JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
Suppose that the function $f(x)=\begin{cases}x^{2}-[x]+2a; & x<-1\\ 4x+5-b; & -1\le x<2\\ 3x^{2}+6[x]-2a; & x\ge2\end{cases}$ is continuous on $(-2,3)$ for some values of $a$ and $b$ where $[x]$ denotes the greatest integer function. Then the value of $8ab$ is
Limits & Continuity
Solution
At $x=-1$: left $=1+2+2a=3+2a$ (since $[x]=-2$), right $=-4+5-b=1-b$, so $2a+b=-2$. At $x=2$: left $=8+5-b=13-b$, right $=12+12-2a=24-2a$, so $2a-b=11$. Adding: $4a=9\Rightarrow a=\tfrac94$, $b=-\tfrac{13}{2}$. $8ab=8\cdot\tfrac94\cdot(-\tfrac{13}{2})=-117$. Option (B).