Limit Continuity And Differentiability Ques 102
- If $f(x)=[x]-\frac{x}{4}, x \in R$ where $[x]$ denotes the greatest integer function, then
(a) $\lim _{x \rightarrow 4+} f(x)$ exists but $\lim _{x \rightarrow 4-} f(x)$ does not exist
(b) $f$ is continuous at $x=4$
(c) Both $\lim _{x \rightarrow 4-} f(x)$ and $\lim _{x \rightarrow 4+} f(x)$ exist but are not equal
(d) $\lim _{x \rightarrow 4-} f(x)$ exists but $\lim _{x \rightarrow 4+} f(x)$ does not exist
Show Answer
Answer:
Correct Answer: 102.(d)
Solution:
- Given, $f(x)=\begin{array}{ll}1+x, & 0 \leq x \leq 2 \ 3-x, & 2<x \leq 3\end{array}$
$$ \begin{alignedat} & \therefore \quad f \circ f(x)=f[f(x)]=\begin{array}{ll} 1+f(x), & 0 \leq f(x) \leq 2 \102-f(x), & 2<f(x) \leq 3 \end{array} \\ & 1+f(x), \quad 0 \leq f(x) \leq 1 \quad 1+(3-x), \quad 2<x \leq 3 \\ & \Rightarrow \quad f \circ f=1+f(x), \quad 1<f(x) \leq 2=1+x, \quad 0 \leq x \leq 1 \\ & 3-f(x), \quad 2<f(x) \leq 3 \quad 3-(1+x), \quad 1<x \leq 2 \\ & 4-x, \quad 2<x < 3 \\ & \Rightarrow \quad(f \circ f)(x)=2+x, \quad 0 \leq x \leq 1 \\ & 2-x, \quad 1<x \leq 2 \end{aligned} $$
Now, $\quad \operatorname{RHL}($ at $x=2$)=$2$ and LHL ($ at $x=2$)=$0$
Also, RHL (at $x=1$) = 1 and LHL (at $x=1$) = 3
Therefore, $f(x)$ is discontinuous at $x=1,2$
$\therefore f[f(x)]$ is discontinuous at $x={1,2}$.
BETA
