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# CAT 2018 Question Paper | Quants Slot 2

###### CAT Previous Year Paper | CAT Functions Questions | Question 30

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Question 30 : Let f(x)= max{5x, 52 - 2x2}, where x is any positive real number.Then the minimum possible value of f(x) is [TITA]

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

Given that f(x) = max{5x , 52 - 2x2}, where x is any positive real number.
The minimum possible value of f(x) has to found.
The minimum possible value is obtained if the two curves intersect or
5x = 52 - 2x2
2x2 + 5x – 52 = 0
2x2 – 8x + 13x – 52 = 0
2x(x - 4) + 13(x - 4) = 0
(2x + 13)(x - 4) = 0
x = $$frac{-13}{2}$ or 4 One of these would be the minimum possible value 5x = 5 × $\frac{-13}{2}$ = $\frac{-65}{2}$ 52 - 2x2 = 52 – 2$$$frac{-13}{2}$)2 = $\frac{-65}{2}$ Similarly when x = 4 5x = 5 × 4 =20 52 - 2x2 = 52 – 2$4)2 = 20
At x = 4 and $$frac{-13}{2}$ the values are equal And when x = $\frac{-13}{2}$ , f$x) = $$frac{-65}{2}$ The minimum possible value of f$x) is $$frac{-65}{2}$ The question is "Let f$x)= max{5x, 52 - 2x2}, where x is any positive real number.Then the minimum possible value of f(x) is [TITA]"

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