If the function f(x)=x3+3(a−7)x2+3a2−9x−1 has positive points of extremum then

# If the function $f\left(x\right)={x}^{3}+3\left(a-7\right){x}^{2}+3\left({a}^{2}-9\right)x-1$ has positive points of extremum then

1. A

$a\in \left(-\mathrm{\infty },-3\right)\cup \left(3,\infty \right)$

2. B

$a\in \left(-\mathrm{\infty },-3\right)\cup \left(3,29/7\right]$

3. C

$\left(-\mathrm{\infty },7\right)$

4. D

$\left(-\mathrm{\infty },29/7\right)$

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### Solution:

We have,

$\begin{array}{r}f\left(x\right)={x}^{3}+3\left(a-7\right){x}^{2}+3\left({a}^{2}-9\right)x-1.\\ ⇒{f}^{\mathrm{\prime }}\left(x\right)=3{x}^{2}+6x\left(a-7\right)+3\left({a}^{2}-9\right)\end{array}$

f the function has a positive point of minimum, then  must have positive roots

is less than the roots of ${f}^{\mathrm{\prime }}\left(x\right)=0$

O and Discriminant

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