CAT 2018 Slot 1 QA Question 25: Let $f(x) = \min\{2x^{2}, 52 - 5x\}$, where $x$ is any positive real number. Then the maximum possible value of $f(x)$ is

Question

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

Accepted Answer

Here $2x^{2}$ is increasing for positive x, while $52 - 5x$ is decreasing.
The maximum possible value of the minimum of the two expressions occurs when they are equal.
$2x^{2} = 52 - 5x$
$2x^{2} + 5x - 52 = 0$
$(2x + 13)(x - 4) = 0$
$x = 4$ since x is positive.
$f(4) = 2 	imes 4^{2} = 32$

Let f(x)=min⁡{2x2,52−5x}f(x) = \min\{2x^{2}, 52 - 5x\}f(x)=min{2x2,52−5x}, where xxx is any positive real number. Then the maximum possible value of f(x)f(x)f(x) is

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