JEE Physics Mathematical Tools MCQs Set 02

Answer: (b) 1

Explanation:
1. \( f(\pi/2)=\cos\dfrac\pi2+\sin\dfrac\pi2=0+1=1 \).

Teacher's Note:
1. Use radians for π/2.
2. Evaluate each term separately.

Answer: (a) 47

Explanation:
1. \( f(2)=4(2)+3=11 \).
2. \( f(11)=4(11)+3=47 \).

Teacher's Note:
1. Apply the function twice, innermost first.
2. f(2) alone is 11.

Answer: (d) \( 2-\sqrt3 \)

Explanation:
1. \( \tan(45^\circ-30^\circ)=\dfrac{1-1/\sqrt3}{1+1/\sqrt3}=\dfrac{\sqrt3-1}{\sqrt3+1} \).
2. Rationalising: \( \dfrac{(\sqrt3-1)^2}{2}=2-\sqrt3\approx0.268 \).

Teacher's Note:
1. Use the difference formula for tan.
2. \( \tan15^\circ\lt 1 \), so the answer must be less than 1.

Answer: (c) \( \dfrac{1-\cos2\theta}{2} \)

Explanation:
1. \( \cos2\theta=1-2\sin^2\theta \).
2. So \( \sin^2\theta=\dfrac{1-\cos2\theta}{2} \).

Teacher's Note:
1. The cosine form with plus is cos²θ.
2. \( \sin^2\theta \) is never negative.

Answer: (b) \( \sin^2A\cos B+\tfrac12\sin2A\sin B \)

Explanation:
1. \( \sin(A+B)=\sin A\cos B+\cos A\sin B \).
2. Multiply by sin A: \( \sin^2A\cos B+\sin A\cos A\sin B \).
3. \( \sin A\cos A=\tfrac12\sin2A \).

Teacher's Note:
1. Expand first, then use the double-angle identity.
2. Test with B = 0: both sides become sin²A.

Answer: (a) (1), (3) and (4)

Explanation:
1. \( \cos(\pi/2+\theta)=-\sin\theta \): yes.
2. \( \cos(\pi/2-\theta)=+\sin\theta \): no.
3. \( \sin(\theta-\pi)=-\sin\theta \) and \( \sin(\pi+\theta)=-\sin\theta \): both yes.

Teacher's Note:
1. Shifting by π flips the sign of sine.
2. Cofunction identities change sin into cos.

Answer: (d) All four

Explanation:
1. \( 8\sin\theta+6\cos\theta=10\left(\tfrac45\sin\theta+\tfrac35\cos\theta\right) \); with \( \sin37^\circ=\tfrac35 \) this is \( 10\sin(\theta+37^\circ) \), whose maximum is 10.
2. \( x_1x_2=48\sin\theta\cos\theta=24\sin2\theta \).
3. \( x_1/x_2=\tfrac86\tan\theta=\tfrac43\tan\theta \).

Teacher's Note:
1. Write a sinθ + b cosθ as R sin(θ+φ).
2. R = √(8²+6²) = 10.

Answer: (c) \( -\dfrac1x \)

Explanation:
1. \( f(f(x))=\dfrac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} \).
2. Numerator \( =\dfrac{-2}{x+1} \), denominator \( =\dfrac{2x}{x+1} \), so the result is \( -\dfrac1x \).

Teacher's Note:
1. Combine the fractions over x + 1.
2. Apply f twice to get −1/x; four times gives back x.

Answer: (b) \( x \)

Explanation:
1. \( f(y)=\dfrac{2y-3}{3y-2} \) with \( y=\dfrac{2x-3}{3x-2} \).
2. Numerator: \( \dfrac{2(2x-3)-3(3x-2)}{3x-2}=\dfrac{-5x}{3x-2} \); denominator: \( \dfrac{3(2x-3)-2(3x-2)}{3x-2}=\dfrac{-5}{3x-2} \).
3. The ratio is x, so f is its own inverse.

Teacher's Note:
1. Substitute y and simplify the double fraction.
2. This function is an involution.

Answer: (a) \( \dfrac{10\sqrt3}{3} \) m

Explanation:
1. The base angles are \( (180^\circ-120^\circ)/2=30^\circ \).
2. Sine rule: \( \dfrac{a}{\sin30^\circ}=\dfrac{10}{\sin120^\circ} \).
3. \( a=\dfrac{10\times\tfrac12}{\sqrt3/2}=\dfrac{10}{\sqrt3}=\dfrac{10\sqrt3}{3} \) m.

Teacher's Note:
1. Use the sine rule with the 120° angle opposite the 10 m side.
2. The equal sides are shorter than 10 m.

Answer: (d) (1) and (3) only

Explanation:
1. Sum-to-product: \( \sin P+\sin Q=2\sin\frac{P+Q}2\cos\frac{P-Q}2 \), so (1) is correct.
2. \( \sin P-\sin Q=2\cos\frac{P+Q}2\sin\frac{P-Q}2 \), so (3) is correct.
3. (2) is false: the sum of sines is not a product of sines.

Teacher's Note:
1. Test (2) with θ₁ = 0 and θ₂ = 90°: the left side is A but the right side is 0.
2. Learn both sum-to-product identities.

Answer: (c) (1), (2) and (4)

Explanation:
1. With \( \sin37^\circ=0.6,\cos37^\circ=0.8 \): (1) 1.4 = 1.4, true.
2. (2) −0.2 = 0.6 − 0.8 = −0.2, true.
3. (3) 1.75 vs 0.333, false.
4. (4) \( \tfrac34\times\tfrac43=1 \), true.

Teacher's Note:
1. 37° and 53° are complementary: sin37 = cos53 and cos37 = sin53.
2. Compute each side numerically.

Answer: (b) \( 2A\cos\dfrac\theta2 \)

Explanation:
1. \( R^2=2A^2(1+\cos\theta)=2A^2\cdot2\cos^2\dfrac\theta2 \).
2. \( R=2A\cos\dfrac\theta2 \).

Teacher's Note:
1. Use \( 1+\cos\theta=2\cos^2(\theta/2) \).
2. Check: at θ = 0, R = 2A.

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

Explanation:
1. \( \dfrac{d}{dx}x^2=2x \), \( \dfrac{d}{dx}x=1 \) and the derivative of a constant is 0.
2. \( \dfrac{dy}{dx}=2x+1 \).

Teacher's Note:
1. The constant 8 disappears.
2. Use the power rule term by term.

Answer: (d) \( \sec^2x-\csc^2x \)

Explanation:
1. \( \dfrac{d}{dx}\tan x=\sec^2x \) and \( \dfrac{d}{dx}\cot x=-\csc^2x \).
2. Adding gives \( \sec^2x-\csc^2x \).

Teacher's Note:
1. The derivative of cot x is negative.
2. Memorise the six trigonometric derivatives.

Answer: (c) \( \cos x-\sin x \) and \( -\sin x-\cos x \)

Explanation:
1. \( y'=\cos x-\sin x \).
2. \( y''=-\sin x-\cos x=-y \).

Teacher's Note:
1. The second derivative equals −y, as for simple harmonic motion.
2. Differentiate the first derivative once more.

Answer: (b) \( \dfrac1x+e^x \) and \( -\dfrac1{x^2}+e^x \)

Explanation:
1. \( y'=\dfrac1x+e^x \).
2. \( y''=-\dfrac1{x^2}+e^x \).

Teacher's Note:
1. Differentiate 1/x as \( x^{-1} \).
2. The derivative of \( e^x \) is \( e^x \) itself.

Answer: (a) \( e^x\left(\ln x+\dfrac1x\right) \)

Explanation:
1. Product rule: \( (uv)'=u'v+uv' \).
2. \( y'=e^x\ln x+e^x\cdot\dfrac1x \).

Teacher's Note:
1. Both factors need differentiating.
2. Check at x = 1: y' = e.

Answer: (d) \( \cos2x \)

Explanation:
1. Product rule: \( \cos x\cos x+\sin x(-\sin x)=\cos^2x-\sin^2x \).
2. This equals \( \cos2x \).

Teacher's Note:
1. Alternatively \( y=\tfrac12\sin2x \).
2. Both routes give the same result.

Answer: (c) \( \dfrac{-19}{(3x-2)^2} \)

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

Teacher's Note:
1. Subtract in the right order: (u'v − uv').
2. The x terms cancel in the numerator.

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

Explanation:
1. Quotient rule: \( \dfrac{\frac1x\cdot x-\ln x\cdot1}{x^2} \).
2. \( =\dfrac{1-\ln x}{x^2} \).

Teacher's Note:
1. The maximum of this function is at x = e.
2. Keep the sign of the \( \ln x \) term.

Answer: (a) 0

Explanation:
1. The product is \( \sec^2x-\tan^2x=1 \), a constant.
2. The derivative of a constant is 0.

Teacher's Note:
1. Simplify before differentiating.
2. Use the identity \( \sec^2x-\tan^2x=1 \).

Answer: (d) 13

Explanation:
1. \( (uv)'=u'v+uv'=(-3)(-1)+(5)(2) \).
2. \( =3+10=13 \).

Teacher's Note:
1. Product rule.
2. Watch the signs of u' and v.

Answer: (c) −7

Explanation:
1. \( \left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}=\dfrac{(-3)(-1)-(5)(2)}{1} \).
2. \( =3-10=-7 \).

Teacher's Note:
1. Quotient rule with \( v^2=1 \).
2. The numerator is u'v − uv'.

Answer: (b) 7/25

Explanation:
1. \( \dfrac{v'u-vu'}{u^2}=\dfrac{(2)(5)-(-1)(-3)}{25} \).
2. \( =\dfrac{10-3}{25}=\dfrac7{25} \).

Teacher's Note:
1. Now v is in the numerator.
2. The denominator is \( u^2=25 \).

Answer: (a) 20

Explanation:
1. \( 7v'-2u'=7(2)-2(-3) \).
2. \( =14+6=20 \).

Teacher's Note:
1. Differentiation is linear.
2. Subtracting a negative derivative adds.

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

Explanation:
1. Chain rule: \( 9(4-3x)^8\times(-3) \).
2. \( =-27(4-3x)^8 \).

Teacher's Note:
1. Multiply by the derivative of the inner function.
2. The inner derivative is −3.

Answer: (c) −1

Explanation:
1. \( 2(x+y)\left(1+\dfrac{dy}{dx}\right)=0 \).
2. Since \( x+y=\pm2\neq0 \), \( 1+\dfrac{dy}{dx}=0 \), so \( \dfrac{dy}{dx}=-1 \).

Teacher's Note:
1. Equivalently x + y = ±2, which are lines of slope −1.
2. Do not divide by x + y without checking it is non-zero.