Question

The number or real roots of the equation $4x^3-6x^2+12x+9=0$ is

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is A $1$

Let $f\left(x\right)=4x^3-6x^2+12x+9$
$\therefore$ $f^{{}^{\prime }}\left(x\right)-12x^2-12x+12$
$=12{(x-\frac{1}{2})}^2+9>0$ for all $x\in R.$
Hence the function is strictly increasing .
$f(-1)=4(-1)-6\left(1\right)+12(-1)+9<0,f\left(1\right)=19>0$
$\therefore$ there is an $\alpha \in (-1,1)$ at which $f\left(\alpha \right)=0$ and there is no other root as f (x) is increasing .