NIMCET 2023 Question Paper with solutions

$\mathit{f}\left(\mathit{x}\right)={\mathit{x}}^3+3\mathit{x}-9\mathit{x}\in \left[-2,3\right]$
Differentiate with respect to $\mathit{x}$
${\mathit{f}}^{\prime }\left(\mathit{x}\right)=3{\mathit{x}}^2+3$
Hence $\mathit{f}\left(\mathit{x}\right)$ is strictly increasing function so its greatest value will be at $\mathit{x}=3$
$\mathit{f}\left(3\right)=3^3+3\times 3-9=27$
$\frac{\mathit{a}}{1-\mathit{r}}=27$
$\Rightarrow \mathit{a}=27-27\mathit{r}$
$\Rightarrow \mathit{a}+27\mathit{r}=27$ ....(i)
${\mathit{f}}^{\prime }\left(0\right)=3$
Also given $\mathit{a}-\mathit{a}\mathit{r}={\mathit{f}}^{\prime }\left(0\right)$
$\Rightarrow \mathit{a}\left(1-\mathit{r}\right)=3$
$\Rightarrow 1-\mathit{r}=\frac{3}{\mathit{a}}$ ....(ii)
From equations, (i) and (ii) we get,
$\mathit{a}+27\left(1-\frac{3}{\mathit{a}}\right)=27$
$\mathit{a}+27-\frac{81}{\mathit{a}}=27$
$\Rightarrow {\mathit{a}}^2=81\Rightarrow \mathit{a}=\pm 9$
$\because$ G.P. is decreasing
$\therefore \mathit{a}=9$
Now, $\frac{9}{1-\mathit{r}}=27\Rightarrow \frac{1}{3}=1-\mathit{r}$
$\mathit{r}=1-\frac{1}{3}$
$\mathit{r}=\frac{2}{3}.$