JEE Mathematics Methods of Differentiation MCQs Set 09

Answer: (a) A-Q, B-R, C-S, D-P

Explanation:
1. A: \( \frac{dx}{dt}=5t^4-15t^2-20=-30 \) and \( \frac{dy}{dt}=12t^2-6t-18=-12 \) at \( t=1 \), so \( \frac{dy}{dx}=\frac25 \) (Q).
2. B: Taylor expansion about 2 gives \( P''(x)=2-12(x-2)+12(x-2)^2 \), so \( P''(3)=2 \) (R).
3. C: with \( y=\frac1x \), \( \frac{dy}{\sqrt{1+y^4}}=-\frac{dx}{\sqrt{1+x^4}} \), so the ratio is \( -1 \) (S).
4. D: the relation forces \( f \) to be linear, so \( f''(0)=0 \) (P).

Teacher's Note:
1. Handle each part independently.
2. Linear functions have zero second derivative.

Answer: (d) 2

Explanation:
1. Expanding the left side as a power series gives \( 1+x+x^3-2x^2+x^4-2x^5+\cdots \), i.e. \( 1+x-2x^2+x^3+x^4-2x^5+\cdots \).
2. Multiply by \( 1+bx+cx^2 \) and compare with \( 1+ax \): the coefficients of \( x \), \( x^2 \), \( x^3 \) give \( a=2 \), \( b=1 \), \( c=1 \).
3. So \( a+b-c=2 \).

Teacher's Note:
1. Series comparison is the safest method for an infinite sum like this.
2. Match coefficients of the first three powers of x.

Answer: (b) 2

Explanation:
1. \( \frac{dy}{dx}=x+\frac12\sqrt{x^2+1}+\frac{x^2}{2\sqrt{x^2+1}}+\frac{1}{2\sqrt{x^2+1}}=x+\sqrt{x^2+1} \).
2. Then \( x\frac{dy}{dx}=x^2+x\sqrt{x^2+1} \), which equals \( 2\left(\frac{x^2}{2}+\frac12x\sqrt{x^2+1}\right) \).
3. And \( \log\frac{dy}{dx}=\log(x+\sqrt{x^2+1})=2\log\sqrt{x+\sqrt{x^2+1}} \).
4. So the left side equals \( 2y \) and \( k=2 \).

Teacher's Note:
1. First compute dy/dx; it simplifies to x + sqrt(x^2 + 1).
2. Compare the two groups of terms with y.

Answer: (d) 6

Explanation:
1. \( \cos3x=4\cos^3x-3\cos x \), so \( \frac{\cos3x}{\cos^3x}=4-3\sec^2x \) and \( \cos^2y=4-3\sec^2x \).
2. Differentiate: \( -2\cos y\sin y\,y'=-6\sec^2x\tan x \), so \( y'=\frac{3\sec^2x\tan x}{\sin y\cos y} \).
3. Squaring and simplifying gives \( (y')^2=\frac{6}{\cos2x+\cos4x} \), so \( k=6 \).

Teacher's Note:
1. Rewrite cos 3x using the triple angle formula first.
2. Check numerically at one value of x to be safe.

Answer: (c) \( (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+n^2y=0 \)

Explanation:
1. Let \( x=\cos\theta \), so \( y=c\cos n\theta \) with \( c=2^{1-n} \) and \( \frac{d^2y}{d\theta^2}=-n^2y \).
2. As in the Chebyshev reduction, \( (1-x^2)y''-xy'=\frac{d^2y}{d\theta^2} \).
3. Hence \( (1-x^2)y''-xy'+n^2y=0 \).

Teacher's Note:
1. The constant factor does not affect a linear differential equation.
2. Use the substitution x = cos theta to reduce it to y'' = -n^2 y.

Answer: (d) 0

Explanation:
1. \( g(x)=\big||x-2|-2\big| \) equals \( |x| \) for \( x\le2 \), \( 4-x \) for \( 2<x\le4 \) and \( x-4 \) for \( x>4 \).
2. The required points are \( -1,1,3,5 \): \( g'(-1)=-1 \), \( g'(1)=1 \), \( g'(3)=-1 \), \( g'(5)=1 \).
3. The sum is \( 0 \).

Teacher's Note:
1. Write the nested modulus piecewise first.
2. Use the slope of the correct piece at each point.

Answer: (c) \( (n+1)!\left[\frac12+\frac13+\cdots+\frac{1}{n+1}\right] \)

Explanation:
1. Take logs and differentiate: \( \frac{1}{x+1}+\cdots+\frac{1}{x+n}=\frac{A_1+2A_2x+\cdots+nA_nx^{n-1}}{A_0+A_1x+\cdots+A_nx^n} \).
2. Put \( x=1 \): the denominator is \( (n+1)! \).
3. So \( A_1+2A_2+\cdots+nA_n=(n+1)!\left[\frac12+\cdots+\frac{1}{n+1}\right] \).

Teacher's Note:
1. Logarithmic differentiation of a product gives the sum of reciprocals.
2. Check with n = 1: the answer is 1.

Answer: (a) \( \frac12 \)

Explanation:
1. \( x\pm2\sqrt{2x-4}=\left(\sqrt{x-2}\pm\sqrt2\right)^2 \).
2. For \( 2<x<4 \), \( f=\sqrt{x-2}+\sqrt2+\sqrt2-\sqrt{x-2}=2\sqrt2 \), so \( f'(3)=0 \).
3. For \( x>4 \), \( f=2\sqrt{x-2} \), so \( f'(6)=\frac{1}{\sqrt{4}}=\frac12 \).
4. Sum: \( \frac12 \).

Teacher's Note:
1. Complete the square under each radical.
2. The modulus changes its sign at x = 4.

Answer: (a) \( -2 \)

Explanation:
1. For \( -1<x\le0 \): \( y=2-x+(x-1)-(x+1)=-x \), so \( y'=-1 \).
2. For \( 0<x\le1 \): \( y=2+x+(x-1)-(x+1)=x \), so \( y'=1 \).
3. For \( x>1 \): \( y=2+x-(x-1)-(x+1)=2-x \), so \( y'=-1 \).
4. Sum: \( -1+1-1-1=-2 \).

Teacher's Note:
1. Split the real line at the critical points -1, 0 and 1.
2. Differentiate each linear piece separately.

Answer: (b) positive for all real \( x \)

Explanation:
1. Put \( x=y=0 \): \( f(0)=a \). The relation gives \( f(x+y)=f(x)+f(y)-a+2xy \).
2. Then \( f'(x)=\frac{f(h)-a}{h}+2x\to f'(0)+2x \), so \( f(x)=x^2+\sqrt{5a-1-a^2}\,x+a \).
3. Discriminant: \( (5a-1-a^2)-4a=-(a^2-a+1)<0 \) for all \( a \), and the leading coefficient is positive, so \( f(x)>0 \).

Teacher's Note:
1. Find f(0) first, then derive f'(x) from the definition.
2. Check the sign of the discriminant of the resulting quadratic.

Answer: (a) \( \frac{1}{\left(\frac{d^2y}{dx^2}\right)^{2/3}}+\frac{1}{\left(\frac{d^2x}{dy^2}\right)^{2/3}} \)

Explanation:
1. Write \( y_1=\frac{dy}{dx} \), \( y_2=\frac{d^2y}{dx^2} \). Then \( R^{2/3}=\frac{1+y_1^2}{y_2^{2/3}}=\frac{1}{y_2^{2/3}}+\frac{y_1^2}{y_2^{2/3}} \).
2. Since \( \frac{d^2x}{dy^2}=-\frac{y_2}{y_1^3} \), \( \left(\frac{d^2x}{dy^2}\right)^{2/3}=\frac{y_2^{2/3}}{y_1^2} \) (real cube roots).
3. So \( \frac{y_1^2}{y_2^{2/3}}=\frac{1}{(d^2x/dy^2)^{2/3}} \), giving the first form.

Teacher's Note:
1. Use the reciprocal-derivative formula for the second derivative of x with respect to y.
2. Cube roots are taken as real cube roots.

Answer: (c) III and IV only

Explanation:
1. On \( [-\frac\pi2,\frac\pi2] \), \( f(x)=x \), so \( f'=1 \) (1st and 4th quadrants).
2. On \( [\frac\pi2,\frac{3\pi}2] \), \( f(x)=\pi-x \), so \( f'=-1 \) (2nd and 3rd quadrants).
3. The pattern repeats with period \( 2\pi \). So III and IV hold; I and II do not.

Teacher's Note:
1. The function is a zig-zag wave of period 2 pi.
2. Its slope alternates between +1 and -1.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( f'(x)=\cos x \), so \( f'(\pi)=-1 \) and \( f'(3\pi)=-1 \); A is true.
2. \( f(\pi)=0=f(3\pi) \); R is true.
3. Equal function values at two points do not by themselves imply equal derivatives; the equality of derivatives follows from periodicity of \( \cos x \). So R is true but is not the explanation of A.

Teacher's Note:
1. Equal values of a function at two points do not imply equal derivatives in general.
2. Here both statements hold because of the periodicity of sine and cosine.

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

Explanation:
1. Since \( y^3 \) takes every real value, \( f(x+t)=f(x)+f(t) \) for all real \( x,t \) (Cauchy equation).
2. A differentiable additive function is \( f(x)=cx \), which is odd, so R is true.
3. Then \( f'(x)=c \) everywhere and \( f'(-2)=f'(2)=a \): the derivative of an odd function is even, so R explains A.

Teacher's Note:
1. The derivative of an odd function is an even function.
2. Additivity plus differentiability forces f(x) to be linear.

Answer: (b) \( f^{(m-1)}(x) \)

Explanation:
1. If \( (x-c)^m \) divides \( f \), then \( f(c)=f'(c)=\cdots=f^{(m-1)}(c)=0 \).
2. The \( m \)-th derivative at \( c \) equals \( m!\times \) (a non-zero constant), so it need not vanish.
3. Hence \( c \) is certainly a root of \( f^{(m-1)}(x) \) (and of all lower derivatives), but not of \( f^{(m)} \) or higher in general.

Teacher's Note:
1. A root of multiplicity m is a common root of f and its first m - 1 derivatives.
2. The m-th derivative does not vanish there.

Answer: (b) 1

Explanation:
1. \( \frac{f'(x)}{f(x)}=\sum_{n=1}^{100}\frac{n(101-n)}{x-n} \).
2. At \( x=101 \): each term is \( \frac{n(101-n)}{101-n}=n \).
3. So \( \frac{f'(101)}{f(101)}=\sum n=5050 \) and \( \frac{f(101)}{f'(101)}=\frac{1}{5050} \), giving \( k=1 \).

Teacher's Note:
1. Logarithmic differentiation of a product turns it into a sum.
2. The factor 101 - n cancels at x = 101.

Answer: (b) \( -4 \)

Explanation:
1. \( g(x+1)=g(x)+\log x \); differentiate twice: \( g''(x+1)=g''(x)-\frac{1}{x^2} \).
2. Apply with \( x=\frac12,\frac32,\ldots,N-\frac12 \) and add: \( g''(N+\frac12)-g''(\frac12)=-\sum_{j=0}^{N-1}\frac{1}{(j+\frac12)^2} \).
3. Each term is \( \frac{4}{(2j+1)^2} \), so the sum equals \( -4\left(1+\frac19+\cdots+\frac{1}{(2N-1)^2}\right) \) and \( k=-4 \).

Teacher's Note:
1. Take the logarithm of the functional equation and differentiate twice.
2. Telescope the resulting difference relation.

Answer: (c) 5

Explanation:
1. \( f(1)=-4+1+1+\frac12+\frac13=-\frac76 \), so \( g\left(-\frac76\right)=1 \).
2. \( f'(x)=2e^{\frac{1-x}{2}}+1+x+x^2 \), so \( f'(1)=2+1+1+1=5 \).
3. \( g'\left(-\frac76\right)=\frac{1}{f'(1)}=\frac15 \), giving \( k=5 \).

Teacher's Note:
1. Find the point where f takes the given value by trial.
2. The inverse derivative is the reciprocal of f' at that point.

Answer: (b) 5

Explanation:
1. Let \( t=2x+\frac1t \), so \( t^2-2xt-1=0 \) and \( t=x+\sqrt{x^2+1} \).
2. \( f(x)=x+\frac1t=x+\sqrt{x^2+1}-x=\sqrt{x^2+1} \).
3. \( f(x)f'(x)=\sqrt{x^2+1}\cdot\frac{x}{\sqrt{x^2+1}}=x \).
4. So the required value is \( \sqrt{\frac{50}{2}}=5 \).

Teacher's Note:
1. Find the repeating unit of the continued fraction first.
2. f times f' simplifies to x.

Answer: (b) 6

Explanation:
1. \( \frac{F'}{F}=\frac{f'}{f}+\frac{g'}{g}+\frac{h'}{h} \) at \( x_0 \).
2. \( 21=4+(-7)+k \), so \( k=24 \).
3. \( \frac k4=6 \).

Teacher's Note:
1. Logarithmic derivatives add for a product.
2. Solve the linear equation for k.

Answer: (a) 9

Explanation:
1. Let \( \psi(x)=f(x)f(-x) \). Then \( \psi'(x)=f'(x)f(-x)-f(x)f'(-x)=0 \) by the given relation.
2. So \( \psi \) is constant: \( \psi(x)=\psi(0)=f(0)^2=9 \).
3. Hence \( f(3)f(-3)=9 \).

Teacher's Note:
1. Look for a combination of the given terms whose derivative is zero.
2. The constant is fixed by the value at 0.

Answer: (b) 0

Explanation:
1. \( y''=2K\cos Kx-K^2x\sin Kx \).
2. \( y''+y=2K\cos Kx+(1-K^2)x\sin Kx \).
3. This equals \( 2K\cos Kx \) for all \( x \) iff \( (1-K^2)x\sin Kx\equiv0 \), i.e. \( K=0 \) or \( K=\pm1 \).
4. The sum of the possible values is \( 0 \).

Teacher's Note:
1. The extra term must vanish identically.
2. K = 0 gives y = 0, which trivially works.

Answer: (a) 3

Explanation:
1. Put \( x=0 \): \( f(0)=2 \), so \( f^{-1}(2)=0 \).
2. Differentiate: \( e^{-x}(f'-f)=\sqrt{x^4+1} \); at \( x=0 \): \( f'(0)-2=1 \), so \( f'(0)=3 \).
3. \( (f^{-1})'(2)=\frac1{f'(0)}=\frac13 \), so \( k=3 \).

Teacher's Note:
1. Use the fundamental theorem of calculus to differentiate the integral.
2. Find the point mapped to 2 first.

Answer: (a) \( -\frac29 \)

Explanation:
1. \( 2x^2-7x+5=(2x-5)(x-1) \), so \( f(x)=\frac{1}{2x-5} \) for \( x\ne1 \); its limit at 1 is \( -\frac13 \), equal to \( f(1) \), so \( f \) equals \( \frac{1}{2x-5} \) everywhere.
2. \( f'(x)=\frac{-2}{(2x-5)^2} \), so \( f'(1)=\frac{-2}{9} \).

Teacher's Note:
1. Factor and cancel to see the removable discontinuity.
2. The given value at x = 1 fills the gap exactly.