The range of the function $f\left(x\right)=\sqrt{-x^2+4x-3}+\sqrt{\sin \frac{\pi }{2}\left(\sin \frac{\pi }{2}(x-1)\right)}$ is $\left[0,k\right] then k=$

The given function is $f\left(x\right)=\sqrt{-x^2+4x-3}+\sqrt{\sin \frac{\pi }{2}\left(\sin \frac{\pi }{2}(x-1)\right)}$

Suppose the given function is, $\mathrm{g}\left(x\right)=\sqrt{-x^2+4x-3}$ and $\mathrm{h}\left(x\right)=\sqrt{\sin \frac{\pi }{2}\left(\sin \frac{\pi }{2}(x-1)\right)}$ 

Since, $-x^2+4x-3=1-(x-2{)}^2$, maximum value of $g=g\left(2\right)=1$

Consider also that , $g\left(1\right)=0$

Therefore, minimum value of $g=g\left(1\right)=0$

Now, $h\left(2\right)=1$ and $h\left(1\right)=0$

Hence, maximum and minimum value of both $g$ and $\mathrm{h}$ are attained at 2 and 1 , respectively.

Further $\mathrm{g}$ and $\mathrm{h}$ are both continuous in $\left[0,2\right]$.

So, Range of $f=\left[f\left(1\right),f\left(2\right)\right]=\left[0,2\right]$

Hence, If the function is $f\left(x\right)=\sqrt{-x^{2}4x-3}\sqrt{\sin \frac{\pi }{2}\left(\sin \frac{\pi }{2}(x-1)\right)}$, then the range of the function is $\left[0,2\right]$.