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)

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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