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  • Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 39 Subquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 39 Subquestion (i) Maths Textbook Solution.

Answers (1)

Answer:

                Continuous

Hint:

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

 

Given:

                f(x)=|x|+|x-1|

Solution:

                f(x)=|x|+|x-1|

We have

(LHL at x=0 )

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

 (RHL at x=0 )

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

Also

f(0)=|0|+|0-1|=0+1=1

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-h-1|]=1 

(RHL at x=1 )

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

Also

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

 

Posted by

infoexpert27

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