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Page No 11.16:

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−1x
(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 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.16:

Question 9:

If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), prove that d2xdθ2=acos θ-θ sin θ,d2ydθ2=asin θ+θ cos θ and d2ydx2=sec3θa θ.

Answer:

We have, x=acosθ+θsinθ
dxdθ=dacosθ+θsinθdθdxdθ=-asinθ+asinθ+aθcosθ=aθcosθ                                   .....id2xdθ2=ddθdxdθ=daθcosθdθd2xdθ2=acosθ-aθsinθ=acosθ-θsinθ

y=asinθ-θcosθdydθ=dasinθ-θcosθdθdydθ=acosθ-acosθ+aθsinθ=aθsinθ                                 .....iid2ydθ2=ddθdydθ=daθsinθdθd2ydθ2=asinθ+aθcosθ=asinθ+θcosθ
From (i) and (ii), we have
dydx=dydθdxdθ=aθsinθaθcosθ=tanθd2ydx2=ddxdydx=dtanθdx=sec2θdθdxd2ydx2=sec2θ1aθcosθd2ydx2=sec3θaθ
Hence proved.

Page No 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.16:

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 11.17:

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 11.17:

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 11.17:

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 11.17:

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 11.17:

Question 20:

If y = (sin−1x)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 11.17:

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 11.17:

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 11.17:

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 11.17:

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 11.17:

Question 25:

If log y = tan−1x, 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 11.17:

Question 26:

If y = tan−1x, 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 11.17:

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 11.17:

Question 28:

If y = (tan−1x)2, then prove that (1 + x2)2y2 + 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 11.17:

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 11.17:

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 11.17:

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 11.17:

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 11.17:

Question 33:

If y = cos−1x, 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 11.17:

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 11.17:

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 11.17:

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 11.17:

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 11.17:

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 11.17:

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 11.18:

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 11.18:

Question 41:

If y = (cot−1x)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 11.18:

Question 42:

If y = cosec−1x, 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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.18:

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 11.22:

Question 1:

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

(a) n2x
(b) −n2x
(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 11.22:

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 11.22:

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 11.22:

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 11.22:

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 11.22:

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 11.22:

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 11.22:

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 11.22:

Question 9:

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

Page No 11.22:

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 11.22:

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 11.23:

Question 12:

If y = a cos (logex) + b sin (logex), then x2y2 + 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 11.23:

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 11.23:

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 11.23:

Question 15:

If y = sin (m sin−1x), 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 11.23:

Question 16:

If y = (sin−1x)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 11.23:

Question 17:

If y = etanx, then (cos2x)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 11.23:

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 11.23:

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 11.23:

Question 20:

If y=logexa+bxx, then x3y2 =

(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 11.23:

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 11.23:

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 11.23:

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 11.23:

Question 24:

If y = xn−1 log x then x2y2 + (3 − 2n) xy1 is equal to

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

Answer:

(a) −(n − 1)2y

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 11.23:

Question 25:

If xy − logey = 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 11.24:

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

Page No 11.24:

Question 1:

  If y = t10 + 1 and x = t8 + 1, then d2ydx2 = ___________________.

Answer:


Given, y = t10 + 1 and x = t8 + 1.

y=t10+1

Differentiating both sides with respect to t, we get

dydt=10t9

x=t8+1

Differentiating both sides with respect to t, we get

dxdt=8t7

dydx=dydtdxdtdydx=10t98t7dydx=5t24

Differentiating both sides with respect to x, we get

ddxdydx=ddx5t24

d2ydx2=54×2tdtdx

d2ydx2=52t×18t7              dxdt=8t7dtdx=18t7

d2ydx2=516t6


If y = t10 + 1 and x = t8 + 1, then d2ydx2 =      516t6     .

Page No 11.24:

Question 2:

If x = a sin θ and y = b cos θ, then d2ydx2 = ______________________.

Answer:


Given, x=asinθ and y=bcosθ.

x=asinθ

Differentiating both sides with respect to θ, we get

dxdθ=acosθ

y=bcosθ

Differentiating both sides with respect to θ, we get

dydθ=-bsinθ

dydx=dydθdxdθ

dydx=-bsinθacosθ

dydx=-btanθa

Differentiating both sides with respect to x, we get

ddxdydx=ddx-btanθa

d2ydx2=-basec2θdθdx

d2ydx2=-basec2θ×1acosθ              dxdθ=acosθdθdx=1acosθ                

d2ydx2=-ba2sec3θ


If x = a sin θ and y = b cos θ, then d2ydx2 =      -ba2sec3θ     .

Page No 11.24:

Question 3:

If y = x + ex, then d2ydx2 = _____________________.

Answer:


y=x+ex

Differentiating both sides with respect to x, we get

dydx=ddxx+ex

dydx=1+ex  

Again differentiating both sides with respect to x, we get

ddxdydx=ddx1+ex

d2ydx2=0+ex=ex

d2ydx2=ex


If y = x + ex, then d2ydx2 =      ex     .

Page No 11.24:

Question 4:

If y=1-x+x22!-x33!+x44! _________________, then d2ydx2 = __________________.

Answer:


y=1-x+x22!-x33!+x44!-...

Differentiating both sides with respect to x, we get

dydx=ddx1-x+x22!-x33!+x44!-...

dydx=ddx1-ddxx+ddxx22!-ddxx33!+ddxx44!-...

dydx=0-1+2x2!-3x23!+4x34!-...

dydx=-1+x1!-x22!+x33!-...

Again differentiating both sides with respect to x, we get

ddxdydx=ddx-1+x1!-x22!+x33!-...

d2ydx2=ddx-1+ddxx1!-ddxx22!+ddxx33!-...

d2ydx2=0+1-2x2!+3x23!-...

d2ydx2=1-x+x22!-x33!+...


If y=1-x+x22!-x33!+x44!- _________________, then d2ydx2 =          1-x+x22!-x33!+...             .

Page No 11.24:

Question 5:

If y = x + ex , then d2xdy2 = ______________.

Answer:


y=x+ex

Differentiating both sides with respect to x, we get

dydx=ddxx+ex

dydx=1+ex

dxdy=11+ex                       dxdy=1dydx

Differentiating both sides with respect to y, we get

ddydxdy=ddy11+ex

d2xdy2=0-1×ddy1+ex1+ex2

d2xdy2=-0+exdxdy1+ex2

d2xdy2=-ex11+ex1+ex2                          dxdy=11+ex

d2xdy2=-ex1+ex3


If y = x + ex , then d2xdy2 =      -ex1+ex3     .

Page No 11.24:

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 11.24:

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 11.24:

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 11.24:

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 11.24:

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 11.24:

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 11.24:

Question 7:

If y = x + ex, find d2xdy2.

Answer:

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



Page No 11.25:

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 11.25:

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



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