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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