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 Continuity exercise 8.2 question 19 maths

Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 19 maths

Answers (1)

Answer:

 Discontinuous at x = 1/2 , 1 , 2

Hint:

 f(x)/g(x) is continuous at every point hen f(x) & g(x) are continuous except 

g(x) \neq 0

Given:

 f(x)=\frac{1}{t^{2}+t-2}, t=\frac{1}{x-1}

Explanation:

\begin{aligned} &f(t)=\frac{1}{t^{2}+t-2} \text { where } t=\frac{1}{x-1}\\ &\text { Clearly } t=\frac{1}{x-1} \text { is discontinuous at } \mathrm{x}=1 \end{aligned}

\begin{aligned} &\text { For } \mathrm{x} \neq 1 \text { we have, }\\ &f(t)=\frac{1}{t^{2}+t-2}=\frac{1}{(t+2)(t-1)} \end{aligned}

\begin{aligned} &\text { This is discontinuous at } x=-2 \text { and } \mathrm{t}=1\\ &\text { For } t=-2, t=\frac{1}{x-1} \Rightarrow x=\frac{1}{2}\\ &\text { For } t=1, t=\frac{1}{x-1} \Rightarrow x=2 \end{aligned}

Hence F is discontinuous at

\begin{aligned} x=\frac{1}{2}, x = 1 \text { and } x = 2 \end{aligned}

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question