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

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