JEE Mathematics Methods of Differentiation MCQs Set 08

Answer: (b) 10

Explanation:
1. Put \( y=1 \): \( f(x)(f(1)-2)=f(1)-2 \); as \( f \) is not constant, \( f(1)=2 \). This is consistent with \( f(x)=1+x^n \), which satisfies the equation.
2. \( f(2)=5 \) gives \( 2^n=4 \), so \( n=2 \).
3. \( f(3)=1+9=10 \).

Teacher's Note:
1. Equations of this form are solved by f(x) = 1 plus or minus a power of x.
2. The minus sign cannot give f(2) = 5.

Answer: (a) \( 2f(x)g(0) \)

Explanation:
1. \( f'(x)=f(x)f'(0) \) from the functional equation.
2. \( f'(0)=\lim_{h\to0}\frac{f(h)-1}{h}=\lim\frac{\sin2h\,g(h)}{h}=2g(0) \).
3. So \( f'(x)=2f(x)g(0) \).

Teacher's Note:
1. Use the definition of derivative at 0.
2. The limit sin 2h over h equals 2.

Answer: (d) B, C, D, A

Explanation:
1. A: \( \sin^{-1}\frac{x}{\sqrt{1+x^2}}=\tan^{-1}x \), so the function is \( 2\tan^{-1}x \) and A \( =\frac{2}{1+0}=2 \).
2. B: \( \sec^{-1}u+\sin^{-1}\frac1u=\frac\pi2 \), a constant, so B \( =0 \).
3. C: \( \tan^{-1}\frac{1+\sin x}{\cos x}=\frac\pi4+\frac x2 \), so C \( =\frac12 \).
4. D: the function is \( 2\tan^{-1}x \), so D \( =\frac{2}{1+1}=1 \).
5. Increasing order of \( 0,\frac12,1,2 \) is B, C, D, A.

Teacher's Note:
1. Evaluate each of the four quantities separately and then compare.
2. Several of them reduce to constants or to 2 tan inverse x.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( f(x)=1+\frac12\left[\cos2x+\cos\left(2x+\frac{2\pi}{3}\right)\right]-\frac12\left[\cos\left(2x+\frac\pi3\right)+\cos\frac\pi3\right] \).
2. The two cosines in the first bracket combine to \( \cos\left(2x+\frac\pi3\right) \), which cancels the term in the second bracket.
3. So \( f(x)=1-\frac14=\frac34 \), a constant, and \( f'(x)=0 \). Both A and R are true and R explains A.

Teacher's Note:
1. Show first that the function is constant by using product-to-sum identities.
2. A constant function has zero derivative, which is exactly the reason given.

Answer: (a) \( -13 \)

Explanation:
1. Expand along the first row: \( f(x)=x^2(x\cos x+3)-(2x+1)(x\sin x+2)+(2x^3-3)(3\sin x-2\cos x) \).
2. Differentiate and put \( x=0 \): the terms give \( 0-(2\cdot2+1\cdot0)+\left[0+(-3)(3\cdot1+2\cdot0)\right] \), i.e. \( -4-9=-13 \).

Teacher's Note:
1. The linear coefficient of a function equals its derivative at zero.
2. Substitute x = 0 only after differentiating the expanded form.

Answer: (b) \( \frac{1}{\sqrt{2ax-x^2}} \)

Explanation:
1. Put \( x=2a\sin^2\theta \): \( \sqrt{2ax-x^2}=2a\sin\theta\cos\theta \) and \( a-x=a\cos2\theta \), so the fraction is \( \tan2\theta \).
2. \( y=2\theta=2\sin^{-1}\sqrt{\frac{x}{2a}} \).
3. \( y'=\frac{2}{\sqrt{1-\frac{x}{2a}}}\cdot\frac{1}{\sqrt{2a}}\cdot\frac{1}{2\sqrt x}=\frac{1}{\sqrt{2ax-x^2}} \).

Teacher's Note:
1. The expression 2ax - x^2 suggests x = 2a sin squared theta.
2. The answer matches the derivative of the angle subtended in a circle.

Answer: (a) \( \frac{\sin2n\theta\cos\theta-2n\sin\theta\cos2n\theta}{2\sin^2\theta} \)

Explanation:
1. Use \( \cos\theta+\cos3\theta+\cdots+\cos(2n-1)\theta=\frac{\sin2n\theta}{2\sin\theta} \).
2. Differentiate with respect to \( \theta \): the left side becomes \( -[\sin\theta+3\sin3\theta+\cdots+(2n-1)\sin(2n-1)\theta] \).
3. The right side derivative is \( \frac{2n\cos2n\theta\sin\theta-\sin2n\theta\cos\theta}{2\sin^2\theta} \); change the sign.

Teacher's Note:
1. Differentiating a trigonometric sum turns cosines into sines with the multiplier equal to the angle coefficient.
2. Remember the overall minus sign.

Answer: (b) \( \frac{d^2y}{dt^2}+y=0 \)

Explanation:
1. \( \frac{dy}{dx}=-\frac{1}{\sin t}\frac{dy}{dt} \).
2. \( \frac{d^2y}{dx^2}=\frac{1}{\sin^2t}\frac{d^2y}{dt^2}-\frac{\cos t}{\sin^3t}\frac{dy}{dt} \).
3. Substituting with \( 1-x^2=\sin^2t \): \( \frac{d^2y}{dt^2}-\cot t\frac{dy}{dt}+\cot t\frac{dy}{dt}+y=0 \), i.e. \( \frac{d^2y}{dt^2}+y=0 \).

Teacher's Note:
1. The first-derivative terms cancel exactly.
2. This is the Chebyshev equation with n = 1.

Answer: (c) 8

Explanation:
1. \( g'(x)=f'(e^x)e^x\,e^{f(x)}+f(e^x)e^{f(x)}f'(x) \).
2. At \( x=0 \): \( f'(1)\cdot1\cdot e^{0}+f(1)e^{0}f'(0)=4+4=8 \).

Teacher's Note:
1. Apply the product rule and chain rule.
2. Evaluate e^0 = 1 and f(0) = 0 before substituting the other values.

Answer: (a) \( -\frac{f''(x)}{(f'(x))^3} \)

Explanation:
1. From \( g(f(x))=x \): \( g'(f(x))f'(x)=1 \).
2. Differentiate again: \( g''(f(x))(f'(x))^2+g'(f(x))f''(x)=0 \).
3. With \( g'(f(x))=\frac{1}{f'(x)} \): \( g''(f(x))=-\frac{f''(x)}{(f'(x))^3} \).

Teacher's Note:
1. Differentiate the identity g(f(x)) = x twice.
2. Substitute the first derivative of g into the second relation.

Answer: (c) \( \pi \)

Explanation:
1. The relation says \( f \) preserves the section formula, so \( f \) is linear: \( f(x)=ax+b \).
2. \( f(0)=3 \) gives \( b=3 \) and \( f'(0)=2 \) gives \( a=2 \).
3. \( \sin(2x+3) \) has period \( \frac{2\pi}{2}=\pi \).

Teacher's Note:
1. Check that the linear function satisfies the relation.
2. The period of sin(ax + b) is 2 pi over |a|.

Answer: (c) \( (0,1)\cup(1,\infty) \)

Explanation:
1. \( \ln f(x)=\frac{1}{\ln x}\ln x=1 \) for \( x>0,\ x\ne1 \).
2. So \( f(x)=e \), a constant.
3. Hence \( f'(x)=0 \) on \( (0,1)\cup(1,\infty) \) (the function is undefined at \( x=1 \)).

Teacher's Note:
1. Simplify before differentiating.
2. Mention the excluded point where ln x = 0.

Answer: (d) \( -22 \)

Explanation:
1. \( y'=-\frac{2x+y}{x+6y} \).
2. \( y''=\frac{-22(x^2+xy+3y^2)}{(x+6y)^3} \) after simplification.
3. Using \( x^2+xy+3y^2=1 \): \( (x+6y)^3y''=-22 \).

Teacher's Note:
1. Differentiate the quotient and combine terms to obtain the original quadratic form.
2. Then use the given equation to replace it by 1.

Answer: (a) \( \frac{2f'(x)}{f(x)} \)

Explanation:
1. \( g=\frac1f \), so \( g'=-\frac{f'}{f^2} \) and \( g''=-\frac{f''}{f^2}+\frac{2f'^2}{f^3} \).
2. \( \frac{g''}{g'}=\frac{f''}{f'}-\frac{2f'}{f} \).
3. Therefore \( \frac{f''}{f'}-\frac{g''}{g'}=\frac{2f'}{f} \).

Teacher's Note:
1. Express g in terms of f and differentiate twice.
2. The ratio g''/g' simplifies neatly.

Answer: (a) \( \frac{488}{125} \)

Explanation:
1. \( t=\frac{x}{1+x^2} \), so \( y=x^2+\frac{x^2}{(1+x^2)^2} \).
2. \( \frac{dy}{dx}=2x+\frac{2x(1+x^2)^2-x^2\cdot2(1+x^2)\cdot2x}{(1+x^2)^4}=2x+\frac{2x(1-x^2)}{(1+x^2)^3} \).
3. At \( x=2 \): \( 4+\frac{4\cdot(-3)}{125}=4-\frac{12}{125}=\frac{488}{125} \).

Teacher's Note:
1. Eliminate t first.
2. Simplify the derivative of the fraction before substituting.

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

Explanation:
1. With \( y=3e^2-x \): \( \ln(x^y)=(3e^2-x)\ln x \).
2. Its derivative is \( -\ln x+\frac{3e^2-x}{x}=0 \), i.e. \( \frac{3e^2}{x}=1+\ln x \).
3. At \( x=e^2 \): left \( =3 \), right \( =1+2=3 \). The left side decreases and the right increases, so this root is unique.

Teacher's Note:
1. Differentiating x^y is the same as differentiating its logarithm and setting it to zero.
2. Test the options after reducing to 3e^2/x = 1 + ln x.

Answer: (c) I, II and III only

Explanation:
1. For \( x>0 \): \( x^2-3x+2=(x-1)(x-2) \) is positive on \( (0,1)\cup(2,\infty) \) and negative on \( (1,2) \), so \( f'=2x-3 \) on the first set (I true).
2. For \( x<0 \): \( x^2+3x+2=(x+1)(x+2) \) is positive on \( (-\infty,-2)\cup(-1,0) \), giving \( f'=2x+3 \) (II true), and negative on \( (-2,-1) \), giving \( f'=-(2x+3)=-2x-3 \) (III true).
3. f is clearly not constant, so IV is false.

Teacher's Note:
1. Remove the modulus by studying the sign of the quadratic inside it.
2. Differentiate separately on each interval.

Answer: (a) \( \frac yx(\log x)^{\log(\log x)}\left(2\log(\log x)+1\right) \)

Explanation:
1. \( \log y=(\log x)^{\log\log x}\cdot\log x \). Let \( L=\log x \), \( M=\log L \); then \( L^M=e^{M^2} \).
2. \( \frac{d}{dx}L^M=L^M\cdot2M\cdot\frac{1}{Lx} \).
3. \( \frac{y'}{y}=\frac{2M\,L^M}{x}+\frac{L^M}{x}=\frac{L^M(2M+1)}{x} \).

Teacher's Note:
1. Write L^M as e^(M^2) to differentiate it simply.
2. The factor L in log x cancels the 1/L from M'.

Answer: (b) \( f_n(x)\,f_{n-1}(x)\cdots f_2(x)\,f_1(x) \)

Explanation:
1. \( \frac{d}{dx}f_n=e^{f_{n-1}}\frac{d}{dx}f_{n-1}=f_n\frac{d}{dx}f_{n-1} \).
2. Repeat: \( f_n f_{n-1}\frac{d}{dx}f_{n-2} \), and so on down to \( f_0=x \).
3. The final derivative is \( 1 \), giving the full product \( f_nf_{n-1}\cdots f_1 \).

Teacher's Note:
1. Apply the chain rule repeatedly.
2. The recursion stops at f_0 = x with derivative 1.

Answer: (b) \( \tan^2\theta \)

Explanation:
1. Multiply by \( x+y \): \( \cos^4\theta+\sin^4\theta+\frac{y\cos^4\theta}{x}+\frac{x\sin^4\theta}{y}=1 \).
2. Using \( 1-2\sin^2\theta\cos^2\theta \) for the first two terms: \( \left(\sqrt{\frac yx}\cos^2\theta-\sqrt{\frac xy}\sin^2\theta\right)^2=0 \).
3. So \( \frac yx=\tan^2\theta \), a constant, and \( \frac{dy}{dx}=\frac yx=\tan^2\theta \).

Teacher's Note:
1. Convert the relation to a perfect square equal to zero.
2. The ratio y/x is therefore constant.

Answer: (b) 12

Explanation:
1. If \( f \) has degree \( n \), comparing degrees of both sides gives \( n=(n-1)+(n-2) \), so \( n=3 \).
2. Let \( f=ax^3+bx^2+cx+d \). Comparing coefficients of \( f(2x)=f'(x)f''(x) \) gives \( 8a=18a^2 \), so \( a=\frac49 \), and \( b=c=d=0 \).
3. Hence \( f(x)=\frac49x^3 \) and \( f(3)=\frac49\cdot27=12 \).

Teacher's Note:
1. Compare the degrees first to find the degree of f.
2. The non-zero polynomial solution is unique.

Answer: (a) three real roots with sum zero

Explanation:
1. With \( f(x)=\frac49x^3 \), \( f(x)=x \) gives \( x\left(\frac49x^2-1\right)=0 \).
2. The roots are \( x=0,\ \pm\frac32 \).
3. Their sum is \( 0 \).

Teacher's Note:
1. Factor out x immediately.
2. The roots are symmetric because the function is odd.

Answer: (a) 4

Explanation:
1. \( f'''(x)=6 \), so \( f'''(3)=6 \). Also \( f'(x)=3x^2+2xf'(1)+f''(2) \) and \( f''(x)=6x+2f'(1) \).
2. Put \( x=1 \) in \( f' \): \( f'(1)=3+2f'(1)+f''(2) \). Put \( x=2 \) in \( f'' \): \( f''(2)=12+2f'(1) \).
3. Solving: \( f'(1)=-5 \) and \( f''(2)=2 \), so \( f(x)=x^3-5x^2+2x+6 \) and \( f(1)=1-5+2+6=4 \).

Teacher's Note:
1. Treat f'(1), f''(2) and f'''(3) as unknown constants.
2. Form two equations by substituting suitable x values.

Answer: (d) many-one and onto

Explanation:
1. From the previous question, \( f(x)=x^3-5x^2+2x+6 \) and \( f'(x)=3x^2-10x+2 \).
2. The discriminant \( 100-24=76>0 \), so \( f' \) changes sign and \( f \) is not monotonic, hence many-one.
3. A cubic takes all real values, so it is onto.

Teacher's Note:
1. A cubic is one-one only if its derivative does not change sign.
2. Every real cubic is onto R.