The number of integers n for which 3x^3-25x+n | vedclass

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(C) Let $f(x) = 3x^3 - 25x + n$.For the cubic equation to have three real roots,the local maximum and local minimum must have opposite signs.$f'(x) =

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(C) Let $f(x) = 3x^3 - 25x + n$.For the cubic equation to have three real roots,the local maximum and local minimum must have opposite signs.$f'(x) = 9x^2 - 25$.Setting $f'(x) = 0$,we get $x^2 = \frac{25}{9}$,so $x = \pm \frac{5}{3}$.Let $x_1 = -\frac{5}{3}$ and $x_2 = \frac{5}{3}$.$f(x_1) = 3(-\frac{125}{27}) - 25(-\frac{5}{3}) + n = -\frac{125}{9} + \frac{125}{3} + n = n + \frac{250}{9}$.$f(x_2) = 3(\frac{125}{27}) - 25(\frac{5}{3}) + n = \frac{125}{9} - \frac{125}{3} + n = n - \frac{250}{9}$.For three real roots,$f(x_1) \cdot f(x_2) This implies $-\frac{250}{9} Since $\frac{250}{9} \approx 27.77$,the integers $n$ range from $-27$ to $27$.The number of such integers is $27 - (-27) + 1 = 55$.

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

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