Differential Calculus> jee king diff, cty...

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

jee king king , 14 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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

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Differential Calculus> jee king diff, cty...

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

jee king king , 14 Years ago

Jitender Singh

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