"The function $$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| 1} \cr } } \right.$$ is :"

Question

"The function $$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$$ is :"

Accepted Answer

"$$f\left( x \right) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {x \in \left[ { - 1,1} \right]} \cr {{1 \over 2}\left( {x - 1} \right),} & {x > 1} \cr {{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \cr } } \right.$$

At x = 1

L.H.L = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$$ = ${{\pi \over 4} + {\pi \over 4}}$ = ${{\pi \over 2}}$

f(1) = ${{\pi \over 4} + {{\tan }^{ - 1}}x}$ = ${{\pi \over 4} + {\pi \over 4}}$ = ${{\pi \over 2}}$

R.H.L = $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {{1 \over 2}\left( {x - 1} \right)} \right)$$ = 0

As L.H.L $\ne$ R.H.L so function is discontinuous $\Rightarrow$ non differentiable.

At x = -1

L.H.L = $$\mathop {\lim }\limits_{x \to - {1^ - }} \left( {{1 \over 2}\left( { - x - 1} \right)} \right)$$ = ${{1 \over 2}\left( { - \left( { - 1} \right) - 1} \right)}$ = 0

f(-1) = ${\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)$ = ${\pi \over 4} - {\pi \over 4}$ = 0

R.H.L = $$\mathop {\lim }\limits_{x \to - {1^ + }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$$

= ${\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)$ = ${\pi \over 4} - {\pi \over 4}$ = 0

As L.H.L = f(-1) = R.H.L so function is continuous.

$$f'\left( x \right) = \left\{ {\matrix{ {{1 \over {1 + {x^2}}},} & {x \in \left[ { - 1,1} \right]} \cr {{1 \over 2},} & {x > 1} \cr { - {1 \over 2},} & {x < - 1} \cr } } \right.$$

For differentiability at x = –1

L.H.D = ${ - {1 \over 2}}$

R.H.D. = ${{1 \over 2}}$

So, non differentiable at x = –1"

The function
$$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$$ is :

Solution

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The function $$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$$ is :

Solution

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More Limits, Continuity and Differentiability questions

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The function
$$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$$ is :