Question

Find the range of values of the parameter $a$ so that $x^3-3x+a=0$ has three real and distinct roots.

• A. $(-1,1)$
• B. $(-2,2)$
• C. $(-3,3)$
• D. $(-4,4)$

Answer 1

Answer: (B)

$(-2,2)$

Let $f\left(x\right)=x^3-3x+a$
$f^{\prime }\left(x\right)=3x^2-3=3(x-1)(x+1)$
$f^{\prime }\left(x\right)=0\Rightarrow x=1,-1$
At $x=1$, $f^{\prime \prime }\left(x\right)>0$, hence $x=1$ is a point of maxima and similarly $x=-1$ is a point of minima.
For real roots, the maxima should have a value $>0$ and the minima should be $<0$.
Now, $f\left(1\right)=a-2$, $f(-1)0=a+2$
Hence,
$a-2<0$ and $a+2>0$
Thus, the given equation would have real and distinct roots, if $a\in (-2,2)$.
Hence, option 'B' is correct.