JEE Mathematics Definite Integration MCQs Set 07

Practice MCQs for JEE Mathematics Definite Integration

Review structured MCQ sets for JEE Mathematics Definite Integration. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: \( \lim_{n\to\infty}\sum_{r=1}^n\frac1{\sqrt{nr}} \) is equal to
(a) 2
(b) 1
(c) 0
(d) none of these
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. Write the sum as \( \frac1n\sum_{r=1}^n\frac1{\sqrt{r/n}} \).
2. As n → ∞ it becomes \( \int_0^1x^{-1/2}dx=2 \).

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Question 2: \( \lim_{n\to\infty}\sum_{r=0}^{n-1}\frac1{\sqrt{n^2-r^2}} \) is
(a) π
(b) \( \frac\pi2 \)
(c) \( \frac\pi4 \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac\pi2 \)

Explanation:
1. Factor n: \( \frac1n\sum\frac1{\sqrt{1-(r/n)^2}} \).
2. The limit is \( \int_0^1\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}1=\frac\pi2 \).

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Question 3: \( \lim_{n\to\infty}\sum_{r=1}^n\frac1ne^{r/n} \) is
(a) e
(b) e − 1
(c) 1 − e
(d) e + 1
Show Answer & Explanation

Answer: (b) e − 1

Explanation:
1. The limit is \( \int_0^1e^xdx=e-1 \).

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Question 4: \( \lim_{n\to\infty}\sum_{r=1}^n\frac1n\sin\frac{r\pi}{2n} \) is
(a) \( \frac\pi2 \)
(b) 2
(c) \( \frac2\pi \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac2\pi \)

Explanation:
1. The limit is \( \int_0^1\sin\frac{\pi x}2dx=\left[-\frac2\pi\cos\frac{\pi x}2\right]_0^1=\frac2\pi \).

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Question 5: \( \lim_{n\to\infty}\sum_{r=1}^{4n}\frac1{n+r} \) is
(a) \( \log_e5 \)
(b) 0
(c) \( \log_e4 \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \log_e5 \)

Explanation:
1. Write \( \frac1{n+r}=\frac1n\cdot\frac1{1+r/n} \) with r/n ranging up to 4.
2. The limit is \( \int_0^4\frac{dx}{1+x}=\ln5 \).

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Question 6: \( \lim_{n\to\infty}\frac1n\sum_{r=1}^{2n}\frac r{\sqrt{n^2+r^2}} \) equals
(a) \( 1+\sqrt5 \)
(b) \( -1+\sqrt5 \)
(c) \( -1+\sqrt2 \)
(d) \( 1+\sqrt2 \)
Show Answer & Explanation

Answer: (b) \( -1+\sqrt5 \)

Explanation:
1. Write \( \frac r{\sqrt{n^2+r^2}}=\frac{r/n}{\sqrt{1+(r/n)^2}} \).
2. The limit is \( \int_0^2\frac{x\,dx}{\sqrt{1+x^2}}=\sqrt{1+x^2}\Big|_0^2=\sqrt5-1 \).

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Question 7: \( \lim_{n\to\infty}\left\{\frac{n!}{(kn)^n}\right\}^{1/n} \), where k ≠ 0 is a constant and n ∈ N, is equal to
(a) ke
(b) \( k^{-1}\cdot e \)
(c) \( ke^{-1} \)
(d) \( k^{-1}\cdot e^{-1} \)
Show Answer & Explanation

Answer: (d) \( k^{-1}\cdot e^{-1} \)

Explanation:
1. Let L be the limit: \( \ln L=\lim\frac1n\sum_{r=1}^n\ln\frac{r}{kn}=\int_0^1\ln\frac xkdx=\int_0^1\ln x\,dx-\ln k=-1-\ln k \).
2. So L = 1/(ke).

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Question 8: \( \lim_{n\to\infty}\frac{2^k+4^k+6^k+\cdots+(2n)^k}{n^{k+1}} \), k ≠ −1, is equal to
(a) \( 2^k \)
(b) \( \frac{2^k}{k+1} \)
(c) \( \frac1{k+1} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac{2^k}{k+1} \)

Explanation:
1. The numerator is \( 2^k\sum r^k \), so the expression is \( 2^k\cdot\frac1n\sum\left(\frac rn\right)^k \).
2. The limit is \( 2^k\int_0^1x^kdx=\frac{2^k}{k+1} \).

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Question 9: Let (a, b) and (λ, μ) be two points on the curve y = f(x). If the slope of the tangent to the curve at (x, y) is φ(x), then \( \int_a^\lambda\varphi(x)\,dx \) is
(a) λ − a
(b) μ − b
(c) λ + μ − a − b
(d) none of these
Show Answer & Explanation

Answer: (b) μ − b

Explanation:
1. φ(x) = f′(x), so the integral is f(λ) − f(a) = μ − b.

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Question 10: If \( g(x)=\int_0^x\cos^4t\,dt \), then g(x + π) equals
(a) g(x) + g(π)
(b) g(x) − g(π)
(c) g(x) g(π)
(d) g(x)/g(π)
Show Answer & Explanation

Answer: (a) g(x) + g(π)

Explanation:
1. \( g(x+\pi)=g(\pi)+\int_\pi^{x+\pi}\cos^4t\,dt \).
2. Since cos⁴t has period π, the last integral equals g(x). So g(x + π) = g(x) + g(π).

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Question 11: If \( af(x)+bf\left(\frac1x\right)=\frac1x-5 \), x ≠ 0, a ≠ b, then \( \int_1^2f(x)\,dx \) equals
(a) \( \frac{(\log2-5)a+\frac{13}2b}{a^2-b^2} \)
(b) \( \frac{(\log2-5)a+\frac72b}{a^2-b^2} \)
(c) \( \frac{(5-\log2)a+\frac72b}{a^2-b^2} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac{(\log2-5)a+\frac72b}{a^2-b^2} \)

Explanation:
1. Replace x by 1/x: \( af(1/x)+bf(x)=x-5 \). Eliminating f(1/x): \( (a^2-b^2)f(x)=a\left(\frac1x-5\right)-b(x-5) \).
2. Integrate from 1 to 2: \( \int_1^2\left(\frac1x-5\right)dx=\ln2-5 \) and \( \int_1^2(x-5)dx=-\frac72 \).
3. So the integral is \( \frac{a(\ln2-5)+\frac72b}{a^2-b^2} \).

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Question 12: \( \int_0^2\sqrt{\frac{2+x}{2-x}}\,dx \) is equal to
(a) π + 1
(b) \( 1+\frac\pi2 \)
(c) \( \pi+\frac32 \)
(d) none of these
Show Answer & Explanation

Answer: (d) none of these

Explanation:
1. Write \( \sqrt{\frac{2+x}{2-x}}=\frac{2+x}{\sqrt{4-x^2}} \).
2. The integral is \( 2\sin^{-1}\frac x2-\sqrt{4-x^2}\Big|_0^2=\pi+2 \), which is not among (a), (b), (c).

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Question 13: If \( a_n=\int_0^{\pi/2}\frac{\sin^2nx}{\sin x}dx \), then \( a_2-a_1,\ a_3-a_2,\ a_4-a_3,\ldots \) are in
(a) AP
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(d) none of these
Show Answer & Explanation

Answer: (c) HP

Explanation:
1. \( a_{n+1}-a_n=\int_0^{\pi/2}\frac{\sin^2(n+1)x-\sin^2nx}{\sin x}dx=\int_0^{\pi/2}\sin(2n+1)x\,dx=\frac1{2n+1} \) (using sin²A − sin²B = sin(A + B) sin(A − B)).
2. The differences are 1/3, 1/5, 1/7, ..., whose reciprocals 3, 5, 7, ... are in AP, so the differences are in HP.

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Multiple Choice Questions (MCQs) for JEE Mathematics Definite Integration

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