| 1. The function is continuous at x = 0. |
| 2. The function is differentiable at x = 0. |
| Which of the above statements is/are correct? |
A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 nor 2
Correct Answer: A
Solution :
| [a] \[f:R\to R,f(x)=\left\{ \begin{matrix} {{x}^{2}}, & x\ge 0 \\ -x, & x<0 \\ \end{matrix} \right.\] |
| For continuity at x = 0 |
| \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,[(0-h)]=\underset{h\to 0}{\mathop{\lim }}\,h=0\] |
| \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)=\underset{h\to 0}{\mathop{\lim }}\,{{(0+h)}^{2}}=0\] |
| and \[f(0)=0\] |
| \[f(x)\] is continuous at x = 0 |
| For differentiability at \[x=0\] |
| \[\underset{h\to 0}{\mathop{\lim }}\,\frac{-(-h)-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\,=\frac{h}{-h}=-1\] |
| and \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\,\,h=0\] |
| \[f(x)\] is not differentiable at x ? 0 |
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