Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 3 maths textbook solution

please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 3 maths textbook solution

Answers (1)

(a)

HINTS: Understand the  definition of differentiability and modulus function/

GIVEN: f(x)=x|x|

SOLUTION:

f(x)=\left\{\begin{matrix} x(-x) & & x<0\\ x(x)& & x\geq 0 \end{matrix}\right.\\

           =\left\{\begin{matrix} -x^{2} & & x<0\\ x^{2}& & x\geq 0 \end{matrix}\right.

For x<0 and x>0 function is differentiable as it is a polynomial function.

Now at x=0

LHD=

\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{f-(-h)^{2}-0}{-h} \\ \end{aligned}

                                     \begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\left(-h^{2}\right)-0}{-h} \\ &=\lim _{h \rightarrow 0}-h \\ &=0 \end{aligned}

RHD  at x=0

\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}-0}{h} \\ &=\lim _{h \rightarrow 0} \mathrm{~h} \\ &=0 \end{aligned}

As LHD at x=0 and RHD at x=0

Hence, f(x) is  differentiable at (-\infty ,\infty )

Posted by

Info Expert 29

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10, 12 compartment 2024 exams today for 2.5 lakh students; exam timings, guidelines
anam-khan

CBSE compartment admit card 2024 out at cbse.gov.in; exams from July 15
anam-khan

Reports saying CBSE cannot conduct board exams twice a year currently ‘totally incorrect’; board denies claims

Latest Question