JEE Main 2013 (Online) 25th AprilQuestion 1
Consider the function : $f(x)=[x]+|1-x|$, $-1\le x\le3$ where $[x]$ is the greatest integer function.
Statement 1: $f$ is not continuous at $x=0,1,2$ and $3$.
Statement 2: $f(x)=\begin{cases}-x,&-1\le x<0\\1-x,&0\le x<1\\1+x,&1\le x<2\\2+x,&2\le x\le3\end{cases}$
Limits & Continuity
Solution
Concept. Open $[x]$ and $|1-x|$ on each unit interval and compare with Statement 2. Formulas used. (1) $|1-x|=1-x$ for $x\le1$, $\ x-1$ for $x\ge1$(2) $[x]=-1,0,1,2$ on $[-1,0),[0,1),[1,2),[2,3)$; $[3]=3$Step 1 - Pieces (1), (2). $[-1,0)$: $-1+1-x=-x$. $[0,1)$: $0+1-x=1-x$. $[1,2)$: $1+x-1=x$. $[2,3)$: $2+x-1=x+1$. $\ f(3)=3+2=5$. Step 2 - Statement 2. It gives $1+x$ on $[1,2)$ and $2+x$ on $[2,3]$, but the correct forms are $x$ and $x+1$. False. Step 3 - Jumps. $x=0$: left $\to0$, $f(0)=1$. $x=1$: left $\to0$, $f(1)=1$. $x=2$: left $\to2$, $f(2)=3$. $x=3$: left $\to4$, $f(3)=5$.Step 4 - Conclude. Discontinuous at $0,1,2,3$: Statement 1 true, Statement 2 false. Answer: A