RD Sharma XII Vol 1 2018 Solutions for Class 12 Science Math Chapter 12 Higher Order Derivatives are provided here with simple step-by-step explanations. These solutions for Higher Order Derivatives are extremely popular among class 12 Science students for Math Higher Order Derivatives Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2018 Book of class 12 Science Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2018 Solutions. All RD Sharma XII Vol 1 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 12.16:

Question 1:

Find the second order derivatives of each of the following functions:

(i) x3 + tan x
(ii) sin (log x)
(iii) log (sin x)
(iv) ex sin 5x
(v) e6x cos 3x
(vi) x3 log x
(vii) tan−1 x
(viii) x cos x
(ix) log (log x)

Answer:

(i) We have,

y=x3+tanxDifferentiating w.r.t. x, we getdydx=3x2+sec2xDifferentiating again w.r.t. x, we getd2ydx2=6x+2sec2x tanx

(ii) We have,

y= sinlogxDifferentiating w.r.t. x, we getdydx=coslogx×1xDifferentiating again w.r.t. x, we getd2ydx2=-sinlogx1x×1x+coslogx×-1x2         =-sinlogx+coslogxx2

(iii) We have,

y= logsinxDifferentiating w.r.t. x, we getdydx=1sinx×cosx =cotxDifferentiating again w.r.t. x, we getd2ydx2=-cosec2x

(iv) We have,

y =exsin5xDifferentiating w.r.t. x, we getdydx=ex sin 5x +ex cos 5x×5Differentiating again w.r.t. x, we getd2ydx2=ex sin 5x +ex cos 5x×5+5ex(-sin5x×5)+5ex cos 5x          =-24ex sin 5x +10ex cos 5x            =2ex5 cos 5x-12 sin 5x

(v) We have,
  y= e6x cos 3xDifferentiating w.r.t. x, we getdydx=e6x×6× cos 3x+e6x(-sin 3x×3)     =6e6x cos3x-3e6xsin 3xDifferentiating again w.r.t. x, we getd2ydx2=6e6x cos3x × 6 -6e6x sin3x×3- 3×6 e6x sin3x-3e6x×3 cos 3x       =27e6x cos3x-36e6x sin3x         =9e6x3 cos3x-4 sin3x

(vi) We have,
y = x3 logxDifferentiating w.r.t. x, we getdydx=3x2 logx+x3×1x       =3x2 logx+x2Differentiating again w.r.t. x, we getd2ydx2=6x logx+3x2×1x+2x         =6x logx +5x

(vii) We have,
y= tan-1xDifferentiating w.r.t. x, we getdydx=11+x2Differentiating again w.r.t. x, we getd2ydx2=-2x×11+x22=-2x1+x22


(viii) We have,
y= x cosxDifferentiating w.r.t. x, we getdydx=cosx -xsinxDifferentiating again w.r.t. x, we getd2ydx2=-sinx-sinx-xcosx       =-2sinx+xcosx

(ix) We have,

y= loglogxDifferentiating w.r.t. x, we getdydx=1logx×1x =1xlogxDifferentiating again w.r.t. x, we getd2ydx2=0-logx+1xlogx2=-1+logxxlogx2

Page No 12.16:

Question 2:

If y = ex cos x, show that d2ydx2=2e-x sin x.

Answer:

Here,

y=e-x cos xDifferentiating w.r.t. x, we getdydx=-e-x sinx -e-x cos x      =-e-x sinx +e-x cos xDifferentiating again w.r.t. x, we getd2ydx2=-e-x cosx -e-x sinx -e-x sinx -e-x cosx        =2e-x sinx

Hence proved.

Page No 12.16:

Question 3:

If y = x + tan x, show that cos2 xd2ydx2-2y+2x=0.

Answer:

Here,

y= x+ tanxDifferentiating w.r.t. x, we getdydx=1+sec2xDifferentiating again w.r.t. x, we getd2ydx2=2sec2xtanxDividing both sides by sec2x, we getcos2x d2ydx2=2tanxcos2x d2ydx2=2(y-x)           y=x +tanxtanx =y-xcos2x d2ydx2-2y+2x=0

Hence proved.

Page No 12.16:

Question 4:

If y = x3 log x, prove that d4ydx4=6x.

Answer:

Here,
y=x3logxDifferentiating w.r.t. x, we getdydx=3x2 logx +x3×1x       =3x2 logx+x2Differentiating again w.r.t. x, we getd2ydx2=6x logx+3x2×1x+2x         =6x logx+5xDifferentiating again w.r.t. x, we getd3ydx3=6logx+6x×1x+5 =6 logx+11Differentiating again w.r.t. x, we getd4ydx4=6x

Hence proved.

Page No 12.16:

Question 5:

If y = log (sin x), prove that d3ydx3=2 cos x cosec3 x.

Answer:

Here,
y=logsinxDifferentiating w.r.t. x, we getdydx=1sinx×cosx=cotxDifferentiating again w.r.t. x, we getd2ydx2=-cosec2xDifferentiating again w.r.t. x, we getd3ydx3=-2cosecx ×-cosecx cotx           = 2cotx cosec2x = 2cosx cosec3x

Hence proved.

Page No 12.16:

Question 6:

If y = 2 sin x + 3 cos x, show that d2ydx2+y=0.

Answer:

Here,
y= 2 sinx+3 cos xDifferentiating w.r.t. x, we getdydx=2cosx-3 sinxDifferentiating again w.r.t. x, we getd2ydx2=-2 sinx-3 cosx d2ydx2=- 2 sinx+3 cos xd2ydx2=-yd2ydx2+y=0

Hence proved.

Page No 12.16:

Question 7:

If y=log xx, show that d2ydx2=2 log x-3x3.

Answer:

Here,

y=logxxDifferentiating w.r.t. x, we getdydx=1-logxx2Differentiating again w.r.t. x, we getd2ydx2=-x-2x1-logxx4       =-x-2x+2xlogxx4       =-3+2logxx3       =2logx-3x3

Hence proved.

Page No 12.16:

Question 8:

If x = a sec θ, y = b tan θ, prove that d2ydx2=-b4a2y3.

Answer:

Here,
x=a secθ and y=b tanθDifferentiating w.r.t. θ, we getdxdθ=asecθ tanθ and dydθ=b sec2θdydx=dydθ×dθdx=b sec2θa secθ tanθ=b cosecθaDifferentiating w.r.t. x, we getd2ydx2=b a×-cosecθ cotθ×dθdx         =-b a×cosecθ cotθ×1asecθ tanθ        =-b a2×cotθ ×1tan2θ        =-b a2×1tan3θ        =-b4a2y3         y=b tanθ

Hence proved.

Page No 12.16:

Question 10:

If y = ex cos x, prove that d2ydx2=2 ex cos x+π2.

Answer:

Here,
y=ex cosx Differentiating w.r.t. x, we getdydx=ex cosx-ex sin x=excosx-sinxDifferentiating again w.r.t. x, we getd2ydx2=ex cosx-sinx+ex -sinx-cosx       =excosx-exsinx-exsinx-excosx       =-2exsinx       =2ex cosx+π2                                     

Hence proved.

Page No 12.16:

Question 11:

If x = a cos θ, y = b sin θ, show that d2ydx2=-b4a2y3.

Answer:

Here,

x=a cosθ and y = b sinθDifferentiating w.r.t. θ, we getdxdθ= -a sinθ and  dydθ= b cosθdydx=b cosθ-a sinθ=-bacotθDifferentiating w.r.t. x, we getd2ydx2=-ba×-cosec2θ dθdx       =ba×cosec2θ×1-a sinθ       =-ba2×1 sin3θ       =-ba2×b3y3        y = b sinθ      =-b4a2y3

Hence proved.

Page No 12.16:

Question 12:

If x = a (1 − cos3 θ), y = a sin3 θ, prove that d2ydx2=3227aat θ=π6.

Answer:

Here,

x=a1-cos3θ,  y =a sin3θDifferentiating w.r.t. θ, we getdxdθ=3a cos2θ sinθ anddydθ=3a sin2θ cosθ dydx=3a sin2θ cosθ3a cos2θ sinθ=tanθDifferentiating w.r.t. x, we getd2ydx2=sec2θ dθdx       =sec2θ3a cos2θ sinθ       =sec4θ3a sinθ  d2ydx2  at θ=π6d2ydx2= secπ6 43a sinπ6=3227a

Page No 12.16:

Question 13:

 If x = a (θ + sin θ), y = a (1 + cos θ), prove that d2ydx2=-ay2.

Answer:

Here,
x=aθ+sinθ and y=a1+cosθDifferentiating w.r.t. θ, we getdxdθ=a+acosθ and dydθ=-a sinθdydx=-a sinθa+a cosθ=- sinθ1+cosθDifferentiating w.r.t. x, we getd2ydx2=-1+cosθcosθ+ sin2θ1+cosθ2dθdx       =-cosθ-cos2θ- sin2θ1+cosθ2×1a+acosθ       =-1+cosθa1+cosθ3       =-1a1+cosθ2       =-ay2        y=a1+cosθ           

Hence proved.

Page No 12.16:

Question 14:

If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find d2ydx2.

Answer:

 Here,

x=aθ-sinθ and y=a1+cosθDifferentiating w.r.t. θ, we getdxdθ=a-acosθ,  dydθ=-a sinθdydx=-a sinθa-a cosθ=- sinθ1-cosθDifferentiating w.r.t. x, we getd2ydx2=-cosθ+cos2θ+ sin2θ1-cosθ2×dθdx      =-cosθ+cos2θ+ sin2θ1-cosθ2×1a-acosθ     =1-cosθa1-cosθ3     =1a1-cosθ2

Page No 12.16:

Question 15:

If x = a(1 − cos θ), y = a(θ + sin θ), prove that d2ydx2=-1aat θ=π2.

Answer:

Here,
x= a1-cosθ and y=aθ+ sinθDifferentiating w.r.t. x, we getdxdθ=a sinθ and dydθ=a+a cosθdydx=a+acosθasinθ=1+cosθsinθDifferentiating w.r.t. x, we getd2ydx2=-sin2θ-1+cosθcosθsin2θdθdx        =-1-1+cosθcosθsin2θ1a sinθ  d2ydx2 at θ=π2 d2ydx2θ=π2=1a-1-01=-1a 

Hence proved.



Page No 12.17:

Question 16:

If x = a (1 + cos θ), y = a(θ + sin θ), prove that d2ydx2=-1aat θ=π2.

Answer:

Here,

 x=a1+cosθ and y =aθ+sinθDifferentiating w.r.t. θ, we getdxdθ=-asinθ and dydθ=a+a cosθdydx=a+acosθ-asinθ=1+cosθ-sinθDifferentiating w.r.t. x, we getd2ydx2=ddθdydxdθdxd2ydx2=--sin2θ-cosθ-cos2θsin2θdθdx       =1+cosθsin2θ×-1a sinθ       =-1+cosθasin3θAt θ=π2: d2ydx2=-1+cosπ2a sinπ23=-1a

Page No 12.17:

Question 17:

If x = cos θ, y = sin3 θ, prove that yd2ydx2+dydx2=3 sin2 θ5 cos2 θ-1.

Answer:

Here,
x= cosθ and y = sin3θDifferentiating w.r.t. θ, we getdxdθ=-sinθ and dydθ=3sin2θ cosθdydx=3sin2θ cosθ-sinθ=-3sinθ cosθDifferentiating w.r.t. x, we getd2ydx2=-3cos2θ+3 sin2θdθdx=-3cos2θ+3 sin2θ-sinθNow,LHS=yd2ydx2+dydx2        = sin3θ×-3cos2θ+3 sin2θ-sinθ+-3sinθ cosθ2        =3sin2θ cos2θ-3 sin4θ+9sin2θ cos2θ         =12sin2θ cos2θ-3 sin4θ         =3sin2θ4cos2θ-sin2θ         =3sin2θ5cos2θ-1              [ cos2+ sin2θ=1]        =RHS
                                                                                                                                                                    
Hence proved.

Page No 12.17:

Question 18:

If y = sin (sin x), prove that d2ydx2+tan x·dydx+y cos2 x=0.

Answer:

Here,
y= sinsinxDifferentiating w.r.t. x, we getdydx=cossinx cosxDifferentiating again w.r.t. x, we getd2ydx2=-sinsinx cos2x-cossinx sinxd2ydx2=-sinsinx cos2x-cossinx cosxtanxd2ydx2=-y cos2x-tanxdydxd2ydx2+tanxdydx+y cos2x=0

Hence proved.

Page No 12.17:

Question 19:

If x = sin t, y = sin pt, prove that 1-x2d2ydx2-xdydx+p2y=0.

Answer:

Here,
x= sint and y =sinptDifferentiating w.r.t. t, we getdxdt=cost and dydt=p cosptdydx=pcosptcostDifferentiating w.r.t. x, we getd2ydx2=-p2sinpt cost+pcosptsintcos2t×dtdxd2ydx2=-p2sinpt cost+pcosptsintcos3td2ydx2=-p2sinpt costcos3t+pcosptsintcos3td2ydx2=-p2ycos2t+xdydxcos2tcos2td2ydx2=-p2y+xdydx1-sin2td2ydx2=-p2y+xdydx1-x2d2ydx2-xdydx+p2y=0.

Hence proved.

Page No 12.17:

Question 20:

If y = (sin−1 x)2, prove that (1 − x2) d2ydx2-xdydx+p2y=0.

Answer:

Here,

y=sin-1x2Now,y1=2 sin-1x 11-x2y2=21-x2+2x sin-1x1-x23/2y2=21-x2+2x sin-1x1-x21-x2y2=21-x2+xy11-x2y21-x2=2+xy1y21-x2-xy1-2=0

Hence proved.

Page No 12.17:

Question 21:

If y=etan-1x, prove that (1 + x2)y2 + (2x − 1)y1 = 0.

Answer:

Here,
y= etan-1xDifferentiating w.r.t. x, we getdydx= etan-1x ×11+x2Differentiating again w.r.t. x, we getd2ydx2=etan-1x11+x22+etan-1x-2x1+x221+x2d2ydx2=etan-1x1+x2-2xetan-1x1+x21+x2d2ydx2=dydx-2xdydx1+x2d2ydx2+2x-1dydx=0

Hence proved.

Page No 12.17:

Question 22:

If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0.

Answer:

Here,
y=3 coslogx+4 sinlogxDifferentiating w.r.t. x, we gety1=-3sinlogx×1x+4 coslogx×1x       =-3sinlogx+4coslogxxDifferentiating again w.r.t. x, we gety2=-3coslogxx-4sinlogxx×x--3sinlogx+4coslogxx2y2=-3coslogx-4sinlogx--3sinlogx+4coslogxx2y2=-3coslogx-4sinlogx--3sinlogx+4coslogxx2y2=-3coslogx-4sinlogxx2--3sinlogx+4coslogxx2y2=-3coslogx+4sinlogxx2--3sinlogx+4coslogxx2y2=-yx2-y1xx2y2=-y-xy1x2y2+y+xy1=0

Hence proved.

Page No 12.17:

Question 23:

If y=e2xax+b, show that y2-4y1+4y=0.

Answer:

Given,

y=e2xax+b

To prove: y2-4y1+4y=0

Proof:

We have,

y=e2xax+b         ...(i)

y1=dydx=ae2x+2e2x(ax+b)                ...(ii)y2 = 2a×e2x+4e2x(ax+b)+2ae2x       =4ae2x + 4e2x(ax+b)                        ...(iii)LHS=y2-4y1+4y=4ae2x+4e2x(ax+b)-4ae2x-8e2x(ax+b)+4e2x(ax+b)= 0=RHS

Page No 12.17:

Question 24:

If x=sin 1alog y, show that (1 − x2)y2xy1a2y = 0.

Answer:

Here,
x= sin1alogy1alogy=sin-1xy=ea sin-1xDifferentiating w.r.t. x, we gety1=ea sin-1x×a1-x2y1=ay1-x2Differentiating again w.r.t. x, we gety2=ay11-x2+xay1-x21-x2y2=ay11-x2+xay1-x21-x2y2=ay11-x2+xay1-x21-x2y2=a2y1-x2+xy11-x21-x2y2-xy1-a2y=0

Page No 12.17:

Question 25:

If log y = tan−1 x, show that (1 + x2)y2 + (2x − 1) y1 = 0

Answer:

Here,
logy= tan-1xDifferentiating w.r.t. x, we get1y×y1=11+x21+x2y1=y1+x2y2+2xy1=y11+x2y2+2xy1-y1=01+x2y2+2x-1y1=0

Hence proved.

Page No 12.17:

Question 26:

If y = tan−1 x, show that 1+x2 d2ydx2+2xdydx=0.

Answer:

Here,
y= tan-1xDifferentiating w.r.t. x, we getdydx=11+x2Differentiating again w.r.t. x, we getd2ydx2=-2x1+x22d2ydx2=-2x1+x2×11+x2d2ydx2=-2xdydx1+x21+x2d2ydx2=-2xdydx1+x2d2ydx2+2xdydx=0

Hence proved.

Page No 12.17:

Question 27:

If y=log x+x2+12, show that 1+x2d2ydx2+xdydx=2.

Answer:

Here,
y=logx+x2+12Differentiating w.r.t. x, we getdydx=2logx+x2+1x+x2+1×1+2x2x2+1dydx=2logx+x2+1x+x2+1×x2+1+xx2+1dydx=2logx+x2+1x2+1Differentiating again w.r.t. x, we getd2ydx2=2-2xlogx+x2+1x2+1x2+1d2ydx2=2-xdydxx2+1x2+1d2ydx2=2-xdydxx2+1d2ydx2+xdydx=2

Page No 12.17:

Question 28:

If y = (tan−1 x)2, then prove that (1 + x2)2 y2 + 2x(1 + x2)y1 = 2.

Answer:

Here,
y=tan-1x2Differentiating w.r.t. x, we gety1= 2 tan-1x1+x2Differentiating again w.r.t. x, we gety2=2-4x tan-1x1+x22y2=21+x22-2 tan-1x ×2x1+x22y2=21+x22-2xy11+x21+x22y2=2-2x1+x2y11+x22y2+2x1+x2y1=2

Hence proved.

Page No 12.17:

Question 29:

If y = cot x show that d2ydx2+2ydydx=0.

Answer:

Here,
y= cotxDifferentiating w.r.t. x, we getdydx=-cosec2xDifferentiating again w.r.t. x, we getd2ydx2=-2 cosecx ×- cosec x cotxd2ydx2=2 cosec2x cotxd2ydx2=-2ydydxd2ydx2+2ydydx=0

Hence proved.

Page No 12.17:

Question 30:

Find d2ydx2, where y=log x2e2.

Answer:

Here,
y=logx2e2Differentiating w.r.t. x, we getdydx=1x2e2×2xe2=2xDifferentiating again w.r.t. x, we getd2ydx2=-2x2

Page No 12.17:

Question 31:

If y = ae2x + bex, show that, d2ydx2-dydx-2y=0.

Answer:

Here,
y= a e2x+b e-xDifferentiating w.r.t. x, we getdydx=2a e2x-b e-xDifferentiating again w.r.t. x, we getd2ydx2=4a e2x+b e-xd2ydx2=2a e2x-b e-x+2a e2x+b e-xd2ydx2=dydx+2y   d2ydx2-dydx-2y=0

Hence proved.

Page No 12.17:

Question 32:

If y = ex (sin x + cos x) prove that d2ydx2-2dydx+2y=0.

Answer:

Here,
y= exsinx+cosxDifferentiating w.r.t. x, we getdydx=exsinx+cosx+excosx-sinx =2 ex cosxDifferentiating w.r.t. x, we getd2ydx2=2ex cosx-2ex sinxNow, LHS=d2ydx2-2dydx+2y=2ex cosx-2ex sinx-4ex cosx+2 exsinx+cosx=0 = RHS

Hence proved.

Page No 12.17:

Question 33:

If y = cos−1 x, find d2ydx2 in terms of y alone.

Answer:

Here,
y= cos-1xDifferentiating w.r.t. x, we getdydx=-11-x2Differentiating again w.r.t. x, we getd2ydx2=-2x21-x23/2=-x1-x23/2Now,y = cos-1xx= cosyd2ydx2=-cosy1-cos2y3/2=-cosysin2y3/2= -coty cosec2y

Page No 12.17:

Question 34:

If y=ea cos-1x, prove that 1-x2d2ydx2-xdydx-a2 y=0.

Answer:

Here,
y= ea cos-1xDifferentiating w.r.t. x, we getdydx= -ea cos-1x×a1-x2Differentiating again w.r.t. x, we getd2ydx2= ea cos-1x×a21-x2+2xa ea cos-1x21-x232d2ydx2=ea cos-1x×a21-x2+xa ea cos-1x1-x21-x2d2ydx2=y×a21-x2-xdydx1-x21-x2d2ydx2=a2y-xdydx1-x2d2ydx2+xdydx-a2y=0

Hence proved.

Page No 12.17:

Question 35:

If y = 500 e7x + 600 e−7x, show that d2ydx2=49y.

Answer:

Here,

y= 500 e7x+600 e-7xDifferentiating w.r.t. x, we getdydx=3500 e7x-4200 e-7x Differentiating again w.r.t. x, we getd2ydx2=24500 e7x+29400 e-7x          =49500 e7x+600 e-7x = 49y

Page No 12.17:

Question 36:

If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find d2ydx2at t=π2.

Answer:

Here,
x= 2 cost -cos2t and y = 2 sint - sin2tDifferentiating w.r.t. t, we getdxdt=-2 sint+2 sin2t  and dydt=2 cost-2 cos2t dydx=2 cost-2 cos2t-2 sint+2 sin2t=cost- cos2t- sint+ sin2tDifferentiating w.r.t. x, we getd2ydx2=-sint+2 sin2t- sint+ sin2t- cost- cos2t- cost+2 cos2t- sint+ sin2t2dtdx          =-sint+2 sin2t- sint+ sin2t- cost- cos2t- cost+2 cos2t- sint+ sin2t2-2 sint+2 sin2tAt t =π2:d2ydx2=-1+0- 1+0- 0+1- 0-2- 1+ 02-2 +0=1+2-2=-32

Page No 12.17:

Question 37:

If x = 4z2 + 5, y = 6z2 + 7z + 3, find d2ydx2.

Answer:

Here,
x= 4z2+5 and y= 6z2+7z+3Differentiating w.r.t. z, we getdxdz=8z and dydz=12z+7 dydx=12z+78zDifferentiating w.r.t. x, we getd2ydx2=12×8z-812z+764z2×dzdx        =96z-96z-56512z3=-764z3

Page No 12.17:

Question 38:

If y log (1 + cos x), prove that d3ydx3+d2ydx2·dydx=0

Answer:

Here,

y=log1+cosxDifferentiating w.r.t. x, we getdydx=-sinx1+cosxDifferentiating again w.r.t. x, we getd2ydx2=-cosx-cos2x-sin2x1+cosx2=-cosx+11+cosx=-11+cosxDifferentiating again w.r.t. x, we getd3ydx3=-sinx1+cosx2d3ydx3+sinx1+cosx2=0d3ydx3+-11+cosx-sinx1+cosx=0d3ydx3+d2ydx2×dydx=0

Page No 12.17:

Question 39:

If y = sin (log x), prove that x2d2ydx2+xdydx+y=0.

Answer:

Here,
y = sinlogxDifferentiating w.r.t. x, we getdydx=coslogxxDifferentiating again w.r.t. x, we getd2ydx2=-sinlogx-coslogxx2d2ydx2=-sinlogxx2-coslogxx2d2ydx2=-yx2-1x×dydxx2d2ydx2+xdydx+y=0
 



Page No 12.18:

Question 40:

If y = 3 e2x + 2 e3x, prove that d2ydx2-5dydx+6y=0

Answer:

Here,
y=3e2x+2 e3xDifferentiating w.r.t. x, we getdydx=6 e2x+6 e3xDifferentiating again w.r.t. x, we getd2ydx2=12 e2x+18 e3xd2ydx2=56 e2x+6 e3x-63e2x+2 e3xd2ydx2=5dydx-6yd2ydx2-5dydx+6y=0

Page No 12.18:

Question 41:

If y = (cot−1 x)2, prove that y2(x2 + 1)2 + 2x (x2 + 1) y1 = 2.

Answer:

Here,
y= cot-1x2Differentiating w.r.t. x, we gety1=2cot-1x×-11+x2=-2cot-1x1+x2Differentiating again w.r.t. x, we gety2=2+4xcot-1x1+x22y2=21+x22+2x×2cot-1x1+x21+x2y2=21+x22-2xy11+x21+x22y2=2-2xy11+x21+x22y2+2xy11+x2=2

Hence proved.

Page No 12.18:

Question 42:

If y = cosec−1 x, x >1, then show that xx2-1d2ydx2+2x2-1dydx=0.

Answer:

Here,
y= cosec-1 xDifferentiating w.r.t. x, we getdydx=-1xx2-1Differentiating again w.r.t. x, we getd2ydx2=x2-1+x2x2-1x2x2-1d2ydx2=x2-1+x2x2x2-1x2-1d2ydx2=2x2-1x2x2-1x2-1d2ydx2=2x2-1x2-1-1x2x2-1x2-1x2-1d2ydx2=2x2-1-1x2x2-1x2-1d2ydx2=-2xdydx+1xdydxxx2-1d2ydx2=-2x2-1dydxxx2-1d2ydx2+2x2-1dydx=0

Hence proved.

Page No 12.18:

Question 43:

If x=cost+log tant2, y=sint, then find the value of d2ydt2 and d2ydx2 at t=π4.

Answer:

We have,x=cost+log tant2   and   y=sintOn differentiating with respect to t, we getdxdt=ddtcost+log tant2=-sint+1tant2×sec2t2×12      =-sint+12sint2cost2=-sint+1sint      =-sin2t+1sint=-sin2t+1sint      =cos2tsintanddydt=ddtsint=costNow, d2ydt2=ddtdydt=ddtcost=-sintd2ydt2t=π4=-sinπ4=-12       ...(1)Also, dydx=dydtdxdt=costcos2tsint=sintcost=tantNow, d2ydx2=ddxdydx=ddxtant                =ddttant×dtdx=sec2t×sintcos2t                =sintcos4td2ydx2t=π4=sinπ4cos4π4=22       ...(2)Hence, at t=π4, d2ydt2=-12 and d2ydx2=22.

Page No 12.18:

Question 44:

If x=a sint and y=acost+log tant2,find d2ydx2.

Answer:

We have,x=a sint and y=acost+log tant2On differentiating with respect to t, we getdxdt=ddta sint=a costanddydt=ddtacost+log tant2=a-sint+1tant2×sec2t2×12       =a-sint+12sint2cost2=a-sint+1sint       =a-sin2t+1sint=acos2tsintNow, dydx=dydtdxdt=acos2tsinta cost=cottTherefore,d2ydx2=ddxdydx=ddxcott        =ddtcott×dtdx=-cosec2t×1a cost        =-1a sin2t costHence, d2ydx2=-1a sin2t cost.

Page No 12.18:

Question 45:

If x=acost+t sint and y=asint-t cost, then find the value of d2ydx2 at t=π4.

Answer:

We have,x=acost+t sint   and   y=asint-t costOn differentiating with respect to t, we getdxdt=ddtacost+t sint=-asint+asint+atcost      =atcostanddydt=ddtasint-t cost=acost-acost+atsint      =atsintNow, dydx=dydtdxdt=atsintatcost=tantd2ydx2=ddxdydx=ddxtant        =ddttant×dtdx=sec2t×1atcost        =1atcos3td2ydx2t=π4=1aπ4cos3π4=82aπHence, at t=π4, d2ydx2=82aπ.

Page No 12.18:

Question 46:

If x=acost+log tant2 and y=asint, evaluate d2ydx2 at t=π3.

Answer:

We have,x=acost+log tant2 and y=a sintOn differentiating with respect to t, we getdxdt=ddtacost+log tant2=a-sint+1tant2×sec2t2×12       =a-sint+12sint2cost2=a-sint+1sint       =a-sin2t+1sint=acos2tsintanddydt=ddta sint=a costNow, dydx=dydtdxdt=a costa cos2tsint=tantTherefore,d2ydx2=ddxdydx=ddxtant       =ddttant×dtdx=sec2t×sinta cos2t       =sinta cos4td2ydx2t=π3=sinπ3a cos4π3=32a116=83aHence, at t=π3, d2ydx2=83a.

Page No 12.18:

Question 47:

If x=acos2t+2t sin2t and y=asin2t-2t cos2t, then find d2ydx2.

Answer:

We have,x=acos2t+2t sin2t and y=asin2t-2t cos2tOn differentiating with respect to t, we getdxdt=ddtacos2t+2t sin2t=a-2sin2t+2sin2t+4t cos2t      =a4t cos2tanddydt=ddtasin2t-2t cos2t=a2cos2t-2cos2t+4t sin2t      =a4t sin2tNow, dydx=dydtdxdt=a4t sin2ta4t cos2t=tan2tTherefore,d2ydx2=ddxdydx=ddxtan2t       =ddttan2t×dtdx=2sec22t×1a4t cos2t       =12at cos32t=12atsec32tHence, d2ydx2=12atsec32t.

Page No 12.18:

Question 48:

If x=3 cot-2cos3t, y=3sint-2sin3t, find d2ydx2.

Answer:

We have,

x=3cos t-2cos3 tdxdt=3-sin t-6cos2 t-sin t=-3sin t+6sin t cos2 t

Also,

y=3sin t-2sin3 tdydt=3cos t-6sin2 t cos t

Now,

dydx=dydtdxdt=3cos t-6sin2 t cos t-3sin t+6sin t cos2t=3cos t1-2sin2 t3sin t-1+2cos2t=cot tcos2tcos2t=cot t

So,

So, d2ydx2=ddxdydx=ddxcot t=-cosec2 t dtdx=-cosec2 tdxdt=-cosec2 t-3sin t+6sin t cos2 t=-cosec2 t-sin t1-2 cos2 t=cosec3 t-cos 2t=-cosec3 tcos 2t

Page No 12.18:

Question 49:

If x=a sint-b cost, y=a cost+b sint, prove that d2ydx2=-x2+y2y3.

Answer:

We have,x=a sint-b cost, y=a cost+b sintOn differentiating with respect to t, we getdxdt=ddta sint-b cost=a cost+b sintanddydt=ddta cost+b sint=-a sint+b costNow, dydx=dydtdxdt=-a sint+b costa cost+b sintTherefore,d2ydx2=ddxdydx=ddx-a sint+b costa cost+b sint       =ddt-a sint+b costa cost+b sint×dtdx       =a cost+b sintddt-a sint+b cost--a sint+b costddta cost+b sinta cost+b sint2×1a cost+b sint       =a cost+b sint-a cost-b sint--a sint+b cost-a sint+b costa cost+b sint3       =-a cost+b sint2--a sint+b cost2a cost+b sint3       =-a cost+b sint2-a sint-b cost2a cost+b sint3       =-y2-x2y3Hence, d2ydx2=-x2+y2y3.

Page No 12.18:

Question 50:

Find A and B so that y=A sin3x+B cos3x satisfies the equationd2ydx2+4dydx+3y=10 cos3x.

Answer:

We have,y=A sin3x+B cos3xdydx=3A cos3x-3B sin3xd2ydx2=-9A sin3x-9B cos3xTherefore,d2ydx2+4dydx+3y=-9A sin3x-9B cos3x+12A cos3x-12B sin3x+3A sin3x+3B cos3x                          =-6A-12B sin3x+-6B+12A cos3xIt is given that,d2ydx2+4dydx+3y=10 cos3xComparing the coefficients of sin3x and cos3x, we get-6A-12B=0           and           -6B+12A=10or A=23           and           B=-13Hence, A=23 and B=-13.

Page No 12.18:

Question 51:

If y=A e-kt cospt+c, prove that d2ydt2+2kdydt+n2y=0, where n2=p2+k2.

Answer:

We have,y=A e-kt cospt+c       ...(1)Differentiating y with respect to t, we getdydt=-kA e-kt cospt+c-pA e-kt sinpt+c      =-ky-pA e-kt sinpt+c         From (1)pA e-kt sinpt+c=-ky-dydt      ...(2)Differentiating dydt with respect to t, we getd2ydt2=-kdydt+pkA e-kt sinpt+c-p2A e-kt cospt+c       =-kdydt+k-ky-dydt-p2y       From (1) and 2       =-kdydt-k2y-kdydt-p2y       =-2kdydt-k2+p2yd2ydt2+2kdydt+k2+p2y=0d2ydt2+2kdydt+n2y=0, where n2=p2+k2.Hence, d2ydt2+2kdydt+n2y=0, where n2=p2+k2.

Page No 12.18:

Question 52:

If y=xna coslogx+b sinlogx, prove that x2d2ydx2+1-2nxdydx+1+n2y=0.

Disclaimer: There is a misprint in the question. It must be x2d2ydx2+1-2nxdydx+1+n2y=0 instead of x2d2ydx2+1-2ndydx+1+n2y=0.

Answer:

We have,y=xna coslogx+b sinlogx       ...(1)Differentiating y with respect to x, we getdydx=nxn-1a coslogx+b sinlogx+xn-a sinlogx×1x+b coslogx×1x      =nxxna coslogx+b sinlogx+xn-1-a sinlogx+b coslogx      =nxy+xn-1-a sinlogx+b coslogx                  From (1)xn-1-a sinlogx+b coslogx=dydx-nxy      ...(2)Differentiating dydx with respect to x, we getd2ydx2=nxdydx-nyx2+n-1xn-2-a sinlogx+b coslogx+xn-1-a coslogx×1x-b sinlogx×1x       =nxdydx-nyx2+n-1xn-1x-a sinlogx+b coslogx-xnx2a coslogx+b sinlogx       =nxdydx-nyx2+n-1xdydx-nxy-yx2                From (1) and 2       =nxdydx-nyx2+n-1xdydx-nn-1yx2-yx2       =dydxn+n-1x-n+n2-n+1yx2       =2n-1xdydx-n2+1yx2x2d2ydx2-x2n-1dydx+n2+1y=0Hence, x2d2ydx2+1-2nxdydx+1+n2y=0.

Page No 12.18:

Question 53:

If y=ax+x2+1n+bx-x2+1-n, prove that x2+1d2ydx2+xdydx-n2y=0.

Disclaimer: There is a misprint in the question, x2+1d2ydx2+xdydx-n2y=0 must be written instead of x2-1d2ydx2+xdydx-n2y=0.

Answer:

We have,y=ax+x2+1n+bx-x2+1-n       ...(1)Differentiating y with respect to x, we getdydx=anx+x2+1n-11+12x2+1×2x-bnx-x2+1-n-11-12x2+1×2x      =anx+x2+1n-11+xx2+1-bnx-x2+1-n-11-xx2+1      =anx+x2+1n-1x2+1+xx2+1-bnx-x2+1-n-1x2+1-xx2+1      =anx+x2+1n-1x+x2+1x2+1+bnx-x2+1-n-1x-x2+1x2+1      =ax+x2+1nnx2+1+bx-x2+1-nnx2+1      =nx2+1y                  From (1)x2+1dydx=nySquaring both sides, we getx2+1dydx2=n2y2          ...(2)Differentiating (2) with respect to x, we getx2+12dydx×d2ydx2+2xdydx2=n22ydydxx2+1d2ydx2+xdydx=n2yx2+1d2ydx2+xdydx-n2y=0Hence,x2+1d2ydx2+xdydx-n2y=0.



Page No 12.22:

Question 1:

If y = a xn + 1 + bxn and x2d2ydx2=λy, then write the value of λ.

Answer:

Here,
y= axn+1+ b x-n and x2d2ydx2=λ yNow,dydx=an+1 xn-bn x-n-1and d2ydx2=ann+1xn-1-bn-n-1 x-n-2Now,  x2d2ydx2=λy    Givenx2ann+1xn-1+bnn+1 x-n-2=λaxn+1+ b x-nann+1 xn+1+ bnn+1x-n=λaxn+1+ b x-nnn+1axn+1+ b x-n=λaxn+1+ b x-nλ=nn+1

Page No 12.22:

Question 2:

If x = a cos ntb sin nt and d2xdt=λx, then find the value of λ.

Answer:

Here,

x=a cos nt-b sin ntNow,dxdt=- an sin nt-bn cos nt d2xdt2=-an2 cos nt+bn2 sin ntAlso,d2xdt2  =λx              Given -an2 cos nt+bn2 sin nt=λa cos nt-b sin nt -n2 a cos nt-b sin nt=λa cos nt-b sin ntλ= -n2

Page No 12.22:

Question 3:

If x = t2 and y = t3, findd2ydx2.

Answer:

Here,
x=t2 and y = t3dxdt=2t and dydt=3t2  dydx=3t2d2ydx2=32dtdx=34t

Page No 12.22:

Question 4:

If x = 2at, y = at2, where a is a constant, then find d2ydx2at x=12.

Answer:

Here,
x= 2at and y =at2Differentiating w.r.t. t, we getdxdt=2a and dydt=2atdydx=2at2a=tDifferentiating again w.r.t. t, we getd2ydx2=1×dtdx=12aNow, d2ydx2x=12=12a

Page No 12.22:

Question 5:

If x = f(t) and y = g(t), then write the value of d2ydx2.

Answer:

Here.
x = f(t) and y = g(t)
dxdt=f't and dydt=g'tdydx=g'tf't


d2ydx2=ddtg'tf't×dtdx               =f'tg''t-g'tf''tf't2×1f't               =f'tg''t-g'tf''tf't3

Page No 12.22:

Question 6:

If y=1-x+x22!-x33!+x44!.....to ∞, then write d2ydx2in terms of y.

Answer:

Here,
y=1-x+x22! -x33!+x44!+...Thus,dydx=-1+2x2!-3x23!+4x34!...        =-1+x-x22!+x33!-...d2ydx2=1-2x2!+3x23!-4x34!+ ...           =1-x+x22! -x33!+...          =y 

Page No 12.22:

Question 7:

If y = x + ex, find d2xdy2.

Answer:

Here,
y= x+ exdydx=1+exdxdy=11+exd2xdy2=-ex1+ex2dxdy=-ex1+ex3

Page No 12.22:

Question 8:

If y = |xx2|, then find d2ydx2.

Answer:

Here,
y=x-x2  =x-x2 if 0<x<1-x+x2  if x>1,x<0dydx=1-2x  if 0<x<1-1+2x  if x>1,x<0d2ydx2=-2   if 0<x<12     if x>1,x<0

Page No 12.22:

Question 9:

If y=loge x, find d2ydx2.

Answer:

Here,

y=logex   =-logex  if 0<x<1          logex if x>1Differentiating  w.r.t. x, we getdydx=-1x  if 0<x<11x if x>1Differentiating again  w.r.t. x, we getd2ydx2=1x2   if 0<x<1-1x2  if x>1

Page No 12.22:

Question 1:

If x = a cos nt b sin nt, then d2xdt2is

(a) n2 x
(b) −n2 x
(c) −nx
(d) nx

Answer:

(b) −n2x

Here,
x=a cosnt-b sinntDifferentiating w.r.t. t, we getdxdt=-an sinnt-bn cosntDifferentiating again w.r.t. t, we getd2xdt2=-an2 cosnt+bn2 sinnt         =-n2acosnt-b sinnt         =-n2x
 
 

Page No 12.22:

Question 2:

If x = at2, y = 2 at, then d2ydx2=

(a) -1t2
(b) 12 at3
(c) -1t3
(d) -12at3

Answer:

  (d) -12at3

Here,
x=at2 and y=2atDifferentiating  w.r.t. t, we getdxdt=2at and dydt=2a dydx=2a2at=1tDifferentiating again w.r.t. t, we getd2ydx2=-1t2dtdx=-12at3



Page No 12.23:

Question 3:

If y = axn+1 + bxn, then x2d2ydx2=

(a) n (n − 1)y
(b) n (n + 1)y
(c) ny
(d) n2y

Answer:

(b) n(n+1)y

Here,
y=axn+1+bx-ndydx=an+1xn-bn x-n-1d2ydx2=ann+1xn-1+bnn+1x-n-2 x2d2ydx2=x2ann+1xn-1+bnn+1x-n-2                   =nn+1axn+1+b x-n                   =nn+1y

Page No 12.23:

Question 4:

d20dx20 2 cos x cos 3 x=

(a) 220 (cos 2 x − 220 cos 4 x)
(b) 220 (cos 2 x + 220 cos 4 x)
(c) 220 (sin 2 x + 220 sin 4 x)
(d) 220 (sin 2 x − 220 sin 4 x)

Answer:

(b) 220(cos2x + 220cos4x)

Here,
  y= 2cosx cos3x=cos3x-x+cos3x+x    =cos2x+ cos4xdydx=-2 sin2x-4 sin4x=-2sin2x+2 sin4xd2ydx2=-4 cos2x -16 cos4x=-22cos2x+ 22cos4xd3ydx3=23sin2x+23sin4xd4ydx4=232cos2x+4×23cos4x=24cos2x+ 24cos4x d20cos2x+ cos4xdx20=220cos2x+ 220cos4x

Page No 12.23:

Question 5:

If x = t2, y = t3, then d2ydx2=

(a) 3/2
(b) 3/4t
(c) 3/2t
(d) 3t/2

Answer:

(b) 3/4t

Here,
x=t2 and y = t3dxdt=2t and dydt=3t2  dydx=3t2d2ydx2=32dtdx=34t

Page No 12.23:

Question 6:

If y = a + bx2, a, b arbitrary constants, then

(a) d2ydx2= 2xy
(b) xd2ydx2=y1
(c) xd2ydx2-dydx+y=0
(d) xd2ydx2=2 xy

Answer:

(b) xd2ydx2=y1

Here,
y= a+bx2y1=2bxy2=2bMultiplying by x on both sides we get, xy2=2bx=y1xd2ydx2=y1

Page No 12.23:

Question 7:

If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to

(a) nn+12
(b) nn+122
(c) -nn+122
(d) none of these

Answer:

(c) -nn+122

Here,
fx=cosx+i sinxcos2x+i sin2x ... cosnx+i sinnxfx=cosx+i sinxcosx+i sinx2... cosx+i sinxnfx=cosx+i sinx1+2+3...........nfx=cosx+i sinxnn+12fx=cosx+i sinxa  where a=nn+12fx=cosax+i sinax        ...1f1=cosa+i sina1=cosa+i sina            ...2      f1=1Differentiating eqn.1, we get,f'x=a-sinax+i cosaxf''x=a2-cosax-i sinaxf''x=-a2cosax+i sinaxf''x=-nn+122cosax+i sinaxf''1=-nn+122cosa+i sinaf''1=-nn+122      Using 2

Page No 12.23:

Question 8:

If y = a sin mx + b cos mx, then d2ydx2 is equal to

(a) −m2y
(b) m2y
(c) −my
(d) my

Answer:

(a) −m2y

Here,

y= a sinmx+b cos mxdydx=am cosmx-bm sinmxd2ydx2=-am2 sinmx-bm2 cosmx               =-m2a sinmx+b cos mx              =-m2y

Page No 12.23:

Question 9:

If fx=sin-1x1-x2, then (1 − x)2 f '' (x) − xf(x) =

(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(a) 1

Here,
fx=sin-1x 1-x2 1-x2 fx=sin-1xDiffferentiating w.r.t. x, we get 1-x2f'x-x fx 1-x2=1 1-x21-x2f'x - xfx=1

DISCLAIMER : In the question instead of (1 − x)2 f '' (x) − xf(x)
                         it should be (1 − x)2 f ' (x) − xf(x)

Page No 12.23:

Question 10:

If y=tan-1 loge e/x2loge ex2+tan-1 3+2 loge x1-6 loge x, then d2ydx2=

(a) 2
(b) 1
(c) 0
(d) −1

Answer:

(c) 0

y=tan-1 loge e/x2loge ex2+tan-1 3+2 loge x1-6 loge xy=tan-1 1-2logex1+2logex+tan-1 3+2 loge x1-6 loge xy=tan-1  1-2logex1+2logex+3+2 loge x1-6 loge x1-1-2logex1+2logex3+2 loge x1-6 loge xy=tan-1  1-2logex1-6 loge x+3+2 loge x1+2logex1+2logex1-6 loge x-1-2logex3+2 loge xy=tan-1 1-8logex+12logex2+3+8logex+4logex21-4logex-12logex2-3+4logex+4logex2y=tan-1 1-8logex+12logex2+3+8logex+4logex21-4logex-12logex2-3+4logex+4logex2y=tan-1 4+16logex2-2-8logex2y=tan-1 41+4logex2-21+4logex2y=tan-1 -2dydx=0d2ydx2=0

Page No 12.23:

Question 11:

Let f(x) be a polynomial. Then, the second order derivative of f(ex) is

(a) f'' (ex) e2x + f'(ex) ex
(b) f'' (ex) ex + f' (ex)
(c) f'' (ex) e2x + f'' (ex) ex
(d) f'' (ex)

Answer:

(a) f''(ex)e2x + f'(ex)ex

Since f(x) is a polynomial,

f'ex=f'ex exf''ex=f''ex (ex)2+f'ex ex           =f''ex e2x+f'ex ex 

Page No 12.23:

Question 12:

If y = a cos (loge x) + b sin (loge x), then x2 y2 + xy1 =

(a) 0
(b) y
(c) −y
(d) none of these

Answer:

(c) −y

Here,
  y= a coslogex+b sinlogexy1=-asinlogex1x+b  coslogex1xy2=-asinlogex+b  coslogexxy2=-acoslogex-b sinlogex--asinlogex+b  coslogexx2x2y2=-acoslogex+b sinlogex--asinlogex+b  coslogex x2y2=-y-xy1 x2y2+xy1=-y    

Page No 12.23:

Question 13:

If x = 2 at, y = at2, where a is a constant, then d2ydx2 at x=12is

(a) 1/2a
(b) 1
(c) 2a
(d) none of these

Answer:

(a) 1/2a

Here,
x= 2at and y =at2Differentiating w.r.t. t, we getdxdt=2a and dydt=2atdydx=2at2a=tDifferentiating w.r.t. x, we getd2ydx2=1×dtdx=12aNow, d2ydx2x=12=12a

Page No 12.23:

Question 14:

If x = f(t) and y = g(t), then d2ydx2is equal to

(a) f' g''-g'f''f'3
(b) f' g''-g'f''f'2
(c) g''f''
(d) f'' g'-g'' f'g'3

Answer:

(a) f' g''-g'f''f'3

Here,
x = f(t) and y = g(t)
dxdt=f't and dydt=g'tdydx=g'tf't


d2ydx2=ddtg'tf't×dtdx               =f'tg''t-g'tf''tf't2×1f't               =f'tg''t-g'tf''tf't3



Page No 12.24:

Question 15:

If y = sin (m sin−1 x), then (1 − x2) y2xy1 is equal to

(a) m2y
(b) my
(c) −m2y
(d) none of these

Answer:

(c)−m2y

Here,
y= sinmsin-1xy1= cosmsin-1xm1-x2y2=- sinmsin-1xm21-x2+mxcosmsin-1x1-x23/2y2=- sinmsin-1xm21-x2+xmcosmsin-1x1-x2×1-x2y2=- sinmsin-1xm21-x2+xy11-x21-x2y2=- ym2+xy11-x2y2-xy1=- m2y

Page No 12.24:

Question 16:

If y = (sin−1 x)2, then (1 − x2)y2 is equal to

(a) xy1 + 2
(b) xy1 − 2
(c) −xy1+2
(d) none of these

Answer:

(a) xy1 + 2

Here,

y=sin-1x2Now,y1=2 sin-1x 11-x2y2=21-x2+2x sin-1x1-x23/2y2=21-x2+2x sin-1x1-x21-x2y2=21-x2+xy11-x2y21-x2=2+xy1

Page No 12.24:

Question 17:

If y = etan x, then (cos2 x)y2 =

(a) (1 − sin 2x) y1
(b) −(1 + sin 2x)y1
(c) (1 + sin 2x)y1
(d) none of these

Answer:

(c) (1 + sin 2x)y1
Here,
y=etanxy1=etanx sec2xy2=etanx sec4x+ etanx ×2secx secx tanxy2=sec2x etanxsec2x+ etanx ×2 tanxcos2xy2=y1+ etanx ×y1sec2x2 tanxcos2xy2=y1+y1×2 sinx cosxcos2xy2=y11+sin2x

Page No 12.24:

Question 18:

If y=2a2-b2tan-1 a-ba+btanx2, a>b>0, then

(a) y1=-1a+b cos x
(b) y2=b sin xa+b cos x2
(c) y1=1a-b cos x
(d) y2=-b sin xa-b cos x2

Answer:

Disclaimer: The question given in the book is wrong.

Page No 12.24:

Question 19:

If y=ax+bx2+c, then (2xy1 + y)y3 =

(a) 3(xy2 + y1)y2
(b) 3(xy1 + y2)y2
(c) 3(xy2 + y1)y1
(d) none of these

Answer:

(a) 3(xy2 + y1)y2

Here,

y=ax+bx2+cx2+cy=ax+bDiffferentiating w.r.t. x, we get2xy+x2+cdydx=aDiffferentiating w.r.t. x, we get2y+2xy1+2xy1+x2+cy2=02y+4xy1+x2+cy2=0Diffferentiating again w.r.t. x, we get2y1+4y1+4xy2+x2+cy3+2xy2=06y1+6xy2+x2+cy3=06y1+6xy2+-2y-4xy1y2y3=0     2y+4xy1+x2+cy2=06y1y2+6xy22-2y-4xy1y3=03y1y2+3xy22-y-2xy1y3=0y1+xy23y2=2xy1+yy3

Page No 12.24:

Question 20:

If y=logexa+bxx, then x3 y2 =

(a) (xy1y)2
(b) (1 + y)2
(c) y-xy1y12
(d) none of these

Answer:

(a) (xy1y)2

Here,
y=logexa+bxxy=x logexa+bx      y1=logexa+bx+x×a+bxx1a+bx-bxa+bx2y1=logexa+bx+aa+bx              ...1y1=yx+aa+bx        y=x logexa+bxxy1-yx=aa+bx          ...2Differentiating 1 we get,y2=a+bxxa+bx-bxa+bx2-baa+bx2y2=axa+bx-baa+bx2y2=aa+bx-abxx a+bx2y2=a2x a+bx2y2=xy1-y2x3               Using 2x3 y2=xy1-y2 

Page No 12.24:

Question 21:

If x = f(t) cos tf' (t) sin t and y = f(t) sin t + f'(t) cos t, then dxdt2+dydt2=

(a) f(t) − f''(t)
(b) {f(t) − f'' (t)}2
(c) {f(t) + f''(t)}2
(d) none of these

Answer:

(c){f(t) + f''(t)}2

Here,
x= ftcost-f't sint and y=ft sint+f'tcostdxdt=f'tcost-ftsint-f''tsint-f'tcost and dydt=f't sint+ftcost+f''tcost-f't sintdxdt=-ftsint-f''tsint and dydt=ftcost+f''tcostThus,dxdt2+dydt2=-ftsint-f''tsint2+ftcost+f''tcost2                      =ftsint+f''tsint2+ftcost+f''tcost2                      =sin2tft+f''t2+cos2tft+f''t2                      =ft+f''t2sin2t+cos2t                      =ft+f''t2

Page No 12.24:

Question 22:

If y1n+y-1n=2x, then find x2-1y2+xy1=

Answer:

c n2yy1n+y-1n=2xDifferentiating the above equation with respect to x1ny1n-1-1ny-1n-1y1=21nyy1n-y-1ny1=2y1n-y-1ny1=2ny     .....1y1n-y-1ny2+y11ny1n-1+1ny-1n-1y1=2ny1nyy1n-y-1ny2+y12y1n+y-1n=2n2yy1Dividing the above equation by y1nyy1y1n-y-1ny2+y1y1n+y-1n=2n2yPutting y1 from equation 1y1n-y-1n22y2+y1y1n+y-1n=2n2y  .....2Now,y1n-y-1n2=y1n+y-1n2-4y1n-y-1n2=4x2-4  .....3Putting the value of 3 in 24x2-1y22+2xy1=2n2yx2-1y2+xy1=n2y

Page No 12.24:

Question 23:

If ddxxn-a1 xn-1+a2 xn-2+...+-1n anex=xn ex, then the value of ar, 0 < rn, is equal to

(a) n!r!
(b) n-r!r!
(c) n!n-r!
(d) none of these

Answer:

(c) n!n-r!

According to the given equation,
ddxxn-a1 xn-1+a2 xn-2+...+-1n anex=xn exddxxn-a1 xn-1+a2 xn-2+...+-1n anex=ddxxn-nxn-1+nn-1xn-2+...+-1n anexComparing the coefficients of the above equation we get,a1=na2=nn-1Similarly,ar=nn-1n-2n-3...n-r+1ar=n!n-r!

Page No 12.24:

Question 24:

If y = xn−1 log x then x2 y2 + (3 − 2n) xy1 is equal to

(a) −(n − 1)2 y
(b) (n − 1)2y
(c) −n2y
(d) n2y

Answer:

(a) −(n − 1)2 y

Here,

y=xn-1 logxy1=n-1xn-2 logx+xn-1xy1=n-1xn-1 logx+xn-1xxy1=n-1y+xn-1xy2+y1=n-1y1+n-1xn-2xy2+y1=n-1y1+n-1xn-1xx2y2+xy1=xn-1y1+n-1xn-1x2y2+xy1=xn-1y1+n-1xy1-n-1yx2y2+xy1=xn-1y1+n-1xy1-n-12yx2y2+xy1=2xn-1y1-n-12yx2y2+xy1-2xn-1y1=-n-12yx2y2+xy11-2n+2=-n-12yx2y2+3-2nxy1=-n-12y

Page No 12.24:

Question 25:

If xy − loge y = 1 satisfies the equation xyy2+y12-y2+λ yy1=0, then λ =

(a) −3
(b) 1
(c) 3
(d) none of these

Answer:

(c) 3

Here,

xy-logey=1xy1+y-y1y=0xyy1+y2-y1=0yy1+xy1y1+xyy2+2yy1-y2=0xy12+yy2-y2+3yy1=0 λ=3

Page No 12.24:

Question 26:

If y2 = ax2 + bx + c, then y3d2ydx2is

(a) a constant
(b) a function of x only
(c) a function of y  only
(d) a function of x and y

Answer:

(a) a constant

Here,

y2=ax2+bx+cNow,2ydydx=2ax+b2yd2ydx2+2dydx2=2a yd2ydx2+dydx2=a yd2ydx2+2ax+b2y2=a      2ydydx=2ax+b4y3d2ydx2+2ax+b2=4ay2y3d2ydx2=4ay2-2ax+b24y3d2ydx2=4aax2+bx+c-2ax+b24      y2=ax2+bx+cy3d2ydx2=4a2x2+4abx+4ac-4a2x2-b2-4axb4y3d2ydx2=4ac-b24=a constant



View NCERT Solutions for all chapters of Class 12