MCQ
$\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to
- A$12$
- B$\frac{19}{3}$
- C$0$
- ✓$19$
$=3 \int \limits_0^1(2+x)^2 d x=27-8=19$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$
$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$
$\mathrm{F}(\mathrm{x})=\int\limits_{1}^{\mathrm{x}} \mathrm{t}^{2} \mathrm{g}(\mathrm{t}) \mathrm{dt},$ where $\mathrm{g}(\mathrm{t})=\int\limits_{1}^{\mathrm{t}} \mathrm{f}(\mathrm{u}) \mathrm{du}$
Then for the function $\mathrm{F}$, the point $\mathrm{x}=1$ is