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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

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										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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago

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