CAT 2018 Question Paper Shift 1 Solved Paper

$\mathit{f}\left(\mathit{x}\right)=\min \left(2{\mathit{x}}^2,52-5\mathit{x}\right)$
The maximum possible value of this function will be attained when $2{\mathit{x}}^2=52-5\mathit{x}$.
$\left(2\mathit{x}+13\right)\left(\mathit{x}-4\right)=0$
$\Rightarrow \mathit{x}=\frac{-13}{2}$ or $\mathit{x}=4$
Since $\mathit{x}$ has to be positive integer, we can discard the case $\mathit{x}=\frac{-13}{2}$.
$\mathit{x}=4$ is the point at which the function attains the maximum value.
putting $\mathit{x}=4$ in the original function, we get, $2{\mathit{x}}^2=2*4^2=32$.
Or the maximum value of $\mathit{f}\left(\mathit{x}\right)=32$