JEE Physics Vectors and Calculus MCQs Set 03

Answer: (b) \( 2x + 1 \)

Explanation:
1. Use the power rule on each term: \( \frac{d}{dx}x^2 = 2x \), \( \frac{d}{dx}x = 1 \), and the derivative of a constant is 0.
2. So \( \frac{dy}{dx} = 2x + 1 \).

Teacher's Note:
1. The derivative of a sum is the sum of the derivatives.
2. Constants vanish on differentiation.

Answer: (d) \( 15t^2 - 15t^4 \)

Explanation:
1. \( \frac{d}{dt}(5t^3) = 15t^2 \) and \( \frac{d}{dt}(3t^5) = 15t^4 \).
2. Hence \( \frac{ds}{dt} = 15t^2 - 15t^4 \).

Teacher's Note:
1. Multiply by the power and reduce the power by one.
2. Watch the minus sign between the terms.

Answer: (a) \( \sin x + x\cos x \)

Explanation:
1. Product rule: \( (uv)' = u'v + uv' \) with \( u = x \) and \( v = \sin x \).
2. \( y' = 1\cdot\sin x + x\cos x \).

Teacher's Note:
1. The product rule gives two terms, both with a plus sign.
2. Do not differentiate each factor separately and multiply.

Answer: (c) \( e^x\ln x + \frac{e^x}{x} \)

Explanation:
1. Product rule with \( u = e^x \), \( v = \ln x \): \( u'v + uv' = e^x\ln x + e^x\cdot\frac1x \).

Teacher's Note:
1. The derivative of \( e^x \) is \( e^x \) and of \( \ln x \) is \( 1/x \).
2. Combine using the product rule.

Answer: (c) \( 3x^2 \)

Explanation:
1. Expand first: \( (x-1)(x^2+x+1) = x^3 - 1 \).
2. So \( \frac{dy}{dx} = 3x^2 \).

Teacher's Note:
1. Simplifying before differentiating saves effort.
2. The product rule would give the same answer.

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

Explanation:
1. \( y = \tan x \), and \( \frac{d}{dx}\tan x = \sec^2 x \).
2. By the quotient rule: \( \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x \).

Teacher's Note:
1. Recognise \( \sin x / \cos x \) as \( \tan x \).
2. The quotient rule confirms the result using the identity \( \sin^{2} + \cos^{2} = 1 \).

Answer: (d) \( \frac{-19}{(3x-2)^2} \)

Explanation:
1. Quotient rule: \( \frac{2(3x-2) - 3(2x+5)}{(3x-2)^2} \).
2. Numerator \( = 6x - 4 - 6x - 15 = -19 \).

Teacher's Note:
1. Quotient rule: (u′v − uv′)/v².
2. The sign of the numerator depends on the order of the subtraction.

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

Explanation:
1. \( y' = \frac{\frac1x\cdot x - \ln x\cdot1}{x^2} = \frac{1 - \ln x}{x^2} \).
2. This equals \( \frac{1}{x^2} - \frac{\ln x}{x^2} \).

Teacher's Note:
1. Use the quotient rule with \( u = \ln x \) and \( v = x \).
2. Split the numerator to match the option format.

Answer: (b) \( 10(2x+1)^4 \)

Explanation:
1. Chain rule: put \( u = 2x + 1 \), so \( y = u^5 \).
2. \( \frac{dy}{dx} = 5u^4\cdot\frac{du}{dx} = 5(2x+1)^4\cdot2 = 10(2x+1)^4 \).

Teacher's Note:
1. Differentiate the outside function, then multiply by the derivative of the inside.
2. Forgetting the inner derivative 2 gives option 2.

Answer: (c) \( -27(4-3x)^8 \)

Explanation:
1. Chain rule: \( 9(4-3x)^8\cdot\frac{d}{dx}(4-3x) = 9(4-3x)^8(-3) \).

Teacher's Note:
1. The inner derivative here is −3, which flips the sign.
2. Multiply 9 by −3 to get −27.

Answer: (a) \( \left(1 - \frac{x}{7}\right)^{-8} \)

Explanation:
1. \( \frac{dy}{dx} = -7\left(1 - \frac{x}{7}\right)^{-8}\cdot\left(-\frac17\right) \).
2. The factors \( -7 \) and \( -\frac17 \) multiply to 1.

Teacher's Note:
1. The outer derivative is \( -7(...)^{-8} \); the inner derivative is −1/7.
2. Their product is \( +1 \).

Answer: (d) \( 5\cos 5x \)

Explanation:
1. Chain rule: \( \frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx} \) with \( u = 5x \).
2. So \( y' = 5\cos5x \).

Teacher's Note:
1. The derivative of sine is cosine, with no sign change.
2. The factor 5 comes from the inner function.

Answer: (b) \( \cos x + \frac2x + 2e^{2x} \)

Explanation:
1. \( \frac{d}{dx}\ln x^2 = \frac{1}{x^2}\cdot2x = \frac2x \).
2. \( \frac{d}{dx}e^{2x} = 2e^{2x} \).
3. Adding with \( \cos x \) gives the answer.

Teacher's Note:
1. Chain rule applies to both \( \ln(x^{2}) \) and \( e^{2x} \).
2. Alternatively \( \ln(x^{2}) = 2 \ln x \), so its derivative is \( 2/x \).

Answer: (d) \( 2\omega\cos(\omega x + \phi) \)

Explanation:
1. Chain rule: the inner derivative is \( \frac{d}{dx}(\omega x + \phi) = \omega \).
2. So \( y' = 2\cos(\omega x + \phi)\cdot\omega \).

Teacher's Note:
1. Treat \( \omega \) and \( \phi \) as constants.
2. The same result is the velocity formula in SHM.

Answer: (a) \( e^x\tan x + e^x\sec^2 x \)

Explanation:
1. Product rule: \( (e^x)'\tan x + e^x(\tan x)' = e^x\tan x + e^x\sec^2 x \).

Teacher's Note:
1. The derivative of \( \tan x \) is \( \sec^{2}x \).
2. Use the product rule for two functions of \( x \).

Answer: (c) \( 48(8x-1)^2 \)

Explanation:
1. \( \frac{dy}{du} = 6u^2 \) and \( \frac{du}{dx} = 8 \).
2. \( \frac{dy}{dx} = 6u^2\cdot8 = 48(8x-1)^2 \).

Teacher's Note:
1. The chain rule multiplies the derivatives dy/du and du/dx.
2. Substitute \( u \) at the end.

Answer: (c) \( 3\cos(3x+1) \)

Explanation:
1. \( \frac{dy}{du} = \cos u \) and \( \frac{du}{dx} = 3 \), so \( \frac{dy}{dx} = 3\cos(3x+1) \).

Teacher's Note:
1. Write the answer in terms of \( x \).
2. Chain rule: \( dy/dx \) = (dy/du)(du/dx).

Answer: (a) \( \frac{-x\sin x - \cos x}{x^2} + \frac{\cos x + x\sin x}{\cos^2 x} \)

Explanation:
1. First term, quotient rule: \( \frac{-x\sin x - \cos x}{x^2} \).
2. Second term, quotient rule: \( \frac{\cos x - x(-\sin x)}{\cos^2 x} = \frac{\cos x + x\sin x}{\cos^2 x} \).
3. Add the two.

Teacher's Note:
1. Apply the quotient rule to each term separately.
2. The minus sign inside the second numerator disappears because the derivative of \( \cos x \) is \( -\sin x \).

Answer: (d) \( 3\sin^2 x\cos x + 3\cos 3x \)

Explanation:
1. \( \frac{d}{dx}\sin^3x = 3\sin^2x\cos x \) by the chain rule.
2. \( \frac{d}{dx}\sin3x = 3\cos3x \).

Teacher's Note:
1. \( \sin^{3}x \) and \( \sin 3x \) are different: one is a power, the other has a multiple angle.
2. Both need the chain rule.

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

Explanation:
1. \( \frac{dq}{dr} = \frac{1}{2\sqrt{2r - r^2}}\cdot(2 - 2r) \).
2. This simplifies to \( \frac{1 - r}{\sqrt{2r - r^2}} \).

Teacher's Note:
1. The factor 2 from the inner derivative cancels the 1/2 of the square root rule.
2. Always multiply by the inner derivative \( 2 - 2r \).

Answer: (b) \( 4\left(\frac{x^2}{8} + x - \frac1x\right)^3\left(\frac x4 + 1 + \frac{1}{x^2}\right) \)

Explanation:
1. Chain rule: outer derivative \( 4(\ldots)^3 \), inner derivative \( \frac{x}{4} + 1 + \frac{1}{x^2} \).
2. The derivative of \( -\frac1x \) is \( +\frac{1}{x^2} \).

Teacher's Note:
1. Differentiate the inside term by term.
2. The derivative of \( -1/x \) is \( +1/x^{2} \).

Answer: (c) \( \frac{1}{2\sqrt x} + \sec^2 x \)

Explanation:
1. \( \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} \) and \( \frac{d}{dx}\tan x = \sec^2x \).

Teacher's Note:
1. Write the root as a power before differentiating.
2. Tan \( x \) gives \( \sec^{2}x \) with a positive sign.

Answer: (a) \( y' = \frac2x + \cos x,\ y'' = -\frac{2}{x^2} - \sin x \)

Explanation:
1. \( y' = \frac{2x}{x^2} + \cos x = \frac2x + \cos x \).
2. \( y'' = -\frac{2}{x^2} - \sin x \).

Teacher's Note:
1. Differentiate twice, carefully with the signs.
2. The derivative of \( 2/x \) is \( -2/x^{2} \).

Answer: (d) \( \frac{-\csc^2 x}{(1 + \cot x)^2} \)

Explanation:
1. Let \( c = \cot x \). Then \( y = \frac{c}{1+c} \) and \( \frac{dy}{dc} = \frac{1}{(1+c)^2} \).
2. With \( \frac{dc}{dx} = -\csc^2x \), \( \frac{dy}{dx} = \frac{-\csc^2 x}{(1+\cot x)^2} \).

Teacher's Note:
1. A substitution simplifies the quotient rule here.
2. The derivative of \( \cot x \) is \( -\csc^{2}x \).

Answer: (b) \( -x^2\sin x \)

Explanation:
1. \( \frac{d}{dx}(x^2\cos x) = 2x\cos x - x^2\sin x \).
2. \( \frac{d}{dx}(-2x\sin x) = -2\sin x - 2x\cos x \) and \( \frac{d}{dx}(-2\cos x) = 2\sin x \).
3. Everything cancels except \( -x^2\sin x \).

Teacher's Note:
1. Look for cancellations after applying the product rule.
2. Such a function is the integral of \( -x^{2} \sin x \).

Answer: (d) \( \cos\theta + \sec^2\theta \)

Explanation:
1. \( (1 + \sec\theta)\sin\theta = \sin\theta + \tan\theta \).
2. Differentiating gives \( \cos\theta + \sec^2\theta \).

Teacher's Note:
1. Simplify first: \( \sec \theta \sin \theta = \tan \theta \).
2. Then each term has a standard derivative.