CAT 2018

Slot 1

QA

Question & Solution

f(x) = min (${2x^{2},52-5x}$)
The maximum possible value of this function will be attained at the point in which $2x^2$ is equal to $52-5x$.

$2x^2 = 52-5x$
$2x^2+5x-52=0$
$(2x+13)(x-4)=0$
=> $x=\frac{-13}{2}$ or $x = 4$
It has been given that $x$ is a positive real number. Therefore, we can eliminate the case $x=\frac{-13}{2}$.
$x=4$ is the point at which the function attains the maximum value. $4$ is not the maximum value of the function. 
Substituting $x=4$ in the original function, we get, $2x^2 = 2*4^2= 32$.
f(x) = $32$.
Therefore, 32 is the right answer.