Minimum Value of Maximum Function

CAT 2006 Slot 1 · Quantitative Ability · Easy · Quantitative Aptitude

This is an easy Quantitative Ability question from the CAT 2006 Slot 1 paper. It tests Quantitative Aptitude. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Let $f(x) = \max(2x + 1, 3 - 4x)$ , where $x$ is any real number. Then the minimum possible value of $f(x)$ is:

A.
1/3
B.
1/2
C.
2/3
D.
4/3
E.
5/3

Answer

5/3

Explanation

The minimum of $\max(2x+1, 3-4x)$ occurs at the point of intersection of $y = 2x+1$ and $y = 3-4x$ . $2x + 1 = 3 - 4x \implies 6x = 2 \implies x = 1/3$ . At $x = 1/3$ , $f(1/3) = 2(1/3) + 1 = 5/3$ .

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Minimum Value of Maximum Function

Explanation

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