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Consider the following statements :
1- The function is discontinuous at $x=3$.
2- The function is not differentiable at $x=0$. What of the above statements is/are correct ?
Consider the function $f(x)=\left\{\begin{array}{ll}x^{2}-5 & x \leq 3 \\ \sqrt{x+13} & x>3\end{array} \quad[2014-I I]\right.$
Options:
1- The function is discontinuous at $x=3$.
2- The function is not differentiable at $x=0$. What of the above statements is/are correct ?
Consider the function $f(x)=\left\{\begin{array}{ll}x^{2}-5 & x \leq 3 \\ \sqrt{x+13} & x>3\end{array} \quad[2014-I I]\right.$
Solution:
1263 Upvotes
Verified Answer
The correct answer is:
Neither 1 nor 2
$1.$ $\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3)$
$\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=4$
Therefore $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=4$
$2.$ Given $f(x)=x^{2}-5 \forall x \leq 3$
$\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}$
$\mathrm{f}^{\prime}(0)=0$
So, $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$ Therefore, neither statement 1 nor 2 is correct.
$\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=4$
Therefore $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=4$
$2.$ Given $f(x)=x^{2}-5 \forall x \leq 3$
$\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}$
$\mathrm{f}^{\prime}(0)=0$
So, $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$ Therefore, neither statement 1 nor 2 is correct.
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