Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (ii)

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (ii)

Answers (1)

Answer:

k=\frac{-2}{\pi}

Hint:

 For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:

                f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.

If  f\left ( x \right ) is continuous at x=1 , then

                \begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\\\ &\lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2}=k \end{aligned}

Putting x-1=y , we get

                \begin{aligned} &\lim _{y \rightarrow 0} y \tan \frac{\pi(y+1)}{2}=k \\\\ &\lim _{y \rightarrow 0} y \tan \left(\frac{\pi y}{2}+\frac{\pi}{2}\right)=k \end{aligned}

                \lim _{y \rightarrow 0} y \tan \left(\frac{\pi}{2}+\frac{\pi y}{2}\right)=k

                -\lim _{y \rightarrow 0} y \cot \left(\frac{\pi y}{2}\right)=k                                                                                   [\because \tan (90+\theta)=-\cot \theta]

                -\frac{2}{\pi} \lim _{y \rightarrow 0} \frac{\frac{\pi y}{2} \cos \left(\frac{\pi y}{2}\right)}{\sin \left(\frac{\pi y}{2}\right)}=k                                                                   \left[\because \cot x=\frac{\cos x}{\sin x}\right]

                -\frac{2}{\pi} \frac{\lim _{y \rightarrow 0} \cos \left(\frac{\pi y}{2}\right)}{\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{\pi y}{2}\right)}{\frac{\pi y}{2}}\right)}=k

                \begin{aligned} & \\\\ &\frac{-2}{\pi} \times \frac{1}{1}=k \\\\ &k=\frac{-2}{\pi} \end{aligned}                                                                                         \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

Posted by

infoexpert27

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question