If $f:[0,3] \rightarrow[0,3]$ is defined by $f(x)$ $=\left\{\begin{array}{ll}1+x, & 0 \leq x \leq 2 \\ 3-x,…

1. Continuous at $x=1$
2. Continuous at $x=2$
3. Discontinuous at $x=1$ and $x=2$
4. Continuous on $[0,3]$

Solution

$ \begin{gathered} g(x)=f(f(x))= \begin{cases}f(1+x) ; & 0 \leq x \leq 2 \\ f(3-x) & ; \quad 2 < x \leq 3\end{cases} \\ = \begin{cases}f(1+x) ; & 0 \leq x \leq 1 \\ f(1+x) ; & 1 \leq x \leq 2 \\ f(3-x) ; & 2 < x \leq 3\end{cases} \end{gathered} $ $ \begin{aligned} & x \in[0,1] \Rightarrow(1+x) \in[1,2] \quad x \in[1,2] \Rightarrow(1+x) \in[2,3] \\ & x \in[2,3] \Rightarrow(3-x) \in[0,1] \end{aligned} $ Hence, $ g(x)= \begin{cases}f(1+x) ; & \text { for } 0 \leq x \leq 1 \Rightarrow 1 \leq x+1 \leq 2 \\ f(1+x) ; & \text { for } 1 \leq x \leq 2 \Rightarrow 2 \leq x+1 \leq 3 \\ f(3-x) ; & \text { for } 2 < x \leq 3 \Rightarrow 0 \leq 3-x \leq 1\end{cases} $ Now, if $(1+x) \in[1,2]$ then,
Using Eqs, (i), (ii), (iii), we get $ g(x)= \begin{cases}2+x & ; \quad 0 \leq x < 1 \\ 2-x ; & 1 < x \leq 2 \\ 4-x ; & 2 < x \leq 3\end{cases} $ Here, as $g(x)$ change the inequality sign at $x=1$ and $x=2$ Thus, to check continuity at $x=1$ and $x=2$ Now, we will check the continuity of $g(x)$ at $ \begin{aligned} & x=1,2 \\ \text { At } x=1, \quad \text { LHL } & =\lim _{x \rightarrow 1^{-}} g(x)=\lim _{x \rightarrow 1^{-}}(2+x)=3 \\ \text { RHL } & =\lim _{x \rightarrow 1^{+}} g(x)=\lim _{x \rightarrow 1^{+}}(2-x)=1 \end{aligned} $ As. LHL $\neq$ RHL $g(x)$ is discontinuous at $x=1$. $ \begin{aligned} \text { At } x=2 \quad \text { LHL } & =\lim _{x \rightarrow 2^{-}} g(x)=\lim _{x \rightarrow 2^{-}}(2-x)=0 \\ \text { RHL } & =\lim _{x \rightarrow 2^{+}} g(x)=\lim _{x \rightarrow 2^{+}}(4-x)=2 \end{aligned} $ As LHL $\neq$ RHL, $g(x)$ is discontinuous at $x=2$ Thus, $g(x)$ is continuous for all $x \in[0,1) \cup(1,2)$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

Practice more Functions questions on Aicharya