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  • Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 16 maths textbook solution

Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 16 maths textbook solution

Answers (1)

Answer:

 No point of discontinuity.

Given:

 f(x)=\left | x \right |-\left | x+1 \right |

Explanation:

f(x)=\left | x \right |-\left | x+1 \right |

\begin{gathered} =\left\{\begin{array}{c} -x+x+1, x<-1 \\ -x-(x+1),-1 \leq x<0 \\ x-(x+1), x \geq 0 \end{array}\right\} \\ =\left\{\begin{array}{c} 1, x<-1 \\ -2 x-1,-1 \leq x<0 \\ -1, x \geq 0 \end{array}\right\} \end{gathered}

Continuity at x = -1

L.H.L

\begin{aligned} &\lim _{x \rightarrow-1^{-}} f(x) \\ &=\lim _{x \rightarrow-1^{-}} 1=1 \end{aligned}

R.H.L

\begin{aligned} &\lim _{x \rightarrow-1^{+}} f(x) \\ &\lim _{x \rightarrow-1^{+}}(-2 x-1) \\ &=-2 \times-1-1=1 \end{aligned}

And

\begin{aligned} &f(-1)=-2 \times-1-1=1 \\ &\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} f(x)=f(-1) \end{aligned}

F(x) is continuous at x=-1

Continuous at x=0

L.H.L

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(-2 x-1) \\ &=-2 \times 0-1=-1 \end{aligned}

R.H.L

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}-1=-1

And

f(0)=-2\times 0-1=-1

L.H.L=R.H.L=f(0)

F(x) is continuous at x=0, hence continuous everywhere

Posted by

Gurleen Kaur

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