Function Extrema Analysis
We are given the function $f(x) = 6x^3-45ax^2+108a^2x+1$, where $a > 0$. We need to find the local maximum and minimum points, $x_1$ and $x_2$, and use the condition $x_1x_2 = 54$ to determine the value of $a + x_1 + x_2$.
Finding Critical Points
Local extrema occur where the first derivative of the function is zero. First, find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(6x^3-45ax^2+108a^2x+1)$
$f'(x) = 18x^2 - 90ax + 108a^2$
Set the derivative to zero to find the critical points $x_1$ and $x_2$:
$18x^2 - 90ax + 108a^2 = 0$
Divide the equation by 18:
$x^2 - 5ax + 6a^2 = 0$
Using Vieta's Formulas
The roots of the quadratic equation $x^2 - 5ax + 6a^2 = 0$ are the locations of the local extrema, $x_1$ and $x_2$. According to Vieta's formulas:
- Sum of roots: $x_1 + x_2 = -(-5a)/1 = 5a$
- Product of roots: $x_1x_2 = 6a^2 / 1 = 6a^2$
Solving for 'a'
We are given the condition that $x_1x_2 = 54$. Using the product of roots from Vieta's formulas:
$6a^2 = 54$
Divide by 6:
$a^2 = 9$
Since we are given that $a > 0$, we take the positive square root:
$a = 3$
Calculating $x_1 + x_2$
Now, use the value of $a$ to find the sum of the roots $x_1 + x_2$ using Vieta's formula:
$x_1 + x_2 = 5a$
$x_1 + x_2 = 5(3)$
$x_1 + x_2 = 15$
Final Calculation
The question asks for the value of $a + x_1 + x_2$. Substitute the values we found:
$a + x_1 + x_2 = 3 + 15$
$a + x_1 + x_2 = 18$
Similar Questions
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
Let $f(x)= \begin{cases} (1+ax)^{1/x} & , x <0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$
be continuous at $x = 0$. Then $e^{abc}$ is equal to:
Let $f (x) = \int x^3\sqrt{3-x^2} \, dx$. If $5f (\sqrt{2}) = -4$, then $f (1)$ is equal to
Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
The area of the region bounded by the curve $y = \max \{|x|, x|x-2|\}$, the x-axis and the lines $x = -2$ and $x = 4$ is equal to
- If the function $f(x) = 2x^3-9ax^2+12a^2x+1$, where $a > 0$, attains its local maximum and local minimum values at p and q, respectively, such that $p^2 = q$, then $f(3)$ is equal to :
Let $f: R\to R$ be a twice differentiable function such that
$(\sin x \cos y) (f(2x+2y)-f(2x-2y)) = (\cos x \sin y) (f(2x+2y)+f(2x-2y))$, for all $x, y \in R$.
If $f'(0) = \frac{1}{2}$, then the value of $24 f'' (\frac{5\pi}{3})$ is:If the area of the region $\{(x, y):|4-x^2|\le y \le x^2, y\le4,x\ge0)$ is $\left(\frac{80\sqrt{2}}{\alpha}-\beta\right)$, $\alpha, \beta\in N$, then $\alpha+\beta$ is equal to _____________.
- Let $f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and $2g(x) - 3g\left(\frac{1}{x}\right) = x, x > 0$. If $\alpha = \int_{1}^{2}f(x) dx$, and $\beta = \int_{1}^{2}g(x) dx$, then the value of $9\alpha + \beta$ is:
Important Questions from Calculus
Let a line passing through the point $(4, 1, 0)$ intersect the line $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then \(\begin{bmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{bmatrix}\) is equal to
If $y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$, $x \in R$, then $\frac{d^2y}{dx^2} + y$ is equal to
Let $f(x)= \begin{cases} (1+ax)^{1/x} & , x <0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$
be continuous at $x = 0$. Then $e^{abc}$ is equal to:
Let $f (x) = \int x^3\sqrt{3-x^2} \, dx$. If $5f (\sqrt{2}) = -4$, then $f (1)$ is equal to
Let $g$ be a differentiable function such that $\int_0^x g(t)dt=x-\int_0^x tg(t)dt$, $x\ge0$ and let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y\tan x = 2(x+1)\sec x g(x)$, $x\in [0,\frac{\pi}{2})$. If $y(0) = 0$, then $y(\frac{\pi}{3})$ is equal to
