Question

Statement I The function
$f(x)=\{\begin{array}{ll}2x,& \text{ if }x<0\\ 0,& \text{ if }0\le x\le 1\\ 4x,& \text{ if }x>1\end{array}$ is continuous everywhere except at $x=1$.
Statement II The function
$f(x)=\{\begin{array}{ll}-2,& \text{ if }x\le -1\\ 2x,& \text{ if }-1<x\le 1\\ 2,& \text{ if }x>1\end{array}$ 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, $f(x)=\{\begin{array}{l}2x,\text{ if }x<0\\ 0,\text{ if }0\le x\le 1\\ 4x,\text{ if }x>1\end{array}$
For $x<0,f(x)=2x;0\le x<1,f(x)=0$ and $x>1$, $f(x)=4x$ 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 $x=0,LHL=\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }(2x)$
Putting $x=0-h$ as $x\to 0^{-},h\to 0$
$\therefore \underset{h\to 0}{\lim }[2(0-h)]=\underset{h\to 0}{\lim }(-2h)=-2\times 0=0$,
$RHL=\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }(0)=0$
Also, $f(0)=0$
$\therefore LHL=RHL=f(0)$
Thus, $f(x)$ is continuous at $x=0$
At $x=1,LHL=\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }(0)=0$,
$RHL=\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(4x)$
Putting $x=1+h$ as $x\to 1^{+}$when $h\to 0$
$\underset{h\to 0}{\lim }[4(1+h)]=\underset{h\to 0}{\lim }(4+4h)=4+4\times 0=4$
$\therefore LHL\ne RHL.$
Thus, $f(x)$ is continuous everywhere except at $x=1$.
II. Here, $f(x)=\{\begin{array}{l}-2,\text{ if }x\le -1\\ 2x,\text{ if }-1<x\le 1\\ 2,\text{ if }x>1\end{array}$
for $x<-1,f(x)=-2;-1<x<1,f(x)=2x$ 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 $x=-1,LHL=\underset{x\to -1^{-}}{\lim }f(x)=\underset{x\to -1^{-}}{\lim }(-2)=-2$
RHL $={\lim }_{x\to -1^{+}}f(x)=\underset{x\to -1^{+}}{\lim }(2x)$
Putting $x=-1+h$ as $x\to -1^{+},h\to 0$
$\therefore \underset{h\to 0}{\lim }2(-1+h)=\underset{h\to 0}{\lim }(-2+2h)=-2+0=-2$
Also, $f(-1)=-2$
$\therefore LHL=RHL=f(-1)$.
Thus, $f(x)$ is continuous at $x=-1$.
At $x=1,LHL=\underset{x\to 1}{\lim }f(x)=\underset{x\to 1}{\lim }(2x)$
Putting $x=1-h$ as $h\to 1^{-},h\to 0$
$\therefore \underset{h\to 0}{\lim }[2(1-h)]=\underset{h\to 0}{\lim }(2-2h)=2-2\times 0=2$
$RHL-\underset{x\to 1^{+}}{\lim }f(x)-\underset{x\to 1^{+}}{\lim }(2)-2$
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 $x$.