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Let $f\left ( x \right )= \frac{1-\tan x}{4x-\pi },x\neq \frac{\pi }{4},x\in \left [ 0,\frac{\pi }{2} \right ].$

$f\left ( x \right )$ is continuous in $\left [ 0,\frac{\pi }{2} \right ],\: then f\left ( \frac{\pi }{4} \right )$ is

Option 1)
$-\frac{1}{2}$

Option 2)
$\frac{1}{2}$

Option 3)
$1$

Option 4)
$-1$

Answers (1)

best_answer

As we learnt in

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow \frac{\pi }{4}}\frac{1-\tan x}{4x-\pi }\, \, \, \, \Rightarrow put\ x=\frac{\pi }{4}-h$

$\lim_{h\rightarrow 0}\frac{1-\tan \left ( \frac{\pi }{4} -h\right )}{-4h}$

$\Rightarrow \lim_{h\rightarrow 0}\frac{1-\frac{1-\tan h}{1+\tan h}}{-4h}$

$\Rightarrow \lim_{h\rightarrow 0}\frac{1+\tan h-1+\tan h}{-4h\left ( 1+\tan h \right )}$

$\Rightarrow \lim_{h\rightarrow 0}\frac{2\tan h}{-4h\left ( 1+\tan h \right )}$

$\Rightarrow -\frac{1}{2}$

Option 1)

$-\frac{1}{2}$

Correct option

Option 2)

$\frac{1}{2}$

Incorrect option

Option 3)

$1$

Incorrect option

Option 4)

$-1$

Incorrect option

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