Animated Solution for Mathematics - Differentiation: Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined by $f(x)=(x-3{)}^{n_1}(x-5{)}^{n_2},n_1,n_2\in \mathbb{N}$. Then, which of the following is NOT true?

Unlocking the Mystery of Extrema

Welcome, warriors of JEE Advanced. Today, we are not just solving a problem; we are exploring the topography of a function.

We are given $f(x)=(x-3{)}^{n_1}(x-5{)}^{n_2}$, a function that seems simple at first glance but hides a subtle trap in its behavior between $x=3$ and $x=5$. Let us embark on this journey to uncover the truth.

The Product Rule Symphony

When we see a product of two functions, our first instinct should be the Product Rule. We are looking for the critical points, the places where the function decides to turn around—the peaks and the valleys.

To find these, we must calculate the derivative $f^{\prime }(x)$. Applying the rule $\frac{d}{dx}[u\cdot v]=u\cdot v^{\prime }+v\cdot u^{\prime }$, we get:

It looks intimidating, doesn't it? A long string of terms. But pause and take a breath. In mathematics, as in life, complexity often masks a simple underlying structure. We don't need to expand this; we need to factor it.

The Art of Factoring

Look closely at the expression. We have common terms: $(x-3)$ and $(x-5)$. We can pull out the lowest powers, $(x-3{)}^{n_1-1}$ and $(x-5{)}^{n_2-1}$.

This is the moment where the fog clears:

Inside the square brackets, we have a linear expression. Let's simplify it: $n_{2}x-3n_2+n_{1}x-5n_1$. Grouping the $x$ terms, we get $(n_1+n_2)x-(5n_1+3n_2)$.

Setting $f^{\prime }(x)=0$ gives us our critical point $\alpha$:

This is the Section Formula! It tells us that $\alpha$ is the weighted average of 3 and 5. It is guaranteed to lie in the interval $(3,5)$, which is the geometric heart of the problem.

The Detective Work (Sign Scheme)

Now, we must determine if $\alpha$ is a local maximum or a local minimum. This depends on how $f^{\prime }(x)$ changes sign as $x$ crosses $\alpha$.

The sign of $f^{\prime }(x)$ is determined by the product of three components: $(x-3{)}^{n_1-1}$, $(x-5{)}^{n_2-1}$, and the linear term $(x-\alpha )$. Let us test Option 3, where $n_1=3$ and $n_2=5$.

Substituting these into our derivative expression:

Here is the trap! Notice the exponents 2 and 4. They are even. For any $x$ in the interval $(3,5)$, $(x-3{)}^2$ is always positive, and $(x-5{)}^4$ is always positive. They do not change sign.

The Verdict

Since the even-powered terms are always positive, the sign of $f^{\prime }(x)$ is dictated entirely by the linear term $(x-3.75)$.

When $x<3.75$, $(x-3.75)$ is negative. When $x>3.75$, $(x-3.75)$ is positive. The derivative changes from negative to positive.

According to the First Derivative Test, this means the function goes from decreasing to increasing. That is the definition of a local minimum.

But look at Option 3 again. It claimed that the function attains a local maxima. We have just proven that it attains a local minima. Therefore, Option 3 is the false statement. We have successfully navigated the trap and found the truth.