JEE MAIN - Mathematics (2004 - No. 5)
Let $$f(x) = {{1 - \tan x} \over {4x - \pi }}$$, $$x \ne {\pi \over 4}$$, $$x \in \left[ {0,{\pi \over 2}} \right]$$.
If $$f(x)$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$, then $$f\left( {{\pi \over 4}} \right)$$ is
If $$f(x)$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$, then $$f\left( {{\pi \over 4}} \right)$$ is
$$-1$$
$${1 \over 2}$$
$$-{1 \over 2}$$
$$1$$
Explanation
$$f\left( x \right) = {{1 - \tan x} \over {4x - \pi }}$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$
$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {{\pi \over 4} + h} \right)$$
$$ = \mathop {\lim }\limits_{h \to 0} {{1 - \tan \left( {{\pi \over 4} + h} \right)} \over {4\left( {{\pi \over 4} + h} \right) - \pi }},h > 0 = \mathop {\lim }\limits_{h \to 0} {{1 - {{1 + \tan \,h} \over {1 - \tan \,h}}} \over {4h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} \,{{ - 2} \over {1 - \tan \,h}}.{{\tan \,h} \over {4h}} = {{ - 2} \over 4} = - {1 \over 2}$$
$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {{\pi \over 4} + h} \right)$$
$$ = \mathop {\lim }\limits_{h \to 0} {{1 - \tan \left( {{\pi \over 4} + h} \right)} \over {4\left( {{\pi \over 4} + h} \right) - \pi }},h > 0 = \mathop {\lim }\limits_{h \to 0} {{1 - {{1 + \tan \,h} \over {1 - \tan \,h}}} \over {4h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} \,{{ - 2} \over {1 - \tan \,h}}.{{\tan \,h} \over {4h}} = {{ - 2} \over 4} = - {1 \over 2}$$
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