JEE Main 2022 (Online) 27th July Morning ShiftQuestion 1
Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined as $f(x)=a\sin\left(\frac{\pi[x]}{2}\right)+[2-x]$, $a\in\mathbb{R}$, where $[t]$ is the greatest integer less than or equal to $t$. If $\lim_{x\to-1}f(x)$ exists, then the value of $\int_0^4f(x)\,dx$ is equal to
Limits & Continuity
Solution
Left limit at $-1$: $[x]=-2$, $\sin(-\pi)=0$, $[2-x]=3$, so $3$. Right limit: $[x]=-1$, $[2-x]=2$, giving $-a+2$. Equal $\Rightarrow a=-1$. For non-integer $x$, $[2-x]=1-[x]$, so $\int_0^4[2-x]dx=4-(0+1+2+3)=-2$. $\int_0^4 -\sin(\pi[x]/2)dx=-(0+1+0-1)=0$. So the integral is $-2$: option (B).