Question

The maximum value of the function defined by $f\left(x\right)=min(e^x,2+e^2-x,8)$ is $\alpha$ then integral value of x satisfying the inequality $\frac{x(x-[\alpha \left]\right)}{x^2-\left[\alpha \right]x+12}<0$, where $\left[\alpha \right]$ is greater integer function of $\alpha$ , is

• A. 1
• B. 3
• C. 5
• D. 6

Answer 1

Answer: (A), (C), (D)

The correct options are
A 1
C 5
D 6

$f\left(x\right)-min(e^x,2+e^2-x,8)$
From graph, maximum value of $f\left(x\right)is,e^2$
$\therefore \left[e^2\right]=7$
$\frac{x(x-7)}{x^2-7x+12}<0\Rightarrow \frac{x(x-7)}{(x-3)(x-4)}<0$