JEE Mathematics Differential Coefficient MCQs Set 04

Answer: (a) 2

Explanation:
1. By the chain rule, \( \frac{dy}{dx}=f'\left(x^2\right)\cdot2x \).
2. \( f'\left(x^2\right)=\sqrt{2x^4-1} \).
3. At \( x=1 \): \( \sqrt{2-1}\cdot2=2 \).

Teacher's Note:
1. Replace x by \( x^2 \) inside \( f' \) and multiply by the derivative of \( x^2 \).

Answer: (a) \( \frac1e \)

Explanation:
1. \( f(x)=\frac{\ln(\ln x)}{\ln x} \).
2. \( f'(x)=\frac{\frac1{x\ln x}\cdot\ln x-\ln(\ln x)\cdot\frac1x}{(\ln x)^2}=\frac{1-\ln\ln x}{x(\ln x)^2} \).
3. At \( x=e \): \( \ln e=1 \) and \( \ln1=0 \), so \( f'(e)=\frac1e \).

Teacher's Note:
1. Change the base to natural logarithms before differentiating.

Answer: (c) 1

Explanation:
1. \( x=t^{-3}+t^{-2} \) gives \( \frac{dx}{dt}=-t^{-4}(3+2t) \), and \( y=\frac32t^{-2}+2t^{-1} \) gives \( \frac{dy}{dt}=-t^{-3}(3+2t) \).
2. So \( \frac{dy}{dx}=t \).
3. Then \( x t^3-t=\frac{(1+t)}{t^3}t^3-t=1 \).

Teacher's Note:
1. Differentiate both x and y with respect to the parameter and divide.

Answer: (d) \( \left(\frac\pi4\right)^{1/\sqrt2}\left(\frac{\sqrt2}2\ln\frac\pi4+\frac{2\sqrt2}\pi\right) \)

Explanation:
1. Near \( x=\frac\pi4 \), \( f(x)=x^{\sin x} \). Taking logs, \( \ln f=\sin x\ln x \).
2. Differentiating: \( \frac{f'}f=\cos x\ln x+\frac{\sin x}x \).
3. At \( \frac\pi4 \): \( \frac{\sqrt2}2\ln\frac\pi4+\frac{\sqrt2}2\cdot\frac4\pi=\frac{\sqrt2}2\ln\frac\pi4+\frac{2\sqrt2}\pi \).
4. Multiplying by \( f\left(\frac\pi4\right)=\left(\frac\pi4\right)^{1/\sqrt2} \) gives option (d).

Teacher's Note:
1. Use logarithmic differentiation for a function raised to a function.

Answer: (d) \( f''\left(e^x\right)e^{2x}+f'\left(e^x\right)e^x \)

Explanation:
1. First derivative: \( \frac d{dx}f\left(e^x\right)=f'\left(e^x\right)e^x \).
2. Second derivative by the product rule: \( f''\left(e^x\right)e^x\cdot e^x+f'\left(e^x\right)e^x \).
3. That is \( f''\left(e^x\right)e^{2x}+f'\left(e^x\right)e^x \).

Teacher's Note:
1. Apply the chain rule twice and remember that the product rule is needed the second time.

Answer: (b) \(-\frac1{(1+x)^2} \)

Explanation:
1. Move one term across and square: \( x^2(1+y)=y^2(1+x) \), i.e. \( (x-y)(x+y+xy)=0 \).
2. The relation requires \( x+y+xy=0 \) (the case \( y=x \) does not satisfy the original equation except at 0), so \( y=-\frac x{1+x} \).
3. Then \( \frac{dy}{dx}=-\frac{(1+x)-x}{(1+x)^2}=-\frac1{(1+x)^2} \).

Teacher's Note:
1. Squaring and factorising gives y explicitly as a function of x.

Answer: (b) \( \frac{1+a^2}{a^{10}} \)

Explanation:
1. For an inverse function, \( g'(y)=\frac1{f'(g(y))} \).
2. With \( g(2)=a \): \( g'(2)=\frac1{f'(a)} \).
3. \( f'(a)=\frac{a^{10}}{1+a^2} \), so \( g'(2)=\frac{1+a^2}{a^{10}} \).

Teacher's Note:
1. The derivative of the inverse is the reciprocal of the derivative of f at the corresponding point.

Answer: (d) \( \frac1{\sqrt{1-x^2}} \)

Explanation:
1. Let \( A=\sin^{-1}x \) and \( B=\sin^{-1}\sqrt x \). Then \( \sin A=x \), \( \cos A=\sqrt{1-x^2} \), \( \sin B=\sqrt x \), \( \cos B=\sqrt{1-x} \).
2. The expression is \( \sin A\cos B+\cos A\sin B=\sin(A+B) \), so \( y=\sin^{-1}x+\sin^{-1}\sqrt x \).
3. Differentiating: \( \frac{dy}{dx}=\frac1{\sqrt{1-x^2}}+\frac1{2\sqrt x\sqrt{1-x}} \), so \( p=\frac1{\sqrt{1-x^2}} \).

Teacher's Note:
1. Recognise the sine addition formula inside the inverse sine.

Answer: (c) 2 and \(-1\)

Explanation:
1. \( 1+x^2+x^4=\left(1+x+x^2\right)\left(1-x+x^2\right) \).
2. So the fraction equals \( 1-x+x^2 \).
3. Its derivative is \( 2x-1 \), so \( a=2 \) and \( b=-1 \).

Teacher's Note:
1. Factorise the numerator as \( (x^2+x+1)(x^2-x+1) \) before differentiating.

Answer: (b) \( \dfrac{2\left(1+x-x^2\right)}{(1+x^2)^2}\sin\left(\dfrac{2x-1}{x^2+1}\right) \)

Explanation:
1. By the chain rule, \( \frac{dy}{dx}=f'(u)\cdot\frac{du}{dx} \) with \( u=\frac{2x-1}{x^2+1} \).
2. \( \frac{du}{dx}=\frac{2(x^2+1)-(2x-1)2x}{(x^2+1)^2}=\frac{2+2x-2x^2}{(x^2+1)^2}=\frac{2(1+x-x^2)}{(1+x^2)^2} \).
3. Since \( f'(u)=\sin u \), the result is option (b).

Teacher's Note:
1. Differentiate the inner function with the quotient rule and multiply by \( f'(u) \).

Answer: (b) \( \frac{x+y}{x-y} \)

Explanation:
1. Use polar form: \( x=r\cos\theta,\ y=r\sin\theta \), where \( \theta=t \). Then the condition says \( r=e^{\theta} \), so \( \frac{dr}{d\theta}=r \).
2. \( \frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}=\frac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta} \).
3. Dividing by r: \( \frac{dy}{dx}=\frac{y+x}{x-y} \).

Teacher's Note:
1. A curve with \( r=e^\theta \) is a logarithmic spiral; differentiate in polar form.

Answer: (d) A straight line with slope 2 and y-intercept 2

Explanation:
1. \( \frac d{dx}\left(f^2+g^2\right)=2ff'+2f'f''=2f'(f+f'')=0 \), so \( h'(x)=f^2+f'^2 \) is a constant k.
2. Hence \( h \) is linear: \( h(x)=kx+h(0)=kx+2 \).
3. Using \( h(1)=4 \): \( k=2 \). So \( y=2x+2 \), a line with slope 2 and y-intercept 2.

Teacher's Note:
1. Show that \( h' \) has zero derivative, so \( h' \) is constant.

Answer: (b) \( \frac1{2(1+x^2)} \) for \( x>0 \) and \(-\frac1{2(1+x^2)} \) for \( x<0 \)

Explanation:
1. Put \( x=\tan\theta \), so \( \sqrt{1+x^2}=\sec\theta \) and the fraction is \( \frac{\sec\theta+1}{2\sec\theta}=\frac{1+\cos\theta}2=\cos^2\frac\theta2 \).
2. So \( y=\cos^{-1}\left|\cos\frac\theta2\right|=\frac{|\theta|}2=\frac{|\tan^{-1}x|}2 \).
3. For \( x>0 \), \( \frac{dy}{dx}=\frac1{2(1+x^2)} \); for \( x<0 \), \( \frac{dy}{dx}=-\frac1{2(1+x^2)} \).

Teacher's Note:
1. The square root gives an absolute value, so the sign of the derivative depends on the sign of x.

Answer: (a) AP

Explanation:
1. Let \( f(x)=px^2+qx+r \). \( f(1)=f(-1) \) gives \( q=0 \), so \( f'(x)=2px \).
2. Then \( f'(a),f'(b),f'(c) \) are \( 2pa,2pb,2pc \).
3. Since a, b, c are in AP, so are \( 2pa,2pb,2pc \).

Teacher's Note:
1. The condition \( f(1)=f(-1) \) removes the linear term, making \( f' \) a linear function through the origin.

Answer: (b) 2

Explanation:
1. The limit is of the form \( \frac00 \) since the integral from 2 to 2 is 0 at \( x=\frac\pi4 \). Apply L'Hopital's rule.
2. The numerator derivative is \( f(\sec^2x)\cdot2\sec^2x\tan x \); the denominator derivative is \( 2x \).
3. At \( x=\frac\pi4 \): \( \frac{f(2)\cdot2\cdot2\cdot1}{\pi/2}=\frac{8}\pi f(2) \). So \( r=2 \).

Teacher's Note:
1. Differentiate the integral with the Leibniz rule, noting \( \sec^2\frac\pi4=2 \).

Answer: (a) 5

Explanation:
1. \( F'(x)=2ff'+2f'f''=2f'(f+f'')=0 \).
2. So F is constant.
3. Therefore \( F(10)=F(5)=5 \).

Teacher's Note:
1. Show that the derivative of \( f^2+f'^2 \) is zero, so the quantity is conserved.

Answer: (b) 1

Explanation:
1. Differentiate \( x^2+y^2=1 \): \( 2x+2yy'=0 \), so \( yy'+x=0 \).
2. Comparing with \( yy'+ax=0 \) gives \( a=1 \).

Teacher's Note:
1. Implicit differentiation of the circle gives the relation directly.