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  • Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 39 subquestion (ii)

Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 39

Answers (1)

Answer:

f\left ( x \right )  is continuous at  x=-1,1

Hint:

f\left ( x \right )  must be defined. The limit of the f\left ( x \right ) approaches the value x must exist.

Given:

                f(x)=|x-1|+|x+1|

Solution:

                f(x)=|x-1|+|x+1|

We have

(LHL at x=-1 )

                \lim _{x \rightarrow-1^{-}} f(x)=\lim _{h \rightarrow 0} f(-1-h)=\lim _{h \rightarrow 0}[|-1-h-1|+|-1-h+1|]=2+0=2 

(RHL at x=-1  )

                \lim _{x \rightarrow-1^{+}} f(x)=\lim _{h \rightarrow 0} f(-1+h)=\lim _{h \rightarrow 0}[|-1+h-1|+|-1+h+1|]=2+0=2 

Also

f(-1)=|-1-1|+|-1+1|=|-2|=2

Now,

(LHL at x=1 )

                \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[|1-h-1|+|1-h+1|]=2 

(RHL at x=1 )

                \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[|1+h-1|+|1+h+1|]=0+2=2 

Also

\begin{aligned} &f(1)=|1+1|+|1-1|=2+0=2 \\ &\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{-}} f(x)=f(-1) \text { and } \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}

Hence f\left ( x \right ) is continuous at x=-1 and x=1 .

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