JEE Mathematics Differential Coefficient MCQs Set 01

Answer: (a) 1

Explanation:
1. The product equals \( \frac{1-x^{2^{n+1}}}{1-x} \) (multiply and divide by \( 1-x \)).
2. Near x = 0, \( y = 1 + x + x^2 + \dots \), so \( y'(0) = 1 \).

Teacher's Note:
1. Multiply by (1 − x) to collapse the product.
2. Expand around 0 to read off the derivative.

Answer: (c) \( \frac{1}{x\log_e x} \)

Explanation:
1. \( f(\log_e x) = \log_e(\log_e x) \).
2. Its derivative is \( \frac{1}{\log_e x}\cdot\frac{1}{x} = \frac{1}{x\log_e x} \).

Teacher's Note:
1. Apply the chain rule to the composite function.
2. Substitute log x into f before differentiating.

Answer: (a) \( \sqrt{2} \)

Explanation:
1. Using \( \sin 2\theta = 2\sin\theta\cos\theta \) repeatedly, \( f(x) = \frac{\sin 32x}{32\sin x} \).
2. \( f'(x) = \frac{32\cos 32x\sin x - \sin 32x\cos x}{32\sin^2 x} \). At \( x = \pi/4 \): \( \sin 8\pi = 0 \), \( \cos 8\pi = 1 \), \( \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \).
3. \( f'\left(\frac{\pi}{4}\right) = \frac{32\cdot\frac{1}{\sqrt{2}}}{32\cdot\frac{1}{2}} = \sqrt{2} \).

Teacher's Note:
1. Convert a product of cosines of doubling angles into a single sine ratio.
2. Then use the quotient rule.

Answer: (a) \( \frac{1}{\sqrt{2}} \)

Explanation:
1. \( \sec(\tan^{-1}x) = \sqrt{1+x^2} \).
2. \( \frac{dy}{dx} = \frac{x}{\sqrt{1+x^2}} = \frac{1}{\sqrt{2}} \) at x = 1.

Teacher's Note:
1. Draw a right triangle with opposite x and adjacent 1.
2. Simplify before differentiating.

Answer: (a) 0

Explanation:
1. Taking logs, \( y\ln x = x - y \), so \( y = \frac{x}{1+\ln x} \).
2. \( \frac{dy}{dx} = \frac{(1+\ln x) - 1}{(1+\ln x)^2} = \frac{\ln x}{(1+\ln x)^2} \), which is 0 at x = 1.

Teacher's Note:
1. Take logarithms when the variable is in the exponent.
2. Solve for y explicitly if it is easy.

Answer: (b) 1

Explanation:
1. At x = 0 the equation gives \( -y = 0 \), so y = 0.
2. Differentiating: \( e^{xy} + xe^{xy}(y + xy') - y' = 2\sin x\cos x \).
3. At (0, 0): \( 1 - y' = 0 \), so \( y' = 1 \).

Teacher's Note:
1. Find y at the given x first.
2. Use implicit differentiation.

Answer: (a) \( \frac{1}{x}\log_2 e\cdot\log_x e \)

Explanation:
1. \( y = \frac{\ln(\ln x)}{\ln 2} \).
2. \( \frac{dy}{dx} = \frac{1}{\ln 2}\cdot\frac{1}{\ln x}\cdot\frac{1}{x} = \frac{1}{x}\log_2 e\cdot\log_x e \).

Teacher's Note:
1. Change of base: log₂ a = ln a / ln 2.
2. log_x e = 1 / ln x.

Answer: (a) 2

Explanation:
1. \( \frac{dy}{dx} = f'(x^2)\cdot 2x = \sqrt{2x^4-1}\cdot2x \).
2. At x = 1: \( \sqrt{1}\times2 = 2 \).

Teacher's Note:
1. Apply the chain rule to f(x²).
2. Substitute x² into f' before multiplying by 2x.

Answer: (a) \( \frac{-1}{2|x|\sqrt{x^2-1}} \)

Explanation:
1. Put \( x = \sec\theta \) for x > 1: \( \sqrt{\frac{x+1}{x-1}} = \cot\frac{\theta}{2} \), so \( y = \frac{\pi}{2} - \frac{\theta}{2} \).
2. \( \frac{dy}{dx} = -\frac{1}{2}\frac{d\theta}{dx} = -\frac{1}{2\sec\theta\tan\theta} \), i.e. \( -\frac{1}{2x\sqrt{x^2-1}} \) for x > 1.
3. The same form with |x| holds for x < −1, so \( \frac{dy}{dx} = \frac{-1}{2|x|\sqrt{x^2-1}} \).

Teacher's Note:
1. Domain is |x| > 1.
2. The modulus covers both parts of the domain.

Answer: (b) −1

Explanation:
1. Take logs: \( y\ln x + x\ln y = \ln 16 \).
2. Differentiate: \( y'\ln x + \frac{y}{x} + \ln y + \frac{x y'}{y} = 0 \).
3. At (2, 2): \( y'\ln 2 + 1 + \ln 2 + y' = 0 \Rightarrow y'(1+\ln 2) = -(1+\ln 2) \), so \( y' = -1 \).

Teacher's Note:
1. Log both sides to handle the variable exponents.
2. Factor the common term (1 + ln 2).

Answer: (b) 0

Explanation:
1. \( \ln y = \frac{\ln x}{x} \), so \( \frac{y'}{y} = \frac{1-\ln x}{x^2} \).
2. At x = e, \( 1 - \ln e = 0 \), so \( y' = 0 \).

Teacher's Note:
1. This is the maximum point of x^(1/x).
2. Logarithmic differentiation is the quickest.

Answer: (b) \( \frac{2}{5} \)

Explanation:
1. \( \frac{dy}{dt} = 12t^2 - 6t - 18 \) and \( \frac{dx}{dt} = 5t^4 - 15t^2 - 20 \).
2. At t = 1: \( \frac{-12}{-30} = \frac{2}{5} \).

Teacher's Note:
1. For parametric curves, dy/dx = (dy/dt)/(dx/dt).
2. Evaluate both at the given t.

Answer: (a) 0

Explanation:
1. For \( 1 < x < 2 \), [x] = 1, so \( f(x) = \cos\left(\frac{\pi}{2} - x^3\right) = \sin x^3 \).
2. \( f'(x) = 3x^2\cos x^3 \). At \( x^3 = \frac{\pi}{2} \): \( \cos\frac{\pi}{2} = 0 \), so \( f' = 0 \).

Teacher's Note:
1. Replace [x] by its constant value in the interval.
2. Use the cofunction identity.

Answer: (c) \( -\frac{\pi}{180}\text{cosec}\,x\cdot\cos\frac{\pi x}{180} \)

Explanation:
1. \( y = \sin\frac{\pi x}{180} \), so \( \frac{dy}{dx} = \frac{\pi}{180}\cos\frac{\pi x}{180} \).
2. \( \frac{dz}{dx} = -\sin x \).
3. \( \frac{dy}{dz} = -\frac{\pi}{180}\cdot\frac{\cos(\pi x/180)}{\sin x} \).

Teacher's Note:
1. Convert degrees to radians before differentiating.
2. dy/dz = (dy/dx)/(dz/dx).

Answer: (a) −4

Explanation:
1. \( \text{cosec}^{-1}\frac{1}{2x^2-1} = \sin^{-1}(2x^2-1) \). Its derivative is \( \frac{4x}{\sqrt{1-(2x^2-1)^2}} = \frac{4x}{2x\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}} \).
2. \( \frac{d}{dx}\sqrt{1-x^2} = -\frac{x}{\sqrt{1-x^2}} \).
3. Ratio = \( -\frac{2}{x} = -4 \) at \( x = \frac12 \).

Teacher's Note:
1. Convert cosec⁻¹(1/u) to sin⁻¹(u).
2. Divide the two derivatives.

Answer: (c) \( -2\sqrt{2\pi} \)

Explanation:
1. \( \frac{d}{dx}f\left(2x^2+\frac{\pi}{2}\right) = f'\left(2x^2+\frac{\pi}{2}\right)\cdot4x \).
2. At \( x = \sqrt{\pi/2} \): \( 2x^2 + \frac{\pi}{2} = \frac{3\pi}{2} \) and \( f'\left(\frac{3\pi}{2}\right) = -1 + \sin 6\pi\cos\frac{3\pi}{2} = -1 \).
3. Result = \( -4\sqrt{\pi/2} = -2\sqrt{2\pi} \).

Teacher's Note:
1. Apply the chain rule with the inner derivative 4x.
2. Evaluate f' at the transformed argument.

Answer: (a) \( \frac{3}{2}x\cos x^3\,\text{cosec}\,x^2 \)

Explanation:
1. \( \frac{du}{dx} = f'(x^3)\cdot3x^2 = 3x^2\cos x^3 \) and \( \frac{dv}{dx} = g'(x^2)\cdot2x = 2x\sin x^2 \).
2. \( \frac{du}{dv} = \frac{3x^2\cos x^3}{2x\sin x^2} = \frac{3}{2}x\cos x^3\,\text{cosec}\,x^2 \).

Teacher's Note:
1. Differentiate both with respect to x, then divide.
2. Note f' is evaluated at x³ and g' at x².

Answer: (a) \( -\frac{1}{2} \)

Explanation:
1. Let \( t = \cos\theta \). Then \( \sqrt{\frac{1-t}{1+t}} = \tan\frac{\theta}{2} \), so \( y = \theta \).
2. \( \sqrt{1+t} = \sqrt2\cos\frac\theta2 \) and \( \sqrt{1-t} = \sqrt2\sin\frac\theta2 \), so \( e^x = \frac{\cos\frac\theta2-\sin\frac\theta2}{\cos\frac\theta2+\sin\frac\theta2} = \tan\left(\frac\pi4-\frac\theta2\right) \).
3. Taking logs, \( x = \ln\tan\left(\frac\pi4-\frac\theta2\right) \), so \( \frac{dx}{d\theta} = -\frac{1}{\cos\theta} \) (using \( \frac{d}{du}\ln\tan u = \frac{2}{\sin 2u} \)). Hence \( \frac{dy}{dx} = -\cos\theta = -t = -\frac12 \) at t = 1/2.

Teacher's Note:
1. Substitute t = cos θ to remove the radicals.
2. The result dy/dx = −t is simple and was confirmed numerically.

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

Explanation:
1. \( t = \frac{x}{1+x^2} \), so \( y = x^2 + \frac{x^2}{(1+x^2)^2} \) and \( \frac{dt}{dx} = \frac{1-x^2}{(1+x^2)^2} \).
2. \( \frac{dy}{dx} = 2x + 2t\frac{dt}{dx} \). At x = 2: \( t = \frac25 \), \( \frac{dt}{dx} = -\frac{3}{25} \).
3. \( \frac{dy}{dx} = 4 - \frac{12}{125} = \frac{488}{125} \).

Teacher's Note:
1. Express t in terms of x first.
2. Then use the chain rule.

Answer: (a) AP

Explanation:
1. \( f(1) = f(-1) \) gives \( f(x) = ax^2 + c \), so \( f'(x) = 2ax \).
2. Then \( f'(a_i) = 2a\,a_i \), a constant multiple of an AP, which is an AP.

Teacher's Note:
1. f(1) = f(−1) removes the linear term.
2. A linear function maps an AP to an AP.

Answer: (b) \( \frac{7}{8} \)

Explanation:
1. Replace x by 1/x and solve the pair: \( 16f(x) = 5(x+2) - 3\left(\frac1x+2\right) \), so \( f(x) = \frac{1}{16}\left(5x+4-\frac3x\right) \).
2. \( f(1) = \frac{6}{16} = \frac38 \) and \( f'(x) = \frac{1}{16}\left(5+\frac{3}{x^2}\right) \), so \( f'(1) = \frac12 \).
3. \( \frac{dy}{dx} = f(1) + 1\cdot f'(1) = \frac38 + \frac12 = \frac78 \).

Teacher's Note:
1. Eliminate f(1/x) using the two equations.
2. Product rule for y = x f(x).

Answer: (b) f'(x) = 0 for all x

Explanation:
1. Fix y: \( f(x)\,[1 + f(y)] = 1 - f(y) \), so \( f(x) = \frac{1-f(y)}{1+f(y)} \), which does not depend on x.
2. Thus f is constant and \( f'(x) = 0 \) for all x.

Teacher's Note:
1. If an expression for f(x) has no x in it, f is constant.
2. A constant function has zero derivative.

Answer: (a) \( \text{cosec}\{g(x)\} \)

Explanation:
1. For an inverse function, \( g'(x) = \frac{1}{f'(g(x))} \).
2. Hence \( g'(x) = \frac{1}{\sin g(x)} = \text{cosec}\{g(x)\} \).

Teacher's Note:
1. Derivative of the inverse is the reciprocal of the derivative at the matching point.
2. Do not forget the g(x) inside.

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

Explanation:
1. From \( y^2 = P \): \( yy' = \frac{P'}{2} \) and \( y'^2 + yy'' = \frac{P''}{2} \).
2. \( y^3y'' = P\left(\frac{P''}{2} - y'^2\right) = \frac{PP''}{2} - \frac{P'^2}{4} \) since \( y'^2 = \frac{P'^2}{4P} \).
3. Differentiate: \( \frac{P'P'' + PP'''}{2} - \frac{P'P''}{2} = \frac{PP'''}{2} \); multiplying by 2 gives \( PP''' \).

Teacher's Note:
1. Differentiate y² = P twice.
2. Cancel the P'P'' terms.

Answer: (d) independent of p

Explanation:
1. Expanding along the first row: \( f(x) = x^3(-p^3) - \sin x\,(6p^3) + \cos x\,(6p^2 + p) \).
2. \( f'''(x) = -6p^3 - 6p^3(-\cos x) + (6p^2+p)\sin x \), so \( f'''(0) = -6p^3 + 6p^3 = 0 \).
3. The value 0 does not depend on p.

Teacher's Note:
1. Expand the determinant and differentiate term by term.
2. The third derivative of x³ is the constant 6.

Answer: (b) 0

Explanation:
1. Each pair of derivatives multiplies by \( -4 \): \( y^{(6)} = -64\sin 2x \).
2. At \( x = \pi/2 \): \( \sin\pi = 0 \), so the value is 0.

Teacher's Note:
1. The nth derivative of sin 2x involves 2ⁿ.
2. Evaluate sin at the point.

Answer: (a) \( \frac{1}{2a} \)

Explanation:
1. Eliminate t: \( t = \frac{x}{2a} \), so \( y = \frac{x^2}{4a} \).
2. \( \frac{d^2y}{dx^2} = \frac{1}{2a} \), a constant.

Teacher's Note:
1. Eliminate the parameter when it is easy.
2. The second derivative of a parabola y = kx² is 2k.

Answer: (b) b = 1 and c = 1

Explanation:
1. \( f'(x) = (a+d+cx)\cos x + (c-b-ax)\sin x \).
2. Comparing with \( x\cos x \): c = 1, a + d = 0, c − b = 0 and a = 0.
3. So a = 0, b = 1, c = 1, d = 0. The correct pair is b = 1 and c = 1.

Teacher's Note:
1. Equate the coefficients of x cos x, cos x, x sin x and sin x.
2. Test each option against the solved values.

Answer: (a) f'(2) = f'(3) and f'(2) = 0 and \( f'\left(\frac12\right) = \frac{16}{5} \)

Explanation:
1. For x > 1, \( \sin^{-1}\frac{2x}{1+x^2} = \pi - 2\tan^{-1}x \), so \( f(x) = \pi \) and f'(2) = f'(3) = 0.
2. For |x| < 1, \( \sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x \), so \( f = 4\tan^{-1}x \) and \( f'\left(\frac12\right) = \frac{4}{1+\frac14} = \frac{16}{5} \).
3. All three statements in the first option are true.

Teacher's Note:
1. The inverse sine of 2x/(1+x²) changes its formula at x = 1.
2. Check the region of each point.

Answer: (a) f(0) + f(2) = f(1) and f(0) + f(3) = 0 and f(1) + f(3) = f(2)

Explanation:
1. Let \( f'(1) = A,\ f''(2) = B \), and \( f''' = 6 \). Then \( f' = 3x^2 + 2Ax + B \) gives \( 3 + 2A + B = A \), and \( f'' = 6x + 2A \) gives \( 12 + 2A = B \).
2. Solving, A = −5, B = 2, so \( f(x) = x^3 - 5x^2 + 2x + 6 \).
3. f(0) = 6, f(1) = 4, f(2) = −2, f(3) = −6: all three relations hold.

Teacher's Note:
1. Treat the derivative values as constants A, B and C.
2. Form equations by substituting x into f', f''.

Answer: (a) f(2) = f(−2) and f'(2) + f'(−2) = 0

Explanation:
1. AP condition: \( 2f(x)f(y) = f(x-y) + f(x+y) \). Put x = 0: \( 2f(0)f(y) = f(-y) + f(y) \), and put y = 0 gives \( f(0) = 1 \) (after cancelling f(0) ≠ 0).
2. Then \( f(-y) = f(y) \): f is even (like cos x), so f(2) = f(−2) and f' is odd, giving f'(2) + f'(−2) = 0.
3. f(3) + f(−3) = 2f(3) need not be 0, and f'(3) = −f'(−3) in general.

Teacher's Note:
1. Setting x = 0 and y = 0 gives evenness.
2. The derivative of an even function is odd.

Answer: (b) an odd function

Explanation:
1. Differentiate f(−x) = f(x): \( -f'(-x) = f'(x) \).
2. So \( f'(-x) = -f'(x) \): f' is odd.

Teacher's Note:
1. Even functions have odd derivatives.
2. Odd functions have even derivatives.

Answer: (a) \( f_n(x)\cdot\frac{d}{dx}\{f_{n-1}(x)\} \) and \( f_n(x)\cdot f_{n-1}(x)\cdots f_2(x)\cdot f_1(x) \)

Explanation:
1. \( \frac{d}{dx}f_n = e^{f_{n-1}}\cdot f_{n-1}' = f_n\cdot f_{n-1}' \).
2. Repeating, \( f_{n-1}' = f_{n-1}f_{n-2}' \), and so on down to \( f_0' = 1 \).
3. So \( f_n' = f_nf_{n-1}\cdots f_1 \), and both expressions in the first option are valid.

Teacher's Note:
1. Apply the chain rule repeatedly.
2. Telescope until f₀' = 1.