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 4 subquestion (v) maths

Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 4 subquestion (v) maths

Answers (1)

Answer:

 a = 3, b = 1

Hint:

 Put LHL = RHL at x = 3 , 5

Given:

 f(x)= \begin{cases}4 & x \leq-1 \\ a x^{2}+b & -1<x<0 \\ \cos x & x \geq 0\end{cases}

Explanation:

At x = -1

\begin{aligned} &f(-1)=4\\ &\text { R.H.L }=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0}(1-h)=\lim _{h \rightarrow 0} a(-1+h)^{2}+b=a+b\\ &\therefore a+b=4 \qquad ....(A) \end{aligned}

Now at x = 0

\begin{aligned} &f(0)=\cos 0=1 \\ &\text { L.H.L }=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} a(-h)^{2}+b \\ &=b \end{aligned}

\begin{aligned} &\therefore f(0)=L . H . L \\ &\Rightarrow b=1 \\ &\text { From }(A) \\ &a=3 \end{aligned}

Hence, a = 3, b = 1

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 practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question