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 3 subquestion (vi) maths

Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 3 subquestion (vi) maths

Answers (1)

Answer:

  x = 0

Hint:

 Use L-Hospital Rule when we find 0/0 form

Given:

 f(x)= \begin{cases}\frac{x^{4}+x^{3}+2 x^{2}}{\tan ^{-1} x} & x \neq 0 \\ 10 & x=0\end{cases}

Explanation:

at x = 0

\begin{aligned} &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{(-h)^{4}+(-h)^{3}+2(-h)^{2}}{\tan ^{-1}(-h)} \\ &=\lim _{h \rightarrow 0} \frac{h^{4}-h^{3}+2 h^{2}}{\tan ^{-1}(h)}=0 \end{aligned}

\begin{aligned} &\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{h^{4}+h^{3}+2 h^{2}}{\tan ^{-1}(h)}=0 \\ &\text { Also } f(0)=10 \\ &\text { Thus L.H.L } =\text { R.H.L } \neq f(0) \end{aligned}

Hence, f(x) is discontinuous at x = 0.

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

CIC asks CBSE to disclose tender process for exam answer sheet procurement under RTI Act
anam-khan

CBSE 10th Second Board Result 2026 LIVE: Class 10 session 2 scores soon at cbse.gov.in; passing marks
anam-khan

Delhi HC refuses to direct reopening of CBSE class 12 re-evaluation portal

Latest Question