Question

The value of $a$ for which $x^3-3x+a=0$ has two distinct roots in $[0,1]$ is given by

• A. $-1$
• B. $1$
• C. $3$
• D. $doesnotexists$

Answer 1

Answer: (D)

$doesnotexists$

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

Let $f\left(x\right)=x^3-3x+a$

$⟹f^{{}^{\prime }}\left(x\right)=3x^2-3=0$

$⟹x^2-1=0$

$⟹x=\pm 1$ are the critical points.

$f^{{}^{\prime \prime }}\left(x\right)=6x$

On substituting the critical points in $f^{{}^{\prime \prime }}\left(x\right)$ we get

$f^{{}^{\prime \prime }}\left(1\right)=6>0⟹x=1\text{is the local minimum}$ and

$f^{{}^{\prime \prime }}(-1)=-6<0⟹x=-1\text{is the local maximum}$

For $f\left(x\right)$ to have two distinct roots we must have either $f\left(1\right)=0$ or $f(-1)=0$. Since, to have real distinct roots, the function should touch the X-axis just once and intersect it once.

$f\left(1\right)=1-3+a=0⟹a=2$ and

$f(-1)=-1+3+a=0⟹a=-2$

Therefore, to have distinct roots, the value of $a$ should be in $[-2,2]$

Let $a=0\in [-2,2]$

Therefore, $f\left(x\right)=x^3-3x+a$

$⟹x^3-3x+0=0$

$⟹x(x^2-3)=0$

$⟹x=0\in [0,1]$ or $x=\sqrt{3}\notin [0,1]$ or $x=-\sqrt{3}\notin [0,1]$

Therefore, only one root is lying in $[0,1]$. Hence, $a$ value doesn't exist to have two roots in the interval $[0,1]$