# CAT 2018 Quant Questions

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

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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.

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