If the point of minima of the function f(x) = | vedclass

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(D) Given $f(x) = 1 + a^2x - x^3$. Find the derivative: $f'(x) = a^2 - 3x^2$. Setting $f'(x) = 0$ gives $3x^2 = a^2$,so $x = \pm \frac{|a|}{\sqrt{3}}$

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$\left( -3\sqrt{3}, -2\sqrt{3} \right) \cup \left( 2\sqrt{3}, 3\sqrt{3} \right)$

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(D) Given $f(x) = 1 + a^2x - x^3$. Find the derivative: $f'(x) = a^2 - 3x^2$. Setting $f'(x) = 0$ gives $3x^2 = a^2$,so $x = \pm \frac{|a|}{\sqrt{3}}$. The second derivative is $f''(x) = -6x$. For a point of minima,$f''(x) > 0$,which implies $-6x > 0$,so $x Thus,the point of minima is $x = -\frac{|a|}{\sqrt{3}}$. Now,solve the inequality $\frac{x^2 + x + 2}{x^2 + 5x + 6} The numerator $x^2 + x + 2$ has a discriminant $D = 1 - 8 = -7 Thus,we need $x^2 + 5x + 6 This holds for $x \in (-3, -2)$. Substituting $x = -\frac{|a|}{\sqrt{3}}$ into the inequality: $-3 Multiplying by $-1$ reverses the inequality: $2 Multiplying by $\sqrt{3}$: $2\sqrt{3} This implies $a \in (-3\sqrt{3}, -2\sqrt{3}) \cup (2\sqrt{3}, 3\sqrt{3})$.

If the point of minima of the function $f(x) = 1 + a^2x - x^3$ satisfies the inequality $\frac{x^2 + x + 2}{x^2 + 5x + 6} < 0$,then $a$ must lie in the interval:

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