JEE Mathematics Methods of Differentiation MCQs Set 01

Answer: (c) \( \frac3{(1-x)^2} \)

Explanation:
1. Solve \( y=\frac{x-1}{x+2} \) for x: \( x=\frac{2y+1}{1-y} \), so \( f^{-1}(x)=\frac{2x+1}{1-x} \).
2. Differentiate: \( \frac{2(1-x)+(2x+1)}{(1-x)^2}=\frac3{(1-x)^2} \).

Teacher's Note:
1. First find the inverse function explicitly, then differentiate it.
2. The alternative is 1/f'(y) evaluated at y = f^(-1)(x), which gives the same result.

Answer: (c) \( e^{2x} \)

Explanation:
1. The determinant gives \( f'(x)^2-f(x)f''(x)=0 \), i.e. \( \frac{d}{dx}\left(\frac{f'}{f}\right)=\frac{f''f-f'^2}{f^2}=0 \).
2. So \( \frac{f'}{f} \) is constant, equal to \( \frac{f'(0)}{f(0)}=2 \).
3. Hence \( f(x)=e^{2x} \).

Teacher's Note:
1. A zero Wronskian-type determinant means f'/f is constant.
2. Use the initial conditions to find the constant.

Answer: (a) 1

Explanation:
1. \( \frac uv=7 \) is a constant function, so its derivative \( q=0 \).
2. \( \frac{p+q}{p-q}=\frac{p}{p}=1 \) (p is non-zero as \( u'=7v' \)).

Teacher's Note:
1. The derivative of a constant function is zero.
2. Check that p is not zero so that the ratio is defined.

Answer: (a) \( -\frac{(3t^2+1)\sin t+6t\cos t}{(3t^2+1)^3} \)

Explanation:
1. \( \frac{dy}{dx}=\frac{\cos t}{3t^2+1} \).
2. \( \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{\cos t}{3t^2+1}\right)\cdot\frac{dt}{dx}=\frac{-\sin t(3t^2+1)-6t\cos t}{(3t^2+1)^2}\cdot\frac1{3t^2+1} \).
3. Hence \( \frac{d^2y}{dx^2}=-\frac{(3t^2+1)\sin t+6t\cos t}{(3t^2+1)^3} \).

Teacher's Note:
1. For parametric equations, differentiate dy/dx with respect to t and divide by dx/dt again.
2. The power of the denominator rises from 2 to 3.

Answer: (a) \( \frac{38}{27} \)

Explanation:
1. Let \( D=2x^2+3x+1 \), \( D'=4x+3 \), \( D''=4 \). Then \( y'=-\frac{D'}{D^2} \).
2. \( y''=\frac{2D'^2-DD''}{D^3} \).
3. At \( x=-2 \): \( D=3,\ D'=-5 \), so \( y''=\frac{2\cdot25-3\cdot4}{27}=\frac{38}{27} \).

Teacher's Note:
1. Write y as D to the power -1 and differentiate twice.
2. For y = 1/D, y'' = (2D'^2 - D D'')/D^3.

Answer: (b) 3

Explanation:
1. Use \( \cos kx\approx1-\frac{k^2x^2}2 \) near 0: \( (\cos2x)^{1/2}\approx1-x^2 \) and \( (\cos3x)^{1/3}\approx1-\frac{3x^2}2 \).
2. \( \cos x\,(\cos2x)^{1/2}(\cos3x)^{1/3}\approx1-\left(\frac12+1+\frac32\right)x^2=1-3x^2 \).
3. So \( f(x)\to3 \) and \( f(0)=3 \).

Teacher's Note:
1. Continuity at 0 means f(0) equals the limit.
2. Expand each cosine factor to second order and add the coefficients.

Answer: (b) \( -\frac yx \)

Explanation:
1. \( \sin(xy)+\cos(xy)=0\Rightarrow\tan(xy)=-1 \), so \( xy=\frac{3\pi}4+k\pi \) is a constant.
2. Differentiate \( xy=\text{constant} \): \( y+x\frac{dy}{dx}=0 \), so \( \frac{dy}{dx}=-\frac yx \).

Teacher's Note:
1. The relation forces the product xy to be constant.
2. Differentiating a constant product gives y + x dy/dx = 0.

Answer: (d) \( \frac1a \)

Explanation:
1. Put \( x=a\sin\theta \): \( f=\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{\cos\frac\theta2+\sin\frac\theta2}{\cos\frac\theta2-\sin\frac\theta2}=\tan\left(\frac\theta2+\frac\pi4\right) \).
2. \( \frac{df}{d\theta}=\frac12\sec^2\left(\frac\theta2+\frac\pi4\right) \), equal to 1 at \( \theta=0 \). Also \( \frac{dx}{d\theta}=a\cos\theta=a \) at 0.
3. So \( f'(0)=\frac1a \).

Teacher's Note:
1. The substitution x = a sin(theta) removes the radicals.
2. Convert back by dividing df/dtheta by dx/dtheta.

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

Explanation:
1. From the relation \( \ln y=\frac1b\cot\frac{x+a}{2b} \). Let \( L=\ln y \). Then \( L'=-\frac{1}{2b^2}\csc^2\frac{x+a}{2b} \) and \( L''=\frac1{2b^3}\csc^2\frac{x+a}{2b}\cot\frac{x+a}{2b}=-L\,L' \).
2. Since \( y'=yL' \) and \( y''=y(L''+L'^2) \): \( yy''=y^2(L'^2-LL')=y'^2-yy'\ln y \).
3. Hence \( yy''+yy'\ln y=y'^2 \).

Teacher's Note:
1. Take the logarithm of y first and differentiate L = ln y twice.
2. The key identity is L'' = - L L'.

Answer: (c) \( P(x)\cdot P'''(x) \)

Explanation:
1. \( 2yy'=P' \) and \( 2yy''+2y'^2=P'' \).
2. So \( y^3y''=y^2\cdot yy''=P\left(\frac{P''}{2}-y'^2\right)=\frac{PP''}2-\frac{P'^2}4 \) (using \( y'^2=\frac{P'^2}{4P} \)).
3. Differentiate: \( \frac{P'P''+PP'''}{2}-\frac{P'P''}{2}=\frac{PP'''}2 \). Multiply by 2: \( PP''' \).

Teacher's Note:
1. Express y y'' and y'^2 through P and its derivatives.
2. The terms with P' P'' cancel.

Answer: (b) 132

Explanation:
1. Put \( u=x-9 \): \( \int\frac{3u+29}{\sqrt u}du=2u^{3/2}+58u^{1/2}+C \).
2. At \( x=10 \) (\( u=1 \)): \( 2+58+C=60 \), so \( C=0 \).
3. At \( x=13 \) (\( u=4 \)): \( 2\cdot8+58\cdot2=16+116=132 \).

Teacher's Note:
1. Substitute u = x - 9 so that the numerator becomes 3u + 29.
2. Use the given value to fix the constant of integration.

Answer: (d) 20

Explanation:
1. Let \( g(x)=f(x^2) \). The limit is \( \lim_{x\to2}\frac{g(2)-g(x)}{2-x}=g'(2) \).
2. \( g'(x)=2xf'(x^2) \), so \( g'(2)=4f'(4)=20 \).

Teacher's Note:
1. The expression is the definition of the derivative of f(x^2) at x = 2.
2. Use the chain rule.

Answer: (a) \( l \) is independent of m and n

Explanation:
1. A power of x beats any power of \( \ln x \): write \( x=e^{-t} \) with \( t\to\infty \) to get \( (-1)^nt^ne^{-mt}\to0 \).
2. So \( l=0 \) for every natural m and n: independent of both.

Teacher's Note:
1. Substitute x = e^(-t) to turn the limit into t^n e^(-mt).
2. Exponential decay beats any polynomial growth.

Answer: (c) 4

Explanation:
1. Expanding the determinant and simplifying gives \( f(x)=-4\sin^3x\cos x \).
2. \( f'(x)=-12\sin^2x\cos^2x+4\sin^4x \).
3. At \( x=\frac\pi2 \): \( f'=0+4=4 \).

Teacher's Note:
1. Expand along the first row and use product-to-sum formulas to simplify.
2. At pi/2, cos x = 0 and sin x = 1.

Answer: (c) \( 2-\sqrt2 \)

Explanation:
1. \( f\left(\frac\pi4\right)=\frac\pi4+\frac1{\sqrt2} \), so \( g\left(\frac\pi4+\frac1{\sqrt2}\right)=\frac\pi4 \).
2. \( g'(f(a))=\frac1{f'(a)} \) with \( f'(x)=1+\cos x \).
3. \( g'=\frac1{1+\frac1{\sqrt2}}=\frac{\sqrt2}{\sqrt2+1}=\sqrt2(\sqrt2-1)=2-\sqrt2 \).

Teacher's Note:
1. Recognise the given argument as f(pi/4).
2. The derivative of an inverse is the reciprocal of the derivative at the corresponding point.

Answer: (a) 19

Explanation:
1. Derivative of \( f(x)-f(2x) \) is \( f'(x)-2f'(2x) \): at 1, \( f'(1)-2f'(2)=5 \); at 2, \( f'(2)-2f'(4)=7 \).
2. Then \( f'(1)=5+2f'(2)=5+2(7+2f'(4))=19+4f'(4) \).
3. Required: \( f'(1)-4f'(4)=19 \).

Teacher's Note:
1. Differentiate f(kx) as k f'(kx).
2. Chain the two given relations to eliminate f'(2).

Answer: (b) \( -\frac{e^x}{(1+e^x)^3} \)

Explanation:
1. \( \frac{dy}{dx}=1+e^x \), so \( \frac{dx}{dy}=\frac1{1+e^x} \).
2. \( \frac{d^2x}{dy^2}=\frac d{dx}\left(\frac1{1+e^x}\right)\cdot\frac{dx}{dy}=\frac{-e^x}{(1+e^x)^2}\cdot\frac1{1+e^x}=-\frac{e^x}{(1+e^x)^3} \).

Teacher's Note:
1. The derivative of the inverse function is 1/(dy/dx).
2. Differentiate again with respect to x and multiply by dx/dy.

Answer: (c) a straight line with slope 2

Explanation:
1. \( \frac{d}{dx}(f^2+g^2)=2ff'+2gg'=2fg+2g(-f)=0 \) because \( g'=f''=-f \).
2. So \( h'(x) \) is a constant \( k \) and \( h(x)=kx+2 \).
3. \( h(1)=4 \) gives \( k=2 \): a straight line of slope 2.

Teacher's Note:
1. Show that f^2 + g^2 is constant by differentiating it.
2. A function with constant derivative is linear.

Answer: (c) \( \frac{b^2+c^2-a^2}{2bc} \)

Explanation:
1. Put \( h=\tfrac1r\to0 \). Numerator \( \approx\left(1-\frac{a^2h^2}2\right)-\left(1-\frac{b^2h^2}2\right)\left(1-\frac{c^2h^2}2\right)\approx\frac{(b^2+c^2-a^2)h^2}2 \).
2. Denominator \( \approx bch^2 \).
3. Limit \( \frac{b^2+c^2-a^2}{2bc} \).

Teacher's Note:
1. Replace 1/r by h and expand to second order.
2. The result looks like the cosine rule.

Answer: (d) \( \frac{a^2-b^2}{3a^2b^2} \)

Explanation:
1. Use \( \arctan u=u-\frac{u^3}3+\cdots \). Then \( a\arctan\frac{\sqrt x}a=\sqrt x-\frac{x^{3/2}}{3a^2}+\cdots \) and similarly for b.
2. The difference is \( -\frac{x^{3/2}}3\left(\frac1{a^2}-\frac1{b^2}\right) \).
3. Dividing by \( x^{3/2} \): \( \frac13\left(\frac1{b^2}-\frac1{a^2}\right)=\frac{a^2-b^2}{3a^2b^2} \).

Teacher's Note:
1. Expand arctan u up to the cubic term; the linear terms cancel.
2. Divide by x^(3/2) to get the finite limit.

Answer: (c) \( \frac{10}3 \)

Explanation:
1. Let \( \cos\alpha=\tfrac2{\sqrt{13}},\ \sin\alpha=\tfrac3{\sqrt{13}} \). Then the first term is \( \cos^{-1}\cos(x+\alpha)=x+\alpha \) and the second is \( \sin^{-1}\sin(x+\beta)=x+\beta \) with \( \sin\beta=\tfrac2{\sqrt{13}},\ \cos\beta=\tfrac3{\sqrt{13}} \) (valid near \( x=\tfrac34 \) as both angles lie in the principal ranges).
2. So \( \frac{df}{dx}=2 \). Also \( \frac{d}{dx}\sqrt{1+x^2}=\frac x{\sqrt{1+x^2}}=\frac{3/4}{5/4}=\frac35 \).
3. Required derivative \( =\frac2{3/5}=\frac{10}3 \).

Teacher's Note:
1. Write the bracketed expressions as cos(x + alpha) and sin(x + beta) with suitable constant angles.
2. Check that the angles lie in the principal value ranges at the given point.

Answer: (d) \( x>0 \)

Explanation:
1. \( f'(x)=5^{2x+1}\ln5 \) and \( g'(x)=5^x\ln5+4\ln5 \).
2. The inequality (divide by \( \ln5>0 \)): \( 5\cdot25^x>5^x+4 \). With \( t=5^x>0 \): \( 5t^2-t-4>0 \), i.e. \( (5t+4)(t-1)>0 \).
3. So \( t>1 \) and \( x>0 \).

Teacher's Note:
1. Differentiate both functions and divide by ln 5, which is positive.
2. Substitute t = 5^x to get a quadratic inequality.

Answer: (d) \( \frac b{ab+2ay} \)

Explanation:
1. The pattern repeats after two steps, so \( y=\frac x{a+\dfrac x{b+y}} \).
2. Cross-multiplying: \( ay(b+y)+xy=x(b+y) \), which simplifies to \( ay(b+y)=xb \), so \( x=\frac{ay(b+y)}b \).
3. \( \frac{dx}{dy}=\frac{a(b+2y)}b \) and \( \frac{dy}{dx}=\frac b{ab+2ay} \).

Teacher's Note:
1. Because the continued fraction repeats, it satisfies a self-referential equation.
2. Express x in terms of y and take the reciprocal of dx/dy.

Answer: (d) A.P.

Explanation:
1. Let \( f(x)=px^2+qx+r \). \( f(1)=f(-1) \) gives \( q=0 \).
2. So \( f'(x)=2px \) and \( f'(a),f'(b),f'(c)=2pa,2pb,2pc \), which are in A.P. because a, b, c are.

Teacher's Note:
1. f(1) = f(-1) removes the linear term.
2. The derivative is then a constant multiple of x.

Answer: (a) \( e^x \)

Explanation:
1. Write the operator as \( (D-m)^2y \) with \( D=\frac d{dx} \).
2. The first part \( (A+Bx)e^{mx} \) is killed by \( (D-m)^2 \).
3. For \( (m-1)^{-2}e^x \): \( (D-m)^2e^x=(1-m)^2e^x \), giving \( e^x \).

Teacher's Note:
1. The left side is the operator (D - m)^2 applied to y.
2. (D - m)^2 annihilates (A + Bx) e^(mx).

Answer: (c) -15

Explanation:
1. Since \( e^{ax} \) and \( e^{bx} \) are independent, each must satisfy \( r^2-2r-15=0 \).
2. Both a and b are roots of this quadratic and \( a\ne b \), so \( ab=-15 \).

Teacher's Note:
1. Substitute e^(rx) to get the characteristic equation.
2. The product of the two distinct roots is the constant term.

Answer: (c) 3

Explanation:
1. \( f'(x)=(k+e^x)h(x)+(kx+e^x)h'(x) \).
2. At 0: \( f'(0)=(k+1)\cdot5+1\cdot(-2)=18 \), so \( 5k+3=18 \) and \( k=3 \).

Teacher's Note:
1. Use the product rule and then substitute x = 0.
2. e^0 = 1.

Answer: (b) \( \cot\frac{5\pi}{12} \)

Explanation:
1. \( x^4-x^2+1=(x^2+1)^2-3x^2=(x^2+\sqrt3x+1)(x^2-\sqrt3x+1) \).
2. So \( y=x^2-\sqrt3x+1 \) and \( y'=2x-\sqrt3 \), giving \( a=2,\ b=-\sqrt3 \).
3. \( a+b=2-\sqrt3=\tan15^\circ=\cot75^\circ=\cot\frac{5\pi}{12} \).

Teacher's Note:
1. Factorise the numerator as a difference of squares.
2. 2 - sqrt3 = tan 15 degrees.

Answer: (b) equal to 1/2

Explanation:
1. The fraction repeats every two levels: \( y=\frac{\sin x}{1+\dfrac{\cos x}{1+y}} \).
2. At \( x=0 \): \( \sin0=0 \), so \( y(0)=0 \). For small x, \( y\approx\frac{x}{1+\frac{1}{1+0}}=\frac x2 \).
3. So \( y'(0)=\tfrac12 \).

Teacher's Note:
1. The continued fraction reproduces itself after two levels.
2. Near x = 0, replace sin x by x and cos x by 1.

Answer: (c) \( \log_23 \)

Explanation:
1. Differentiate: \( 3^x\ln3-2^y\ln2\cdot y'=0 \), so \( y'=\frac{3^x\ln3}{2^y\ln2} \).
2. As \( x\to\infty \), \( 2^y=3^x-1\sim3^x \), so \( y'\to\frac{\ln3}{\ln2}=\log_23 \).

Teacher's Note:
1. Replace 2^y by 3^x - 1 using the curve equation.
2. Ratio of logarithms in different bases is a logarithm to base 2.