If the functions f(x) = x³/3 + 2bx + ax²/2 and g(x) = x³/3 + ax + bx², a ≠ 2b have a common extreme

Q: Wait, why are we setting the derivatives to zero? Shouldn't we set $f(x) = g(x)$?

That is a classic trap! Setting $f(x) = g(x)$ finds the intersection points of the functions themselves. But the problem asks for a common extreme point. An extreme point is defined by the slope, not the position. We need the derivatives to be zero at the same $x$.

Q: Why does the problem explicitly state $a \neq 2b$?

Great catch! That condition is your safety net. When we subtract the equations, we get $(a - 2b)x = a - 2b$. If $a = 2b$, then $0 = 0$, which is true for all $x$, meaning the equations are identical. The constraint ensures that the common root is unique and solvable.

Q: Is there a geometric interpretation of the subtraction we performed?

Absolutely! By subtracting the two parabolas, we are essentially finding the line that passes through their intersection points. Since they share a root, that line must pass through the x-axis at that exact root, allowing us to isolate $x$ linearly.

Question

If the functions $f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$ and $g(x) = \frac{x^3}{3} + ax + bx^2, a 
e 2b$ have a common extreme point, then $a + 2b + 7$ is equal to

Options:

(A) 4

(B) $\frac{3}{2}$

(C) 3

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a rollercoaster. The track is defined by a cubic function, $f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$. As you ride, you experience moments of pure stillness—the peaks and the valleys.

In the language of calculus, these are your extreme points. Today, we are going to solve a beautiful puzzle: what happens when two different rollercoasters, $f(x)$ and $g(x) = \frac{x^3}{3} + ax + bx^2$, share the exact same moment of stillness?

### The Calculus of Stillness

To find these moments of stillness, we must look at the rate of change. An extreme point is where the slope of the tangent line is zero. Mathematically, this means we need to find the first derivative of our functions and set them to zero.

Let us differentiate $f(x)$ with respect to $x$. Using the power rule, we obtain:

$$f'(x) = x^2 + ax + 2b$$

Similarly, for $g(x)$, we find:

$$g'(x) = x^2 + 2bx + a$$

These two expressions, $f'(x)$ and $g'(x)$, are parabolas. When we set them to zero, we are looking for the roots of these parabolas—the points where they kiss the $x$-axis.

### The Algebraic Dance

The problem tells us that these two functions share a common extreme point. This means there exists some value $x$ such that both $f'(x) = 0$ and $g'(x) = 0$ are satisfied simultaneously. We have a system of two quadratic equations:

$$x^2 + ax + 2b = 0$$

$$x^2 + 2bx + a = 0$$

Many students would immediately reach for the quadratic formula. But stop! Take a breath and look at the symmetry. We have two equations with the same leading term, $x^2$. This is a gift.

If we subtract the second equation from the first, the $x^2$ term vanishes entirely. This leaves us with:

$$(x^2 + ax + 2b) - (x^2 + 2bx + a) = 0$$

Simplifying this, we get:

$$(a - 2b)x + (2b - a) = 0$$

### The Elegant Cancellation

This is where the magic happens. We can rewrite the equation as:

$$(a - 2b)x = a - 2b$$

The problem gives us a crucial constraint: $a 
eq 2b$. This is our green light. Because $a - 2b$ is not zero, we can divide both sides by it without fear.

The variables $a$ and $b$ vanish, leaving us with the elegant result:

$$x = 1$$

We have found the common root! It is not a complex, messy number; it is simply unity.

### The Final Reveal

Now that we know the common extreme point occurs at $x = 1$, we can substitute this back into our derivative equation $f'(x) = 0$. This gives us:

$$1^2 + a(1) + 2b = 0$$

This simplifies to $1 + a + 2b = 0$, or:

$$a + 2b = -1$$

The question asks us to find the value of $a + 2b + 7$. Since we know $a + 2b = -1$, we simply substitute this value into the expression:

$$-1 + 7 = 6$$

And there it is. Through the power of derivatives and the elegance of algebraic symmetry, we have arrived at our answer: **6**.

Remember, in JEE Advanced, the math is rarely about brute force. It is about finding the path of least resistance, the symmetry that simplifies the complex, and the beauty hidden within the equations.

Animated Solution for Mathematics - Differentiation: If the functions $f (x) = \frac{x ^{3}}{3} + 2 b x + \frac{a x ^{2}}{2}$ and $g (x) = \frac{x ^{3}}{3} + a x + b x^{2}, a \neq = 2 b$ have a common extreme point, then $a + 2 b + 7$ is equal to

Select Answer:

Visualized Solution

Understanding Extreme Points

Differentiating $f (x)$

Differentiating $g (x)$

The Common Root Condition

Subtracting the Equations

Finding the Common Root

Substituting the Root

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Calculus of Stillness

The Algebraic Dance

The Elegant Cancellation

The Final Reveal

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