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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $x = \pm \frac{|a|}{\sqrt{3}}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

Let $f(x)=(x-2)^{17}(x+5)^{24}$. Then

The perimeter of a triangle is $10 \text{ cm}$. If one of its sides is $4 \text{ cm}$,then the remaining sides of the triangle,when the area of the triangle is maximum,are

If $p(x)$ is a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2$,then $p(0)$ is equal to:

The minimum value of the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$ on the set $A = \{x \mid x^2 + 20 \le 9 x\}$ is

What is the minimum value of the sum of a real number $x$ and its reciprocal?

Vedclass Test Series

Exam Paper Generator

Online Exam Module