Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 32

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 32

Answers (1)

Answer:

                k=-4

Hint:

f\left ( x \right ) must be defined. The limit of the  f\left ( x \right ) approaches the value x must exist.

Given:

                f(x)=\left(\begin{array}{cc} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} & , x \neq 0 \\\\ k & , x=0 \end{array}\right)

Solution:

                f(x)=\left(\begin{array}{cc} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} & , x \neq 0 \\\\ k & , x=0 \end{array}\right)

If f\left ( x \right )  is continuous at x=0 , then

                \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1}=k \end{aligned}

                \lim _{x \rightarrow 0} \frac{1-\sin ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1}=k                                                   \left[\because \cos ^{2} x=1-\sin ^{2} x\right]

                \lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x}{\sqrt{x^{2}+1}-1}=k

                \lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x\left(\sqrt{x^{2}+1}+1\right)}{\left(\sqrt{x^{2}+1}-1\right)\left(\sqrt{x^{2}+1}+1\right)}=k  [Multiplying and dividing by \sqrt{x^{2}+1}+1  ]

                \lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x\left(\sqrt{x^{2}+1}+1\right)}{x^{2}}=k                                                     \left[\because a^{2}-b^{2}=(a+b)(a-b)\right]

                \begin{aligned} &-2 \lim _{x \rightarrow 0} \frac{\left(\sin ^{2} x\right)\left(\sqrt{x^{2}+1}+1\right)}{x^{2}}=k \\\\ &-2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2} \lim _{x \rightarrow 0}\left(\sqrt{x^{2}+1}+1\right)=k \end{aligned}

                -2 \times 1 \times 1(1+1)=k                                                                              \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                k=-4

 

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

CBSE Class 10, 12 compartment 2024 exams today for 2.5 lakh students; exam timings, guidelines
anam-khan

CBSE compartment admit card 2024 out at cbse.gov.in; exams from July 15
anam-khan

Reports saying CBSE cannot conduct board exams twice a year currently ‘totally incorrect’; board denies claims

Latest Question