CAT 2018 Set-1 | Question: 97

Question

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

Answer 1

Given that, $ f(x) = \min \{2x^{2}, 52 – 5x\} \quad \longrightarrow (1)$

For maximum value of $f(x):$

$ 2x^{2} = 52 – 5x $

$ \Rightarrow 2x^{2} + 5x – 52 = 0 $

$ \Rightarrow 2x^{2} + 13x – 8x – 52 = 0 $

$ \Rightarrow x(2x+13) – 4(2x+13) = 0 $

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

$ \Rightarrow x-4=0, 2x+13=0 $

$ \Rightarrow x=4, x= \frac{-13}{2} (\text {rejected, because $x$ is not positive real number})$

$ \Rightarrow \boxed{x = 4}$

Now, $f(x) = \min \{2x^{2}, 52 – 5x\}$

$ \Rightarrow f(x) = \min \{2(4)^{2}, 52 – 20\}$

$ \Rightarrow f(x) = \min \{32,32\}$

$ \Rightarrow \boxed{f(x) = 32}$

$\therefore$ The maximum possible valve of $ f(x) = 32 $

Correct Answer $:32 $

Comments

Why is maximum value of f(x) only possible when both 2x^2 and 52-x are equal !  could you please explain …