- ✓$1: 1$
- B$1: 2$
- C$2: 1$
- D$\sqrt{2}: 1$
Surface area of solid i.e. $\quad S=2 \pi r^2+2 \pi r h+\pi r^2$
$S=3 \pi r^2+2 \pi r h$
$h=\frac{S-3 \pi r^2}{2 \pi r}$
Volume of solid
$\text { i.e. } \quad V=\pi r^2 h+\frac{2}{3} \pi r^3$
$\Rightarrow \quad V=\pi r^2\left(\frac{S-3 \pi r^2}{2 \pi r}\right)+\frac{2}{3} \pi r^3$
$\Rightarrow \quad \quad V=\frac{1}{2}\left(S r-3 \pi r^3\right)+\frac{2}{3} \pi r^3$
$\Rightarrow \quad \frac{d V}{d r}=\frac{S}{2}-\frac{9 \pi r^2}{2}+2 \pi r^2$
$\text { For maximum or minimum } \frac{d V}{d r}=0$
$\therefore \quad S=5 \pi r^2$
$\Rightarrow \quad h=\frac{S-3 \pi r^2-5 \pi r^2-3 \pi r^2}{2 \pi r}=r$
$\therefore \quad h=r \Rightarrow h: r=1: 1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Match the conditions / expressions in Column $I$ with statements in Column $II$ and indicate your answers by darkening the appropriate bubbles in $4 \times 4$ matrix given in the $ORS$.
| Column $I$ | Column $II$ |
| $(A)$ If $-1 < x < 1$, then $f$ ( $x$ ) satisfies | $(p)$ $ 0 < $ f (x) $ < 1$ |
| $(B)$ If $1 < x < 2$, then $f(x)$ satisfies | $(q)$ $\mathrm{f}(\mathrm{x}) < 0$ |
| $(C)$ If $3 < x < 5$, then $f(x)$ satisfies | $(r)$ $ \mathrm{f}(\mathrm{x}) > 0$ |
| $(D)$ If $x > 5$, then $f(x)$ satisfies | $(s)$ $ f (\mathrm{x}) < 1$ |
$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$