Question

Let the function $f\left(x\right)=\{\begin{array}{cc}{\sin }^{-1}x,& -1\le x\le 1\\ a-(x-1{)}^2,& x>1\end{array}$ has a point of local maxima at $x=1,$ then the minimum value of $\frac{\pi }{a}(a>0)$ is

Answer 1

Given : $f\left(x\right)=\{\begin{array}{cc}{\sin }^{-1}x,& -1\le x\le 1\\ a-(x-1{)}^2,& x>1\end{array}$
Plotting the graph of $f\left(x\right):$

For point of maxima at $x=1,f(1-h)<f\left(1\right)>f(1+h)$ as $h\to 0^{+}$
From graph it is clear that $f(1-h)<f\left(1\right)$ and for $f\left(1\right)>f(1+h),a$ should be less than or equal to $\frac{\pi }{2}.$
$\Rightarrow a\le \frac{\pi }{2}$
$\Rightarrow \frac{1}{a}\ge \frac{2}{\pi }$
$\Rightarrow \frac{\pi }{a}\ge 2$
$\therefore$ minimum value of $\frac{\pi }{a}$ is $2.$