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
v]`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`
Reference: