BITSAT 2014
- Exam
- BITSAT
- Year
- 2014
- Subject
- Mathematics
- Chapter
- Continuity and Differentiability
BITSAT 2014 · Mathematics · Continuity and Differentiability
Let f(x)=\ array c (x-1) (1)/(x-1) if x ≠ 1 0 if x=1 array . Then...
Let f(x)=\left\{\right. Then which one of the following is true?
Options
- A is differentiable at and
- B is differentiable at but not at
- C is differentiable at but not at
- D is neither differentiable at nor at
Correct answer
B. is differentiable at but not at
Step-by-step solution
We have; f(x)=\left\{\right.R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}
which does not exist.
is not differentiable at Also \left.f^{\prime}(0)=\sin \frac{1}{(x-1)}-\frac{x-1}{(x-1)^{2}} \cos \left(\frac{1}{x-1}\right)\right]_{x=0}=-\sin 1+\cos 1\therefore fx=0