Question

Number of intergers in the range of 'a' so that the equation $x^3-3x+a=0$ has all its roots real and distinct,is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

Given equation is,

$x^3-3x+a=0$

DIfferentiate w.r.t. x, we get,

$f^{\prime }\left(x\right)=3x^2-3$

To find extreme values of x, put $f^{\prime }\left(x\right)=0$

$\therefore 3x^2-3=0$

$\therefore 3x^2=3$

$\therefore x^2=1$

$\therefore x=\pm 1$

$\therefore x_1=1$ and $\therefore x_2=-1$

For all roots to be real, $f\left(x_1\right).f\left(x_2\right)<0$

$\therefore [{\left(1\right)}^3-3\left(1\right)+a][{(-1)}^3-3(-1)+a]<0$

$\therefore [1-3+a][-1+3+a]<0$

$\therefore (a-2)(a+2)<0$

To satisfy this condition,

$(a-2)<0$ and $(a+2)>0$

$\therefore a<2$ and $a>-2$

$\therefore -2<a<2$

Thus, there are 3 integers between -2 and 2