For a 0,if the function f(x) = 2x^3 - 9ax^2 | vedclass

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

Question

(C) Given,$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. To find the critical points,we calculate the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$. Setting $f'(x) =

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given,$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. To find the critical points,we calculate the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$. Setting $f'(x) = 0$,we get $6(x^2 - 3ax + 2a^2) = 0$,which factors as $6(x - a)(x - 2a) = 0$. Thus,the critical points are $x = a$ and $x = 2a$. We find the second derivative: $f''(x) = 12x - 18a$. For $x = a$,$f''(a) = 12a - 18a = -6a$. Since $a > 0$,$f''(a) For $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a$. Since $a > 0$,$f''(2a) > 0$,so $f(x)$ has a minimum at $q = 2a$. Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$. Since $a > 0$,we can divide by $a$ to get $a = 2$.

For $a > 0$,if the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ attains its maximum value at $p$ and minimum value at $q$ such that $p^2 = q$,then $a =$

Similar Questions

If $f(x) = x + \frac{1}{x}$ for $x > 0$,then its greatest value is:

For the function $f(x) = \int_{0}^{x} \frac{\sin t}{t} dt$,where $x > 0$,which of the following is true?

If the function $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2}=q$,then $a$ equals

Let $f(x) = \begin{cases} |x| & \text{for } 0 < |x| \leqslant 2 \\ 1 & \text{for } x = 0 \end{cases}$. What is the nature of $f$ at $x = 0$?

$A$ stone thrown upwards has its equation of motion $s = 490t - 4.9t^2$. Then the maximum height reached by it is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module