Applyo - College Application Platform

CAT 2018 Slot 2 QA Question & Solution

AlgebraMedium

Question

Let f(x)= $\max(5x, 52-2x^2)$, where x is any positive real number. Then the minimum possible value of f(x) 

Solution

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$x$ = $52 - 2x^2$
or, $2x^2 + 5x - 52$ = 0
On solving, we get $x$ = 4 or $-\dfrac{13}{2}$
But, it is given that $x$ is a positive number.
So, $x$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.