Minimum Value of Maximum Function

CAT 2006 Slot 1 · QA · Easy · Quantitative Aptitude

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

  1. A.

    1/3

  2. B.

    1/2

  3. C.

    2/3

  4. D.

    4/3

  5. E.

    5/3

Answer

5/3

Explanation

The minimum of max⁡(2x+1,3−4x)\max(2x+1, 3-4x) occurs at the point of intersection of y=2x+1y = 2x+1 and y=3−4xy = 3-4x. 2x+1=3−4x  ⟹  6x=2  ⟹  x=1/32x + 1 = 3 - 4x \implies 6x = 2 \implies x = 1/3. At x=1/3x = 1/3, f(1/3)=2(1/3)+1=5/3f(1/3) = 2(1/3) + 1 = 5/3.

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