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Q.
Let $f( x )=\operatorname{Min} \cdot\left(\frac{1}{2}-\frac{3 x ^2}{4}, \frac{5 x ^2}{4}\right)$ for $0 \leq x \leq 1$. The maximum value of $f( x )$ is
Application of Derivatives
Solution:
Sodving, $\frac{1}{2}-\frac{3 x^2}{4}=\frac{5 x^2}{4}$
$\therefore 2-3 x^2=5 x^2 \Rightarrow 8 x^2=2 \Rightarrow x=\frac{1}{2} \text { or }-\frac{1}{2}$
$\therefore f$ is maximum when $x =1 / 2$
$f_{\max }=\frac{5}{4} \cdot \frac{1}{4}=\frac{5}{16} $
