Let f be a function defined on R such that f'(x) = 2010(x - 2009)(x - 2010)²(x - 2011)³(x - 2012)⁴ f

Q: Wait, why does the exponential term $e^{f(x)}$ not affect the local maxima?

That is a fantastic question! The derivative of $g(x)$ is $g'(x) = e^{f(x)} \cdot f'(x)$. Since the exponential function $e^{f(x)}$ is strictly positive for all real $x$, it acts like a positive multiplier. It scales the derivative but never changes its sign. Therefore, the sign of $g'(x)$ is entirely determined by the sign of $f'(x)$. We can safely ignore the exponential term when looking for where the derivative changes sign!

Q: Why do we ignore the even powers in the Wavy Curve method?

Think of it this way: an even power, like $(x - 2010)^2$, is always non-negative. Whether $x$ is slightly larger or slightly smaller than $2010$, the term $(x - 2010)^2$ remains positive. Because it doesn't change sign as we cross $2010$, it doesn't contribute to a sign flip for the entire expression. That is why the curve 'bounces' off the axis instead of crossing it.

Q: How do I know if a point is a local maximum or a local minimum?

It is all about the direction of the sign change! If the derivative $g'(x)$ changes from positive to negative, the function $g(x)$ goes from increasing to decreasing, which creates a peak—a local maximum. If it changes from negative to positive, the function goes from decreasing to increasing, creating a valley—a local minimum. Just track the sign of $f'(x)$ as you move from left to right!

Question

Let $f$ be a function defined on $\mathbb{R}$ such that $f'(x) = 2010(x - 2009)(x - 2010)^2(x - 2011)^3(x - 2012)^4$ for all $x \in \mathbb{R}$. If $g$ is a function defined on $\mathbb{R}$ with values in $(0, \infty)$ such that $f(x) = \ln(g(x))$, then the number of points in $\mathbb{R}$ at which $g$ has a local maximum is

Sigma Percentile · Accepted Answer

### The Hidden Elegance of Functions

My dear student, I know that when you first look at a problem like this, with its large numbers like $2009, 2010, 2011,$ and $2012$, it feels like a mountain of arithmetic. But take a deep breath.

In the world of JEE Advanced, these numbers are often just labels, designed to test your conceptual clarity, not your ability to perform massive calculations. Let us peel back the layers of this problem together.

### Unmasking the Function $g(x)$

We are given $f(x) = \ln(g(x))$. Our goal is to find the local maxima of $g(x)$.

By exponentiating both sides, we get $g(x) = e^{f(x)}$. Now, here is the first piece of wisdom: the exponential function $e^{f(x)}$ is a beacon of positivity.

It is always greater than zero, no matter what $f(x)$ is. This means $g(x)$ is always positive, which is a relief!

### The Derivative Dance

To find local extrema, we must look at the derivative, $g'(x)$. Using the chain rule, we find:

$$g'(x) = e^{f(x)} \cdot f'(x)$$

This is the moment of truth. Since $e^{f(x)}$ is always positive, it cannot change the sign of $g'(x)$.

The sign of our derivative is entirely dictated by $f'(x)$. We have effectively reduced a complex-looking problem into a simple sign analysis of the polynomial:

$$f'(x) = 2010(x - 2009)(x - 2010)^2(x - 2011)^3(x - 2012)^4$$

### The Wavy Curve Method: A Map of the Terrain

Now, let us plot our critical points on the number line: $2009, 2010, 2011,$ and $2012$. We start from the far right, where $x > 2012$.

If you pick a number like $3000$, every single factor in $f'(x)$ is positive. So, the curve starts above the axis.

As we move left, we encounter $x = 2012$. The factor $(x - 2012)^4$ has an even power, so the sign does not change—the curve bounces off the axis and stays positive.

Next, we hit $x = 2011$. The factor $(x - 2011)^3$ has an odd power, so the sign flips from positive to negative.

Then, at $x = 2010$, the factor $(x - 2010)^2$ has an even power, so the sign stays negative. Finally, at $x = 2009$, the factor $(x - 2009)^1$ has an odd power, and the sign flips back to positive.

### The Final Verdict

A local maximum occurs when the derivative changes from positive to negative. Looking at our sign scheme, at $x = 2009$, the sign changes from positive to negative.

That is our **local maximum**! At $x = 2011$, the sign changes from negative to positive, which is a local minimum.

The other points are not extrema. Thus, we have found exactly one point of local maximum. You see? With the right perspective, even the most intimidating problems reveal their simple, elegant core.

Enter Numerical Value:

Visualized Solution

Relationship between $f (x)$ and $g (x)$

Differentiating $g (x)$

Sign Analysis of $g^{'} (x)$

Critical Points on Number Line

Wavy Curve: $x > 2012$

Crossing $x = 2012$ (Even Power)

Crossing $x = 2011$ (Odd Power)

Crossing $x = 2010$ (Even Power)

Crossing $x = 2009$ (Odd Power)

Condition for Local Maximum

Conclusion

The Sigma Insight: Maxima and Minima

The Hidden Elegance of Functions

Unmasking the Function $g (x)$

The Derivative Dance

The Wavy Curve Method

The Final Verdict

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Enter Numerical Value:

Visualized Solution

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

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

Sign Analysis of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)

Critical Points on Number Line

Wavy Curve: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>2012

Crossing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2012 (Even Power)

Crossing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2011 (Odd Power)

Crossing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2010 (Even Power)

Crossing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2009 (Odd Power)