^{-1} ≤ft( 1+x²/2x ) is

Correct Answer is : continuous but not differentiable at x=1x=1x=1

continuous but not differentiable at x=1

neither continuous nor differentiable atx=1

Correct Answer is : continuous but not differentiable at x=1x=1x=1

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CGPET 2011 Solved Paper

Section: Mathematics

Question : 123 of 150

Marks: +1, -0

\sin^{-1} \left( \frac{1+x^2}{2x} \right)

Solution:

Given function

\sin^{-1} \left( \frac{1+x^2}{2x} \right)

x=1, \text{LHL}

\lim\limits_{h \to 0^{-}} \sin^{-1} \left[ \frac{1+(1-h)^2}{2(1-h)} \right]

= \sin^{-1} \left( \frac{1+1}{2} \right) = \frac{\pi}{2}

and RHL

\lim\limits_{h \to 0^{+}} \sin^{-1} \left[ \frac{1+(1+h)^2}{2(1+h)} \right]

= \sin^{-1} \left( \frac{2}{2} \right) = \frac{\pi}{2}

and

f(1) = \sin^{-1} \left( \frac{2}{2} \right) = \frac{\pi}{2}

Thus, given function is continuous at

x=1

Now,

\frac{d}{dx} \left\{ \sin^{-1} \left( \frac{1+x^2}{2x} \right) \right\}

= \frac{1}{\sqrt{1- \left( \frac{1+x^2}{2x} \right)^2}} \cdot \left[ \frac{2x(2x) - (1+x^2) \cdot 2}{(2x)^2} \right]

= \frac{2x}{\sqrt{4x^2 - (1 + x^4 + 2x^2)}} \cdot \frac{2x^2 - 1}{(2x)^2}

= \frac{2x^2 - 1}{\sqrt{-(1 + x^4 - 2x^2)} \cdot (2x)}

= \frac{2x^2 - 1}{2x \sqrt{-(1 - x^2)^2}}

which does not exist at

x=1

Hence, given function is not differentiable at

x=1

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