sinx=limh→0sink+h=limh→0sinkcosh+cosksinh=sinkHere,atx=k,LHL=RHL=gk.Hence,thefunctiongiscontinuousforallrealnumbers.

Lethx=cosxLetkbeanyrealnumber.Atx=k,hk=coskLHL=limx→k-hx=limx→k-cosx=limh→0cosk-h=limh→0coskcosh+sinksinh=coskRHL=limx→k+hx=limx→k+cosx=limh→0cosk+h=limh→0coskcosh-sinksinh=coskHere,atx=k,LHL=RHL=hk.

Hence,thefunctionhiscontinuousforallrealnumbers.

Weknowthatifgandharetwocontinuousfunctions,thenthefunctions gh,h≠0,1h,h≠0and1g,g≠0arecontinuousfunctions.

Therefore,cosec x=1sin x,sin x≠0iscontinuous⇒x≠nπn∈Ziscontinuous.

Hence,cosecxiscontinuousexceptx=nπn∈Z.secx=1cosx,cosx≠0iscontinuous.⇒x≠2n+1π2n∈Ziscontinuous.Hence,secxiscontinuousexceptx=2n+1π2n∈Z.cotx=cosxsinx,sinx≠0iscontinuous.⇒x≠nπn∈Ziscontinuous.Hence,cotxiscontinuousexceptx=nπn∈Z.

Q.18. Findallpointsofdiscontinuityoff,wherefx=sin xx,Ifx<0x+1,Ifx≥0.




Solution: Givenfunctionisdefinedbyfx=sin xx,Ifx<0x+1,Ifx≥0Letkbeanyrealnumber.Accordingtoquestion,k<0ork=0ork>0

Firstcase:Ifk<0fk=sin kkandlimx→k fx=limx→k sin xx=sin kk.Here,limx→k fx=fkHence,thefunctionfiscontinuousforallrealnumberslessthan0.

Secondcase:Ifk=0f0=0+1=1LHL=limx→0-fx=limx→0-x+1=0+1=1RHL=limx→0+ fx=limx→0+ x+1=0+1=1,Hereatx=0,LHL=RHL=f0.Hence,thefunctionfiscontinuousatx=0.

Thirdcase:Ifk>0fk=k+1andlimx→kfx=limx→kx+1=k+1.Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumbersgreaterthan0.

Therefore,thefunctionfiscontinuousforallrealnumbers.

Q.19. Determineiffdefinedbyfx=x2sin 1x,Ifx≠00,Ifx=0isacontinuousfunction?

Solution: Givenfunctionisdefinedbyfx=x2sin 1x,Ifx≠00,Ifx=0Letkbeanyrealnumber.Accordingtoquestion,k≠0ork=0

Firstcase:Ifk≠0fk=k2sin 1kandlimx→k fx=limx→k x2sin 1x=k2sin 1k.Here,limx→k fx=fkHence,thefunctionfiscontinuousfork≠0.

Secondcase:If,k=0,f0=0LHL=limx→0-fx=limx→0-x2sin1x=limx→0x2sin1x

Weknowthat,-1≤sin 1x≤1,x≠0⇒-x2≤x2sin 1x≤x2

⇒limx→0-x2≤limx→0x2sin1x≤limx→0x2⇒0≤limx→0x2sin1x≤0⇒limx→0x2sin1x=0⇒limx→0-fx=0Similarly,RHL=limx→0+fx=limx→0+x2sin1x=limx→0x2sin1x=0Here,atx=0,LHL=RHL=f0Hence,atx=0,fiscontinuous.Therefore,thefunctionfiscontinuousforallrealnumbers.

Q.20. Examinethecontinuityoff,wherefisdefinedbyfx=sinx-cosx,Ifx≠0-1,Ifx=0.


Givenfunctionisdefinedbyfx=sin x-cosx,Ifx≠0-1,Ifx=0

Letkbeanyrealnumber.Accordingtoquestion,k≠0ork=0Heref(x)=sinx-cosxforx<0andx>0.And,f(0)=-1.LHL=limk→0-fx=limk→0-sinx-cosx=0-1=-1RHL=limk→0+fx=limk→0+sinx-cosx=0-1=-1Hence,forx=0,LHL=RHL=fxHence,thefunctionfiscontinuousatx=0.

Q.21. Findthevaluesofksothatthefunctionfiscontinuousattheindicatedpoint.fx=kcosxπ-2x,Ifx≠π23,Ifx=π2atx=π2

Solution: Given:Thefunctionisdefinedbyfx=kcosxπ-2x,Ifx≠π23,Ifx=π2iscontinuousatx=π2


Giventhatthefunctioniscontinuousatx=π2.Therefore,LHL=RHL=fπ2

⇒limx→π2-fx=limx→π+2fx=fπ2⇒limx→π2-kcosxπ-2x=limx→π+2kcosxπ-2x=3⇒limh→0-kcosπ2-hπ-2π2-h=limh→0+kcosπ2+hπ-2π2+h=3⇒limh→0-ksinh2h=limh→0+-ksinh-2h=3⇒k2=k2=3∵limh→0sinhh=1

⇒k=6

Hence,fork=6,thegivenfunctionfiscontinuousattheindicatedpoint.

Q.22. Findthevaluesofksothatthefunctionfiscontinuousattheindicatedpoint.fx=kx2,Ifx≤23,Ifx>2atx=2.

Solution: Givenfunctionisdefinedbyfx=kx2,Ifx≤23,Ifx>2atx=2


Giventhatthefunctioniscontinuousatx=2.Therefore,LHL=RHL=f2⇒limx→2-fx=limx→2+fx=f2⇒limx→2-kx2=limx→2+3=k22⇒4k=3=4k⇒k=34Hence,fork=34,thegivenfunctionfiscontinuousattheindicatedpoint.

Q.23. Findthevaluesofksothatthefunctionfiscontinuousattheindicatedpoint.fx=kx+1,Ifx≤πcosx,Ifx>πatx=π.

Solution: Givenfunctionisfx=kx+1,Ifx≤πcosx,Ifx>πatx=π


Giventhatthefunctioniscontinuousatx=π,Therefore,LHL=RHL=fπ⇒limx→π-fx=limx→π+fx=fπ⇒limx→π-kx+1=limx→π+cosx=kπ+1⇒kπ+1=cosπ=kπ+1⇒kπ+1=-1=kπ+1⇒πk=-2⇒k=-2πHence,fork=-2π,thegivenfunctionfiscontinuousattheindicatedpoint.

Q.24. Findthevaluesofksothatthefunctionfiscontinuousattheindicatedpoint.

fx=kx+1,Ifx≤53x-5,Ifx>5atx=5

Solution: Givenfunctionisdefinedbyfx=kx+1,Ifx≤53x-5,Ifx>5atx=5


Giventhatthefunctioniscontinuousatx=5.Therefore,LHL=RHL=f5⇒limx→5-fx=limx→5+fx=f5⇒limx→5-kx+1=limx→5+3x-5=5k+1⇒5k+1=15-5=5k+1⇒5k=9⇒k=95Hence,fork=95,thegivenfunctionfiscontinuousattheindicatedpoint.

Q.25. Examinethefollowingfunctionforcontinuity:fx=x-5.




Solution: Givenfunctionfx=x-5.

Toexamine:Continuityofthegivenfunction.

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).Letkbeanyrealnumber.Atx=k,fk=k-5.LHL=limx→k-fx=limx→k-x-5=k-5RHL=limx→k+fx=limx→k+x-5=k-5Here,atx=k,LHL=RHL=fk=k-5Hence,thefunctionfiscontinuousforallrealnumbers.

Q.26. Examinethefollowingfunctionforcontinuity:fx=1x-5,x≠5.

Solution: Givenfunctionfx=1x-5,x≠5

Toexamine:Continuityofthegivenfunction.

Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).Letkbeanyrealnumberk≠5.Atx=k,fk=1k-5LHL=limx→k-fx=limx→k-1x-5=1k-5RHL=limx→k+fx=limx→k+1x-5=1k-5Here,atx=k,LHL=RHL=fk=1k-5Hence,thefunctionfiscontinuousforallrealnumbersexcept5.

Q.27. Examinethefollowingfunctionforcontinuity:fx=x2-25x+5,x≠-5

Solution: Givenfunctionfx=x2-25x+5,x≠-5


Letkk≠-5beanyrealnumber.Atx=k,fk=k2-25k+5=k+5k-5k+5=k-5LHL=limx→k-fx=limx→k-x2-25x+5=limx→k-x+5x-5x+5=limx→k-x-5=k-5RHL=limx→k+fx=limx→k+x2-25x+5=limx→k+k+5k-5k+5=limx→k-x-5=k-5Here,atx=k,LHL=RHL=fk=k-5.Hence,thefunctionfiscontinuousforallrealnumbers(except-5).

Q.28. Examinethefollowingfunctionforcontinuity:fx=x-5

Solution: Givenfunctionisfx=x-5=5-x,x<5x-5,x≥5


Letkbeanyrealnumber.Accordingtoquestion,kcanbek<5ork=5ork>5.Firstcase:Ifk<5,fk=5-kandlimx→kfx=limx→k5-x=5-k.Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumberslessthan5.

Secondcase:Ifk=5,

fk=k-5andlimx→k fx=limx→k x-5=k-5.Here,limx→k fx=fk

Hence,thefunctionfiscontinuousatx=5.

Thirdcase:Ifk>5,

fk=k-5andlimx→k fx=limx→k x-5=k-5.Here,limx→k fx=fk

Hence,thefunctionfiscontinuousforallrealnumbersgreaterthan5.Hence,thefunctionfiscontinuousforallrealnumbers.

Q.29. Findthevaluesofaandbsuchthatthefunctiondefinedbyfx=5,Ifx≤2ax+b,If2<x<1021,Ifx≥10isacontinuousfunction.

Solution: Givenfunctionisfx=5,Ifx≤2ax+b,If2<x<1021,Ifx≥10


Giventhatthefunctioniscontinuousatx=2.Therefore,LHL=RHL=f2⇒limx→2-fx=limx→2+fx=f2⇒limx→2-5=limx→2+ax+b=5⇒2a+b=5…(i)

Giventhatthefunctioniscontinuousatx=10.Therefore,LHL=RHL=f10

⇒limx→10-fx=limx→10+ fx=f10⇒limx→10-ax+b=limx→10+ 21=21⇒10a+b=21…(ii)

Solvingtheequation(i)and(ii),wegeta=2b=1

Q.30. Showthatthefunctiondefinedbyfx=cosx2isacontinuousfunction.

Solution: Givenfunctionisdefinedbyfx=cosx2

Assumingthatthefunctionsarewelldefinedforallrealnumbers,wecanwritethegivenfunction finthecombinationofgandhf=goh.

Where,gx=cosxandh(x)=x2.Ifgandhbotharecontinuousfunctionthenfisalsocontinuous.∵gohx=ghx=gx2=cosx2

LetthefunctiongxbecosxLetkbeanyrealnumber.Atx=k,gk=cosk

limx→k gx=limx→k cosx=limh→0 cosk+h=limh→0 coskcosh-sin ksin h=cosk

Here,limx→kgx=gk.Hence,thefunctiongiscontinuousforallrealnumbers.

Andletthefunctionhxbex2Letkbeanyrealnumber.Atx=k,hk=k2

limx→k hx=limx→k x2=k2

Here,limx→khx=hk.Hence,thefunctionhiscontinuousforallrealnumbersTherefore,gandhbotharecontinuousfunction.Hence,fiscontinuous.

Q.31. Showthatthefunctiondefinedbyfx=cosxisacontinuousfunction.




Solution: Giventhatthefunctionisdefinedbyfx=cosx

Assumingthatthefunctionsarewell-definedforallrealnumbers,wecanwritethegivenfunction finthecombinationofgandhf=goh.

Where,gx=xandhx=cosx.Ifgandhbotharecontinuousfunctionthenfisalsocontinuous.∵gohx=ghx=gcosx=cosx

LetthefunctiongxbexRearrangingthefunctiong,wegetgx=-x,Ifx<0x,Ifx≥0Letkbeanyrealnumber.Accordingtoquestion,k<0ork=0ork>0Firstcase:Ifk<0,gk=0andlimx→k gx=limx→k -x=0,here,limx→k gx=gkHence,thefunctiongiscontinuousforallrealnumberslessthan0.

Secondcase:Ifk=0,g0=0=0LHL=limx→0-gx=limx→0--x=0RHL=limx→0+ gx=limx→0+ x=0,Hereatx=0,LHL=RHL=g0Hence,thefunctiongiscontinuousatx=0.

Thirdcase:Ifk>0,gk=0andlimx→k gx=limx→k x=0.Here,limx→k gx=gkHence,thefunctiongiscontinuousforallrealnumbersgreaterthan0.

Hence,thefunctiongiscontinuousforallrealnumbers.AndletthefunctionhxbecosxLetkbeanyrealnumber.Atx=k,hk=cosklimx→khx=limx→kcosx=coskHere,limx→khx=hk.Hence,thefunctionhiscontinuousforallrealnumbers.

Therefore,gandhbotharecontinuousfunction.Hence,fiscontinuous.

Q.32. Examinethatsin xisacontinuousfunction.

Solution: Letthegivenfunctionbefx=sinxAssumingthatthefunctionsarewelldefinedforallrealnumbers,wecanwritethegivenfunctionfinthecombinationofgandhf=hog.

Where,hx=sin xandgx=x.

Ifgandhbotharecontinuousfunctionthenfisalsocontinuous.∵hogx=hgx=hx=sinx

Functionhx=sinxLetkbeanyrealnumber.

Atx=k,hk=sin klimx→k hx=limx→k sin x=sin kHere,limx→k hx=hk.

Hence,thefunctionhiscontinuousforallrealnumbers.

Functiongx=xRedefiningthefunctiong,weget,gx=-x,Ifx<0x,Ifx≥0


Accordingtoquestion,k<0ork=0ork>0Firstcase:Ifk<0,gk=-kandlimx→kgx=limx→k-x=-k.Here,limx→kgx=gkHence,thefunctiongiscontinuousforallrealnumberslessthan0.

Secondcase:Ifk=0,g0=0LHL=limx→0-gx=limx→0--x=0RHL=limx→0+ gx=limx→0+ x=0Hereatx=0,LHL=RHL=g0Henceatx=0,thefunctiongiscontinuous.

Thirdcase:Ifk>0,gk=kandlimx→kgx=limx→kx=limx→kx=k.Here,limx→kgx=gkHence,thefunctiongiscontinuousforallrealnumbersgreaterthan0.

Hence,thefunctiongiscontinuousforallrealnumbers.Therefore,gandhbotharecontinuousfunctions.Hence,fiscontinuous.

Q.33. Findallthepointsofdiscontinuityoffdefinedbyfx=x-x+1.




Solution: Giventhatthefunctionisdefinedbyfx=x-x+1Assumingthatthefunctionsarewelldefinedforallrealnumbers,wecanwritethegivenfunctionfinthecombinationofgandhf=g-h,where,gx=xandhx=x+1.Ifgandhbotharecontinuousfunctionthenfalsobecontinuous.Functiongx=xRedefiningthefunctiong,weget,gx=-x,Ifx<0x,Ifx≥0Letkbeanyrealnumber.Accordingtoquestion,k<0ork=0ork>0Firstcase:Ifk<0,gk=-kandlimx→k gx=limx→k -x=-kHere,limx→k gx=gkHence,thefunctiongiscontinuousforallrealnumberslessthan0.

Secondcase:Ifk=0,g0=0=0LHL=limx→0-gx=limx→0--x=0andRHL=limx→0+ gx=limx→0+ x=0Here,atx=0,LHL=RHL=g0Hence,thefunctiongiscontinuousatx=0.

Thirdcase:Ifk>0,gk=kandlimx→kgx=limx→kx=k.Here,limx→kgx=gkHence,thefunctiongiscontinuousforallrealnumbersmorethan0.

Hence,thefunctionkiscontinuousforallrealnumbers.

Functionhx=x+1Redefiningthefunctionh,wegethx=-x+1,Ifx<-1x+1,Ifx≥-1Letkbeanyrealnumber.Accordingtoquestion,k<-1ork=-1ork>-1Firstcase:Ifk<-1,hk=-k+1andlimx→k hx=limx→k -x+1=-k+1.Here,limx→k hx=hkHence,thefunctiongiscontinuousforallrealnumberslessthan-1.

Secondcase:Ifk=-1,h-1=-1+1=0LHL=limx→-1-hx=limx→-1--x+1=--1+1=0RHL=limx→-1+ hx=limx→-1+ x+1=-1+1=0Hereatx=-1,LHL=RHL=h-1Hence,thefunctionhiscontinuousatx=-1.

Thirdcase:Ifk>-1hk=k+1andlimx→k hx=limx→k k+1=k+1.Here,limx→k hx=hkHence,thefunctiongiscontinuousforallrealnumbersgreaterthan-1.Hence,thefunctionhiscontinuousforallrealnumbers.Therefore,gandhbotharecontinuousfunction.Hence,fiscontinuous.

Q.34. Provethatthefunctionfx=xn,iscontinuousatx=n,wherenisapositiveinteger

Solution: Givenfunctionisfx=xn.


Atx=n,fn=nnAgain,limx→nfx=limx→nxn=nnHere,atx=n,limx→nfx=fn=nnHence,thefunctionfiscontinuousatx=n,wherenispositiveinteger.

Q.35. Isthefunctionfdefinedbyfx=x,x≤15,x>1continuousatx=0?Atx=1?Atx=2?

Solution: Givenfunctionisfx=x,x≤15,x>1


Atx=0,f0=0limx→0fx=limx→0x=0Hereatx=0,limx→0fx=f0=0Hence,thefunctionfiscontinuousatx=0.

Atx=1,f1=1

LHL=limx→1-fx=limx→1-x=1

RHL=limx→1+fx=limx→1+5=5Here,atx=1,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=1.

Atx=2,f2=5

limx→2 fx=limx→2 5=5

Here,atx=2,limx→2fx=f2=5Hence,thefunctionfiscontinuousatx=2.

Q.36. Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=2x+3,Ifx≤22x-3,Ifx>2

Solution: Given:fx=2x+3,Ifx≤22x-3,Ifx>2


Letkbeanyrealnumber.Accordingtoquestion,k<2ork=2ork>2Firstcase:k<2fk=2k+3andlimx→kfx=limx→k2x+3=2k+3.Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumberssmallerthan2.

Secondcase:Ifk=2,f2=2k+3

LHL=limx→2-fx=limx→2-2x+3=7

RHL=limx→2+fx=limx→2+2x-3=1Here,atx=2,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=2.

Thirdcase:Ifk>2,

fk=2k-3andlimx→k fx=limx→k 2x-3=2k-3.Here,limx→k fx=fk

Therefore,thefunctionfiscontinuousforallrealnumbersgreaterthan2.Hence,thefunctionfisdiscontinuousonlyatx=2.

Q.37. Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=x+3,Ifx≤-3-2x,If-3<x<36x+2,Ifx≥3.




Solution: Given:fx=x+3,Ifx≤-3-2x,If-3<x<36x+2,Ifx≥3


Letkbeanyrealnumber.Accordingtoquestion,k<-3ork=-3or-3<k<3ork=3ork>3

Firstcase:Ifk<-3,

fx=-k+3andlimx→k fx=limx→k -x+3=-k+3.Here,limx→k fx=fk

Hence,thefunctionfiscontinuousforallrealnumberslessthan-3.

Secondcase:Ifk=-3,f-3=--3+3=6

LHL=limx→-3-fx=limx→-3--x+3=--3+3=6

RHL=limx→-3+fx=limx→-3+-2x=-2-3=6.Here,limx→kfx=fkHence,thefunctionfiscontinuousatx=-3.

Thirdcase:If-3<k<3,

fk=-2kandlimx→k fx=limx→k f-2x=-2k.Here,limx→k fx=fk

Hence,thefunctionfiscontinuousat-3<x<3.

Fourthcase:Ifk=3,

LHL=limx→k-fx=limx→k--2x=-2k

RHL=limx→k+fx=limx→k+6x+2=6k+2,Here,atx=3,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=3.

Fifthcase:Ifk>3,

fk=6k+2andlimx→k fx=limx→k 6x+2=6k+2.Here,limx→k fx=fk

Hence,thefunctionfiscontinuousforallnumbersgreaterthan3.Hence,thefunctionfisdiscontinuousonlyatx=3.

Q.38. Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=xx,Ifx≠00,Ifx=0.

Solution: Givenfunction:fx=xx,Ifx≠00,Ifx=0

Afterredefiningthefunctionf,weget

fx=-xx=-1,Ifx<00,Ifx=0xx=1,Ifx>0Letkbeanyrealnumber.Afunctionf(x)iscontinuousatapointk,iflimx→k+f(x)=limx→k-f(x)=f(k).Accordingtoquestion,k<0ork=0ork>0.

Firstcase:Ifk<0,

fk=-kk=-1andlimx→k fx=limx→k -xx=limx→k -1=-1.

Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumberssmallerthan0.

Secondcase:If,k=0,f0=0

LHL=limx→k-fx=limx→k--xx=-1andRHL=limx→k+ fx=limx→k+ xx=1.

Here,atx=0,LHL≠RHL.Hence,thefunctionfisdiscontinuousatx=0.

Thirdcase:Ifk>0,

fk=kk=1andlimx→k fx=limx→k xx=limx→k 1=1.

Here,limx→kfx=fkHence,thefunctionfiscontinuousforallrealnumbersgreaterthan0.Therefore,thefunctionfisdiscontinuousonlyatx=0.

Q.39. Findallpointsofdiscontinuityoff,wherefisdefinedbyfx=xx,Ifx<0-1,Ifx≥0.

Solution: Givenfunction:fx=xx,Ifx<0-1,Ifx≥0

Weknowthat,forx<0,x=-x.

Therefore,f(x)=xx=x-x=-1.And,giventhatf(x)=-1ifx>0.So,f(x)=-1,foranyvalueofx

Now,letkbeanyvalueofx,whereweneedtocheckthecontinuityofthefunction.

Nowwhenk=0,

LHL=limx→k-f(x)=limx→0-xx=limx→0-x-x=limx→0--1=-1RHL=limx→k+f(x)=limx→0+-1=-1So,hereLHL=RHL=f(k)forallrealvaluesofk.Hence,thefunctionfiscontinuousforallrealnumbers.




Exercise5.2

Q.1. Differentiatethefunctionwithrespecttox.sin x2+5

Solution: Givenfunction:sinx2+5

Lety=sin x2+5

Differentiatingbothsideswithrespecttoxandbyusingchainruleweget,dydx=ddxsinx2+5=cosx2+5·ddxx2+5∵ddxf(g(x))=f'(g(x))·g'(x)=cosx2+5·2x∵ddxxn=nxn-1Hence,dsinx2+5dx=2xcosx2+5

Q.2. Provethatthegreatestintegerfunctiondefinedbyfx=x,0<x<3,isnotdifferentiableatx=1andx=2.

Solution: Givenfunction:fx=x,0<x<3

Toprove:Thegreatestintegerfunctiondefinedbyfxisnotdifferentiableatx=1andx=2.

Atx=1,Afunctionisdifferentiableatapointifleft-handderivativeandright-handderivativeareequalatthatpoint. Forafunctionf(x),thederivativeatapointkisgivenby,f'(x)=limx→kf(x+h)-f(x)h,whereh=x-k.Here,LHD=limh→0f1-h-f1-h=limh→01-h-1-h=limh→00-1-h=∞RHD=limh→0f1+h-f1h=limh→01+h-1h=limh→01-1h=0Here,LHD≠RHD.Therefore,thefunctionfx=x,0<x<3,isnotdifferentiableatx=1.

Atx=2,LHD=limh→0f2-h-f2-h=limh→02-h-2-h=limh→01-2-h=∞

RHD=limh→0f2+h-f2h=limh→0 2+h-2h=limh→0 2-2h=0

Here,LHD≠RHD.Therefore,thefunctionfx=x,0<x<3,isnotdifferentiableatx=2.

Q.3. Differentiatethefunctionwithrespecttoxcossin x

Solution: Givenfunction:cossinx

Lety=cossin x

Differentiatingbothsideswithrespecttoxweget,dydx=dcossinxdsinx·dsinxdx∵ddxf(g(x))=f'(g(x))·g'(x)⇒dydx=-sinsinx·ddxsinx=-sinsinx·cosxHence,dcossinxdx=-sinsinx·cosx

Q.4. Differentiatethefunctionwithrespecttox.sin ax+b

Solution: Lety=sinax+b

Differentiatingbothsideswithrespecttoxbyapplyingchainruleweget,

dydx=cosax+b·ddxax+b=cosax+b·aHence,dsinax+bdx=cosax+b·a

Q.5. Differentiatethefunctionwithrespecttox.sectan x

Solution: Givenfunction:sec(tan(x))

Lety=sectan x

Differentiatingbothsideswithrespecttoxbyapplyingchainruleweget, dydx=sectanxtantanx·ddxtanx∵ddxf(g(x))=f'(g(x))·g'(x)Applyingchainruleagainweget,=sectanxtantanx·sec2xddxx=sectanxtantanx·sec2x12x∵ddxxn=nxn-1wheren=12=sectanxtantanxsec2x2xTherefore,dsectanxdx=sectanxtantanxsec2x2x.

Q.6. Differentiatethefunctionwithrespecttox.

cos(x)

Solution: Lety=cos(x)

Weneedtofindderivativeofyw.r.t.x

i.e.y'=ddxcosx⇒dydx=d(cosx)dxApplyingchainruleweget,=-sinx·dxdx∵ddxf(g(x))=f'(g(x))g'(x)=-sinx·12x∵ddxxn=nxn-1=-sinx2xTherefore,thederivativeofcos(x)is-sinx2x.

Q.7. Differentiatethefunctionwithrespecttox.cosx3·sin2 x5

Solution: Lety=cosx3·sin2x5


dydx=cosx3·ddxsin2x5+sin2x5·ddxcosx3∵(uv)'=u'v+uv'ApplyingchainruleofdifferentiationforboththetermsonRHSweget,=cosx3·2sinx5cosx5·ddxx5+sin2x5-sinx3·ddxx3∵ddxf(g(x))=f'(g(x))·g'(x)=cosx3·2sinx5cosx5·5x4-sin2x5sinx3·3x2∵ddxxn=nxn-1=cosx3·sin2x5·5x4-sin2x5sinx3·3x2Therefore,dcosx3·sin2x5dx=cosx3·sin2x5·5x4-sin2x5sinx3·3x2

Q.8. Differentiatethefunctionwithrespecttox.2cot x2

Solution: Givenfunction:2cotx2

Lety=2cot x2

Differentiatingbothsideswithrespecttoxandapplyingchainruleweget,dydx=2·12cotx2·ddxcotx2∵ddxf(g(x))=f'(g(x))·g'(x)Againapplyingchainruleweget,=1cotx2·-cossec2x2·ddxx2=1cotx2·-cossec2x2.2x∵ddxxn=nxn-1=-2xcosec2x2cotx2Therefore,d2cotx2dx=-2xcosec2x2cotx2.

Q.9. Differentiatethefunctionswithrespecttox:

cos(x)




Solution: Givenfunction:cos(x)

Lety=cos(x).

Differentiatingbothsideswithrespecttoxweget,dydx=d(cosx)dxByapplyingchainruleweget, dydx=-sinx·dxdx∵ddxf(g(x))=f'(g(x))·g'(x)=-sinx·12x∵ddxxn=nxn-1,wheren=12=-sinx2xTherefore,thederivativeofcos(x)withrespecttoxis-sinx2x.

Q.10. Provethatthefunctionfgivenbyfx=x-1,x∈R,isnotdifferentiableatx=1.

Solution: Givenfunction:f(x)=x-1

Toprove:Provethatthefunctionfgivenbyfxisnotdifferentiableatx=1.

Afunctionisdifferentiableatapointifleft-handderivativeandright-handderivativeareequalatthatpoint. Forafunctionf(x),thederivativeatapointkisgivenby,f'(x)=limx→kf(x+h)-f(x)h,whereh=x-k.Here,Atx=1,LHD=limh→0-f1-h-f1-h=limh→0-1-h-1-1-1-h=limh→0-h-h=-1RHD=limh→0+f1+h-f1h=limh→0+1+h-1-1-1h=limh→0+hh=1Here,LHD≠RHD.Therefore,thefunctionfx=x-1,x∈R,isnotdifferentiableatx=1.(Henceproved)




Exercise5.3

Q.1. Finddydxinthefollowing:2x+3y=sin x.

Solution: Givenequationis2x+3y=sinx

Differentiatingbothsideswithrespecttox,weget

ddx2x+3y=ddxsinx⇒ddx2x+ddx3y=ddxsinx∵ddxf(x)+g(x)=ddxf(x)+ddxg(x)⇒2+3dydx=cosx∵ddxxn=nxn-1andddxk·f(x)=kddxf(x)⇒dydx=cosx-23Therefore,thevaluefordydxiscosx-23.

Q.2. Finddydxinthefollowing:y=tan-1 3x-x31-3x2,-13<x<13.

Solution: Givenequationisy=tan-13x-x31-3x2Letx=tanθ

Therefore,y=tan-1 3tan θ-tan3 θ1-3tan2 θ

=tan-1tan3θ=3θ=3tan-1x⇒y=3tan-1xDifferentiatingbothsideswithrespecttox,weget,dydx=31+x2∵dtan-1pdp=11+p2

Q.3. Finddydxinthefollowing:y=cos-1 1-x21+x2,0<x<1.

Solution: Givenequationisy=cos-11-x21+x2Letx=tanθTherefore,y=cos-11-tan2θ1+tan2θ

=cos-1 cos2θ=2θ=2tan-1 x

⇒y=2tan-1xDifferentiatingbothsideswithrespecttox,wegetdydx=21+x2

Q.4. Finddydxinthefollowing:y=sin-1 1-x21+x2,0<x<1.

Solution: Givenequationisy=sin-11-x21+x2Letx=tanθ

Therefore,

y=sin-11-tan2θ1+tan2θ=sin-1cos2θ=sin-1sinπ2-2θ=π2-2θ=π2-2tan-1x⇒y=π2-2tan-1xDifferentiatingbothsideswithrespecttox,wegetdydx=0-21+x2=-21+x2

Q.5. Finddydxinthefollowing:y=cos-1 2x1+x2,-1<x<1.

Solution: Givenequationisy=cos-12x1+x2Letx=tanθ

Therefore,y=cos-1 2tan θ1+tan2 θ

=cos-1sin2θ=cos-1cosπ2-2θ=π2-2θ=π2-2tan-1x⇒y=π2-2tan-1xDifferentiatingbothsideswithrespecttox,wegetdydx=0-21+x2=-21+x2

Q.6. Finddydxinthefollowing:y=sin-1 2x1-x2,-12<x<12.

Solution: Givenequationisy=sin-12x1-x2

Letx=sin θ

Therefore,y=sin-12sinθ1-sin2θ=sin-12sinθcosθ=sin-1sin2θ=2θ=2sin-1x⇒y=2sin-1xDifferentiatingbothsideswithrespecttox,wegetdydx=21-x2

Q.7. Finddydxinthefollowing:y=sec-1 12x2-1,0<x<12

Solution: Givenequationisy=sec-112x2-1Letx=cosθTherefore,y=sec-112cos2θ-1=sec-11cos2θ

=sec-1 sec2θ=2θ=2cos-1 x

⇒y=2cos-1xDifferentiatingbothsideswithrespecttox,wegetdydx=-21-x2

Q.8. Finddydxinthefollowing:2x+3y=sin y.

Solution: Givenequationis2x+3y=siny

Differentiatingbothsideswithrespecttox,weget,

ddx2x+ddx3y=ddxsiny⇒2+3dydx=cosydydx⇒dydxcosy-3=2⇒dydx=2cosy-3Therefore,thevalueofdydxis2cosy-3.

Q.9. Finddydxinthefollowing:ax+by2=cosy.

Solution: Givenequationisax+by2=cosy

Differentiatingbothsideswithrespecttoxbyusingtheadditionruleofdifferentiationweget,

ddxax+ddxby2=ddxcosy⇒a+2bydydx=-sinydydx∵ddxxn=nxn-1⇒dydx2by+siny=-a⇒dydx=-a2by+sinyTherefore,thevalueofdydxis-a2by+siny.

Q.10. Finddydxinthefollowing:xy+y2=tan x+y.

Solution: Givenequationisxy+y2=tanx+y

Differentiatingbothsideswithrespecttox,weget

ddxxy+ddxy2=ddxtanx+dydxApplyingproductrule:(uv)'=u'v+uv'andusingtheformula:ddxxn=nxn-1weget,⇒xdydx+y+2ydydx=sec2x+dydx⇒dydxx+2y-1=sec2x-y⇒dydx=sec2x-yx+2y-1Therefore,thevalueofdydxissec2x-yx+2y-1.

Q.11. Finddydxinthefollowing:x2+xy+y2=100.




Solution: Givenequationisx2+xy+y2=100


ddxx2+ddxxy+ddxy2=ddx100Applyingproductrule:(uv)'=u'v+uv'andusingtheformula:ddxxn=nxn-1weget,⇒2x+xdydx+y+2ydydx=0⇒dydxx+2y=-(2x+y)⇒dydx=-2x+yx+2yTherefore,thevalueofdydxis-2x+yx+2y.

Q.12. Finddydxinthefollowing:x3+x2y+xy2+y3=81.

Solution: Givenequationisx3+x2y+xy2+y3=81


ddxx3+ddxx2y+ddxxy2+ddxy3=ddx81⇒3x2+x2dydx+y·2x+x·2ydydx+y2·1+3y2dydx=0⇒dydxx2+2xy+3y2=-3x2+2xy+y2⇒dydx=-3x2+2xy+y2x2+2xy+3y2

Q.13. Finddydxinthefollowing:sin2 y+cosxy=k.

Solution: Givenequationissin2y+cosxy=k


ddxsin2y+ddxcosxy=ddxk⇒2sinycosydydx-sinxyxdydx+y=0⇒sin2ydydx-xsinxydydx-ysinxy=0⇒sin2y-xsinxydydx=ysinxy⇒dydx=ysinxysin2y-xsinxy

Q.14. Finddydxinthefollowing:sin2 x+cos2 y=1.

Solution: Givenequationissin2x+cos2y=1


ddxsin2x+ddxcos2y=ddx1⇒2sinxcosx+2cosy-sinydydx=0⇒sin2x-sin2ydydx=0⇒dydx=sin2xsin2y

Q.15. Finddydxinthefollowing:y=sin-1 2x1+x2.

Solution: Givenequationisy=sin-12x1+x2Letx=tanθ

Therefore,y=sin-1 2tan θ1+tan2 θ=sin-1 sin 2θ=2θ=2tan-1 x

⇒y=2tan-1xDifferentiatingbothsideswithrespecttox,weget,dydx=21+x2∵dtan-1pdp=11+p2




Exercise5.4

Q.1. Differentiatethefollowingw.r.t.x:exsin x

Solution: GivenexpressionisexsinxLety=exsinx

Byapplyingthequotientruleanddifferentiatingbothsideswithrespecttoxweget,

dydx=ex·ddxsinx-sinxddxexsin2x=excosx-sinx·exsin2x∵ddxuv=uv'-vu'v2=excosx-sinxsin2xTherefore,derivativeofthefunctionexsinxis=excosx-sinxsin2x.

Q.2. Differentiatethefollowingw.r.t.x:coslogx+ex

Solution: Expressioniscoslogx+ex

Lety=coslogx+ex

Byapplyingchainruleanddifferentiatingbothsideswithrespecttoxweget,dydx=ddxcos(logx+ex)=-sinlogx+exddx(logx+ex)=-sinlogx+ex·1x+exTherefore,thederivativeofcoslogx+exis-sinlogx+ex·1x+ex.

Q.3. Differentiatethefollowingw.r.t.x:esin-1 x

Solution: Givenexpressionisesin-1xLety=esin-1x

Byusingchainruleanddifferentiatingbothsideswithrespecttoxweget,dydx=esin-1 x·ddxsin-1 x=esin-1 x·11-x2=esin-1 x1-x2

Therefore,derivativeofesin-1xisesin-1x1-x2.

Q.4. Differentiatethefollowingw.r.t.x:ex3

Solution: Givenexpressionisex3Lety=ex3

Byusingchainruleanddifferentiatingbothsideswithrespecttoxweget,

dydx=ex3·ddxx3=ex3·3x2=3x2ex3Therefore,derivativeofex3is3x2ex3.

Q.5. Differentiatethefollowingw.r.t.x:sin tan-1 e-x

Solution: Expressionissintan-1e-xLety=sintan-1e-x

Byusingchainruleanddifferentiatingwithrespecttoxweget,

dydx=costan-1e-x·ddxtan-1e-x=costan-1e-x·11+e-x2·ddxe-x=costan-1e-x·11+e-2x·-e-x=-e-xcostan-1e-x1+e-2x.Therefore,thederivativeofsintan-1e-xis-e-xcostan-1e-x1+e-2x.

Q.6. Differentiatethefollowingw.r.t.x:logcosex

Solution: Expressionislogcosex

Lety=logcosex,

Byusingchainruleanddifferentiatingwithrespecttoxweget,dydx=1cosex·ddxcosex=1cosex-sinexddxex=-tanex·exTherefore,thederivativeoflogcosexis-tanex·ex.

Q.7. Differentiatethefollowingw.r.t.x:ex+ex2+…+ex5

Solution: Expressionisex+ex2+…+ex5Lety=ex+ex2+ex3+ex4+ex5


dydx=ex+ex2ddxx2+ex3ddxx3+ex4ddxx4+ex5ddxx5∵ddxf(x)±g(x)±h(x)=ddxf(x)±ddxg(x)±ddxh(x)andddxef(x)=ef(x)ddxf(x)=ex+ex2·2x+ex3·3x2+ex4·4x3+ex5·5x4∵ddxxn=nxn-1=ex+2xex2+3x2ex3+4x3ex4+5x4ex5

Q.8. Differentiatethefollowingw.r.t.x:ex,x>0

Solution: Expressionisex,x>0Lety=ex

Weknowthat,

df(g(h(x)))dx=df(g(h(x)))dg(h(x))×dg(h(x))dh(x)×dh(x)dxtherefore,dydx=12exddxex=12ex·ex·ddxx=12ex·ex·12x=ex4x

Q.9. Differentiatethefollowingw.r.t.x:loglogx,x>1

Solution: Expressionisloglogx,x>1.

Byapplyingchainruleanddifferentiatingwithrespecttoxweget,

dydx=1logx·ddxlogx=1logx·1x=1xlogxTherefore,thederivativeoflog(logx)is1xlogx.




Q.10. Differentiatethefollowingw.r.t.x:cosxlogx,x>0

Solution: Givenexpressioniscosxlogx,x>0Lety=cosxlogx

Byapplyingquotientruleanddifferentiatingbothsideswithrespecttoxweget,

dydx=logxddxcosx-cosxddxlogxlogx2=logx·-sinx-cosx·1xlogx2=-xsinxlogx+cosxxlogx2∵uv'=u'v-uv'v2Therefore,dydx=-xsinxlogx+cosxxlogx2.




Exercise5.5

Q.1. Differentiatethefunctionsgiven:cosx·cos2x·cos3x.

Solution: Functioniscosx·cos2x·cos3x

Lety=cosx·cos2x·cos3x,takinglogonboththesides,

logy=logcosx+logcos2x+logcos3x[∵log(ab)=loga+logb]Differentiatingbothsidesbyxandapplyingchainruleforeachtermweget,1ydydx=1cosx·ddxcosx+1cos2x·ddxcos2x+1cos3x·ddxcos3x∵ddxlogx=1x⇒dydx=y1cosx·-sinx+1cos2x·-sin2x·2+1cos3x·-sin3x.3=cosx·cos2x·cos3x-tanx-2tan2x-3tan3xTherefore,thederivativeofcosx·cos2x·cos3xiscosx·cos2x·cos3x-tanx-2tan2x-3tan3x.

Q.2. Differentiatethefunctiongiven:xxcosx+x2+1x2-1.

Solution: Givenfunctionisxxcosx+x2+1x2-1

Letu=xxcosxandv=x2+1x2-1andy=u+v.

Differentiatingbothsideswithrespecttox,wegetdydx=dudx+dvdx…(i)

u=xxcosx,takinglogonboththesides

logu=xlogx,

Differentiatingbothsideswithrespecttox,weget1ududx=xcosx·ddxlogx+logx·ddxxcosx=xcosx·1x+logx·-x·sinx+cosx1ududx=cosx-xsinxlogx+cosxlogxdudx=ucosx-xsinxlogx+cosxlogx=xxcosxcosx-xsinxlogx+cosxlogx…(ii)

v=x2+1x2-1,takinglogonboththesides

logv=logx2+1-logx2-1,

Differentiatingbothsideswithrespecttox,weget1vdvdx=1x2+1·2x-1x2-1·2x=2x(x2-1)-2x(x2+1)(x2+1)(x2-1)=-4x(x2+1)(x2-1)dvdx=v-4xx2+1x2-1=x2+1x2-1-4xx2+1x2-1=-4x(x2-1)2…(iii)Puttingthevalueofdudxfrom(ii)anddvdxfrom(iii)inequation(i),wegetdydx=xxcosxcosx-xsinxlogx+cosxlogx-4xx2-12

Q.3. Differentiatethefunctionsgiven:xcosxx+xsin x1x.

Solution: Functionisxcosxx+xsinx1x

Letu=xcosxxandv=xsin x1x,andy=u+v.

Differentiatingwithrespecttox,weget,dydx=dudx+dvdx…(i)

u=xcosxx,takinglogonboththesides

logu=xlogxcosx,

Differentiatingwithrespecttox,weget,1ududx=x·ddxlogxcosx+logxcosx·ddxx=x·1xcosx-xsinx+cosx+logxcosx·1=-xtanx+1+logxcosxdudx=xcosxx1-xtanx+logxcosx=xcosxx1-xtanx+logxcosx…(ii)

v=xsinx1x,takinglogonboththesides

logv=1xlogxsin x,

Differentiatingwithrespecttox,weget,1vdvdx=1x·ddxlogxsinx+logxsinx·ddx1x=1x·1xsinxxcosx+sinx+logxsinx-1x2dvdx=vxcotx+1-logxsinxx2=xsinx1xxcotx+1-logxsinxx2…(iii)

Puttingthevalueofdudxfrom(ii)anddvdxfrom(iii)inequation(i),weget

dydx=xcosxx1-xtan x+logxcosx+xsin x1xxcot x+1-logxsin xx2

Q.4. Finddydxofthefunctiongiven:xy+yx=1.

Solution: Functionisxy+yx=1

Letu=xyandv=yx,therefore,u+v=1

Differentiatingwithrespecttox,wegetdudx+dvdx=0…(i)

u=xy,takinglogonboththesides,

logu=ylogx,

Differentiatingwithrespecttox,weget1ududx=y·ddxlogx+logx·ddxy=y·1x+logx·dydxdudx=xyyx+logx·dydx…(ii)

Andv=yx,takinglogonboththesides

logv=xlogy,

Differentiatingwithrespecttox,weget1vdvdx=x·ddxlogy+logy·ddxx=x·1ydydx+logy·1dvdx=vxydydx+logy=yxxydydx+logy…(iii)


xyyx+logx·dydx+yxxydydx+logy=0

⇒yxy-1+xylogxdydx+xyx-1dydx+yxlogy=0⇒dydxxylogx+xyx-1=-yxlogy+yxy-1⇒dydx=-yxlogy+yxy-1xylogx+xyx-1

Q.5. Finddydxofthefunctionsgiven:yx=xy.

Solution: Functionisyx=xy

Takinglogonboththesides,xlogy=ylogx,

x·ddxlogy+logy·ddxx=y·ddxlogx+logx·ddxy⇒x·1ydydx+logy·1=y·1x+logx·dydx⇒dydxxy-logx=yx-logydydxx-ylogxy=y-xlogyx⇒dydx=y(y-xlogy)x(x-ylogx)

Q.6. Finddydxofthefunctionsgiven:cosxy=cosyx.




Solution: Functioniscosxy=cosyx

Takinglogonboththesides,ylogcosx=xlogcosy,

Differentiatingbothsideswithrespecttoxweget,y·ddxlogcosx+logcosx·ddxy=x·ddxlogcosy+logcosy·ddxx⇒y1cosx-sinx+logcosx·dydx=x·1cosy-sinydydx+logcosy·1dydxlogcosx+xtany=cosy+ytanx⇒dydx=cosy+ytanxlogcosx+xtany

Q.7. Finddydxofthefunctiongiven:xy=ex-y.

Solution: Functionisxy=ex-y

Takinglogonboththesides,

logx+logy=x-yloge⇒logx+logy=x-y,Differentiatingbothsideswithrespecttoxweget,1x+1y·dydx=1-dydxdydx1y+1=1-1x⇒dydx1+yy=x-1x⇒dydx=yx-1xy+1

Q.8. Findthederivativeofthefunctiongivenbyfx=1+x1+x21+x41+x8,andhencefindf'1.

Solution: Functionisfx=1+x1+x21+x41+x8

Takinglogonboththesides,

logfx=log1+x+log1+x2+log1+x4+log1+x8,

1fx·ddxfx=11+x+11+x2·ddxx2+11+x4·ddxx4+11+x8·ddxx8

⇒1fx·f'x=11+x+11+x2·2x+11+x4·4x3+11+x8·8x7

⇒f'(x)=fx11+x+2x1+x2+4x31+x4+8x71+x8⇒f'(x)=1+x1+x21+x41+x811+x+2x1+x2+4x31+x4+8x71+x8

f'(1)=1+11+11+11+111+1+21+1+41+1+81+1

⇒f'1=222212+22+42+82=16152=120

Q.9. Differentiatex2-5x+8x3+7x+9inthewaymentionedbelow:

byusingtheproductrule

Solution: Givenfunction:x2-5x+8x3+7x+9

dydx=x2-5x+8ddxx3+7x+9+x3+7x+9ddxx2-5x+8∵(uv)'=u'v+v'u

=x2-5x+83x2+7+x3+7x+92x-5=3x4+7x2-15x3-35x+24x2+56+2x4-5x3+14x2-35x+18x-45=5x4-20x3+45x2-52x+11


byexpandingtheproducttoobtainasinglepolynomial.

Solution: y=x2-5x+8x3+7x+9

=x5+7x3+9x2-5x4-35x2-45x+8x3+56x+72

=x5-5x4+15x3-26x2+11x+72Now,differentiatingbothsideswithrespecttoxweget,dydx=ddxx5-5ddxx4+15ddxx3-26ddxx2+11ddxx+ddx72=5x4-20x3+45x2-52x+11


bylogarithmicdifferentiation.

Solution: y=x2-5x+8x3+7x+9


logy=logx2-5x+8+logx3+7x+9

1y·dydx=1x2-5x+8·ddxx2-5x+8+1x3+7x+9·ddxx3+7x+9

1y·dydx=1x2-5x+8·2x-5+1x3+7x+9·3x2+7

dydx=y2x-5x3+7x+9+3x2+7x2-5x+8x2-5x+8x3+7x+9

=y2x4+14x2+18x-5x3-35x-45+3x4-15x3+24x2+7x2-35x+56x2-5x+8x3+7x+9

dydx=x2-5x+8x3+7x+95x5-20x3+45x2-52x+11x2-5x+8x3+7x+9⇒dydx=5x4-20x3+45x2-52x+11

Q.12. Ifu,vandwarefunctionsofx,thenshowthat

ddxu·v·w=dudxv·w+u·dvdx·w+u·vdwdx

intwoways–firstbyrepeatedapplicationofproductrule,secondbylogarithmicdifferentiation.

Solution: Giventhatu,vandwarefunctionsofx

Lety=u.v.w=u.v.w

Differentiationbyrepeatedapplicationofproductruledydx=u·ddxv·w+v·w·ddxu=uvddxw+wddxv+v·w·dudx

dydx=u·v·dwdx+u·w·dvdx+v·w·dudx

Differentiationusinglogarithmic

Lety=u·v·wTakinglogonboththesides,logy=logu+logv+logw1y·dydx=1u·dudx+1v·dvdx+1w·dwdx⇒dydx=y1u·dudx+1v·dvdx+1w·dwdx

dydx=u·v·w1u·dudx+1v·dvdx+1w·dwdx

⇒dydx=u·v·wu·dudx+u·v·wv·dvdx+u·v·ww·dwdx

⇒dydx=v·w·dudx+u·w·dvdx+u·v·dwdx

Q.13. Differentiatethefunctionsgiven:x-1x-2x-3x-4x-5.




Solution: Functionisx-1x-2x-3x-4x-5

Lety=x-1x-2x-3x-4x-5=x-1x-2x-3x-4x-512

Takinglogonboththesidesweget,logy=logx-1x-2x-3x-4x-512=12logx-1+logx-2-logx-3-logx-4-logx-5∵logAm=mlogAandlogABC=logA+logB-logCDifferentiatingbothsideswithrespecttoxweget,1ydydx=121x-1+1x-2-1x-3-1x-4-1x-5∴dydx=12x-1x-2x-3x-4x-51x-1+1x-2-1x-3-1x-4-1x-5

Q.14. Differentiatethefunctionsgiven:logxcosx.

Solution: Givenfunctionislogxcosx

Lety=logxcosx,takinglogonboththesides,

logy=loglogxcosx=cosx·log(logx)Differentiatingbothsideswithrespecttoxweget,1ydydx=cosx·ddxloglogx+loglogx·ddxcosxdydx=ycosx·1logx.1x+loglogx·-sinx⇒dydx=logxcosxcosx-sinxloglogxxlogx

Q.15. Differentiatethefunctionsgiven:xx-2sin x.

Solution: Givenfunctionisxx-2sinx

Letu=xxandv=2sin xtherefore,y=u-v

Differentiatingwithrespecttoxonbothsidesdydx=dudx-dvdx…(i)Here,u=xx,takinglogonboththesideslogu=xlogx,therefore,1ududx=x·ddxlogx+logx·ddxx=x.1x+logx·1=1+logxdudx=u1+logx=xx1+logx…(ii)

Andv=2sinx,takinglogonboththesides

logv=sin xlog2,therefore,

1vdvdx=log2⋅ddxsinx=log2⋅cosxdvdx=v[cosxlog2]=2sinx[cosxlog2]…(iii)Therefore,dydx=dudx-dvdx=xx1+logx-2sinx[cosxlog2]

Q.16. Differentiatethefunctiongiven:x+32·x+43·x+54

Solution: Functionisx+32·x+43·x+54Lety=x+32·x+43·x+54,takinglogonboththesideslogy=2logx+3+3logx+4+4logx+5


1ydydx=2·1x+3+3·1x+4+4·1x+5

⇒dydx=y2x+4x+5+3x+3x+5+4x+3x+4x+3x+4x+5⇒dydx=y2x2+9x+20+3x2+8x+15+4x2+7x+12(x+3)(x+4)(x+5)dydx=(x+3)2·(x+4)3·(x+5)49x2+70x+133(x+3)(x+4)(x+5)⇒dydx=(x+3)·(x+4)2⋅(x+5)39x2+70x+133

Q.17. Differentiatethefunctiongiven:x+1xx+x1+1x.

Solution: Givenfunctionisx+1xx+x1+1xLetu=x+1xxandv=x1+1x,

therefore,y=u+vDifferentiatingwithrespecttox,weget,dydx=dudx+dvdx…(i)

u=x+1xx,takinglogonboththesideslogu=xlogx+1x,Differentiatingwithrespecttox,weget,1ududx=x⋅ddxlogx+1x+logx+1x⋅ddxx=x⋅1x+1x⋅1-1x2+logx+1x⋅1=x2x2+1⋅x2-1x2+logx+1xdudx=x+1xxx2-1x2+1+logx+1x…(ii)

v=x1+1x,

takinglogonboththesideslogv=1+1xlogx,

Differentiatingwithrespecttox,weget,1vdvdx=1+1x·ddxlogx+logx·ddx1+1x=1+1x·1x+logx·-1x2dvdx=vx+1x2-logxx2=x1+1xx+1-logxx2…(iii)Puttingthevalueofdudxfrom(ii)anddvdxfrom(iii)inequation(i),weget,dydx=x+1xxx2-1x2+1+logx+1x+x1+1xx+1-logxx2

Q.18. Differentiatethefunctionsgivenlogxx+xlogx

Solution: Functionislogxx+xlogxLetu=logxxandv=xlogx,therefore,y=u+vDifferentiatingwithrespecttox,wegetdydx=dudx+dvdx…(i)

u=logxx,takinglogonboththesideslogu=xloglogx1ududx=x.ddxloglogx+loglogx.ddxx=x.1logx.1x+loglogx.1=1logx+loglogxdudx=logxx1+logx.loglogxlogx=logxx-11+logx.loglogx…(ii)

v=xlogx,takinglogonboththesideslogv=logxlogx,therefore,1vdvdx=logx.ddxlogx+logx.ddxlogx=logx.1x+logx.1xdvdx=v2logxx=xlogx2logxx=xlogx-12logx…(iii)

Puttingthevalueofdudxfrom(ii)anddvdxfrom(iii)inequation(i),wegetdydx=logxx-11+logx.loglogx+xlogx-12logx

Q.19. Differentiatethefunctiongiven:sin xx+sin-1 x.




Solution: Functionissinxx+sin-1x

Letu=sin xxandv=sin-1 x,therefore,y=u+v


u=sinxx,takinglogonboththesides

logu=xlogsin x,therefore,

1ududx=x·ddxlogsinx+logsinx·ddxx=x·1sinx·cosx+logsinx·1=xcotx+logsinxdudx=sinxxxcotx+logsinx…(ii)

v=sin-1x,therefore,

dvdx=ddxsin-1x

⇒dvdx=11-x·ddxx=11-x·12x=12x-x2…(iii)Puttingthevalueofdudxfrom(ii)anddvdxfrom(iii)inequation(i),wegetdydx=sinxxxcotx+logsinx+12x-x2

Q.20. Differentiatethefunctionsgiven:xsin x+sin xcosx.

Solution: Functionisxsinx+sinxcosx

Letu=xsin xandv=sin xcosxandy=u+v


u=xsinx,takinglogonboththesides

logu=sin xlogx,

Differentiatingwithrespecttox,weget,1ududx=sinx·ddxlogx+logx·ddxsinx=sinx·1x+logx·cosx=sinxx+logxcosxdudx=xsinxsinxx+logxcosx=xsinx-1sinx+xlogxcosx…(ii)

v=sinxcosx,takinglogonboththesides,

logv=cosxlogsin x,

Differentiatingwithrespecttox,weget,1vdvdx=cosx·ddxlogsinx+logsinx·ddxcosx=cosx·1sinxcosx+logsinx-sinxdvdx=vcosxcotx-sinxlogsinx=sinxcosxcosxcotx-sinxlogsinx…(iii)


dydx=xsin x-1sin x+xlogxcosx+sin xcosxcosxcot x-sin xlogsin x




Exercise5.6

Q.1. Ifxandyareconnectedparametricallybytheequationsgiveninthisquestion,withouteliminatingtheparameter.Finddydx.

x=2at2,y=at4

Solution: Giventhatxandyareconnectedparametricallyandhere,x=2at2,y=at4.

Therefore,dxdt=2a2tanddydt=a4t3

dydx=dydtdxdt=4at34at=t2


x=acosθ+θsinθ,y=asinθ-θcosθ

Solution: Giventhatxandyareconnectedparametricallyandhere,

x=acosθ+θsinθ,y=asinθ-θcosθ

Weknowthatddxcosx=-sinxandddxsinx=cosxApplyingproductrule:ddxuv=udvdx+vdudxdxdθ=a-sinθ+θcosθ+sinθ=aθcosθanddydθ=acosθ--θsinθ+cosθ=aθsinθdydx=dydθdxdθ=aθsinθaθcosθ=tanθ

Q.3. Ifx=asin-1 t,y=acos-1 t,showthatdydx=-yx


x=asin-1 t,y=acos-1 t

dxdt=12asin-1t·ddtasin-1t=12asin-1t·asin-1t·loga11-t2=12x·x2·loga11-t2=xloga21-t2dydt=12acos-1t·ddtacos-1t=12acos-1t·acos-1t·loga-11-t2=12y·y2·loga11-t2=-yloga21-t2dydx=dydtdxdt=-yloga21-t2xloga21-t2=-yx(Henceproved)

Q.4. Ifxandyareconnectedparametricallybytheequationsgiveninthisquestion,withouteliminatingtheparameter.Finddydx.x=acosθ,y=bcosθ

Solution: Giventhatxandyareconnectedparametricallyandhere,x=acosθ,y=bcosθ.

Differentiatingbothxandywithrespecttoθweget,

dxdθ=a-sinθanddydθ=b-sinθdydx=dydθdxdθ=-bsinθ-asinθ=ba


x=sin t,y=cos2t

Solution: Giventhatxandyareconnectedparametricallyandhere,x=sint,y=cos2t

Therefore,dxdt=costanddydt=-sin 2t·2

dydx=dydtdxdt=-2sin2tcott=-22sintcostcost=-4sint

Q.6. Ifxandyareconnectedparametricallybytheequationsgiveninthisquestion,withouteliminatingtheparameter.Finddydx.x=4t,y=4t

Solution: Giventhatxandyareconnectedparametricallyandhere,x=4t,y=4tTherefore,dxdt=4anddydt=-4t2

Therefore,dydx=dydtdxdt=-4t24=-1t2


x=cosθ-cos2θ,y=sin θ-sin 2θ


x=cosθ-cos2θ,y=sin θ-sin 2θ

Therefore,dxdθ=-sinθ+2sin2θanddydθ=cosθ-2cos2θHence,dydx=dydθdxdθ=cosθ-2cos2θ-sinθ+2sin2θ


x=aθ-sin θ,y=a1+cosθ

Solution: Giventhatxandyareconnectedparametricallyandhere,x=aθ-sinθ,y=a1+cosθ

Therefore,dxdθ=a1-cosθanddydθ=a0-sin θ

Hence,dydx=dydθdxdθ=-asin θa1-cosθ=-2sin θ2cosθ22sin2 θ2=-cot θ2

Q.9. Ifxandyareconnectedparametricallybytheequationsgiveninthisquestion,withouteliminatingtheparameter.Finddydx.x=sin3 tcos2t,y=cos3 tcos2t




Solution: Giventhatxandyareconnectedparametricallyandhere,x=sin3 tcos2t,y=cos3 tcos2tTherefore,dxdt=sin3 tddtcos2t-cos2tddtsin3 tcos2t2=sin3 t·12cos2t·-sin 2t·2-cos2t·3sin2 tcostcos2t=-sin3 t·sin 2t-3cos2t·sin2 tcostcos2tcos2t

dydt=cos3 tddtcos2t-cos2tddtcos3 tcos2t2=cos3 t·12cos2t-sin 2t·2-cos2t·3cos2 t-sin tcos2t=-cos3 t·sin 2t+3cos2t·cos2 tsin tcos2tcos2t

dydx=dydtdxdt=-cos3t·sin2t+3cos2t·cos2tsint-sin3t·sin2t-3cos2t·sin2tcost=-cos3t·2sintcost+3cos2t·cos2tsint-sin3t·2sintcost-3cos2t·sin2tcost=cos2tsint-2cos2t+3cos2tsin2tcost-2sin2t-3cos2t

dydx=cost-2cos2 t+32cos2 -1sin t-2sin2 t-31-2sin2 t=cost-2cos2 t+6cos2 t-3sin t-2sin2 t-3+6sin2 tdydx=cost4cos2 t-3sin t-3+4sin2 t=-4cos3 t-3cost3sin t-4sin3 t=-cos3tsin 3t=-cot 3t

Q.10. Ifxandyareconnectedparametricallybytheequationsgiveninthisquestion,withouteliminatingtheparameter.Finddydx.x=acost+logtan t2y=asin t


x=acost+logtan t2,y=asin t

dxdt=a-sint+1tant2·sec2t2·12=a-sint+cost2sint2·1cos2t2·12=a-sint+12sint2cost2=a-sint+1sint=a-sin2t+1sint=acos2tsintdydt=acostdydx=dydtdxdt=acostacos2tsint=sintcost=tant

Q.11. Ifxandyareconnectedparametricallybytheequationsgiveninthisquestion,withouteliminatingtheparameter.Finddydx.x=asecθ,y=btan θ

Solution: Giventhatxandyareconnectedparametricallyandhere,x=asecθ,y=btanθ

dxdθ=asecθtan θ

dydθ=bsec2θdydx=dydθdxdθ=bsec2θasecθtanθ=bsecθatanθ=b1cosθasinθcosθ=bacosecθ




Exercise5.7

Q.1. Findthesecond-orderderivativesofthefunctiongiven

x2+3x+2

Solution: Givenfunctionisx2+3x+2

Lety=x2+3x+2,therefore,

dydx=ddxx2+3x+2=2x+3d2ydx2=ddx2x+3=2Secondorderderivativeofx2+3x+2=2

Q.2. Findthesecond-orderderivativeofthefunctiongiven:

sin logx

Solution: Givenfunctionissinlogx

Lety=sin logx,

Differentiatingwithrespecttoxweget,dydx=ddxsinlogx=coslogx·1x=coslogxxDifferentiatingagainwithrespecttoxweget,Secondorderderivative=d2ydx2=ddxdydx=x-sinlogx·1x-coslogx·1x2=-sinlogx-coslogxx2

Q.3. Ify=5cosx-3sin x,provethatd2ydx2+y=0

Solution: Giventhaty=5cosx-3sinx,


dydx=ddx5cosx-3sinx=-5sinx-3cosx

Differentiatingbothsidesagainwithrespecttoxweget,

d2ydx2=ddx-5sin x-3cosx

=-5cosx+3sinx=-5cosx-3sinx=-y⇒d2ydx2+y=0

Q.4. Ify=cos-1 x,findd2ydx2intermsofyalone.

Solution: Giventhaty=cos-1x⇒cosy=x,


-sinydydx=1⇒dydx=-1siny=-cosecy


d2ydx2=-cosec ycot y·dydx=-cosec ycot y·-cosec y

=cosec2 ycot y

Q.5. Ify=3coslogx+4sin logx,showthatx2y2+xy1+y=0.

Solution: Giventhaty=3coslogx+4sinlogx,therefore,

dydx=ddx3coslogx+4sin logx

=-3sinlogx·1x+4coslogx·1xxdydx=-3sinlogx+4coslogx⇒xd2ydx2+dydx·ddxx=ddx-3sinlogx+4coslogx=-3coslogx·1x-4sinlogx·1x=-1x3coslogx+4sinlogx=-1x·y⇒xd2ydx2+dydx=-1xy⇒x2d2ydx2+xdydx=-y⇒x2d2ydx2+xdydx+y=0⇒x2y2+xy1+y=0

Q.6. Ify=Aemx+Benx,showthatd2ydx2-m+ndydx+mny=0.

Solution: Giventhaty=Aemx+Benx,therefore,

dydx=ddxAemx+Benx=mAemx+nBenx

⇒d2ydx2=ddxmAemx+nBenx=m2Aemx+n2BenxPuttingthevalueofd2ydx2anddydxind2ydx2-m+ndydx+mny,wegetLHS=m2Aemx+n2Benx-m+nmAemx+nBenx+mny=m2Aemx+n2Benx-m2Aemx+mnBenx+mnAemx+n2Benx+mny=-mnAemx+mnBenx+mny=-mnAemx+Benx+mny=-mny+mny=0=RHS

Q.7. Ify=500e7x+600e-7x,showthatd2ydx2=49y.

Solution: Giventhaty=500e7x+600e-7x,therefore,

dydx=ddx500e7x+600e-7x=500e7x·7+600e-7x·-7=7500e7x-600e-7x

Again,d2ydx2=ddx7500e7x-600e-7x=7500e7x·7-600e-7x·-7=49500e7x+600e-7x=49y⇒d2ydx2=49y

Q.8. Ifeyx+1=1,showthatd2ydx2=dydx2.

Solution: Giventhateyx+1=1,therefore,

eydydxx+1+x+1ddxey=ddx1

⇒ey+x+1eydydx=0⇒dydx=-1x+1

d2ydx2=ddx-1x+1=-x+1·ddx1-1·ddxx+1x+12=-0-1x+12=1x+12

⇒d2ydx2=-1x+12

⇒d2ydx2=dydx2

Q.9. Ify=tan-1 x2,showthatx2+12y2+2xx2+1y1=2.




Solution: Given,

y=tan-1 x2

Differentiatingbothsideswithrespecttox,wegetdydx=ddxtan-1x2⇒dydx=2tan-1x·11+x2[Usingchainruleinright-handside,ddxf(g(x))=ddx(f(g(x))·ddx(g(x))]⇒dydx=2tan-1x1+x2⇒1+x2dydx=2tan-1x

Again,differentiatingbothsideswithrespecttox,weget

1+x2d2ydx2+dydx·ddx1+x2=ddx2tan-1 x

[UsingProductruleinLeft-handside,ddxf(x)·g(x)=f(x)·ddxg(x)+g(x)·ddxf(x)]⇒1+x2·d2ydx2+dydx·2x=21+x2⇒1+x22·d2ydx2+2x·1+x2dydx=2⇒x2+12y2+2xx2+1y1=2[Whered2ydx2=y2,dydx=y1]HenceProved

Q.10. Findthesecond-orderderivativeofthefunctiongiven:x20.

Solution: Givenfunctionisx20Lety=x20,therefore,

Byapplyingtheformuladdxxn=nxn-1,weget,y'=dydx=ddxx20=20x19

Tofindthesecond-orderderivative,differentiatingy'withrespecttoxweget,y''=d2ydx2=dy'dx=ddx20x19=20ddxx19=20×19×x18=380x18Therefore,thesecond-orderderivativeofthegivenfunctionis380x18.


x·cosx

Solution: Lety=x·cosx,therefore,

dydx=ddxx·cosx=x·ddxcosx+cosx·ddxx=-xsin x+cosx

d2ydx2=ddx-xsinx+cosx=-xddxsinx+sinxddxx-sinx

=-xcosx-sin x-sin x=-xcosx+2sin x


logx

Solution: Givenfunctionislogx

Lety=logx,

Differentiatingbothsideswithrespecttoxweget,dydx=ddxlogx=1xDifferentiatingagainbothsideswithrespecttoxweget,d2ydx2=ddx1x=-1x2

Q.13. Findthesecondorderderivativesofthefunctiongivenx3logx

Solution: Givenfunctionisx3logx

Lety=x3logx,therefore,

dydx=ddxx3logx=x3·ddxlogx+logx·ddxx3=x3·1x+logx·3x2=x2+3x2logx

d2ydx2=ddxx2+3x2logx=2x+3x2ddxlogx+logxddxx2

=2x+3x2·1x+logx·2x=2x+3x+6xlogx=5x+6xlogx=x5+6logx


exsin 5x

Solution: Givenfunctionisexsin5x

Lety=exsin 5x,

Differentiatingbothsideswithrespecttoxweget,dydx=ddxexsin5x=ex·ddxsin5x+sin5x·ddxex=ex·cos5x·5+sin5x·ex


d2ydx2=ddx5excos5x+exsin 5x

=5ex·ddxcos5x+cos5x·ddxex+ex·ddxsin5x+sin5x·ddxex=5ex·-sin5x·5+cos5x·ex+ex·cos5x·5+sin5x·ex=ex-25sin5x+5cos5x+5cos5x+sin5x=ex10cos5x-24sin5x


e6xcos3x

Solution: Givenfunctionise6xcos3x

Lety=e6xcos3x,therefore,

dydx=ddxe6xcos3x=e6x·ddxcos3x+cos3x·ddxe6x=e6x·-sin3x·3+cos3x·e6x·6=3e6x-sin3x+2cos3x

⇒d2ydx2=ddx3e6x-sin3x+2cos3x=3e6x·ddx-sin3x+2cos3x+-sin3x+2cos3x·ddx3e6x

=3e6x·-3cos3x-6sin 3x+-sin 3x+2cos3x·18e6x

=e6x-9cos3x-18sin3x-18sin3x+36cos3x=e6x27cos3x-36sin3x=9e6x3cos3x-4sin3x

Q.16. Findthesecondorderderivativesofthefunctiongiventan-1 x




Solution: Lety=tan-1 x,therefore,dydx=ddxtan-1 x=11+x2

d2ydx2=ddx11+x2=1+x2ddx1-1.ddx1+x21+x22=0-2x1+x22=-2x1+x22

Q.17. Findthesecond-orderderivativeofthefunctiongiven

loglogx

Solution: Givenfunctionisloglogx

Lety=loglogx,

Differentiatingwithrespecttox,wegetdydx=ddxloglogx=1logx·1x=1xlogx

Differentiatingagainwithrespecttox,weget

d2ydx2=ddx1xlogx=xlogxddx1-1·ddxxlogxxlogx2

=0-x·1x+logxxlogx2=-1+logxxlogx2




Exercise5.8

Q.1. VerifyRolle’sTheoremforthefunctionfx=x2+2x-8,x∈-4,2.

Solution: Givenfunctionisfx=x2+2x-8,x∈-4,2

(i)Functionfisapolynomialfunction,soitiscontinuousincloseinterval-4,2.

(ii)f'x=2x+2Hence,thefunctionfisdifferentiableinopeninterval-4,2.(iii)f-4=-42+2-4-8=16-8-8=0andf2=22+22-8=4+4-8=0⇒f-4=f2Here,allthethreeconditionsofRolle’sTheoremissatisfied.Therefore,theremustbeanumberc∈-4,2suchthatf'c=0.⇒f'c=2c+2=0⇒c=-1∈-4,2Hence,theRolle’sTheoremisverifiedforthefunction fx=x2+2x-8,x∈-4,2.

Q.2. ExamineifRolle'sTheoremisapplicabletoanyofthefollowingfunctions.CanyousaysomethingabouttheconverseofRolle'sTheoremfromtheseexample?(i)fx=xforx∈5,9(ii)fx=xforx∈-2,2(iii)fx=x2-1forx∈1,2

Solution: Rolle’sTheoremisapplicabletofunctionf:a,b→RthefollowingthreeconditionsofRolle’sTheoremissatisfied.

(i)Functionfiscontinuousinclosedintervala,b.

(ii)Functionfisdifferentiableinopenintervala,b.(iii)fa=fbSoconverseofRolle'stheoremstatesthat,Iff'(c)=0,inanyintervalcontainingcinwhichfiscontinuousanddifferentiable,thereexista,bsuchthatf(a)=f(b).(i)fx=xforx∈5,9Thegreatestintegerfunctionfisneithercontinuousinclosedinterval5,9nordifferentiableinopeninterval5,9.Alsof5≠f9.Hence,theRolle’sTheoremisnotapplicabletofx=xforx∈5,9.

(ii)fx=xforx∈-2,2

Thegreatestintegerfunctionfisneithercontinuousinclosedinterval-2,2nordifferentiableinopeninterval2,2.Alsof-2≠f2.

Hence,theRolle’sTheoremisnotapplicabletofx=xforx∈-2,2.

(iii)fx=x2-1forx∈1,2

Thefunctionfisapolynomialfunction,soitiscontinuousintheclosedinterval1,2.

f'x=2x,hence,thefunctionfisdifferentiableintheopeninterval1,2.f1=12-1=0andf2=22-1=3,⇒f1≠f2Hence,Rolle’sTheoremisnotapplicabletothefunctionfx=x2-1forx∈1,2.

Q.3. Iff:-5,5→Risadifferentiablefunctionandiff'xdoesnotvanishanywhere,thenprovethatf-5≠f5.

Solution: f:-5,5→Risadifferentiablefunction,hence

(i)Thefunctionfiscontinuousinclosedinterval-5,5.

(ii)Thefunctionfiscontinuousinopeninterval-5,5.

AccordingtoMeanValueTheorem,thereexistsavaluec∈-5,5,suchthat,

f'c=f5-f-55--5

Butitisgiventhatf'xdoesnotvanishanywhere,hencef'c=f5-f-55--5≠0⇒f5-f-5≠0⇒f5≠f-5

Q.4. VerifyMeanValueTheorem,iffx=x2-4x-3intheintervala,b,wherea=1andb=4.

Solution: Givenfunctionisfx=x2-4x-3,x∈1,4

(i)Thefunctionfisapolynomialfunction,henceitiscontinuousintheclosedinterval1,4.

(ii)f'x=2x-4

Hence,thefunctionfisdifferentiableinopeninterval1,4.

AccordingtoMeanValueTheorem,thereexistsavaluec∈1,4,suchthat

f'c=f4-f14-1⇒2c-4=42-44-3-12-41-33⇒2c-4=-3--63=33=1⇒2c=5⇒c=52∈1,4Hence,forthefunctionfx=x2-4x-3,x∈1,4,theMeanValueTheoremisverified

Q.5. VerifyMeanValueTheorem,iffx=x3-5x2-3xintheintervala,b,wherea=1andb=3.Findallc∈1,3forwhichf'c=0.

Solution: Givenfunctionisfx=x3-5x2-3x,x∈1,3

(i)Functionfisapolynomialfunction,henceitiscontinuousinclosedinterval1,3.

(ii)f'x=3x2-10x-3Hence,thefunctionfisdifferentiableinopeninterval1,3.

AccordingtoMeanValueTheorem,thereexistsavaluec∈1,3,suchthat

f'c=f3-f13-1

⇒3c2-10c-3=33-532-33-13-512-312⇒3c2-10c-3=27-54-1-82=-27+72=-10⇒3c2-10c+7=0

⇒3c2-3c-7c+7=0

⇒3cc-1-7c-1=0

⇒c-13c-7=0⇒c-1=0or3c-7=0⇒c=1orc=73⇒c=73∈1,3Hence,forthefunctionfx=x3-5x2-3x,x∈1,3,theMeanValueTheoremisverified.Forthevalueofc=73thefunctionf'c=0.

Q.6. ExaminetheapplicabilityofMeanValueTheoremforallthreefunctions

(i)fx=xforx∈5,9(ii)fx=xforx∈-2,2(iii)fx=x2-1forx∈1,2




Solution: MeanValueTheoremisapplicabletofunctionf:a,b→R,ifthefollowingtwoconditionsofMeanValueTheoremissatisfied.

(i)Functionfiscontinuousintheclosedintervala,b.

(ii)Functionfisdifferentiableintheopenintervala,b.(i)fx=xforx∈5,9Thegreatestintegerfunctionfisneithercontinuousinthecloseinterval5,9nordifferentiableintheopeninterval5,9.Hence,theMeanValueTheoremisnotapplicabletofx=xforx∈5,9.

(ii)fx=xforx∈-2,2.

Thegreatestintegerfunctionfisneithercontinuousincloseinterval-2,2nordifferentiableinopeninterval-2,2.

Hence,theMeanValueTheoremisnotapplicabletofx=xforx∈-2,2.

(iii)fx=x2-1forx∈1,2

Thefunctionfisapolynomialfunction,soitiscontinuousintheclosedinterval1,2.

f'x=2x,hence,thefunctionfisdifferentiableintheopeninterval1,2.Hence,theMeanValueTheoremisapplicabletofx=x2-1forx∈1,2.




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