JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 19)
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 :
$$f(x) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr {{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr } } \right.$$ is :
continuous on R–{–1} and differentiable on R–{–1, 1}
both continuous and differentiable on R–{1}
both continuous and differentiable on R–{–1}
continuous on R–{1} and differentiable on R–{–1, 1}
Explanation
$$f\left( x \right) = \left\{ {\matrix{
{{\pi \over 4} + {{\tan }^{ - 1}}x,} & {x \in \left[ { - 1,1} \right]} \cr
{{1 \over 2}\left( {x - 1} \right),} & {x > 1} \cr
{{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \cr
} } \right.$$
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
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
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