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 multiple choice question 8 maths

Explain solution RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 8 maths

Answers (1)

Answer:

 The correct option is (b)

Hint:

 Use the given formula:

 \text { (i) Standard limit } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1

(ii) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text { iii }) \lim _{x \rightarrow a}\{f(x) \pm g(x)\}=1 \pm m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}

Given:

f(x)=\left\{\begin{array}{l} \frac{\log (1+a x)-\log (1-b x)}{x} \\ k \end{array}\right.        \begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}

And f(x) is continuous at x = 0

Solution:

\because f(x) is continuous at x = 0

\Rightarrow \lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}=k

Using formula (ii)

\Rightarrow a \lim _{x \rightarrow 0} \frac{\log (1+a x)}{a x}+b \lim _{x \rightarrow 0} \frac{\log (1-b x)}{-b x}=k

Using formula (i)

\Rightarrow a+b=k

So, the correct option is (b)

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 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