Question

The number of integers $n$ for which $3x^3-25x+n=0$ has three real roots is$?$

• A. $1$
• B. $25$
• C. $55$
• D. Infinite

Answer 1

Answer: (C)

$55$

Given cubic is $f\left(x\right)=3x^3-25x+n$.

Using the fact that, between two roots of derivative, a function has at most 1 root.

$f^{\prime }\left(x\right)=9x^2-25$

$\Rightarrow f^{\prime }\left(x\right)=0$ for $x=\pm \frac{5}{3}$

$f\left(\frac{5}{3}\right)=3\times (\frac{5}{3}{)}^3-25\times \frac{5}{3}+n=n-\frac{250}{9}$

$f(-\frac{5}{3})=n+\frac{250}{9}$

For the cubic to have $3$ real roots, $f(-\frac{5}{3})>0\text{and}f\left(\frac{5}{3}\right)<0$

i.e., $-\frac{250}{9}<n<\frac{250}{9}$

There are $55$ such $n$ possible. i.e., $n\in [-27,27]$