Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 8

Need solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 8

Answers (1)

Answer:

 \frac{1}{2}

Given:

f(x)= \begin{cases}\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x} & x \neq \pi / 4 \\ k & x=\pi / 4\end{cases}

Hint:

 Apply L-Hospital Rule when you get 0/0 from.

Explanation:

At x = \pi/4

\begin{aligned} &\text { L.H.L } =\text { R.H.L }=f\left(\frac{\pi}{4}\right) \quad \ldots(A) \\ &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right) \end{aligned}

=\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}+h\right)}{\cot \left(\frac{\pi}{4}-h\right)}=\lim _{h \rightarrow 0} \frac{\tanh }{\tan 2 h}

=\lim _{h \rightarrow 0} \frac{\frac{\tanh }{h}}{\frac{\tan 2 h}{2 h} \times 2}=\frac{1}{2}

Hence F (x) will be continious on

[0, \frac{\pi }{2}]

If

f(\frac{\pi }{4})=\frac{1}{2}

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 Maths paper 2025 'moderate but lengthy', questions from PYQ, NCERT
anam-khan

CBSE Board Exam 2025 Live: Class 12 mathematics paper 'moderate but lengthy'; memory-based questions
anam-khan

CBSE Class 12 Chemistry Exam Analysis 2025: Paper 'moderately difficult' but well balanced

Latest Question