Get Answers to all your Questions

header-bg qa

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 41

Answers (1)

Answer:

                k  can be any real number.

Hint:

 f\left ( x \right ) must be defined. The limit of the f\left ( x \right )  approaches the value x must exist.

Given:

                f(x)=\left\{\begin{array}{l} 2 x^{2}+k, \text { if } x \geq 0 \\ -2 x^{2}+k, \text { if } x<0 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{l} 2 x^{2}+k, \text { if } x \geq 0 \\ -2 x^{2}+k, \text { if } x<0 \end{array}\right.

We have

(LHL at x=0)

                \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}- 2-(h)^{2}+k=k

(RHL at x=0 )

                \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} 2(h)^{2}+k=k

If f\left ( x \right ) is continuous at x=0 .

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=k \end{aligned}

k can be any real number.

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

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