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

A
$a \in (- \infty, - 3) \cup (3, \infty)$
B
$a \in (- \infty, - 3) \cup (3, 29 / 7]$
C
$(- \infty, 7)$
D
$(- \infty, 29 / 7)$

Solution:

We have,

$\begin{aligned} f (x) = x^{3} + 3 (a - 7) x^{2} + 3 (a^{2} - 9) x - 1 . \\ \Rightarrow f^{'} (x) = 3 x^{2} + 6 x (a - 7) + 3 (a^{2} - 9) \end{aligned}$

f the function $f (x)$ has a positive point of minimum, then $f' (x) = 0$ must have positive roots

$\Rightarrow x = 0$ is less than the roots of $f^{'} (x) = 0$

$\Rightarrow f^{'} (0) > 0$ O and Discriminant $\geq 0$

$\Rightarrow 4 (a - 7)^{2} - 4 (a^{2} - 9) \geq 0 and 3 (a^{2} - 9) > 0$

$\begin{array}{l} \Rightarrow 7 a - 29 \leq 0 and a^{2} - 9 > 0 \\ \Rightarrow a \leq \frac{29}{7} and (a < - 3 or a > 3) \Rightarrow a \in (- \infty, - 3) \cup (3, 29 / 7] \end{array}$

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

Solution:

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