JEE Physics Vectors and Calculus MCQs Set 04

Answer: (b) \( \frac{x^4}{2} - \frac{5x^2}{2} + 7x + C \)

Explanation:
1. Use \( \int x^n dx = \frac{x^{n+1}}{n+1} \) term by term.
2. \( \frac{2x^4}{4} - \frac{5x^2}{2} + 7x + C \).

Teacher's Note:
1. Integration raises the power by one and divides by the new power.
2. Always add the constant of integration for an indefinite integral.

Answer: (d) \( \frac23 x^{3/2} + \frac34 x^{4/3} + C \)

Explanation:
1. \( \int x^{1/2}dx = \frac{x^{3/2}}{3/2} = \frac23x^{3/2} \).
2. \( \int x^{1/3}dx = \frac{x^{4/3}}{4/3} = \frac34x^{4/3} \).

Teacher's Note:
1. Dividing by 3/2 means multiplying by 2/3.
2. Do not confuse integration with differentiation: the powers go up.

Answer: (a) \( -\frac1x - \frac{1}{2x^2} + C \)

Explanation:
1. Expand: \( x^{-3}(x + 1) = x^{-2} + x^{-3} \).
2. \( \int x^{-2}dx = -x^{-1} \) and \( \int x^{-3}dx = -\frac{x^{-2}}{2} \).

Teacher's Note:
1. Expand before integrating.
2. Negative powers integrate in the same way, except for \( x^{-1} \).

Answer: (c) \( -\cos\theta + \theta + C \)

Explanation:
1. \( \cos\theta\tan\theta = \sin\theta \) and \( \cos\theta\sec\theta = 1 \).
2. So the integral is \( \int(\sin\theta + 1)d\theta = -\cos\theta + \theta + C \).

Teacher's Note:
1. Simplify using trigonometric identities first.
2. The integral of \( \sin \) is \( -\cos \).

Answer: (c) \( \frac{3\pi^2}{2} \)

Explanation:
1. \( \left[\frac{\theta^2}{2}\right]_\pi^{2\pi} = \frac{4\pi^2 - \pi^2}{2} = \frac{3\pi^2}{2} \).

Teacher's Note:
1. Substitute the upper limit and subtract the lower.
2. Square the limits before subtracting.

Answer: (a) 0

Explanation:
1. \( [\sin x]_0^{\pi} = \sin\pi - \sin0 = 0 \).

Teacher's Note:
1. Positive and negative areas of \( \cos x \) cancel over 0 to π.
2. The integral is net area, not total area.

Answer: (d) \( \frac13\ln\frac52 \)

Explanation:
1. \( \int\frac{dx}{3x+2} = \frac13\ln|3x+2| \).
2. Limits: \( \frac13(\ln5 - \ln2) = \frac13\ln\frac52 \).

Teacher's Note:
1. The factor 1/3 comes from the inner derivative of \( 3x + 2 \).
2. Difference of logarithms becomes the \( \log \) of a quotient.

Answer: (b) \( b^3 \)

Explanation:
1. Area \( = \int_0^b 3x^2dx = [x^3]_0^b = b^3 \).

Teacher's Note:
1. The area under a curve is the definite integral.
2. Check: at \( b = 1 \) the area is 1.

Answer: (b) \( \frac{b^2}{4} + b \)

Explanation:
1. \( \int_0^b\left(\frac x2 + 1\right)dx = \left[\frac{x^2}{4} + x\right]_0^b = \frac{b^2}{4} + b \).

Teacher's Note:
1. Integrate each term of the straight line.
2. This is a trapezium of parallel sides 1 and \( b/2 + 1 \).

Answer: (c) 24

Explanation:
1. \( \left[\frac{r^2}{2}\right]_{\sqrt2}^{5\sqrt2} = \frac{50 - 2}{2} = 24 \).

Teacher's Note:
1. The squares of the limits are 50 and 2.
2. Divide the difference by 2.

Answer: (a) \( e - 1 \)

Explanation:
1. \( [e^x]_0^1 = e - e^0 = e - 1 \).

Teacher's Note:
1. The integral of \( e^x \) is \( e^x \) itself.
2. Do not forget to subtract the lower-limit value \( e^{0} = 1 \).

Answer: (d) \( -\frac{2}{t^2} - \frac{2}{3t^{3/2}} + C \)

Explanation:
1. Split: \( 4t^{-3} + t^{-5/2} \).
2. \( \int4t^{-3}dt = -2t^{-2} \) and \( \int t^{-5/2}dt = \frac{t^{-3/2}}{-3/2} = -\frac23t^{-3/2} \).

Teacher's Note:
1. Write every term as a power of \( t \).
2. Dividing by −3/2 flips the sign.

Answer: (b) \( 2\sqrt t - \frac{2}{\sqrt t} + C \)

Explanation:
1. The integrand is \( t^{-1/2} + t^{-3/2} \).
2. \( \int t^{-1/2}dt = 2t^{1/2} \) and \( \int t^{-3/2}dt = -2t^{-1/2} \).

Teacher's Note:
1. Simplify the fraction to powers of \( t \) before integrating.
2. Check by differentiating the answer.

Answer: (d) 8

Explanation:
1. \( v = \frac{dx}{dt} = -2t + 4 = 0 \) gives \( t = 2 s \).
2. \( \frac{d^2x}{dt^2} = -2 < 0 \), so it is a maximum: \( x_{\max} = -4 + 8 + 4 = 8 \).

Teacher's Note:
1. A maximum occurs where the first derivative is zero and the second is negative.
2. Substitute the time back into \( x \).

Answer: (a) 39 and 38

Explanation:
1. \( f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3) \), zero at \( x = 2 \) and 3.
2. \( f''(x) = 12x - 30 \) is negative at 2 (maximum) and positive at 3 (minimum).
3. \( f(2) = 16 - 60 + 72 + 11 = 39 \) and \( f(3) = 54 - 135 + 108 + 11 = 38 \).

Teacher's Note:
1. A local maximum can be smaller than a local minimum elsewhere only for a cubic like this where the maximum occurs first.
2. Use the second derivative test to classify the points.

Answer: (c) \( \frac78 \)

Explanation:
1. \( y' = 4x - 1 = 0 \) gives \( x = \frac14 \), and \( y'' = 4 > 0 \).
2. \( y_{\min} = \frac{2}{16} - \frac14 + 1 = \frac78 \).

Teacher's Note:
1. For a quadratic ax² + bx \( + c \) with \( a \gt 0 \), the minimum is \( c - b^{2}/4a \).
2. Here that is \( 1 - 1/8 = 7/8 \).

Answer: (c) At t = 1, S is a maximum

Explanation:
1. \( \frac{dS}{dt} = t^2 - 4t + 3 = (t-1)(t-3) \), zero at \( t = 1 \) and 3.
2. \( \frac{d^2S}{dt^2} = 2t - 4 \) is −2 at \( t = 1 \) (maximum) and \( +2 \) at \( t = 3 \) (minimum).

Teacher's Note:
1. The sign of the second derivative decides between maximum and minimum.
2. \( t = 2 \) is not a stationary point.

Answer: (a) 6

Explanation:
1. \( y' = 8x - 4 = 0 \) gives \( x = \frac12 \).
2. \( y_{\min} = 1 - 2 + 7 = 6 \).

Teacher's Note:
1. Complete the square: \( 4(x - 1/2)^{2} + 6 \).
2. The square is zero at the minimum.

Answer: (d) 2 and 1

Explanation:
1. \( y' = 3x^2 - 3 = 0 \) gives \( x = \pm1 \); \( y'' = 6x \).
2. At \( x = -1 \), \( y'' < 0 \), so it is a maximum, with \( y = -1 + 3 = 2 \). At \( x = 1 \), \( y'' > 0 \), so \( y \) is a minimum, with \( y = -2 \).

Teacher's Note:
1. Test both stationary points with y″.
2. The local maximum occurs at the negative \( x \) value.

Answer: (b) 900

Explanation:
1. \( P = x(60 - x) \), \( \frac{dP}{dx} = 60 - 2x = 0 \) gives \( x = 30 \).
2. \( \frac{d^2P}{dx^2} = -2 < 0 \), so \( P_{\max} = 30\times30 = 900 \).

Teacher's Note:
1. For a fixed sum, the product is greatest when the numbers are equal.
2. That is the AM-GM rule.

Answer: (b) \( \sqrt{\frac{40}{3}} \) m

Explanation:
1. Let the base side be \( x \) and height \( h \): \( x^2 + 4xh = 40 \), so \( h = \frac{40 - x^2}{4x} \).
2. \( V = x^2h = \frac{x(40 - x^2)}{4} \) and \( \frac{dV}{dx} = \frac{40 - 3x^2}{4} = 0 \) gives \( x = \sqrt{40/3} \).
3. \( \frac{d^2V}{dx^2} = -\frac{3x}{2} < 0 \), so it is a maximum (and \( h = x/2 \)).

Teacher's Note:
1. Eliminate \( h \) using the area constraint before differentiating.
2. An open tank has one square base and four sides, so the area is \( x^{2} \) + 4xh.

Answer: (c) 9 m/s

Explanation:
1. \( v = \frac{dx}{dt} = 4t + 1 \).
2. At \( t = 2 \): \( v = 9 m/s \).

Teacher's Note:
1. Velocity is the first derivative of position.
2. Substitute the time after differentiating.

Answer: (d) 8 m/s\(^2\)

Explanation:
1. \( a = \frac{dv}{dt} = 4t \).
2. At \( t = 2 s \), \( a = 8 m/s ^2 \).

Teacher's Note:
1. Acceleration is the time derivative of velocity.
2. Substituting \( t = 2 \) into \( v \) instead would give 6, which is velocity.

Answer: (b) 30 N

Explanation:
1. \( F = \frac{dp}{dt} = 6t^2 + 2t + 2 \).
2. At \( t = 2 \): \( F = 24 + 4 + 2 = 30 N \).

Teacher's Note:
1. Force is the rate of change of momentum.
2. Substitute \( t = 2 \) after differentiating.

Answer: (d) 23 A

Explanation:
1. \( i = \frac{dq}{dt} = 4t + 3 \).
2. At \( t = 5 \): \( i = 23 A \).

Teacher's Note:
1. Current is the rate of flow of charge.
2. The derivative is evaluated at the instant asked.