CAT 2018 Quant Questions

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

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Correct Answer: 20

The minimum value of the function will occur when the expressions inside the function are equal.

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

or, \(2{x^2} + 5x - 52\) = 0

On solving, we get \(x\) = 4 or \( - \frac{{13}}{2}\)

But, it is given that \(x\) is a positive number.

So, \(x\) = 4

And the minimum value = 5*4 = 20

Hence, 20 is the correct answer.

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CAT 2018 Quant Questions
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