Question

If $a$ is a real root of $2x^3-3x^2+6x+6=0$, then $[\propto ]$ equals

• A. 0
• B. -1
• C. 1
• D. -2

Answer 1

The correct option is B -1

Let $f\left(x\right)=2x^2-3x^2+6x+6$
$f\left(x\right)=6x^2-6x+6=6(x^2-x+1)>0$ for all $x\in R$
$f\left(x\right)$ is increase on R
$f\left(x\right)$ has only real root.
$\Rightarrow f\left(0\right)=6$ and $f(-1)=-5$
So, Real root $a$ of $f\left(x\right)$ lies in interval $(-1,0)$.
Hence $\left[a\right]=-1$