If maximum and minimum value of $|\arcsin x|+|\arccos x| $ is represented by $M$ and $m$, respectively, then find $M+m$.
Since $ -\frac{\pi}{2} \leq \arcsin x \leq \frac{\pi}{2}$, we have$$0 \leq |\arcsin x| \leq \frac{\pi}{2} .$$
Also, $ \pi \leq \arccos x \leq 0$, so$$ \frac{\pi}{2} \leq |\arccos x| \leq 0 .$$
Can we proceed in this way... please guide... the correct method thanks..
Yes. For the $sin^{-1}(x)$, use the chain rule to take the derivative of $|sin^{-1}(x)|$:
$$\frac{d}{dx}|sin^{-1}(x)|=\frac{|sin^{-1}(x)|}{sin^{-1}(x)}\frac{1}{\sqrt{1-x^2}}$$
Find the critical points b finding when the derivative equals 0, and you'll find that it equals 0 when x = 0. Since the only other two critical points are the boundaries, which are both $\pi/2$, 0 must be the minimum, and $\pi/2$ must be the maximum.
The same can be done for $cos^{-1}(x)$, but you'll find that $cos^{-1}(x)$ has no critical points in its domain. This means that the minimum and maximum are simply the endpoints of its co-domain (also known as range).