A) \[f(x)=({{x}^{2}}-1)\left| (x-1)(x-2) \right|\]
B) \[f(x)=\sin (\left| x-1 \right|)-\left| x-1 \right|\]
C) \[f(x)=\tan (\left| x-1 \right|)+\left| x-1 \right|\]
D) None of these
Correct Answer: C
Solution :
| [c] \[f(x)=({{x}^{2}}-1)\left| (x-1)(x-2) \right|\] |
| \[f'({{1}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{({{(1+h)}^{2}}-1)\left| h\cdot (1+h-2) \right|-0}{h}=0,f'({{1}^{-}})\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{({{(1-h)}^{2}}-1)\left| -h\cdot (1-h-2) \right|-0}{-h}=0\] |
| Hence, it is differentiable at x = 0 |
| For, \[f(x)=sin(\left| x-1 \right|)-\left| x-1 \right|\] |
| \[f'({{0}^{+}})=\underset{h\to 0}{\mathop{lim}}\,\frac{\sin \,h-h-0}{h}=0,f'({{0}^{-}})\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \left| -h \right|-\left| -h \right|}{-h}=0\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h-h}{-h}=0\] |
| Hence, \[f(x)\] is differentiable at \[x=0\] |
| For \[f(x)=\tan (\left| x-1 \right|)+\left| x-1 \right|\] |
| \[f'({{0}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{\tan \,h+h-0}{h}=2\], |
| \[f'({{0}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{\tan \left| -h \right|+\left| -h \right|}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\tan \,h+h}{-h}=-2\] |
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