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] \[f(x)=({{x}^{2}}-1)\left| (x-1)(x-2) \right|\] \[=({{x}^{2}}-1)\left| (x-1)(x-2) \right|\] \[=(x+1)[(x-1)\left| x-1 \right|]\left| x-2 \right|\] Which is differentiable at x=1. For \[f(x)=sin(\left| x-1 \right|)-\left| x-1 \right|,\] \[f'({{1}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h-h-0}{h}=0\] \[f'({{1}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h\left| -h \right|-\left| -h \right|}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h-h}{-\,h}=0\]Hence, \[f(x)\]is differentiable at x=1. For\[f(x)=tan(\left| x-1 \right|)+\left| x-1 \right|\], \[f'({{1}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{tan\,h+h-0}{h}=2\] \[f'({{1}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h\left| -h \right|-\left| -h \right|}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\tan \,h+h}{-h}=-2\] Hence, f(x) is non-differentiable at x=1.
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