| 1. \[\underset{x\to 1}{\mathop{\lim \,f}}\,(x)\] does not exist. |
| 2. f(x) is differentiable at x = 0 |
| 3. f(x) is continuous at x = 0 |
| Which of the above statements is/are correct? |
A) 1 only
B) 3 only
C) 2 and 3 only
D) 1 and 3 only
Correct Answer: D
Solution :
| [d] For \[x\ge 0\] |
| \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,2+x=2+1=3\] |
| For \[x<0\] |
| \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,2-x=2-1=1\] |
| So, \[\underset{x\to 1}{\mathop{\lim }}\,f(x)\] does not exist. |
| At \[x=0\] |
| \[RHL:\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,f(0+h)=\underset{h\to 0}{\mathop{\lim }}\,2+h=2\]\[LHL:\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,2-h=2\] |
| \[f(0)=2+0=2.\] |
| So, RHL = LHL = f(0) |
| \[\Rightarrow f(x)\] is continuous at \[x=0\] |
| Differentiability at \[x=0\] |
| \[LHD:\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{2+h-2}{-h}\] |
| \[=\frac{-h}{h}=-1\] |
| \[RHD:\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2+h-2}{h}=1\] |
| Since \[LHD\ne RHD\] |
| So, \[f(x)\] is not differentiable at \[x=0\] |
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