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jee king king Grade: 12
        

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)








 
7 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										
Ans:

f(x) = |sin^{-1}(x)|




241-2077_arcsin_graph.gif


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})|






241-1468_arccos_graph.gif




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

3 years ago
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