Let $f(x)=\left\{\begin{matrix}\sin^{-1}α+x^2,&0<x<1\\2x,&x≥1\end{matrix}\right.$, f(x) can have a minimum at x = 1 then the value of a can be:

None of these

$f(x)=\left\{\begin{matrix}\sin^{-1}α+x^2,&0<x<1\\2x,&x≥1\end{matrix}\right.$

Clearly 2x has its minimum value at x = 1 for x ≥ 1 and that value is 2.

Now if $\sin^{-1}α+x^2> 2$, then minimum of f(x) will be at x = 1

$0<x<1, 0<x^2<1$

$\sin^{-1}α<\sin^{-1}α+x^2<1+\sin^{-1}α⇒min(\sin^{-1}α+x^2)≥2⇒\sin^{-1}α≥2$