Question

Statement I The function
$f(x)=\begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}$ is continuous everywhere except at $x=1$.
Statement II The function
$f(x)= \begin{cases}-2, & \text { if } x \leq-1 \\ 2 x, & \text { if }-1 < x \leq 1 \\ 2, & \text { if } x>1\end{cases}$ is continuous for every value of $x$.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (B)

l. Here, $f(x)=\begin{cases}2 x, \text { if } x<0 \\ 0, \text { if } 0 \leq x \leq 1 \\ 4 x, \text { if } x>1\end{cases}$
For $x<0, f(x)=2 x ; 0 \leq x<1, f(x)=0$ and $x>1$, $f(x)=4 x$ are polynomial and constant function, so it is continuous in the given interval. So, we have to check the continuity at $x=0$ and 1 .
At $x=0, LHL =\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{x \rightarrow 0^{-}}(2 x)$
Putting $x=0-h$ as $x \rightarrow 0^{-}, h \rightarrow 0$
$\therefore \displaystyle\lim _{h \rightarrow 0}[2(0-h)]=\displaystyle\lim _{h \rightarrow 0}(-2 h)=-2 \times 0=0$,
$R H L=\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{x \rightarrow 0^{+}}(0)=0$
Also, $f(0)=0$
$\therefore LHL = RHL =f(0)$
Thus, $f(x)$ is continuous at $x=0$
At $x=1, LHL =\displaystyle\lim _{x \rightarrow 1^{-}} f(x)=\displaystyle\lim _{x \rightarrow 1^{-}}(0)=0$,
$RHL =\displaystyle\lim _{x \rightarrow 1^{+}} f(x)=\displaystyle\lim _{x \rightarrow 1^{+}}(4 x)$
Putting $x=1+h$ as $x \rightarrow 1^{+}$when $h \rightarrow 0$
$\displaystyle \lim _{h \rightarrow 0}[4(1+h)]=\displaystyle\lim _{h \rightarrow 0}(4+4 h)=4+4 \times 0=4 $
$ \therefore LHL \neq RHL .$
Thus, $f(x)$ is continuous everywhere except at $x=1$.
II. Here, $f(x)=\begin{cases}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-1 < x \leq 1 \\ 2, \text { if } x>1\end{cases}$
for $x < -1, f(x)=-2 ;-1 < x < 1, f(x)=2 x$ and for $x>1, f(x)=2, a$ constant and polynomial function, thus continuous.
We have to check the continuity only at $x=-1$ and 1 .
At $x=-1, LHL =\displaystyle\lim _{x \rightarrow-1^{-}} f(x)=\displaystyle\lim _{x \rightarrow-1^{-}}(-2)=-2$
RHL $=\lim _{x \rightarrow-1^{+}} f(x)=\displaystyle\lim _{x \rightarrow-1^{+}}(2 x)$
Putting $x=-1+h$ as $x \rightarrow-1^{+}, h \rightarrow 0$
$\therefore \displaystyle \lim _{h \rightarrow 0} 2(-1+h)=\displaystyle\lim _{h \rightarrow 0}(-2+2 h)=-2+0=-2$
Also, $ f(-1)=-2$
$\therefore LHL = RHL =f(-1)$.
Thus, $f(x)$ is continuous at $x=-1$.
At $x=1, LHL =\displaystyle\lim _{x \rightarrow 1} f(x)=\displaystyle\lim _{x \rightarrow 1}(2 x)$
Putting $x=1-h$ as $h \rightarrow 1^{-}, h \rightarrow 0$
$ \therefore \displaystyle\lim _{h \rightarrow 0}[2(1-h)]=\displaystyle\lim _{h \rightarrow 0}(2-2 h)=2-2 \times 0=2$
$ RHL -\displaystyle\lim _{x \rightarrow 1^{+}} f(x)-\displaystyle\lim _{x \rightarrow 1^{+}}(2)-2$
Also, $f(1) = 2\times 1 = 2 \,\,[\because f(x) = 2x]$
$\therefore LHL = RHL = f(1)$
Thus, $f(x)$ is continuous at $x=1$.
Hence, $f(x)$ is continuous for every value of $x$.