Mathematics - Functions Question with Solution | TestHub

${\sin }^{-1}x\ge 0\Rightarrow 0\le x\le 1$ and $2^x+2^{1-x}\le 3\Rightarrow 2^x+2\cdot 2^{-x}-3\le 0$. Put $2^x=t$, then $t^2-3t+2\le 0\Rightarrow (t-2)(t-1)\le 0$ $\Rightarrow 1\le t\le 2$ i.e. $1\le 2^x\le 2\Rightarrow 0\le x\le 1$.