# CAT 2018 Question Paper | Quants Slot 1

###### CAT Previous Year Paper | CAT Functions Questions | Question 33

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Question 33 : Let f(x)=min{2x2, 52 - 5x}, where x is any positive real number.Then the maximum possible value of f(x) is [TITA]

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##### Method of solving this CAT Question from Functions

Given f(x) = min {2x2, 52 − 5x}
From graph, we see that f(x) increases initially and then decreases after intersection
So, maximum value occurs when 2x2 = 52 − 5x
2x2 + 5x – 52 = 0
2x2 - 8x + 13x - 52 = 0
2x(x-4) + 13(x-4) = 0
(2x+13) (x-4) = 0
Since x is a Positive real number, x = 4
min{2x2,52-5x} = min {32,32} = 32 = max f(x)

The question is " Let f(x)=min{2x2, 52 - 5x}, where x is any positive real number.Then the maximum possible value of f(x) is [TITA]"

##### Hence, the answer is 32

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