Let a 0. If the function f(x) = 6x^3 - 45ax | vedclass

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

Question

(B) Given $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$. To find the local maximum and minimum,we find the derivative $f'(x)$: $f'(x) = 18x^2 - 90ax + 108a^2$.

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) Given $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$. To find the local maximum and minimum,we find the derivative $f'(x)$: $f'(x) = 18x^2 - 90ax + 108a^2$. Setting $f'(x) = 0$ for critical points: $18(x^2 - 5ax + 6a^2) = 0$ $18(x - 2a)(x - 3a) = 0$. Thus,the critical points are $x = 2a$ and $x = 3a$. Since $f''(x) = 36x - 90a$,we check the nature of the points: $f''(2a) = 36(2a) - 90a = -18a $f''(3a) = 36(3a) - 90a = 18a > 0$ (Local minimum at $x_2 = 3a$). Given $x_1x_2 = 54$,we have $(2a)(3a) = 54$,so $6a^2 = 54$,which gives $a^2 = 9$. Since $a > 0$,we have $a = 3$. Then $x_1 = 2(3) = 6$ and $x_2 = 3(3) = 9$. Finally,$a + x_1 + x_2 = 3 + 6 + 9 = 18$.

Let $a > 0$. If the function $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ respectively such that $x_1x_2 = 54$,then $a + x_1 + x_2$ is equal to:

Similar Questions

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$.

The figure shown below is the graph of the derivative of some function $y=f(x)$. Then,

The lengths of the sides of a triangle are $10+x^2$,$10+x^2$ and $20-2x^2$. If for $x=k$,the area of the triangle is maximum,then $3k^2$ is equal to

$x^x$ has a stationary point at

Find the point where $f(x) = 2\sin x + \cos 2x$ has a maximum value in the interval $0 \leq x \leq 2\pi$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module