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Question

The domain of the function $$f(x)=\begin{cases} \left( { x }^{ 2 }-9 \right) /\left( x-3 \right) ,if\quad x\neq 3 \\ 6,\quad if\quad x=3 \end{cases}$$ is


A
(0,3)
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B
(,3)
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C
(,)
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D
(3,)
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E
(3,3)
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Solution

The correct option is C $$\left( -\infty ,\infty \right) $$
Given $$f(x)=\begin{cases} \cfrac { { x }^{ 2 }-9 }{ x-3 } ,x\neq 3 \\ 6,x=3 \end{cases}$$ at $$x=3$$
$$\quad LHL=\lim _{ x\rightarrow { 3 }^{ - } }{ \cfrac { { x }^{ 2 }-9 }{ x-3 }  } =\lim _{ x\rightarrow { 3 }^{ - } }{ (x+3) } =\lim _{ h\rightarrow 0 }{ (3-h+3) } =6$$
$$RHL=\lim _{ x\rightarrow { 3 }^{ + } }{ \cfrac { { x }^{ 2 }-9 }{ x-3 }  } =\lim _{ x\rightarrow { 3 }^{ +- } }{ (x+3) } =\lim _{ h\rightarrow 0 }{ (3+h+3) } =6\quad $$
$$\Rightarrow f(3)=6\quad $$
$$\therefore$$ LHL=RHL
Hence, $$f(x)$$ is continuous at $$x=3$$
$$\therefore$$ Domain of $$f(x)=\left( -\infty ,\infty  \right) $$

Mathematics

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