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  • Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 43

Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 43

Answers (1)

Answer:

                k=2

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}{l} 2 x+1, \text { if } x<2 \\ k, \text { if } x=2 \\ 3 x-1, \text { if } x>2 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{l} 2 x+1, \text { if } x<2 \\ k, \text { if } x=2 \\ 3 x-1, \text { if } x>2 \end{array}\right.

We have

(LHL at  x=2 )

                \lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[2(2-h)+1]=[4+1]=5

(RHL at x=2  )

                \lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}[3(2+h)-1]=[6-1]=5

Also  f\left ( 2 \right )=k

If f\left ( x \right )  is continuous at  x=2 .

\begin{aligned} &\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(2) \\\\ &5=5=k \\\\ &k=5 \end{aligned}

Since  LHS =RHS ,f\left ( x \right )   is continuous

Thus,f\left ( x \right )  is continuous at x=2 .

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