Continuity & Differentiability NCERT Solutions

Miscellaneous Ex Q12-23

Question: (12) Find $(dy)/dx$ if $y= 12(1 cos t), x= 10 (t- sin t), (-pi)/2 < t < pi/2$

Solution:

Here $x=10 (t - sin t)$ and $y = 12(1- cos t)$

Differentiating both sides w.r.t. $x$ , we have

$(dx)/(dt)= d/(dt)[10(t-sin t)]$ and $(dy)/(dt) = d/(dt)$

$= d/(dt) [12(1- cos t)]$

$= 10 (1-cos t)$

$=12 sin t$

Now $(dy)/(dx) = (dy//dt)/(dx//dt)$

$(12 sin t)/(10 (1- cos t))$

$(12xx2 sin \ t/2 cos \ t/2)/(10 xx 2 sin ^2 \ t/2)$

$= 6/5 cot \ t/2$

Question: (13)Find $(dy)/(dx)$ , if $y= sin^-1 x + sin ^-1 sqrt (1-x^2), -1<= x<= 1$

Solution:

Here $y = sin^-1 x + sin ^-1 sqrt (1-x^2)$

$= sin^-1[xsqrt (1-(1-x^2))+ sqrt(1-x^2). sqrt(1-x^2)]$

$= sin^-1[x sqrt(1-((1-x^2) + sqrt (1-x^2) . sqrt (1-x^2)]$

$[∵ sin^-1 x+ sin ^-1 y= sin^-1 (xsqrt(1-y^2) +ysqrt (1-x^2))]$

$=sin^-1[x^2+1- x^2]=sin^-1(1) = pi/2$

$y= pi/2$

Differentiating both sides w.r.t. $x$ , we have

$(dy)/(dx)= d/(dx)(pi/2) = 0$

Question: (14) If $x sqrt(1+y)+ ysqrt(1+x) = 0, for-1< x < 1$ , prove that

Solution:

Here $xsqrt (1+y) + y sqrt (1+x)=0$

$x sqrt(1+y) = - y sqrt(1+x)$

squaring both sides, we have

$x^2 (1+y)$

$= y^2 (1+x)$

$=>x^2+ x^2y$

$= xy(y-x)$

$=>x+y =-xy$

$=> y+xy=-x$

$=> y= -x/(1+x)$

Differentiating both sides w.r.t. $x$ , we have

$(dy)/(dx)= d/(dx)[-x/(1+x)]$

$((1+x)d/dx (-x)-(-x)d/dx(1+x))/((1+x)^2)$

$(-1-x+x)/(1+x^2)= -1/((1+x)^2)$

Question: (15) If $(x-a)^2 + (y-b)^2 = c^2, for some c > 0$ , prove that

$[1+(dy/dx)^2]^(3/2)/((d^2y)/(dx^2)$ is a constant independent of $a$ and $b$

Solution:

Here $(x-a)^2 +(y-b)^2 =c^2$

Differentiating both sides w.r.t. $x$ , we have

$d/dx [(x-a)^2]+(y-b)^2]$

$=d/dx(c^2)$

$2(x-a)d/dx(x-a)$ $+2(y-b).d/dx(y-b)=0$

$2(x-a)+ 2(y-b)(dy)/(dx)=0$

$(x-a)+ (y-b)(dy)/(dx)=0$

Again differentiating both sides w.r.t $x$ , we have

$d/dx [(x-a)+ (y- b)(dy)/(dx)]=d/dx(0)$

$1+(y-b).d/dx(dy/dx)$ $+ dy/dx.d/dx(y-b)=0$

$1+(y-b).(d^2y)/(dx^2)+ (dy/dx)^2 =0$

$(y-b)(d^2y)/(dx^2)= -(1+ (dy/dx)^2)$

$(y-b)=-[(1+(dy/dx)^2)/((d^2y)/(dx^2))]$ ----(iii)

Putting value of $(y-b)$ in (ii), we have

$(x-a)-[(1+(dy/dx)^2)/((d^2y)/(dx^2))](dy)/(dx)=0$

$(x-a)=[(1+(dy/dx)^2)/((d^2y)/(dx^2))](dy)/(dx)$

Putting value of $(x-a)$ and $(y-b)$ in (i), we get

$[(1+(dy/dx)^2)/((d^2y)/(dx^2))](dy/dx)^2$ $+[(1+(dy/dx)^2)/((d^2y)/(dx^2))]=c^2$

$[(1+(dy/dx)^2)]^2 [1+(dy/dx)^2]=c^2((d^2y)/(dx^2))^2$

$[1+(dy/dx)^2]^3 = c^2(d^2y/dx^2)^2$

Taking square root on both sides, we have

$[(1+(dy/dx)^2)]^(3/2) =c((d^2y)/(dx^2))$

$c= ([1+(dy/dx)^2]^(3/2))/((d^2y)/(dx^2))$

Question: (16) If $cos y = xcos(a+y)$ with $cos a!= +-1$ , prove that $(dy)/(dx) =cos^2 \ ((a+y))/(sin a)$

Solution:

Here $cos y=x cos(a+y)$

$:. x= (cos y)/(cos (a+y))$

Differentiating both sides w.r.t. $x$ , we have

$dx/dy =d/dy[(cos y)/(cos(a+y))]$

$=(cos (a+y)d/dy (cos y)- cos y d/dy (cos(a+y)))/(cos^2(a+y)$

$=-(cos(a+y)sin y+cosy sin(a+y))/(cos^2(a+y))$

$= (sin[a+y-y])/(cos^2(a+y))$

$= (sin a)/(cos^2(a+y)$

$:. dy/dx = (cos^2(a+y))/(sin a)$

Question: (17) If $x = a(cos t+ t sin t)$ and $y=a(sin t-t cos t)$ , find $(d^2y)/(dx^2)$

Solution:

Here $x= a (cos t + t)$ and $y = a (sin t - t cos t)$

Differentiating both sides w.r.t. $x$ , we have

$dx/dy =d/dt[a(cos t + t sin t)]$

$a[-sin t + t d/dt(sin t)+ sin td/dt (t)]$

$a[ -sin t+ t cos t + sin t]= at cos t$

$dy/dt = d/dt [a (sin t- t cos t)]$

$a[cos t-{t \ d/dt(cost t)+ cos t\ d/dt(t)}]$

$a[ cos t + tsin t - cos t] = at sin t$

Now, $dy/dx =(dy/dt)/(dx/dt)$

$(at sin t)/(at cos t)=tan t$

Differentiating both sides w.r.t. $x$ , we have

$d/dx(dy/dt)=d/dx (tan t)$

$(d^2y)/(dx^2)=sec^2t (dt)/(dx)$

$sec^2 t. 1/ (at cost)$

$= 1/at sec^3 t.$

Question: (18) If $f(x)= |x|^3$ , show that $t f'(x)$ exists for all real $x$ and find it.

Solution:

Here $f(x)=|x|^3$

When $x>= 0$

$f(x)=|x|^3= x^3$

Differentiating both sides w.r.t. $x$ , we have

$d/dx[f(x)]=d/dx (x^3)$

Again differentiating both sides w.r.t. $x$ , we have

$d/dx [f(x)]= d/dx (3x^2)$

$f(x) =6x$

When $x<0$

$f|x| = |x|^3$

$= (-x)^3=-x^3$

Differentiating both sides w.r.t. $x$ , we have

$d/dx [f(x)]= d/dx(-x^3)$

$f(x) =-3x^2$

Again differentiating both sides w.r.t. $x$ , we have

$d/dx [f(x)]= d/dx(-3x^2)$

$f'(x) =6x$

Question: (19) Using mathematical induction prove that $d/dx(x^n)=nx^n-1$ for all positive integers n.

Solution:

Here $P(n): d/dx(x^n)=nx^(n-1)$

For $n=1$

$p(1): d/dx(x^1)= 1x^1 -1$

which is true.

Suppose it is trus for $n=k$

$P(k):d/dx(x^k)=kx^(k-1)$ is true.

Let $n=k+1$

$P(k+1): d/dx(x^(k+1))$

$=(k+1)^(k+1-1)$

$=(k+1)x^k$

$d/dx(x^(k+1))$

$= d/dx (x^k.x)$

$=x^k. d/dx (x)$ $+x.d/dx(x^k)$

$= x^k. 1+ x. kx^(k-1)$

$x^k+kx^k$

$=(k+1)x^(k+1)$

So it is true for $n=k+1$ , whenever it is true for $n=k$

Thus by principal of mathematical induction it is true for all values of $k$ .

Question: (20) Using the fact that $sin(A+B)=sin A cos B+cos A sin B$ and the differentation , obtain the sum formula for cosine.

Solution:

Here $sin (A+B)= sin A cos B+ cos A sin B$

Let A and B be functions of t, then differetiating both sides w.r.t. $t$ we have

$d/dt[sin(A+B)]$

$=d/dt[ sin A cos B + cos A sin B]$

$cos (A+B)[(dA)/(dt)+ (dB)/(dt)]$

$sin A d/dt cos B + cos B d/dt sin A + cos A d/dt sin B + sin B d/dt cos A$

$=- sin A cos B . (dB)/dt + cos A cos B (dA)/dt + cos A cos B (dB)/ dt -sin A sin B (dA)/dt$

$cos (A+B)[(dA)/dt + (dB)/dt]$

$cos A cos B [(dA)/dt +(dB)/dt]- sin A sin B [(dA)/dt +(dB)/dt]$

$(cos A cos B - sin A sin B)[(dA)/dt+ (dB)/dt]$

$:. cos (A+B)=cos A cos B -sin A sin B$

Question: (21) Does there exist a function which is continuous everywhere but not differentiable at exactly two points? justify your answer.

Solution:

If we take $f(x)=|x|+|x-1|$

Then $f(x)$ is continuous everywhere, but it is not differentiable at $x=0$ and $x=1$

Question: (22) If $y=|(f(x), g(x), h(x)), (l, m, n), (a, b, c)|$ ,prove that $dy/dx= |(f'(x), g'(x), h'(x)), (l, m, n), (a, b, c)|$

Solution:

Here $y=|(f(x), g(x), h(x)), (l, m, n), (a, b, c)|$

Differentiating both sides w.r.t. $x$ we have

$(dy)/dx =d/dx |(f'(x), g'(x), h'(x)), (l, m, n), (a, b, c)|$

$=|(f'(x), g'(x), h'(x)), (l, m, n), (a, b, c)|$ $+|(f(x), g(x), h(x)), (0, 0, 0), (a, b, c)|$ $+|(f(x), g(x), h(x)), (l, m, n), (0, 0, 0)|$

$=|(f(x), g(x), h(x)), (l, m, n), (a, b, c)|$

Question: (23) If $y= e^(a cos -1 x), -1<=x <= 1$ , show that $(1=x^2)(d^2y)/(dx^2) -x dy/dx - a^2 y =0$

Solution:

Here $y= e^(a cos -1x)$

Differentiating both sides w.r.t. $x$ we have

$(dy)/dx =d/dx (e^(a cos-1 x))$

$=e^(a cos-1 x). d/dx (a cos ^-1x)$

$e^(a cos-1 x). -a/sqrt(1-x^2)$

$:.sqrt (1-x^2 dy/dx$

$= -a e^(a cos-1 x)$

Again differentiating both sides w.r.t. $x$ we have

$d/dx[sqrt(1-x^2).(dy)/dx]$

$=d/dx [-ae^a cos-1 x]$

$sqrt(1-x^2).d/dx (dy/dx)+ dy/dx. d/dx (sqrt(1-x^2))$

$=-a.e^(acos-1x). d/dx (a cos ^-1x)$

$sqrt(1-x^2)(d^2y)/(dx^2)$ $+ dy/dx .1/2 sqrt(1-x^2) . -2x$

$-a.e^(a cos-1x). -a/ sqrt(1-x^2)$

$sqrt (1-x^2)(d^2y)/(dx^2)- x/ sqrt (1-x^2) (dy)/dx$

$-a^2/(sqrt(1-x^2)) e^(acos-1x)$

Multiplying both sides by $sqrt(1-x^2)$ we get

$(1-x^2)(d^2y)/(dx^2)-x (dy)/dx$

$=a^2y[:. y=e^(a cos -1x)]$

$(1-x^2)(d^2y)/(dx^2)-x(dy)/(dx) -a^2y =0$

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