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Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 28 Maths Textbook Solution.

Answers (1)

Answer:

                a=1, b=-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)=\left\{\begin{array}{r} \frac{x-4}{|x-4|}+a, \text { if } x<4 \\\\ a+b, \text { if } x=4 \\\\ \frac{x-4}{|x-4|}+b, \text { if } x>4 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{r} \frac{x-4}{|x-4|}+a, \text { if } x<4 \\\\ a+b, \text { if } x=4 \\\\ \frac{x-4}{|x-4|}+b, \text { if } x>4 \end{array}\right.

We observe

(LHL at x=4 )

\begin{aligned} \lim _{x \rightarrow 4^{-}} f(x) &=\lim _{h \rightarrow 0} f(4-h) \\\\ &=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right)=\lim _{h \rightarrow 0}\left(\frac{-h}{|-h|}+a\right)=a-1 \end{aligned}            

 (RHL at x=4 )

\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h) \\\\ &\quad=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right)=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right)=b+1 \\\\ &f(4)=a+b \end{aligned}

If f\left ( x \right ) is continuous at  x=4

           \begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4) \\ &a-1=b+1=a+b \\ &a-1=a+b, b+1=a+b \\ &b=-1, a=1 \end{aligned}

          

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