#### need solution for RD Sharma maths class 12 chapter 8 Continuity exercise Fill in the blanks question 3

Hint: You must know about the concept of continuous function
Given:

$f(x)=\left\{\begin{array}{l} a x^{2}-b, 0 \leq x<1 \\ 2, x=1 \\ x+1,1
Solution:

If $f(x)$ is continuous at $x=1$, then

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ &\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}

$\lim _{h \rightarrow 0} a(1-h)^{2}-b=\lim _{h \rightarrow 0}(1+h)+1=\lim _{x \rightarrow 1} 2$

\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

Hint: You must know about the concept of continuous function
Given:

$f(x)=\left\{\begin{array}{l} a x^{2}-b, 0 \leq x<1 \\ 2, x=1 \\ x+1,1
Solution:

If $f(x)$ is continuous at $x=1$, then

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ &\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}

\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}