Question

Assertion :Let $f\left(x\right)=\{\begin{array}{cc}1+x& x<0\\ 1+\left[x\right]+\sin x& 0\le x\le \pi /2\\ 3& x\ge \pi /2\end{array}$
f is continuous on R $-\left\{1\right\}$ Reason: The greatest integer function is discontinuous at every integer.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

$f\left(0\right)=1=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(1+x)=1$
$=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(1+[x]+\sin x)=1$
So f is continuous at $x=0$.
Since [x] is right continuous but not left continuous at $x=1$ so also is f.
$f\left(\pi /2\right)=3=\underset{x\to \pi /2^{+}}{lim}f\left(x\right)$
$\underset{x\to \pi /2^{-}}{lim}f\left(x\right)=\underset{x\to \pi /2^{-}}{lim}(1+[x]+\sin x)$
$=1+\underset{x\to \pi /2^{-}}{lim}\left[x\right]+\underset{x\to \pi /2^{-}}{lim}\sin x$
$=1+1+1=3$
So f is continuous at $x=\pi /2$. Thus f is continuous on $R\sim \left\{1\right\}$. Also [x] is not continuous at every $x\in I$