JEE Mathematics Definite Integration MCQs Set 34

Answer: (a) \( e^2 \)

Explanation:
1. \( d/dt (t e^{t^2+t}) = e^{t^2+t}(1 + t(2t+1)) = e^{t^2+t}(2t^2 + t + 1) \).
2. The integral is \( [te^{t^2+t}]_0^1 = e^2 \).

Teacher's Note:
1. The integrand is an exact derivative.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac38 \)

Explanation:
1. The limit is \( \int_0^1\cos^4\frac{\pi x}2dx = \frac2\pi\int_0^{\pi/2}\cos^4t\,dt = \frac2\pi\cdot\frac{3\pi}{16} = \frac38 \).

Teacher's Note:
1. Riemann sum of \( cos^4 \) on \( [0, 1] \) with argument pi \( x / 2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( 2 \)

Explanation:
1. \( x = \ln(1/y) \) means \( y = e^{-x} \). The two curves meet at \( (0, 1); y = \ln(x + e) \) meets the x-axis at \( x = 1 - e \).
2. Area = \( \int_{1-e}^0\ln(x+e)dx + \int_0^\infty e^{-x}dx = \int_1^e\ln u\,du + 1 = 1 + 1 = 2 \).

Teacher's Note:
1. Left piece under the logarithm is \( 1 \), right piece under \( e^{-x} \) is \( 1 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 1 \)

Explanation:
1. Let \( h(x) = \int_0^x\frac{t^2}{1+t^4}dt - 2x + 1 \). Then \( h(0) = 1 \gt 0 \) and \( h(1) \lt 1/3 - 1 \lt 0 \).
2. \( h'(x) = \frac{x^2}{1+x^4} - 2 \lt 0 \), so \( h \) is strictly decreasing and has exactly one zero.

Teacher's Note:
1. Use monotonicity and the intermediate value theorem.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac52 \)

Explanation:
1. For each \( x \) the region is between \( y = x^2/4 \) and the smaller of \( 3 - x \) and \( 1 + \sqrt x \); these two meet at \( x = 1 \).
2. Area = \( \int_0^1\left(1 + \sqrt x - \frac{x^2}4\right)dx + \int_1^2\left(3 - x - \frac{x^2}4\right)dx = \left(1 + \frac23 - \frac1{12}\right) + \left(3 - \frac32 - \frac{7}{12}\right) = \frac{19}{12} + \frac{11}{12} = \frac52 \).

Teacher's Note:
1. The upper boundary changes at \( x = 1 \) and the region ends at \( x = 2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 1 \)

Explanation:
1. Expand: \( sin^2x + sin^2(ax) + 2sin x sin(ax) \). The average of each square is \( 1/2 \) and the cross term averages \( 0 \) when \( |a| \ne 1 \) and a \( \ne 0 \).
2. \( L = 1/2 + 1/2 = 1 \).

Teacher's Note:
1. The cross term has zero mean.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac32 \)

Explanation:
1. Put \( u = x + 3: y \ge \sqrt{|u|}, y \le (u+6)/5, u \le 9 \). The line meets \( y = \sqrt{-u} \) at \( u = -1 \) and \( y = \sqrt u \) at \( u = 4 \); between these the line lies above the curve.
2. Area = \( \int_{-1}^4\left(\frac{u+6}5 - \sqrt{|u|}\right)du = \frac{37.5}5 - \frac23 - \frac{16}3 = 7.5 - 6 = \frac32 \).

Teacher's Note:
1. Find the intersections of the line with both branches.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac{3\sqrt5}{\sqrt8} \)

Explanation:
1. Let \( S \) be the sum of the two roots. Then \( S^2 = 4x + 2\sqrt{4x^2-5(4x-5)} = 4x + 2|2x-5| \).
2. On \( [3/2, 9/4], 2x - 5 \lt 0 \), so \( S^2 = 4x + 10 - 4x = 10 \) and \( S = \sqrt{10} \).
3. The integral is \( \sqrt{10}(9/4 - 3/2) = 3\sqrt{10}/4 = 3\sqrt5/\sqrt8 \).

Teacher's Note:
1. The sum of the roots is the constant \( sqrt10 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 3 \)

Explanation:
1. Differentiate: \( -\sin^2x\,f(\sin x)\cos x = -\cos x \), so \( f(\sin x) = 1/\sin^2x \).
2. Hence \( f(t) = 1/t^2 \) and \( f(1/\sqrt3) = 3 \).

Teacher's Note:
1. Differentiating gives \( f(t) = 1/t^2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) All of (i), (ii), (iii) and (iv)

Explanation:
1. \( f'(x) = e^{x^2}(x-2)(x-3) \) is positive for \( x \lt 2 \), negative on \( (2, 3) \) and positive for \( x \gt 3 \).
2. So \( f \) has a local maximum at \( 2 \), decreases on \( (2, 3) \) and has a local minimum at \( 3 \). Since \( f'(2) = f'(3) = 0 \), Rolle's theorem gives \( f''(c) = 0 \) for some \( c \) in \( (2, 3) \).

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (c) \( 2R_1=R_2 \)

Explanation:
1. \( f \) is symmetric about \( x = 1/2 \) on \( [-1, 2] \), so \( \int_{-1}^2xf(x)dx = \frac12\int_{-1}^2f(x)dx (use x \to 1 - x \) and \( average) \).
2. \( R_2 = \int_{-1}^2f(x)dx \), so \( R_1 = R_2/2 \).

Teacher's Note:
1. Symmetry about \( x = 1/2 \) gives the mean position \( 1/2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac\pi4 \)

Explanation:
1. The sum is \( 1 - 1/3 + 1/5 - 1/7 + \) ..., the Leibniz series, which equals \( \tan^{-1}1 = \pi/4 \).
2. Equivalently \( \sum\int_0^1(x^{4n-4} - x^{4n-2})dx = \int_0^1\frac{1-x^2}{1-x^4}dx = \int_0^1\frac{dx}{1+x^2} \).

Teacher's Note:
1. Express each term as an integral and sum the geometric series.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 2015 \)

Explanation:
1. The \( sin x |x| \) part is odd and integrates to \( 0 \). For the other part, \( \cos x \to 1 \), so \( \int_{-1/n}^{1/n}2015\cos x\,|x|dx \approx 2015\cdot2\cdot\frac{1}{2n^2} = \frac{2015}{n^2} \).
2. Multiplying by \( n^2 \) gives \( 2015 \).

Teacher's Note:
1. Separate odd and even parts.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{20\sqrt2}3 \)

Explanation:
1. For a given \( y, x \) runs between \( |x| = \sqrt y/2 \) and \( |x| = 3\sqrt y \), so the width on both sides is \( 2(3\sqrt y - \tfrac12\sqrt y) = 5\sqrt y \).
2. Area = \( \int_0^25\sqrt y\,dy = 5\cdot\frac23\cdot2\sqrt2 = \frac{20\sqrt2}3 \).

Teacher's Note:
1. Integrate with respect to \( y \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \int_0^{\log(1+\sqrt2)}2(e^u+e^{-u})^{16}du \)

Explanation:
1. Put \( \cot x = \sinh u \), so \( u \) runs from \( 0 (x = \pi/2) \) to \( \ln(1+\sqrt2) (x = \pi/4) \). Then \( \csc x = \cosh u \) and \( dx = -du/\cosh u \) in absolute value, so \( \csc^{17}x\,dx = \cosh^{16}u\,du \).
2. \( (2\csc x)^{17}dx = 2^{17}\cosh^{16}u\,du = 2(e^u + e^{-u})^{16}du \).

Teacher's Note:
1. The hyperbolic substitution reduces the power by one.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( \frac\pi4\ln(2+\sqrt3)+\frac{\pi^2}{12}-\frac\pi{\sqrt3} \)

Explanation:
1. With \( x = \tan\theta, \cos^{-1}\frac{2x}{1+x^2} = \frac\pi2 - 2\tan^{-1}x \) on \( |x| \le 1 \). The \( tan^{-1}x \) part gives an odd integrand, so it vanishes.
2. The rest is \( \frac\pi2\cdot2\int_0^{1/\sqrt3}\frac{x^4}{1-x^4}dx = \pi\int_0^{1/\sqrt3}\left(-1 + \frac{1}{1-x^4}\right)dx \), and \( \int\frac{dx}{1-x^4} = \frac12\left(\tan^{-1}x + \tanh^{-1}x\right) \).
3. Evaluating at \( 1/\sqrt3: \pi\left(-\frac1{\sqrt3} + \frac\pi{12} + \frac14\ln(2+\sqrt3)\right) \).

Teacher's Note:
1. Odd parts cancel; the rest is a rational integral.
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( 19 \)

Explanation:
1. Split \( [-2, 4] \) into unit intervals with \( n = -2, -1, 0, 1, 2, 3 \).
2. Sum of \( n^2 = 4 + 1 + 0 + 1 + 4 + 9 = 19 \).

Teacher's Note:
1. Add \( n^2 \) for \( n = -2,...,3 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 12-8\ln2 \)

Explanation:
1. \( |x - 2y| + |x + 2y| = 2\max(|x|, 2|y|) \), so the first condition is \( |x| \le 4, |y| \le 2 (a rectangle) \). The condition xy \( \ge 2 \) selects parts in the first and third quadrants.
2. First quadrant: \( \int_1^4\left(2 - \frac2x\right)dx = 6 - 4\ln2 \). Doubling for the third quadrant: \( 12 - 8\ln2 \).

Teacher's Note:
1. Rewrite the first condition as a rectangle.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) Only (ii)

Explanation:
1. The curve touches \( y = -1 \) at \( x = \pm1 \) and lies above it between them, so the area is \( \int_{-1}^1(x^2 - 2|x| + 1)dx = 2\int_0^1(1 - x)^2dx = 2/3 \).
2. \( 2\int_0^1(2x - x^2)dx = 4/3 \), and \( (2/3)(2) = 4/3 \), so \( (i) \) and \( (iii) \) are false.

Teacher's Note:
1. Examine each statement separately before choosing the combination.
2. Test doubtful statements with a specific function or number.

Answer: (d) \( 5-\sqrt2-\sqrt3 \)

Explanation:
1. \( [x^2] = 0, 1, 2, 3 \) on \( [0,1), [1,\sqrt2), [\sqrt2,\sqrt3), [\sqrt3,2) \) respectively.
2. The integral is \( 1(\sqrt2 - 1) + 2(\sqrt3 - \sqrt2) + 3(2 - \sqrt3) \).
3. = \( 5 - \sqrt2 - \sqrt3 \).

Teacher's Note:
1. Find the points where \( x^2 \) crosses \( 1, 2 \) and \( 3 \).

Answer: (a) \( \frac\pi4 \)

Explanation:
1. By the property \( \theta \to \pi/2 - \theta \), the integral equals \( \frac12\int_0^{\pi/2}(\sin\theta + \cos\theta)\sqrt{\sin2\theta}\,d\theta \).
2. Put \( t = \sin\theta - \cos\theta: \sin2\theta = 1 - t^2 \), so I = \( \frac12\int_{-1}^1\sqrt{1-t^2}dt = \frac12\cdot\frac\pi2 = \pi/4 \).

Teacher's Note:
1. A half-circle area appears after the substitution.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( \frac{81}{2} \)

Explanation:
1. \( x = 9y - y^3 = y(3 - y)(3 + y) \) meets \( x = 0 \) at \( y = 0, \pm3 \), and the curve is odd in \( y \).
2. Area = \( 2\int_0^3(9y - y^3)dy = 2(81/2 - 81/4) = 81/2 \).

Teacher's Note:
1. Two symmetric lobes.
2. Verify the final value with a quick numerical or limiting check.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. The region lies in the upper half-disc of radius \( \sqrt2 \). Its lower boundary is \( y = \sqrt x \) for \( x \ge 0 \) and \( y = x^2 \) for \( x \lt 0 \); both meet the circle at \( x = \pm1 \).
2. Area = \( \int_{-1}^1\sqrt{2-x^2}dx - \int_0^1\sqrt x\,dx - \int_{-1}^0x^2dx = (1 + \pi/2) - 2/3 - 1/3 = \pi/2 \).

Teacher's Note:
1. Subtract the two lower curves from the circle's upper arc.
2. Verify the final value with a quick numerical or limiting check.

Answer: (d) \( 2 \)

Explanation:
1. For a = \( 1 \) the integrand is \( 4\sin^2x \), whose average value is \( 4\cdot\frac12 = 2 \).

Teacher's Note:
1. Average of \( 4 sin^2 x \) is \( 2 \).
2. Verify the final value with a quick numerical or limiting check.

Answer: (a) \( \frac14 \)

Explanation:
1. The logarithm of the limit is \( \lim\frac1n\sum_{r=1}^{2n-1}\ln\sin\frac{r\pi}{2n} = 2\int_0^1\ln\sin\frac{\pi x}{2}dx = \frac4\pi\int_0^{\pi/2}\ln\sin t\,dt \).
2. \( \int_0^{\pi/2}\ln\sin t\,dt = -\frac\pi2\ln2 \), so the logarithm is \( -2\ln2 \) and the limit is \( 1/4 \).

Teacher's Note:
1. Riemann sum of \( ln sin \), with \( 2n - 1 \) factors.
2. Verify the final value with a quick numerical or limiting check.

Answer: (b) \( 0 \)

Explanation:
1. By the fundamental theorem, \( F'(x) = e^{x^2/2}(1 - x^2) \).
2. At \( x = 1 \) this equals \( 0 \).

Teacher's Note:
1. Apply the fundamental theorem of calculus.
2. Verify the final value with a quick numerical or limiting check.