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NEET-UG Biology

Integral as the limit of a sum — practice questions

16 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · difficulty L3 · understanding

    Let [][\cdot] denote the greatest integer function and f(x)=limn1n3k=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 12j=1f(i)12 \sum\limits_{j=1}^{\infty} f(i) is equal to ________.

  2. Question 2 · difficulty L3 · understanding

     If limn(n+1)k1nk+1[(nk+1)+(nk+2)++(nk+n)]=33limn1nk+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, limn2n(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(233)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

    limn3n{4+(2+1n)2+(2+2n)2++(31n)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=limnk=1n2nn2+k2a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} and f(x)=1cosx1+cosxf(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

    limn12n(1112n+1122n+1132n+...+112n12n)\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

    limnr=1nr2r27rn+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. loge(32){\log _e}\left( {{{\sqrt 3 } \over 2}} \right)
    • B. loge(334){\log _e}\left( {{{3\sqrt 3 } \over 4}} \right)
    • C. loge(274){\log _e}\left( {{{27} \over 4}} \right)
    • D. loge(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 limn(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}}}

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