If the function f(x) = 2x^3 - 9ax^2 + 12a^2x | vedclass

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

Question

(D) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ with $a > 0$. First,find the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$. Set $f'(x) = 0$ to f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ with $a > 0$. First,find the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$. Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0$,which factors to $6(x - a)(x - 2a) = 0$. Thus,the critical points are $x = a$ and $x = 2a$. Find the second derivative: $f''(x) = 12x - 18a$. At $x = a$,$f''(a) = 12a - 18a = -6a  0$),so $x = a$ is a point of local maximum. Thus,$p = a$. At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a > 0$,so $x = 2a$ is a point of local minimum. Thus,$q = 2a$. Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$. Since $a > 0$,we can divide by $a$ to get $a = 2$.

If the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$,where $a > 0$,attains its maximum and minimum values at $p$ and $q$ respectively,such that $p^2 = q$,then what is the value of $a$?

Similar Questions

The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five observations are $2, 4, 10, 12, 14,$ then the absolute difference of the remaining two observations is

In a chloroplast,chlorophyll is found in:

The angle between the lines joining the origin to the points of intersection of the line $x + 2y + 1 = 0$ and the curve $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ is

If $f(x)=|x|+|sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ is

The foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide. Then the value of $b^{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module