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

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Question

(c)

We have, $3 x^3-25 x+n=0$

Let $f(x)=3 x^3-25 x+n$

$f^{\prime}(x) =9 x^2-25$

$	ext { Put } f^{\prime}(x) =0$

$9 x^2-25 =0$

$x =\

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(c)

We have, $3 x^3-25 x+n=0$

Let $f(x)=3 x^3-25 x+n$

$f^{\prime}(x) =9 x^2-25$

$	ext { Put } f^{\prime}(x) =0$

$9 x^2-25 =0$

$x =\pm \frac{5}{3} \Rightarrow x_1=\frac{-5}{3}, x_2=\frac{5}{3}$

$f(x)$ has three real roots.

$	herefore\left(x_1ight) f\left(x_2ight) < 0$

$	herefore\left(3\left(\frac{-5}{3}ight)^3-25 \left(\frac{-5}{3}ight)+night)$

$\left(3\left(\frac{5}{3}ight)^3-25\left(\frac{5}{3}ight)+night) < 0$

$\left(\frac{-125}{9}+\frac{125}{3}+night)\left(\frac{125}{9}-\frac{125}{3}+night) < 0$

$\left(n+\frac{250}{9}ight)\left(n-\frac{250}{9}ight) < 0$

$n \in\left(\frac{-250}{9}, \frac{250}{9}ight]$

Hence, $n \in I$

$	herefore$ Total integer is $55$

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

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