Let f(x) = x²⁰²⁵ - x²⁰⁰⁰, x ∈ [0, 1] and the minimum value of the function f(x) in the interval [0,

Q: Why do we factor out $x^{1999}$ instead of just dividing by it?

Great question! If you divide by $x^{1999}$, you implicitly assume $x \neq 0$. While $x=0$ is a boundary point, it is a critical part of the domain. Factoring ensures we don't lose the root at $x=0$.

Q: Why is the minimum value negative? I thought $x^{2025}$ would be larger than $x^{2000}$?

Remember that for $x \in (0, 1)$, a higher power makes the number smaller. Since $x^{2025} < x^{2000}$ in this interval, the difference $x^{2025} - x^{2000}$ is guaranteed to be negative.

Q: Can I just use the second derivative test to confirm it's a minimum?

You certainly could, but in a competitive exam like JEE, time is your currency. Since the function is continuous on a closed interval $[0, 1]$ and $f(0)=f(1)=0$, any critical point found between them must be the global minimum.

Q: How did you know to rewrite $x^{2000}$ as $(x^{25})^{80}$?

It's all about pattern matching! We found $x^{25} = \frac{80}{81}$. By expressing $x^{2000}$ as a power of $x^{25}$, we can substitute the fraction directly without ever needing to calculate the actual value of $x$.

Q: Is it possible for $n$ to be negative in the expression $(n)^{-81}$?

Yes! Since the exponent $81$ is odd, the negative sign is preserved. $(-81)^{-81}$ is equivalent to $-\frac{1}{81^{81}}$, which perfectly matches our derived result.

Question

Let $f(x) = x^{2025} - x^{2000}, x \in [0, 1]$ and the minimum value of the function $f(x)$ in the interval $[0, 1]$ be $(80)^{80}(n)^{-81}$. Then $n$ is equal to

Options:

(A) -40

(B) -41

(C) -80

(D) -81

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast, flat plain defined by the interval $[0, 1]$. You are tasked with tracking the trajectory of a function, $f(x) = x^{2025} - x^{2000}$.

At first glance, this looks like a daunting, high-degree polynomial. But let us look at the soul of this function. When $x=0$, $f(0)=0$. When $x=1$, $f(1) = 1^{2025} - 1^{2000} = 0$.

Between these two points, the function must dip into the negative realm. For any $x$ between $0$ and $1$, $x^{2025}$ is strictly smaller than $x^{2000}$. Our mission is to find the deepest point of this valley—the global minimum.

### The Search for the Critical Point

To find the bottom of the valley, we must find where the slope of the function vanishes. We invoke the power rule of calculus:

$$f'(x) = 2025x^{2024} - 2000x^{1999}$$

Setting this derivative to zero is our key to unlocking the mystery. We factor out the common term, $x^{1999}$, to reveal the structure:

$$f'(x) = x^{1999}(2025x^{25} - 2000) = 0$$

This gives us two paths. One is the boundary $x=0$. The other, more interesting path, is $2025x^{25} = 2000$. Solving this, we find the critical value:

$$x^{25} = \frac{2000}{2025} = \frac{80}{81}$$

This is the "golden ratio" for our specific function. We do not need to find $x$ itself; we only need this power.

### The Elegant Substitution

Now, we return to our original function. Instead of getting lost in the weeds of high-degree exponents, we rewrite $f(x)$ to highlight the $x^{25}$ term we just discovered:

$$f(x) = x^{2000}(x^{25} - 1)$$

Since $x^{2000} = (x^{25})^{80}$, we can substitute our value $\frac{80}{81}$ directly into the expression:

$$f_{min} = \left(\frac{80}{81}\right)^{80} \left(\frac{80}{81} - 1\right)$$

The bracket simplifies beautifully to $-\frac{1}{81}$. Thus, our minimum value becomes:

$$f_{min} = \left(\frac{80}{81}\right)^{80} \left(-\frac{1}{81}\right) = -\frac{80^{80}}{81^{80} \cdot 81} = -\frac{80^{80}}{81^{81}}$$

### The Final Revelation

We are told the minimum value is given in the form $(80)^{80}(n)^{-81}$. By equating our result to this form:

$$(80)^{80} \cdot n^{-81} = -\frac{80^{80}}{81^{81}}$$

We see the $(80)^{80}$ terms cancel out, leaving us with:

$$n^{-81} = -\frac{1}{81^{81}} = -\left(\frac{1}{81}\right)^{81} = (-81)^{-81}$$

Because the exponent is odd, the negative sign can be brought inside the base. Therefore, **$n = -81$**. You have successfully navigated the complexity of high-degree polynomials by focusing on the underlying structure rather than the raw numbers.

Animated Solution for Mathematics - Differentiation: Let $f (x) = x^{2025} - x^{2000}, x \in [0, 1]$ and the minimum value of the function $f (x)$ in the interval $[0, 1]$ be $(80)^{80} (n)^{- 81}$ . Then $n$ is equal to

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

The Condition for Minimum

Differentiating $f (x)$

Factoring the Derivative

Finding Critical Points

Solving for $x^{25}$

Preparing to Find the Minimum Value

Substituting the Critical Value

Simplifying the Minimum Value

Equating with the Given Form

Solving for $n$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Search for the Critical Point

The Elegant Substitution

The Final Revelation

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