Question

Number of integers k for which the equation $x^3–27x+k=0$ has at least two distinct integer roots is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let $f\left(x\right)=x^3–27x+k$
$f^{\prime }\left(x\right)=3x^2-27=0$
$\Rightarrow 3x^2=27$
$\Rightarrow x^2=9$
$\Rightarrow x=\pm 3$ and
$f^{\prime \prime }\left(x\right)=6x$
$\Rightarrow f^{\prime \prime }\left(3\right)>0$
$\Rightarrow f^{\prime \prime }(-3)<0$
So $x=3$ is local minima and $x=-3$ is local maxima.
For the function to have exactly two roots either $f(-3)=0$ or $f\left(3\right)=0$
$f\left(3\right)=0$
$\Rightarrow (3{)}^3–27(3)+k=0$
$\Rightarrow \left(27\right)-81+k=0$
$\Rightarrow -54+k=0$
$\Rightarrow k=54$
$f(-3)=0$
$\Rightarrow (-3{)}^3–27(-3)+k=0$
$\Rightarrow -27+81+k=0$
$\Rightarrow 54+k=0$
$\Rightarrow k=-54$
Suppose that the value of the function at local maximum is lesser than 0 or if the value of the function at local minimum is greater than 0, then the curve of the function would not touch the $x$-axis at all and would intersect the $x$-axis only once.
The equation will have one real and two complex roots.
For $k<-54$ and $k>54$, equation will have complex roots.
So, to have at least two distinct real roots $k\in [-54,54]$
Therefore, number of integral values of $k$ is $109$.