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  • Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 2

Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 2

Answers (1)

Answer:

                x=3

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,

f(x)=\left(\begin{array}{cc} \frac{x^{2}-x-6}{x-3} & , x \neq 3 \\\\ 5 & , x=3 \end{array}\right)   

We observe,

[LHL at x=3 ]

\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)

\lim _{h \rightarrow 0} \frac{(3-h)^{2}-(3-h)-6}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}-6 h-3+h-6}{-h}                     \left[\because(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)\right]  

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h} \\\\ &=\lim _{h \rightarrow 0}(5+h)=5 \end{aligned}

[RHL at x=3 ]

\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)

\lim _{h \rightarrow 0} \frac{(3+h)^{2}-(3+h)-6}{(3+h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}+6 h-3-h-6}{h}                    \left [ \because \left ( a+b \right )^{2}=\left ( a^{2}+b^{2}+2ab \right ) \right ]

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h} \\\\ &=\lim _{h \rightarrow 0}(5+h)=5 \end{aligned}

Also, f\left ( 3 \right )=5

\therefore           \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)=f(3)

Hence, f\left ( x \right ) is continuous at x=3.

Posted by

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