JEE Mathematics Differentiation MCQs Set 01

Answer: (b) 2

Explanation:
1. Add and subtract \( 2f(2) \) in the numerator: \( \dfrac{xf(2)-2f(2)+2f(2)-2f(x)}{x-2}=f(2)-2\cdot\dfrac{f(x)-f(2)}{x-2} \).
2. As \( x\to2 \) the limit is \( f(2)-2f'(2)=4-2=2 \).

Teacher's Note:
1. Add and subtract a suitable term so that the first-principle limit \( \frac{f(x)-f(a)}{x-a} \) appears.
2. L'Hospital's rule gives the same result: \( f(2)-2f'(2) \).

Answer: (c) 6

Explanation:
1. Put \( x=5,y=0 \): \( f(5)=f(5)f(0) \), so \( f(0)=1 \).
2. \( f'(5)=\displaystyle\lim_{h\to0}\dfrac{f(5)f(h)-f(5)}{h}=f(5)\lim_{h\to0}\dfrac{f(h)-f(0)}{h}=f(5)f'(0) \).
3. \( f'(5)=2\times3=6 \).

Teacher's Note:
1. The functional equation \( f(x+y)=f(x)f(y) \) gives \( f'(x)=f(x)f'(0) \).
2. Always find \( f(0) \) first.

Answer: (b) \( \dfrac78 \)

Explanation:
1. Replace x by \( \dfrac1x \): \( 5f\!\left(\frac1x\right)+3f(x)=\dfrac1x+2 \). Solving with the original: \( 16f(x)=5x-\dfrac3x+4 \).
2. \( 16f'(x)=5+\dfrac3{x^2} \), so \( f(1)=\dfrac{6}{16} \) and \( f'(1)=\dfrac{8}{16} \).
3. \( \dfrac{dy}{dx}=f+xf' \) gives at \( x=1 \): \( \dfrac6{16}+\dfrac8{16}=\dfrac78 \).

Teacher's Note:
1. Eliminate \( f(1/x) \) by solving two simultaneous equations.
2. Use the product rule on \( xf(x) \) after finding f explicitly.

Answer: (a) 0

Explanation:
1. \( f(x)=x^3 \) for \( x\ge0 \) and \( -x^3 \) for \( x<0 \).
2. \( f'(x)=3x^2 \) for \( x>0 \) and \( -3x^2 \) for \( x<0 \).
3. Right and left derivatives at 0 are both 0, so \( f'(0)=0 \).

Teacher's Note:
1. At a point where a modulus changes sign, compare the left and right derivatives.
2. Powers higher than 1 smooth out the corner.

Answer: (b) \( -1 \)

Explanation:
1. \( \sin2x\cos2x=\dfrac12\sin4x \) and \( \log_22^{x+3}=x+3 \). So \( f(x)=\dfrac12\sin4x\cos3x+x+3 \).
2. \( f'(x)=\dfrac12\left(4\cos4x\cos3x-3\sin4x\sin3x\right)+1 \).
3. At \( x=\pi \): \( \dfrac12(4\cdot1\cdot(-1)-0)+1=-2+1=-1 \).

Teacher's Note:
1. Simplify the function using identities before differentiating.
2. \( \log_a a^k=k \) removes the logarithm completely.

Answer: (c) \( \dfrac12(\sqrt3-1) \)

Explanation:
1. Near \( x=\dfrac{2\pi}3 \), \( \cos x<0 \) and \( \sin x>0 \), so \( y=-\cos x+\sin x \).
2. \( \dfrac{dy}{dx}=\sin x+\cos x \).
3. At \( \dfrac{2\pi}3 \): \( \dfrac{\sqrt3}2-\dfrac12=\dfrac12(\sqrt3-1) \).

Teacher's Note:
1. Remove the modulus signs by checking the sign of each expression near the given point.
2. Differentiate only after the modulus is removed.

Answer: (b) \( \dfrac1e \)

Explanation:
1. \( f(x)=\dfrac{\ln(\ln x)}{\ln x} \).
2. \( f'(x)=\dfrac{\frac1{x\ln x}\cdot\ln x-\frac{\ln(\ln x)}{x}}{(\ln x)^2} \).
3. At \( x=e \): \( \ln e=1,\ \ln(\ln e)=0 \), so \( f'(e)=\dfrac{1/e-0}{1}=\dfrac1e \).

Teacher's Note:
1. Change the base to the natural logarithm first.
2. Then apply the quotient rule.

Answer: (d) none of these

Explanation:
1. \( f(x)=-\log x \) for \( 0<x<1 \) and \( \log x \) for \( x\ge1 \).
2. So \( f'(x)=-\dfrac1x \) for \( 0<x<1 \) and \( \dfrac1x \) for \( x>1 \).
3. No single option (a), (b) or (c) is true for all \( x\ne1 \), so the answer is none of these.

Teacher's Note:
1. A function defined with a modulus has different derivatives on each side of the change point.
2. Check each option over the whole domain, not at one value only.

Answer: (a) 0

Explanation:
1. \( \sec^{-1}u=\cos^{-1}\dfrac1u \), so the first term is \( \cos^{-1}\dfrac{\sqrt x-1}{\sqrt x+1} \).
2. Then \( y=\cos^{-1}t+\sin^{-1}t=\dfrac\pi2 \), where \( t=\dfrac{\sqrt x-1}{\sqrt x+1} \).
3. A constant has derivative 0.

Teacher's Note:
1. Use \( \sin^{-1}t+\cos^{-1}t=\dfrac\pi2 \) whenever both inverse functions have the same argument.
2. Convert secant inverse to cosine inverse of the reciprocal.

Answer: (d) \( -1 \)

Explanation:
1. Divide numerator and denominator by \( \cos x \): \( \dfrac{1-\tan x}{1+\tan x}=\tan\left(\dfrac\pi4-x\right) \).
2. So the expression is \( \dfrac\pi4-x \) (in the principal range).
3. Its derivative is \( -1 \).

Teacher's Note:
1. The ratio \( \frac{1-\tan x}{1+\tan x} \) is the tangent of \( \frac\pi4-x \).
2. Simplify with a trigonometric identity before differentiating.

Answer: (b) \( \dfrac12 \)

Explanation:
1. Put \( x=\cos\theta \): \( \sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}=\tan\dfrac\theta2 \).
2. \( \cot^{-1}\tan\dfrac\theta2=\dfrac\pi2-\dfrac\theta2 \), so \( y=\sin^2\left(\dfrac\pi2-\dfrac\theta2\right)=\cos^2\dfrac\theta2=\dfrac{1+\cos\theta}{2}=\dfrac{1+x}{2} \).
3. \( \dfrac{dy}{dx}=\dfrac12 \).

Teacher's Note:
1. The substitution \( x=\cos\theta \) works well with \( \sqrt{\frac{1-x}{1+x}} \).
2. Use the half-angle formula to simplify.

Answer: (a) 1

Explanation:
1. Let \( \cos\alpha=\dfrac5{13} \), \( \sin\alpha=\dfrac{12}{13} \). Then \( y=\cos^{-1}\cos(x+\alpha) \).
2. Since \( x+\alpha \) lies in \( (0,\pi) \) for the given x, \( y=x+\alpha \).
3. \( \dfrac{dy}{dx}=1 \).

Teacher's Note:
1. Write \( a\cos x+b\sin x \) in the form \( R\cos(x\mp\alpha) \).
2. Check that the angle lies in the principal range \( [0,\pi] \) of \( \cos^{-1} \).

Answer: (c) \( \sec^2x \)

Explanation:
1. \( \tan^22x-\tan^2x=(\tan2x-\tan x)(\tan2x+\tan x) \) and \( 1-\tan^22x\tan^2x=(1+\tan2x\tan x)(1-\tan2x\tan x) \).
2. The fraction \( =\tan(2x-x)\tan(2x+x)=\tan x\tan3x \).
3. Multiplying by \( \cot3x \) leaves \( \tan x \), whose derivative is \( \sec^2x \).

Teacher's Note:
1. Factorise the numerator and denominator to use \( \tan(A\pm B) \).
2. Simplify completely before differentiating.

Answer: (a) \( -1 \)

Explanation:
1. Put \( x^x=\tan\theta \): \( \dfrac{\tan^2\theta-1}{2\tan\theta}=-\cot2\theta \). So \( f=\cot^{-1}(-\cot2\theta)=\pi-2\theta \).
2. \( f(x)=\pi-2\tan^{-1}(x^x) \), so \( f'(x)=-\dfrac{2x^x(1+\log x)}{1+x^{2x}} \).
3. At \( x=1 \): \( -\dfrac{2\cdot1\cdot1}{2}=-1 \).

Teacher's Note:
1. The expression \( \frac{t-1/t}{2} \) with \( t=\tan\theta \) reduces to \( -\cot2\theta \).
2. Remember that \( x^x \) has derivative \( x^x(1+\log x) \).

Answer: (a) 1

Explanation:
1. Multiply and divide by \( 1-x \): \( y=\dfrac{1-x^{2^{n+1}}}{1-x} \).
2. \( \dfrac{dy}{dx}=\dfrac{-2^{n+1}x^{2^{n+1}-1}(1-x)+(1-x^{2^{n+1}})}{(1-x)^2} \).
3. At \( x=0 \): \( \dfrac{0+1}{1}=1 \).

Teacher's Note:
1. The product telescopes after multiplying by \( 1-x \).
2. Alternatively the coefficient of x in the product is simply 1.

Answer: (c) 1

Explanation:
1. At \( x=0 \): \( 0=y+0 \), so \( y=0 \).
2. Differentiating: \( e^{xy}+xe^{xy}\left(x\dfrac{dy}{dx}+y\right)=\dfrac{dy}{dx}+2\sin x\cos x \).
3. Putting \( x=0,y=0 \): \( 1=\dfrac{dy}{dx} \).

Teacher's Note:
1. Find the value of y at the given x before substituting into the differentiated equation.
2. Implicit differentiation needs the product rule on \( xy \).

Answer: (a) 1

Explanation:
1. At \( x=0 \): \( \ln y=0 \), so \( y=1 \).
2. Differentiate: \( \dfrac{1+y'}{x+y}=2(xy'+y) \).
3. At \( (0,1) \): \( \dfrac{1+y'}{1}=2(0+1)=2 \), so \( y'(0)=1 \).

Teacher's Note:
1. Use the equation itself to find the y-value at \( x=0 \).
2. Substitute only after differentiating.

Answer: (c) \( \dfrac yx \)

Explanation:
1. Take logarithms: \( m\log x+n\log y=\log2+(m+n)\log(x+y) \).
2. Differentiate: \( \dfrac mx+\dfrac ny\dfrac{dy}{dx}=\dfrac{m+n}{x+y}\left(1+\dfrac{dy}{dx}\right) \).
3. Solving for \( \dfrac{dy}{dx} \) gives \( \dfrac{dy}{dx}=\dfrac yx \).

Teacher's Note:
1. Logarithmic differentiation simplifies powers and products.
2. The result says y is proportional to x along the curve.

Answer: (a) \( (\sin x)^{\tan x}(1+\sec^2x\log\sin x) \)

Explanation:
1. \( \log y=\tan x\log\sin x \).
2. \( \dfrac1y\dfrac{dy}{dx}=\tan x\cot x+\sec^2x\log\sin x=1+\sec^2x\log\sin x \).
3. \( \dfrac{dy}{dx}=(\sin x)^{\tan x}\left[1+\sec^2x\log\sin x\right] \).

Teacher's Note:
1. For \( y=[f(x)]^{g(x)} \) take logarithms and differentiate both factors.
2. Do not use the power rule alone: the exponent also depends on x.

Answer: (b) \( \tan\theta \)

Explanation:
1. \( \dfrac{dx}{d\theta}=a(-\sin\theta+\sin\theta+\theta\cos\theta)=a\theta\cos\theta \).
2. \( \dfrac{dy}{d\theta}=a(\cos\theta-\cos\theta+\theta\sin\theta)=a\theta\sin\theta \).
3. \( \dfrac{dy}{dx}=\dfrac{a\theta\sin\theta}{a\theta\cos\theta}=\tan\theta \).

Teacher's Note:
1. For parametric equations use \( \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} \).
2. Look for terms cancelling after applying the product rule.

Answer: (d) 1

Explanation:
1. With \( t=\tan\phi \) (principal values), \( \cos x=\cos\phi \) and \( \sin y=\sin\phi \).
2. So \( x=\tan^{-1}t \) and \( y=\tan^{-1}t \), giving \( y=x \).
3. Hence \( \dfrac{dy}{dx}=1 \).

Teacher's Note:
1. Identify the right-hand sides as \( \cos\phi \) and \( \sin\phi \) of the same angle.
2. Eliminating the parameter is sometimes quicker than the parametric formula.

Answer: (b) \( y^2 \)

Explanation:
1. \( y=x^y \), so \( \log y=y\log x \).
2. \( \dfrac1y\dfrac{dy}{dx}=\dfrac yx+\log x\dfrac{dy}{dx} \).
3. \( \left(\dfrac1y-\log x\right)\dfrac{dy}{dx}=\dfrac yx \), so \( x(1-y\log x)\dfrac{dy}{dx}=y^2 \).

Teacher's Note:
1. A power tower repeats itself, so \( y=x^y \).
2. Take logarithms before differentiating.

Answer: (a) \( \dfrac{2xy}{2y-x^2} \)

Explanation:
1. \( y=x^2+\dfrac1y \), so \( y^2=x^2y+1 \).
2. \( 2y\dfrac{dy}{dx}=2xy+x^2\dfrac{dy}{dx} \).
3. \( \dfrac{dy}{dx}(2y-x^2)=2xy \), so \( \dfrac{dy}{dx}=\dfrac{2xy}{2y-x^2} \).

Teacher's Note:
1. In a continued fraction the tail equals the whole expression again.
2. Collect the \( \frac{dy}{dx} \) terms on one side.

Answer: (b) 1

Explanation:
1. For \( x>20 \): \( f(x)=x-2>0 \).
2. \( g(x)=f(x-2)=|x-4|=x-4 \).
3. \( g'(x)=1 \).

Teacher's Note:
1. Evaluate the inner function first and remove the modulus using the given range of x.
2. A composite of linear pieces is linear on that range.

Answer: (a) \( \dfrac34 \)

Explanation:
1. \( \cos^{-1}(\sin u)=\dfrac\pi2-u \) for \( u=\sqrt{\frac{1+x}2} \) (here \( u\in[0,\pi/2] \) near \( x=1 \)).
2. \( f'(x)=-\dfrac1{\sqrt2}\cdot\dfrac1{2\sqrt{1+x}}+x^x(1+\log x) \).
3. At \( x=1 \): \( -\dfrac1{\sqrt2\cdot2\sqrt2}+1=-\dfrac14+1=\dfrac34 \).

Teacher's Note:
1. Use \( \cos^{-1}(\sin u)=\frac\pi2-u \) for u in the principal range.
2. The derivative of \( x^x \) is \( x^x(1+\log x) \).

Answer: (a) \( n^2y \)

Explanation:
1. \( \dfrac{dy}{dx}=\dfrac{ny}{\sqrt{1+x^2}} \), so \( \sqrt{1+x^2}\,y'=ny \).
2. Differentiate again: \( \sqrt{1+x^2}\,y''+\dfrac x{\sqrt{1+x^2}}y'=ny'=\dfrac{n^2y}{\sqrt{1+x^2}} \).
3. Multiply by \( \sqrt{1+x^2} \): \( (1+x^2)y''+xy'=n^2y \).

Teacher's Note:
1. Find a simple first-order relation first, then differentiate it once more.
2. This is a standard identity for \( (x+\sqrt{1+x^2})^n \).