Question

The real values of 'a' for the equation $x^3-3x+a=0$ has three real and distinct roots, where ${}^{\prime }{x}^{\prime }$ lies $[-1,1]$ is

• A. $-2<a<2$
• B. $a\le -2$
• C. $a\ge -2$
• D. $-2<a<\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

$-2<a<2$

Given equation

$x^3-3x+a=0$ ……. (1)

Consider put $x=2\cos \theta$ where $-1\le \cos \theta \le 1$ then $0\le \theta \le \pi$

Now, By equation (1)

$8{\cos }^3\theta -6\cos \theta +a=0$

$8{\cos }^3\theta -6\cos \theta =-a$

$2(4{\cos }^3\theta -3\cos \theta )=-a$

$(4{\cos }^3\theta -3\cos \theta )=\frac{-a}{2}$

$\cos 3\theta =\frac{-a}{2}$

Now,

$-1\le \cos 3\theta \le 1$

So, $-1\le \frac{-a}{2}\le 1$

Then, $-2\le -a\le 2$

Hence, we get the value of $a=\{-2,-1,0,1,2\}$

Hence option $\left(a\right)$ is correct.