Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\mathrm{\text { If } f(x)=\left\{\begin{array}{ll} 5 & ,-3<x<-1 \\ 5+2 x & ,-1 \leq x<0 \\ 5-x & , 0 \leq x<2 \\ x^2+x-3 & , 2 \leq x<3 \end{array} \text {,

\mathrm{\text { If } f(x)=\left\{\begin{array}{ll} 5 & ,-3<x<-1 \\ 5+2 x & ,-1 \leq x<0 \\ 5-x & , 0 \leq x<2 \\ x^2+x-3 & , 2 \leq x<3 \end{array} \text {, then } f(|x|)\right. \text { is }}

Option: 1

 differentiable but not continuous in (-3,3)


Option: 2

 continuous but not differentiable in (-3,3)


Option: 3

 continuous as well as differentiable in (-3,3)


Option: 4

neither continuous nor differentiable in (-3,3)


Answers (1)

best_answer

\mathrm{\text { We have, } f(x)= \begin{cases}5 & ,-3<x<-1 \\ 5+2 x & ,-1 \leq x<0 \\ 5-x & , 0 \leq x<2 \\ x^2+x-3,2 \leq x<3\end{cases}}

\mathrm{\begin{aligned} & \therefore f(|x|)= \begin{cases}5-|x| & \text {, if } 0 \leq|x|<2 \\ |x|^2+|x|-3 & \text {, if } 2 \leq|x|<3\end{cases} \\ & \Rightarrow f(|x|)= \begin{cases}5+x & \text {, if }-2<x<0 \\ 5-x & \text {, if } 0 \leq x<2 \\ x^2-x-3 & \text {, if }-3<x \leq-2 \\ x^2+x-3 & \text {, if } 2 \leq x<3\end{cases} \end{aligned}}

Clearly, f(x) is continuous in (–3, 3). But, it is not
differentiable at x = 2, –2, 0.
Hence f(|x|) is continuous but not differentiable in (–3, 3).

Posted by

Gunjita

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 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in
anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

Latest Question