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)

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.

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Test Series JEE Main April 2022

Take chapter-wise, subject-wise and Complete syllabus mock tests and get an in-depth analysis of your test..

₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions