Mathematics · 2021
JEE Main · 3 August 2021, Shift 1 · Q66
Suppose that the function f(x)= {x^2-[x]+2 a; x<-1; 4 x+5-b; -1 ≤ x<2; 3 x^2+6[x]-2 a; x ≥ 2} is continuous on (-2,3) for some values of a and b…
Suppose that the function
$$f(x)= \begin{cases}x^2-[x]+2 a ; & x<-1 \\ 4 x+5-b ; & -1 \leq x<2 \\ 3 x^2+6[x]-2 a ; & x \geq 2\end{cases}
$$
is continuous on $\displaystyle (-2,3)$ for some values of $\displaystyle a$ and $\displaystyle b$ where $\displaystyle [x]$ denotes the greatest integer function. Then the value of $\displaystyle 8 a b$ is
Official answer
(2)
$\displaystyle -117$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.