Q116 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics

Given that $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. If $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=x^2+2$, then $$ \lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]= $$

  • Correct answer
    A
    55
  • B
    57
  • C
    104
  • D
    7

Solution

Given expression is $$ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p}+\left(\frac{r}{n}\right)=\int_0^p+(x) d x \text {. } $$ Take, $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{x}\right)+f\left(\frac{14}{x}\right)+f\left(\frac{21}{n}\right)+\ldots+f\left(\frac{7 n}{n}\right)\right]$ $$ \begin{aligned} \Rightarrow & 3 \lim _{x \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{7 \times \frac{n}{7}} \mathrm{f}\left(\frac{7 r}{n}\right)=3 \lim _{x \rightarrow \infty} \frac{1}{7 \times\left(\frac{n}{7}\right)} \sum_{r=1}^{7 \times\left(\frac{n}{7}\right)} \mathrm{f}\left(\frac{\mathrm{r}}{\left(\frac{\mathrm{n}}{7}\right)}\right) \\ & \text { Here, } \mathrm{p}=7 \end{aligned} $$ Here, $\mathrm{p}=7$ $$ \begin{aligned} & \text { So, } \frac{3}{7} \int_0^{\mathrm{p}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{3}{7} \int_0^7\left(\mathrm{x}^2+2\right) \mathrm{dx} \\ & =\frac{3}{7}\left[\frac{\mathrm{x}^3}{3}+2 \mathrm{x}\right]_0^7=\frac{3}{7}\left[\frac{7 \times 7 \times 7}{3}+7 \times 2\right] \\ & =\frac{3}{7}\left[\frac{49 \times 7+7 \times 6}{3}\right]=\frac{3}{7} \times \frac{7}{3}[49+6]=55 \end{aligned} $$ So, option (a) is correct.

Hence, the correct answer is option: 'A'.

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