Let a 0. If the function f(x) = 6x³ - 45ax² + 108a²x + 1 attains its local maximum and minimum val

Q: Wait, why is the local maximum at $x_1 = 2a$ and the local minimum at $x_2 = 3a$? How do we know which is which?

Great question! For a cubic function $f(x) = Ax^3 + Bx^2 + Cx + D$ with $A > 0$, the graph increases, reaches a local maximum, decreases to a local minimum, and then increases again. Therefore, the local maximum must occur at the smaller root of the derivative, and the local minimum at the larger root. Since $a > 0$, we know $2a < 3a$, so $x_1 = 2a$ and $x_2 = 3a$.

Q: Can I use Vieta's formulas instead of factoring the quadratic?

Absolutely! Using Vieta's formulas is a brilliant way to save time. For the quadratic $x^2 - 5ax + 6a^2 = 0$, the product of the roots $x_1x_2$ is simply the constant term, which is $6a^2$. Setting $6a^2 = 54$ immediately gives $a^2 = 9$. It is a very elegant approach.

Q: Why do we reject the solution $a = -3$?

The problem explicitly states that $a > 0$. In mathematics, whenever we solve an equation like $a^2 = 9$, we get two potential roots, $3$ and $-3$. However, we must always check these against the constraints provided in the problem statement. Since the condition $a > 0$ is given, we must discard the negative value.

Question

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 :-

Options:

(A) 15

(B) 18

(C) 24

(D) 13

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we look at a function like $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$, we are not just looking at a collection of terms; we are looking at a landscape.

Because the coefficient of the $x^3$ term is positive, this function behaves like a mountain range that rises, dips, and rises again. This 'N' shape is the signature of a cubic function with a positive leading coefficient.

The peaks and valleys of this landscape are the local maximum and minimum, and they are the keys to unlocking the secrets of this problem.

### The Power of the Derivative

To find these peaks and valleys, we turn to the most powerful tool in our calculus arsenal: the derivative. At the exact moment the function reaches a local maximum or minimum, the tangent line is perfectly horizontal.

This means the slope of the function, which is the derivative $f'(x)$, must be zero. Let us perform the differentiation.

Applying the power rule to $f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$, we get:

$$f'(x) = 18x^2 - 90ax + 108a^2$$

This is the velocity of our function's growth. When we set this to zero, we are asking: "Where does the function stop climbing and start falling, or vice versa?"

### The Algebraic Elegance

Now, let us simplify this expression. Notice that $18$, $90$, and $108$ are all divisible by $18$. Factoring out $18$, we get:

$$f'(x) = 18(x^2 - 5ax + 6a^2)$$

This quadratic $x^2 - 5ax + 6a^2$ is a beautiful piece of algebra. We need two numbers that multiply to $6a^2$ and add to $-5a$. Those numbers are $-2a$ and $-3a$.

Thus, the derivative factorizes into:

$$f'(x) = 18(x - 2a)(x - 3a)$$

The roots are $x = 2a$ and $x = 3a$. Since $a > 0$, we know $2a < 3a$. Following the geometry of our cubic curve, the local maximum $x_1$ must be the smaller root, $2a$, and the local minimum $x_2$ must be the larger root, $3a$.

### The Final Synthesis

We are given the condition $x_1x_2 = 54$. Substituting our values, we have:

$$(2a)(3a) = 54$$

This simplifies to $6a^2 = 54$. Dividing by $6$, we find $a^2 = 9$. Given $a > 0$, we conclude **$a = 3$**.

Now, we simply find the coordinates: $x_1 = 2(3) = 6$ and $x_2 = 3(3) = 9$. The question asks for the sum $a + x_1 + x_2$.

Plugging in our values, we get $3 + 6 + 9 = 18$. It is a perfect, clean result.

Remember, math is not just about calculation; it is about understanding the structure of the problem. You have navigated the geometry, the calculus, and the algebra with grace. The final answer is **$18$**.

Animated Solution for Mathematics - Differentiation: Let $a > 0$ . If the function $f (x) = 6 x^{3} - 45 a x^{2} + 108 a^{2} x + 1$ attains its local maximum and minimum values at the points $x_{1}$ and $x_{2}$ respectively such that $x_{1} x_{2} = 54$ , then $a + x_{1} + x_{2}$ is equal to :-

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Visualized Solution

Visualizing the Cubic Function $f (x)$

Identifying Critical Points

Differentiating $f (x)$

The First Derivative $f^{'} (x)$

Simplifying the Derivative

Factorizing the Quadratic

Identifying $x_{1}$ and $x_{2}$

Applying the Product Condition

Solving for $a$

Finding the Value of $a$

Calculating $x_{1}$ and $x_{2}$

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Power of the Derivative

The Algebraic Elegance

The Final Synthesis

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Visualized Solution

Visualizing the Cubic Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Identifying Critical Points

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Simplifying the Derivative

Factorizing the Quadratic

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​

Applying the Product Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​