PYQs Homework Notes Mocks Board Daily Homework JEE Mains Maths Question type Limit of a sum evaluated as a definite integral This type of question, asked 6 times: 2021 Paper 5, 14 Jun 2022, 26 Jul 2022, 30 Jan 2023, 30 Jan 2024, 9 Apr 2024. All 6 questions below, oldest first.
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Question 1 2021 Paper 5
82 Let f : ( 0 , 2 ) → R f:(0,2)\rightarrow\mathbb{R} f : ( 0 , 2 ) → R be defined as f ( x ) = log 2 ( 1 + tan ( π x 4 ) ) f(x)=\log_2\left(1+\tan\left(\frac{\pi x}{4}\right)\right) f ( x ) = log 2 ( 1 + tan ( 4 π x ) ) . Then lim n → ∞ 2 n ( f ( 1 n ) + f ( 2 n ) + … + f ( n n ) ) \lim_{n\rightarrow\infty}\frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\ldots+f\left(\frac{n}{n}\right)\right) lim n → ∞ n 2 ( f ( n 1 ) + f ( n 2 ) + … + f ( n n ) ) is equal to ______.
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Question 2 14 Jun 2022
157 lim n → ∞ ( n 2 ( n 2 + 1 ) ( n + 1 ) + n 2 ( n 2 + 4 ) ( n + 2 ) + n 2 ( n 2 + 9 ) ( n + 3 ) + … + n 2 ( n 2 + n 2 ) ( n + n ) ) \lim_{n \rightarrow \infty} \left( \frac{n^{2}}{\left( n^{2}+ 1 \right)(n + 1)}+\frac{n^{2}}{\left( n^{2}+ 4 \right)(n + 2)}+\frac{n^{2}}{\left( n^{2}+ 9 \right)(n + 3)}+ \ldots +\frac{n^{2}}{\left( n^{2}+n^{2} \right)(n + n)} \right) lim n → ∞ ( ( n 2 + 1 ) ( n + 1 ) n 2 + ( n 2 + 4 ) ( n + 2 ) n 2 + ( n 2 + 9 ) ( n + 3 ) n 2 + … + ( n 2 + n 2 ) ( n + n ) n 2 ) is equal to A. π 8 + 1 4 log e 2 \frac{\pi}{8} + \frac{1}{4}\log_{e}2 8 π + 4 1 log e 2 B. π 4 + 1 8 log e 2 \frac{\pi}{4} + \frac{1}{8}\log_{e}2 4 π + 8 1 log e 2 C. π 4 − 1 8 log e 2 \frac{\pi}{4} - \frac{1}{8}\log_{e}2 4 π − 8 1 log e 2 D. π 8 + log e 2 \frac{\pi}{8} + \log_{e}\sqrt{2} 8 π + log e 2 Tap an option to check your answer.
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Question 3 26 Jul 2022
70 If a = lim n → ∞ ∑ k = 1 n 2 n n 2 + k 2 a =\lim_{n \rightarrow \infty} \sum_{k = 1}^{n} \frac{2n}{n^{2}+k^{2}} a = lim n → ∞ ∑ k = 1 n n 2 + k 2 2 n and f ( x ) = 1 − cos x 1 + cos x , x ∈ ( 0 , 1 ) f\left( x \right)=\sqrt{\frac{1 - \cos x}{1 + \cos x}},x \in (0,1) f ( x ) = 1 + c o s x 1 − c o s x , x ∈ ( 0 , 1 ) , then: A. 2 2 f ( a 2 ) = f ′ ( a 2 ) 2\sqrt{2}f\left( \frac{a}{2} \right) = f'\left( \frac{a}{2} \right) 2 2 f ( 2 a ) = f ′ ( 2 a ) B. f ( a 2 ) f ′ ( a 2 ) = 2 f\left( \frac{a}{2} \right)f'\left( \frac{a}{2} \right) = \sqrt{2} f ( 2 a ) f ′ ( 2 a ) = 2 C. 2 f ( a 2 ) = f ′ ( a 2 ) \sqrt{2}f\left( \frac{a}{2} \right) = f'\left( \frac{a}{2} \right) 2 f ( 2 a ) = f ′ ( 2 a ) D. f ( a 2 ) = 2 f ′ ( a 2 ) f\left( \frac{a}{2} \right) = \sqrt{2}f'\left( \frac{a}{2} \right) f ( 2 a ) = 2 f ′ ( 2 a ) Tap an option to check your answer.
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Question 4 30 Jan 2023
163 lim n → ∞ 3 n { 4 + ( 2 + 1 n ) 2 + ( 2 + 2 n ) 2 + ⋯ + ( 3 − 1 n ) 2 } \lim_{n \rightarrow \infty} \frac{3}{n}\left\{ 4 +\left( 2 +\frac{1}{n} \right)^{2}+\left( 2 +\frac{2}{n} \right)^{2}+ \cdots +\left( 3 -\frac{1}{n} \right)^{2} \right\} lim n → ∞ n 3 { 4 + ( 2 + n 1 ) 2 + ( 2 + n 2 ) 2 + ⋯ + ( 3 − n 1 ) 2 } is equal to A. 12
B. C. 0
D. 19
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Question 5 30 Jan 2024
66 The value of lim n → ∞ ∑ k = 1 n n 3 ( n 2 + k 2 ) ( n 2 + 3 k 2 ) \lim_{n \rightarrow \infty} \sum_{k = 1}^{n} \frac{n^{3}}{\left( n^{2}+k^{2} \right)\left( n^{2}+ 3k^{2} \right)} lim n → ∞ ∑ k = 1 n ( n 2 + k 2 ) ( n 2 + 3 k 2 ) n 3 is: A. ( 2 3 + 3 ) π 24 \frac{(2\sqrt{3} + 3)\pi}{24} 24 ( 2 3 + 3 ) π B. 13 π 8 ( 4 3 + 3 ) \frac{13\pi}{8(4\sqrt{3} + 3)} 8 ( 4 3 + 3 ) 13 π C. 13 ( 2 3 − 3 ) π 8 \frac{13(2\sqrt{3} - 3)\pi}{8} 8 13 ( 2 3 − 3 ) π D. π 8 ( 2 3 + 3 ) \frac{\pi}{8(2\sqrt{3} + 3)} 8 ( 2 3 + 3 ) π Tap an option to check your answer.
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Question 6 9 Apr 2024
84 Let lim n → ∞ ( n n 4 + 1 − 2 n ( n 2 + 1 ) n 4 + 1 + n n 4 + 16 − 8 n ( n 2 + 4 ) n 4 + 16 \lim_{n \rightarrow \infty} \left( \frac{n}{\sqrt{n^{4}+ 1}}-\frac{2n}{\left( n^{2}+ 1 \right)\sqrt{n^{4}+ 1}}+\frac{n}{\sqrt{n^{4}+ 16}}-\frac{8n}{\left( n^{2}+ 4 \right)\sqrt{n^{4}+ 16}} \right. n → ∞ lim ( n 4 + 1 n − ( n 2 + 1 ) n 4 + 1 2 n + n 4 + 16 n − ( n 2 + 4 ) n 4 + 16 8 n + … … + n n 4 + n 4 − 2 n ⋅ n 2 ( n 2 + n 2 ) n 4 + n 4 + \ldots\ldots +\frac{n}{\sqrt{n^{4}+n^{4}}}-\frac{2n \cdot n^{2}}{\left( n^{2}+n^{2} \right)\sqrt{n^{4}+n^{4}}} + …… + n 4 + n 4 n − ( n 2 + n 2 ) n 4 + n 4 2 n ⋅ n 2 be π k \frac{\pi}{k} k π , using only the principal values of the inverse trigonometric functions. Then k 2 k^{2} k 2 is equal to............
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