CAT 2018 Slot 1 - QUANT Question 7

Question 7

Let f(x) = min ($${2x^{2},52-5x}$$) where x is any positive real number. Then the maximum possible value of f(x) is


Correct Answer: 32

Solution

f(x) = min ($${2x^{2},52-5x}$$)
The maximum possible value of this function will be attained at the point in which $$2x^2$$ is equal to $$52-5x$$.

$$2x^2 = 52-5x$$
$$2x^2+5x-52=0$$
$$(2x+13)(x-4)=0$$
=> $$x=\frac{-13}{2}$$ or $$x = 4$$
It has been given that $$x$$ is a positive real number. Therefore, we can eliminate the case $$x=\frac{-13}{2}$$.
$$x=4$$ is the point at which the function attains the maximum value. $$4$$ is not the maximum value of the function. 
Substituting $$x=4$$ in the original function, we get, $$2x^2 = 2*4^2= 32$$.
f(x) = $$32$$.
Therefore, 32 is the right answer.


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