JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 06

Practice MCQs for JEE Mathematics Properties and Applications of Definite Integrals

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Multiple Choice Questions

Question 1: \( \int_{-1}^{1}(x-[2x])\,dx \) is equal to
(a) 0
(b) 2
(c) 1
(d) 4
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. The integral of x over [-1, 1] is 0
2. [2x] is -2, -1, 0, 1 on [-1,-1/2), [-1/2,0), [0,1/2), [1/2,1), giving -1 - 1/2 + 0 + 1/2 = -1
3. Result = 0 - (-1) = 1

Teacher's Note:
1. Break [2x] into steps of width 1/2.

Question 2: If f(x) = |x| + 1 for \( -1\le x\lt0 \) and f(x) = 1 + x^2 for \( 0\le x\le1 \), then \( \int_{-1}^{1}f(x)\,dx \) is equal to
(a) \( -1/6 \)
(b) \( 17/6 \)
(c) \( -17/6 \)
(d) \( 5/6 \)
Show Answer & Explanation

Answer: (b) \( 17/6 \)

Explanation:
1. From -1 to 0: integral of (1 - x) = 1 + 1/2 = 3/2
2. From 0 to 1: integral of (1 + x^2) = 4/3
3. Total = 3/2 + 4/3 = 17/6

Teacher's Note:
1. Integrate each branch separately.

Question 3: Let f(x) = maximum {x + |x|, -[x]}, where [x] is the greatest integer not exceeding x. Then \( \int_{-2}^{2}f(x)\,dx \) is equal to
(a) 3
(b) 2
(c) 1
(d) 7
Show Answer & Explanation

Answer: (d) 7

Explanation:
1. On [-2, -1): -[x] = 2 and x +
2. x
3. = 0, so f = 2. On [-1, 0): f = 1
4. On [0, 2]: f = 2x (since -[x] is at most 0 on [0,1) and negative later)
5. Integral = 2 + 1 + 4 = 7

Teacher's Note:
1. Evaluate each interval for both expressions.

Question 4: \( \int_0^3|x^3-3x^2+2x|\,dx \) is equal to
(a) \( 11/4 \)
(b) \( 3/4 \)
(c) \( 7/4 \)
(d) \( 13/4 \)
Show Answer & Explanation

Answer: (a) \( 11/4 \)

Explanation:
1. x^3 - 3x^2 + 2x = x(x - 1)(x - 2), positive on (0,1), negative on (1,2), positive on (2,3)
2. The areas are 1/4, 1/4 and 9/4
3. Total = 11/4

Teacher's Note:
1. Split at the roots 0, 1, 2.

Question 5: Let f be continuous with f(a - x) + f(x) = 0 on [0, a]. Then \( \int_0^a\dfrac{dx}{1+e^{f(x)}} \) is equal to
(a) a
(b) f(a)
(c) \( a/2 \)
(d) \( f(a)/2 \)
Show Answer & Explanation

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

Explanation:
1. Replacing x by a - x gives f(a - x) = -f(x)
2. Adding the two forms, the integrands sum to 1, so 2I = a
3. I = a/2

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

Question 6: \( \int_{-a}^{a}\log_e\left(x+\sqrt{1+x^2}\right)dx \) is equal to
(a) \( 2\log_e a \)
(b) \( \log_e2+\log a \)
(c) a
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. Let g(x) = log(x + sqrt(1 + x^2)). Then g(-x) = log(sqrt(1 + x^2) - x) = -g(x)
2. The function is odd, so the integral is 0

Teacher's Note:
1. This is the inverse hyperbolic sine, an odd function.

Question 7: Let \( f(x)=\dfrac{e^x+1}{e^x-1} \) and \( \int_0^1\dfrac{e^x+1}{e^x-1}x\,dx=\lambda \). Then \( \int_{-1}^{1}tf(t)\,dt \) is equal to
(a) 0
(b) \( 2\lambda \)
(c) \( \lambda \)
(d) \( 4\lambda \)
Show Answer & Explanation

Answer: (b) \( 2\lambda \)

Explanation:
1. f(-x) = -f(x), so f is odd
2. Then t f(t) is even
3. The integral over [-1, 1] is twice that over [0, 1], i.e. 2 lambda

Teacher's Note:
1. Odd times odd is even.

Question 8: Let f be continuous with f(x) + f(y) = f(x + y) for all real x, y. If \( \int_0^3 f(x)\,dx=k \), then \( \int_{-3}^{3}f(x)\,dx \) is equal to
(a) 0
(b) 2k
(c) \( k/2 \)
(d) -2k
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. The functional equation gives f(0) = 0 and f(-x) = -f(x), so f is odd
2. The integral over a symmetric interval is 0

Teacher's Note:
1. Additive functions are odd.

Question 9: Let \( I=\int_{-a}^{a}(p\tan^3x+q\cos^2x+r\sin x)\,dx \), where p, q, r are arbitrary constants. The value of I depends on
(a) p, q and r
(b) q, r and a
(c) q and a
(d) p, r and a
Show Answer & Explanation

Answer: (c) q and a

Explanation:
1. tan^3 x and sin x are odd and integrate to 0
2. cos^2 x is even, giving 2q times the integral of cos^2 x from 0 to a
3. So I depends on q and a

Teacher's Note:
1. Drop the odd terms.

Question 10: \( \int_0^{100\pi}\sqrt{1+\cos 2x}\,dx \) is equal to
(a) 0
(b) \( 200\sqrt2 \)
(c) \( 100\sqrt2 \)
(d) 100
Show Answer & Explanation

Answer: (b) \( 200\sqrt2 \)

Explanation:
1. sqrt(1 + cos 2x) = sqrt2
2. cos x
3.
4. The integral of
5. cos x
6. over one period of pi is 2, and there are 100 such periods
7. I = 200 sqrt2

Teacher's Note:
1. Use the half-angle identity and periodicity.

Question 11: If \( \int_0^a f(x)dx=\lambda \) and \( \int_0^a f(2a-x)dx=\mu \), then \( \int_0^{2a}f(x)dx \) is equal to
(a) \( \lambda-\mu \)
(b) \( 2\lambda-\mu \)
(c) \( \lambda-2\mu \)
(d) \( \lambda+\mu \)
Show Answer & Explanation

Answer: (d) \( \lambda+\mu \)

Explanation:
1. Split: \( \int_0^{2a}f=\int_0^af+\int_a^{2a}f \)
2. In the second part put x = 2a - t: it becomes \( \int_0^af(2a-t)dt=\mu \)
3. Total = lambda + mu

Teacher's Note:
1. Split at a and substitute in the second half.

Question 12: If f(-x) + f(x) = 0, then \( \int_0^x f(t)\,dt \) is
(a) an even function
(b) an odd function
(c) a periodic function
(d) none of these
Show Answer & Explanation

Answer: (a) an even function

Explanation:
1. f is odd
2. Let F(x) be the integral. F(-x) = integral from 0 to -x of f(t)dt; put t = -u: this equals the integral from 0 to x of f(-u)(-du) = F(x)
3. So F is even

Teacher's Note:
1. The integral of an odd function is even.

Question 13: Let f and g be continuous on [0, a] with f(x) = f(a - x) and g(x) + g(a - x) = 2. Then \( \int_0^a f(x)g(x)\,dx \) is equal to
(a) \( \int_0^ag(x)dx \)
(b) 2a
(c) \( \int_0^af(x)dx \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \int_0^af(x)dx \)

Explanation:
1. Replace x by a - x: \( I=\int_0^af(x)g(a-x)dx \)
2. Add: \( 2I=\int_0^af(x)\{g(x)+g(a-x)\}dx=2\int_0^af \)
3. So I is the integral of f

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

Question 14: Let f(x + k) = f(x) for all real x. Then \( \int_a^{a+k}f(x)\,dx \) depends for its value on
(a) a only
(b) both a and k
(c) neither a nor k
(d) k only
Show Answer & Explanation

Answer: (d) k only

Explanation:
1. For a function of period k, the integral over any interval of length k is the same
2. So the integral does not change with a

Teacher's Note:
1. Integral over one period is independent of the starting point.

Question 15: \( \int_{\pi/4}^{3\pi/4}\dfrac{x}{1+\sin x}dx \) is equal to
(a) \( (\sqrt2+1)\pi \)
(b) \( (\sqrt2-1)\pi \)
(c) \( \pi \)
(d) \( 2\pi \)
Show Answer & Explanation

Answer: (b) \( (\sqrt2-1)\pi \)

Explanation:
1. Put x to \( \pi-x \) and add: \( 2I=\pi\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1+\sin x} \)
2. An antiderivative of 1/(1 + sin x) is tan x - sec x, so the integral equals \( 2(\sqrt2-1) \)
3. \( I=(\sqrt2-1)\pi \)

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

Question 16: The value of \( \int_0^{\pi/2}\dfrac{dx}{1+\tan^3x} \) is
(a) \( \pi/4 \)
(b) \( \pi/2 \)
(c) \( \pi \)
(d) none of these
Show Answer & Explanation

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

Explanation:
1. Replace x by pi/2 - x: tan becomes cot, so the form becomes tan^3 x/(1 + tan^3 x)
2. Adding gives 1, so 2I = pi/2
3. I = pi/4

Teacher's Note:
1. King's property.

Question 17: If f and g are continuous, then \( \int_{-\pi/2}^{\pi/2}\{f(x)+f(-x)\}\{g(x)-g(-x)\}dx \) is equal to
(a) \( \pi \)
(b) 1
(c) 0
(d) -1
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. f(x) + f(-x) is even and g(x) - g(-x) is odd
2. The product is odd
3. The integral over a symmetric interval is 0

Teacher's Note:
1. Identify even and odd parts.

Question 18: Let \( \int_a^bf(x)dx=p \) and \( \int_a^b|f(x)|dx=q \). Then
(a) \( p>q \)
(b) \( |p|\le q \)
(c) \( p+q=0 \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( |p|\le q \)

Explanation:
1. Since \(
2. \int f
3. \le\int
4. f
5. \), we get
6. p
7. is at most q

Teacher's Note:
1. Triangle inequality for integrals.

Question 19: If f(2a - x) = f(x) and \( \int_0^af(x)dx=\lambda \), then \( \int_0^{2a}f(x)dx \) is
(a) \( \lambda \)
(b) 0
(c) none of these
(d) \( 2\lambda \)
Show Answer & Explanation

Answer: (d) \( 2\lambda \)

Explanation:
1. f is symmetric about x = a, so the integrals over [0, a] and [a, 2a] are equal
2. Total = 2 lambda

Teacher's Note:
1. Symmetric about the midpoint.

Question 20: The function \( f(x)=\int_1^x t(e^t-1)(t-2)^3(t-3)^5dt \) has a local minimum at x equal to
(a) 3
(b) 0
(c) 1
(d) 2
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. f'(x) = x(e^x - 1)(x - 2)^3 (x - 3)^5
2. Near x = 0 the factor x(e^x - 1) is non-negative and does not change sign, so there is no extremum
3. At x = 2, f' changes from + to -: a maximum. At x = 3, f' changes from - to +: a minimum

Teacher's Note:
1. Study the sign change of the integrand.

Question 21: The value of \( \int_0^{\pi}\dfrac{\sin nx}{\sin x}dx \), n a natural number, is
(a) \( \pi \) if n is even
(b) 0 if n is odd
(c) 0 if n is even
(d) \( \pi \) for all n
Show Answer & Explanation

Answer: (c) 0 if n is even

Explanation:
1. The result is pi for odd n and 0 for even n
2. Check n = 1 gives pi and n = 2: sin 2x/sin x = 2 cos x, which integrates to 0

Teacher's Note:
1. Test small values of n.

Question 22: Let \( f(x)=\int_0^x|t-1|\,dt \), x \( \ge \) 0. Consider f'(x): I. continuous at x = 1; II. continuous at x = 2; III. differentiable at x = 1; IV. differentiable at x = 2. Which are true?
(a) I and II only
(b) II and III only
(c) III and IV only
(d) I, II and IV
Show Answer & Explanation

Answer: (d) I, II and IV

Explanation:
1. f'(x) =
2. x - 1
3. by the fundamental theorem
4. It is continuous everywhere, so I and II hold
5.
6. x - 1
7. has a corner at x = 1 so III fails, but it is smooth at x = 2 so IV holds

Teacher's Note:
1. Corner points of a modulus function.

Question 23: Let f(a) \( \gt \) 0 and f be a non-decreasing continuous function on [a, b]. Then \( \frac{1}{b-a}\int_a^bf(x)dx \) has: I. maximum value f(b); II. minimum value f(a); III. maximum value b f(b); IV. minimum value f(a)/(b - a). Which are true?
(a) II and III
(b) I and II
(c) I and IV
(d) III and IV
Show Answer & Explanation

Answer: (b) I and II

Explanation:
1. The mean value lies between the minimum and maximum of f
2. For a non-decreasing function these are f(a) and f(b), so I and II are correct

Teacher's Note:
1. The average of a function lies between its extremes.

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