CAT 2018 [SLOT 1] Quant Question with Solution 31

f(x) = min (${2x^{2},52-5x}$)

The maximum possible value of this function will be attained when $2x^{2}=52-5x$.

$2x^2+5x-52=0$

$(2x+13)(x-4)=0$

=> $x=\frac{-13}{2}$ or $x = 4$

Since x has to be positive integer, we can discard the case $x=\frac{-13}{2}$.

$x=4$ is the point at which the function attains the maximum value.

putting $x=4$ in the original function, we get, $2x^2 = 2*4^2= 32$.

Or the maximum value of f(x) = $32$.