CAT 1993 — QA Question 32

So maximum possible value will be at the point of intersection of the two graphs.

$$\begin{gathered} \therefore \frac{1}{2}-\frac{3x^2}{4}=\frac{5x^2}{4} \\ \Rightarrow x^2=\frac{1}{4} \end{gathered}$$

Hence, required maximum value = $\frac{5x^2}{4}=\frac{5}{4}\times \frac{1}{4}=\frac{5}{16}.$

Hence, option (d).