Skip to content

JEE Main Mathematics · Class 12

Integral as the limit of a sum — NEET & JEE Questions

Integral as the limit of a sum is a topic in Integral Calculus, a Class 12 Mathematics chapter of the JEE Main syllabus. TestBudy holds 16 published JEE Main questions on Integral as the limit of a sum, each with the marking scheme of the real paper.

questions
16
shown here
10
Test me on this topic

Five free tests with an account. Your score comes with the result; where you stand is Pro.

Integral as the limit of a sum MCQs with solutions

10 of the 16 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L3 · understanding

    Let [⋅][\cdot] denote the greatest integer function and f(x)=lim⁡n→∞1n3∑k=1n[k23x]f(x) = \lim\limits_{n \to \infty} \frac{1}{n^{3}} \sum\limits_{k=1}^n \left[ \frac{k^2}{3^x} \right]. Then 12∑j=1∞f(i)12 \sum\limits_{j=1}^{\infty} f(i) is equal to ________.

  2. Question 2 · difficulty L3 · understanding

     If lim⁡n→∞(n+1)k−1nk+1[(nk+1)+(nk+2)+…+(nk+n)]=33⋅lim⁡n→∞1nk+1⋅[1k+2k+3k+…+nk] \begin{aligned} &\text { If } \lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)] \\ &=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right] \end{aligned}, then the integral value of k\mathrm{k} is equal to _____________

  3. Question 3 · difficulty L3 · understanding

    Let f : (0, 2) → \to R be defined as f(x) = log 2 (1+tan⁡(πx4))\left( {1 + \tan \left( {{{\pi x} \over 4}} \right)} \right). Then, lim⁡n→∞2n(f(1n)+f(2n)+...+f(1))\mathop {\lim }\limits_{n \to \infty } {2 \over n}\left( {f\left( {{1 \over n}} \right) + f\left( {{2 \over n}} \right) + ... + f(1)} \right) is equal to ___________.

  4. Question 4 · difficulty L3 · understanding

    The value of \lim _\limits{n \rightarrow \infty} \sum_\limits{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)} is :

    • A. π8(23+3)\frac{\pi}{8(2 \sqrt{3}+3)}
    • B. (23+3)π24\frac{(2 \sqrt{3}+3) \pi}{24}
    • C. 13π8(43+3)\frac{13 \pi}{8(4 \sqrt{3}+3)}
    • D. 13(23−3)π8\frac{13(2 \sqrt{3}-3) \pi}{8}
  5. Question 5 · difficulty L3 · understanding

    Among (S1): \lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1 (S2) : \lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}

    • A. Only (S1) is true
    • B. Both (S1) and (S2) are true
    • C. Both (S1) and (S2) are false
    • D. Only (S2) is true
  6. Question 6 · difficulty L3 · understanding

    lim⁡n→∞3n{4+(2+1n)2+(2+2n)2+…+(3−1n)2}\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\} is equal to :

    • A. 0
    • B. 193\frac{19}{3}
    • C. 19
    • D. 12
  7. Question 7 · difficulty L3 · understanding

    If a=lim⁡n→∞∑k=1n2nn2+k2a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} and f(x)=1−cos⁡x1+cos⁡xf(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} , x∈(0,1)x \in (0,1), then :

    • A. 22f(a2)=f′(a2)2\sqrt 2 f\left( {{a \over 2}} \right) = f'\left( {{a \over 2}} \right)
    • B. f(a2)f′(a2)=2f\left( {{a \over 2}} \right)f'\left( {{a \over 2}} \right) = \sqrt 2
    • C. 2f(a2)=f′(a2)\sqrt 2 f\left( {{a \over 2}} \right) = f'\left( {{a \over 2}} \right)
    • D. f(a2)=2f′(a2)f\left( {{a \over 2}} \right) = \sqrt 2 f'\left( {{a \over 2}} \right)
  8. Question 8 · difficulty L3 · understanding

    lim⁡n→∞12n(11−12n+11−22n+11−32n+  ...  +  11−2n−12n)\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right) is equal to

    • A. 12\frac{1}{2}
    • B. 1
    • C. 2
    • D. −-2
  9. Question 9 · difficulty L3 · understanding

    lim⁡n→∞∑r=1nr2r2−7rn+6n2\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} is equal to :

    • A. log⁡e(32){\log _e}\left( {{{\sqrt 3 } \over 2}} \right)
    • B. log⁡e(334){\log _e}\left( {{{3\sqrt 3 } \over 4}} \right)
    • C. log⁡e(274){\log _e}\left( {{{27} \over 4}} \right)
    • D. log⁡e(43){\log _e}\left( {{4 \over 3}} \right)
  10. Question 10 · difficulty L3 · understanding

    If Un=(1+1n2)(1+22n2)2.....(1+n2n2)n{U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n, then lim⁡n→∞(Un)−4n2\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}} is equal to :

    • A. e216{{{e^2}} \over {16}}
    • B. 4e{4 \over e}
    • C. 16e2{{16} \over {{e^2}}}
    • D. 4e2{4 \over {{e^2}}}

Practise this chapter

Reading a question and answering it under a clock are different things. A chapter-length test measures whether the topic went in.