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  • Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 42

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 42

Answers (1)

Answer:

                For any value  \lambda, f(x)  is Continuous at x=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}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.

If  f\left ( x \right ) is continuous at  x=0, then

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right) \end{aligned}

Putting  x=0-h \text { as } x \rightarrow 0^{-}

When,h\rightarrow 0

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \lambda\left((0-h)^{2}-2(0-h)\right) \\\\ &=\lim _{h \rightarrow 0} \lambda\left(h^{2}+2 h\right) \\\\ &=0 \end{aligned}

At x=0 , RHL = \lim _{x \rightarrow 0^{+}} f(x)

=\lim _{x \rightarrow 0^{+}} 4 x+1

Putting x=0+h \text { as } x \rightarrow 0^{+}

When,  h\rightarrow 0

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}[4(0+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4 h+1) \\\\ &=1 \end{aligned}

LHL \neq RHL. Thus, f\left ( x \right )  is discontinuous at x=0 for any value \lambda .

At  x=1 , LHL=    \lim _{x \rightarrow 1^{-}} f(x)

=\lim _{x \rightarrow 1^{-}} 4 x+1

Putting x=1-\text {has } x \rightarrow 1^{-}

When,  h\rightarrow 0

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

At  x=1, RHL =  \lim _{x \rightarrow 1^{+}} f(x)
 

=\lim _{x \rightarrow 1^{+}} 4 x+1

Putting  x=1+h  as x\rightarrow 1^{+}

When,h\rightarrow 0

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

LHL =  RHL

Therefore, for any value of  \lambda  for which f\left ( x \right ) is continuous at  x=1

Posted by

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