# pls giv detailed soln with graph for the following: C) f(x)=mod sine inverse x is R) Differentiable in (0,1) D) f(x) = mod cos inverse (x-1/2) is S) not differentiable at least at one point in (-1,1) T) not continuous at least at one point in(-1,1)

Jitender Singh IIT Delhi
8 years ago
Ans:
$f(x) = |sin^{-1}(x)|$
This is the graph of sin-1x. For |sin-1x|, the portion from -1 to 0 will be above y- axis & symmetric to other portion.
So, it is clear from the graph that f(x) is continuous in (0, 1) & the curve is smooth. So f(x) is differentiable in (0, 1).
$f(x) = |cos^{-1}(x-\frac{1}{2})|$
This is the graph of |cos-1x|. For |cos-1(x-1/2)|, domain would be
$-1\leq x - \frac{1}{2}\leq 1$
$\frac{-1}{2}\leq x\leq \frac{3}{2}$
So, in the interval(-1, -1/2) the graph is not continuous & not differentiable.
Thanks & Regards
Jitender Singh
IIT Delhi