JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 05

Answer: (b) \( \int_{-a}^{a} f(x)\,dx \)

Explanation:
1. Split the integral: \( \int_0^a f(-x)dx \) becomes \( \int_{-a}^0 f(x)dx \) by putting x = -t
2. Adding the two parts gives the integral from -a to a

Teacher's Note:
1. Splitting at 0 turns two half-range integrals into one full-range integral.

Answer: (d) \( \pi/4 \)

Explanation:
1. Let I be the integral. Replace x by \( \pi/2-x \) to get the companion integral
2. Adding, \( 2I=\int_0^{\pi/2}1\,dx=\pi/2 \), so \( I=\pi/4 \)

Teacher's Note:
1. Use the property of replacing x by (a + b - x).

Answer: (a) \( \pi/20 \)

Explanation:
1. Here a + b = \( \pi/5+3\pi/10=\pi/2 \), so x and \( \pi/2-x \) give the sine form
2. \( 2I=\int_{\pi/5}^{3\pi/10}1dx=\pi/10 \), so \( I=\pi/20 \)

Teacher's Note:
1. The sum of cos and sin forms is 1, so the integral is half the length of the interval.

Answer: (c) \( \lambda \) and \( \nu \)

Explanation:
1. The term \( \mu\sin x/(1+\cos x) \) is odd, so it integrates to 0 over a symmetric interval
2. The other terms are even: \( 2[\lambda(1-\tfrac{1}{\sqrt2})+\nu\pi/4]=0 \)
3. This involves only \( \lambda \) and \( \nu \)

Teacher's Note:
1. Odd functions vanish over symmetric limits.

Answer: (d) 2

Explanation:
1. f(x) is the fractional part, with period 1
2. Each unit interval gives 1/2, and there are 4 unit intervals in [-2, 2]
3. Total = 4 x 1/2 = 2

Teacher's Note:
1. Fractional part has average 1/2 per period.

Answer: (b) 0

Explanation:
1. Replace x by \( \pi-x \): \( \sin x \) is unchanged and \( \cos^5x \) changes sign
2. So I = -I, giving I = 0

Teacher's Note:
1. Odd power of cos about x = pi/2 gives zero.

Answer: (a) 4

Explanation:
1. The integrand is even, so I = 2 times the integral from 0 to 2
2. From 0 to 1: 1 - x^2 gives 2/3. From 1 to 2: x^2 - 1 gives 4/3
3. I = 2(2/3 + 4/3) = 4

Teacher's Note:
1. Split at the points where the modulus changes sign.

Answer: (c) \( \pi^2 \)

Explanation:
1. Use \( x\to2\pi-x \) and add: \( 2I=2\pi\int_0^{2\pi}\dfrac{\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx \)
2. That integral equals \( 4\int_0^{\pi/2}\cdots=4\cdot\pi/4=\pi \)
3. So \( I=\pi\cdot\pi=\pi^2 \).

Teacher's Note:
1. Combine the x-weight trick with the sin/cos symmetry.

Answer: (b) \( \log_e 2 \)

Explanation:
1. For x \( \gt \) 0, \( 2e^{-x}\lt 2 \), so the bracket is 1 when \( 2e^{-x}\ge 1 \), i.e. \( x\le\ln2 \), and 0 afterwards
2. The integral is the length of [0, ln 2] = ln 2

Teacher's Note:
1. Greatest integer function: find where the value changes.

Answer: (d) \( 17/3 \)

Explanation:
1. x(x-1) is positive for x \( \lt \) 0 and x \( \gt \) 1, negative in (0, 1)
2. From -2 to 0: 14/3. From 0 to 1: 1/6. From 1 to 2: 5/6
3. Total = 14/3 + 1 = 17/3

Teacher's Note:
1. Split at the roots 0 and 1.

Answer: (a) 0

Explanation:
1. For x in (-2, 0), [x/pi] = -1 so the denominator is -1/2. For x in (0, 2), [x/pi] = 0 so it is 1/2
2. The integrand is therefore odd, since the numerator is even
3. The integral over a symmetric interval is 0

Teacher's Note:
1. Check whether the integrand is odd.

Answer: (c) \( -\pi/2 \)

Explanation:
1. On (0, pi/2], cos x lies in [0, 1), so [cos x] = 0
2. On (pi/2, pi), cos x lies in (-1, 0), so [cos x] = -1
3. Integral = -(pi - pi/2) = -pi/2

Teacher's Note:
1. Find where cos x is above and below 0.

Answer: (d) 2 : 1

Explanation:
1. Integrating by parts gives the known result \( a_n=\dfrac{\pi}{2^{n+1}} \)
2. So \( a_n/a_{n+1}=2 \), i.e. 2 : 1

Teacher's Note:
1. Reduction formula for the product of powers of cos and cos nx.

Answer: (b) \( 1/2 \)

Explanation:
1. Differentiate both sides: \( f(x)=1-xf(x) \)
2. So \( f(x)=\dfrac{1}{1+x} \) and \( f(1)=1/2 \)

Teacher's Note:
1. Differentiate under the integral using the fundamental theorem.

Answer: (a) 9/2

Explanation:
1. On [-1, 0] the maximum is 2 - x. On [0, 1] the maximum is 2
2. Integral = 2 + 1/2 + 2 = 9/2

Teacher's Note:
1. Compare the three expressions on each sub-interval.

Answer: (c) \( 1/2 \)

Explanation:
1. Replace x by \( 1-x \): \( x(1-x) \) is unchanged
2. So \( I_1=\int(1-x)f\,dx \), and adding gives \( 2I_1=I_2 \)
3. Hence \( I_1/I_2=1/2 \)

Teacher's Note:
1. Use the symmetry of the limits about 1/2.

Answer: (b) \( -n\pi \)

Explanation:
1. Over each full period, [sin x] = 0 on (0, pi) and -1 on (pi, 2pi), giving -pi
2. n periods give -n pi
3. On (2n pi, 2n pi + pi), [sin x] = 0, so nothing more is added

Teacher's Note:
1. Count full periods first.

Answer: (d) \( \pi/2 \)

Explanation:
1. Replace x by \( \pi-x \): \( \cos x\to-\cos x \)
2. Adding, \( 2I=\int_0^\pi\left(\dfrac{1}{1+3^{c}}+\dfrac{3^{c}}{3^{c}+1}\right)dx=\pi \)
3. So \( I=\pi/2 \)

Teacher's Note:
1. The two forms add to 1.

Answer: (a) \( -\pi/2 \)

Explanation:
1. [2 sin x] = 1 on [pi/2, 5pi/6], 0 on (5pi/6, pi], -1 on (pi, 7pi/6] and -2 on (7pi/6, 3pi/2)
2. Integral = pi/3 - pi/6 - 2(pi/3) = -pi/2

Teacher's Note:
1. Find the sub-intervals where 2 sin x crosses integers.

Answer: (c) \( \pi \)

Explanation:
1. Replace x by \( 2\pi-x \): sin x changes sign
2. Adding the two forms gives 1, so \( 2I=2\pi \) and \( I=\pi \)

Teacher's Note:
1. Use the a + b - x property.

Answer: (d) \( 1-\dfrac{1}{\sqrt2} \)

Explanation:
1. On (0, pi/4), [x] = 0, so d(x - [x]) = dx
2. I = integral of sin x from 0 to pi/4 = 1 - cos(pi/4) = 1 - 1/sqrt2

Teacher's Note:
1. The greatest integer function is constant on this interval.

Answer: (b) \( \tfrac13(\alpha^3+\beta^3) \)

Explanation:
1. For x \( \lt \) 0, x
2. x
3. = -x^2, so the part from alpha to 0 is \( \alpha^3/3 \)
4. For x \( \gt \) 0 the part from 0 to beta is \( \beta^3/3 \)
5. Sum = (alpha^3 + beta^3)/3

Teacher's Note:
1. Split at 0 because of the modulus.

Answer: (a) \( \dfrac{\pi}{8}\log_e 2 \)

Explanation:
1. Replace x by \( \pi/4-x \): \( 1+\tan(\pi/4-x)=\dfrac{2}{1+\tan x} \)
2. So \( 2I=\int_0^{\pi/4}\log2\,dx=\dfrac{\pi}{4}\log2 \)
3. \( I=\dfrac{\pi}{8}\log2 \)

Teacher's Note:
1. A standard integral using x to a - x.