CAT 2018 Slot 2 - QUANT Question 2

Question 2

Let f(x)= $$\max(5x, 52-2x^2)$$, where x is any positive real number. Then the minimum possible value of f(x) 


Correct Answer: 20

Solution

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$$x$$ = $$52 - 2x^2$$
or, $$2x^2 + 5x - 52$$ = 0
On solving, we get $$x$$ = 4 or $$-\dfrac{13}{2}$$
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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