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The value of k for which the function

is continuous at, $x=\frac{\pi }{2}$ is:

Option 1)
$\frac{17}{20}$

Option 2)
$\frac{2}{5}$

Option 3)
$\frac{3}{5}$

Option 4)
$-\frac{2}{5}$

Answers (1)

best_answer

As we learnt in

Result of 1 to the power of infinity Form -

$\lim_{x\rightarrow \infty}\left(1+\frac{1}{x} \right )^{x}=e$

$\lim_{x\rightarrow 0}\:\;\:(1+x)^{\frac{1}{x}}=e$

$\lim_{x\rightarrow 0}\:\;\:(1+\lambda x)^{\frac{1}{x}}=e^{\lambda }$

$\lim_{x\rightarrow \infty}\left(1+\frac{\lambda}{x} \right )^{x}=e^{\lambda}$

-

$\lim_{x\rightarrow \frac{\pi^-}{2}}\left ( \frac{4}{5} \right )^{\frac{\tan 4x}{\tan 5x}}$

Let $y= \lim_{x\rightarrow \frac{\pi^{-} }{2}}\left ( \frac{4}{5} \right )^{\frac{\tan 4x}{\tan 5x}}$

$log_{e}^{y}= \lim_{x\rightarrow \frac{\pi^{-} }{2}}\frac{\tan 4x}{\tan 5x}log\frac{4}{5}$

Put $x=\frac{\pi }{2}-h$

$log_{e}^{y}=\lim_{h\rightarrow 0}- \frac{\tan 4h}{\cot 5h}log\frac{4}{5}$

$= \lim_{h\rightarrow 0}-\tan 4h\cdot \tan 5h\:log\frac{4}{5}$

$= 0$

$\therefore y=1$

So that,

$k+\frac{2}{5}= 1$

$k= 1-\frac{2}{5}= \frac{3}{5}$

Option 1)

$\frac{17}{20}$

This option is incorrect.

Option 2)

$\frac{2}{5}$

This option is incorrect.

Option 3)

$\frac{3}{5}$

This option is correct.

Option 4)

$-\frac{2}{5}$

This option is incorrect.

Posted by

prateek

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