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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)


6 years ago


Answers : (1)


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

askIITians Faculty

2 years ago

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