CAT 2018 Question Paper Shift 2 Solved Paper

Question : 92 of 100

Marks: +1, -0

Let

f(x)=\max\{5x, 52-2x^2\}

, where x is any positive real number. Then the minimum possible value of f(x) is

Solution:

Given

x

is positive real number. The minimum value of the maximum

\{5 x, 52-2 x^{2}\}

will occur
when both the graphs intersect. i.e., when

5 x=52-2 x^{2}

2 x^{2}+5 x-52=0

2 x^{2}+13 x-8 x-52=0

x(2 x+13)-4(2 x+13)=0

(x-4)(2 x+13)=0

x=4

\;{-13}/{2}

When

x=4, f(x)=20

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