Minimum possible value of maximum function

CAT 2018 Slot 2 · QA · Medium · Functions

Let f(x)=max{5x,522x2}f(x) = \max\{5x, 52 - 2x^2\}, where xx is any positive real number. Then the minimum possible value of f(x)f(x) is

  1. A.

    20

  2. B.

    25

  3. C.

    15

  4. D.

    30

Answer

A

Explanation

The function 5x5x is strictly increasing for x>0x > 0. The function 522x252 - 2x^2 is strictly decreasing for x>0x > 0.

The minimum of max{5x,522x2}\max\{5x, 52 - 2x^2\} occurs at the point of intersection: 5x=522x25x = 52 - 2x^2 2x2+5x52=02x^2 + 5x - 52 = 0 (2x+13)(x4)=0(2x + 13)(x - 4) = 0 Since x>0x > 0, we have x=4x = 4.

At x=4x = 4, f(4)=5(4)=20f(4) = 5(4) = 20. So the minimum possible value of f(x)f(x) is 20.

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