retagged by
734 views

1 Answer

Best answer
2 votes
2 votes
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 $
selected by
Answer:

Related questions

2 votes
2 votes
1 answer
1
go_editor asked Mar 20, 2020
657 views
Let $f\left (x \right ) = \max\left \{5x, 52 – 2x^{2}\right \}$ , where $x$ is any positive real numbers. Then the minimum possible value of $f(x)$ is ________
2 votes
2 votes
1 answer
4
go_editor asked Mar 19, 2020
731 views
While multiplying three real numbers, Ashok took one of the numbers as $73$ instead of $37$. As a result, the product went up by $720$. Then the minimum possible value of...
2 votes
2 votes
1 answer
5