Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

\text { Let } \begin{aligned} \mathrm{f(x)}=\left\{ \begin{matrix} \mathrm{(x-1)^2 \cos \frac{1}{x-1}-|x|, x \neq 1 }\\\mathrm{ -1, x=1 }\\ \end{matrix}\right.\end{aligned}
The set of points where \mathrm{f(x)} is not differentiable is

Option: 1

{1}


Option: 2

{0,1}


Option: 3

{0}


Option: 4

none of these


Answers (1)

best_answer

The doubtful points are \mathrm{x=0,1} because these are the turning points of the definition. \mathrm{|x|} is not differentiable at \mathrm{x=0}, while \mathrm{(x-1)^2 \cos \frac{1}{x-1}} is differentiable. The algebraic sum of a differentiable function and a nondifferentiable function is nondifferentiable. So, \mathrm{f(x)} is not differentiable at \mathrm{x=0}. Now,

\begin{aligned} \mathrm{f^{\prime}(1+0) }& =\mathrm{\lim _{h \rightarrow 0} \frac{(1+h-1)^2 \cos \frac{1}{1+h-1}-|1+h|-(-1)}{h} }\\ & =\mathrm{\lim _{h \rightarrow 0} \frac{h^2 \cos \frac{1}{h}-h}{h}=\lim _{h \rightarrow 0}\left(h \cos \frac{1}{h}-1\right)=0-1=-1 .} \end{aligned}

Similarly, \mathrm{f^{\prime}(1-0)=-1}. So, \mathrm{f(x)} is differentiable at \mathrm{x=1}.

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question