Question

The set of values of 'a'for which the equation $x^3-3x+a=0$ has three distinct real roots , is

• A. ( -¥, ¥ )
• B. (-2, 2)
• C. ( -1, 1)
• D. none of these

Answer 1

The correct option is B (-2, 2)


Solution : $f\left(x\right)=x^3-3x+a$
$f^{\prime }\left(x\right)=3x^2-3$
$f^{\prime }\left(x\right)=0$ at $x=-1$
Coefficient of $x^3$ is positive
$f\left(x\right)$ has three distinct real root is
$i.e.f(-1)<0$ and $f\left(1\right)>0$
$-1+3+a>0$ and $1-3+a<0$
$a=-2$ and $a=2$
$a\in (-2,2)$