How to find minimum of a function on closed interval?

Find critical points by setting f'(x) = 0. Evaluate f at critical points and endpoints. The smallest value is the minimum.

What is the derivative of x^n?

d/dx(x^n) = nx^(n-1). Apply this formula to each term separately.

How to solve x^2025 - x^2000 minimum?

Find f'(x) = 2025x^2024 - 2000x^1999 = 0. Factor out x^1999 to get x^1999(2025x^25 - 2000) = 0.

What are critical points?

Points where f'(x) = 0 or f'(x) undefined. Include endpoints when finding extrema on closed intervals.

How to factor polynomials?

Look for common factors first. For x^1999(2025x^25 - 2000) = 0, either x^1999 = 0 or 2025x^25 = 2000.

On closed intervals [a,b], extrema occur at critical points OR endpoints. Must evaluate f at all three types of points.

How to compare values?

Calculate f(0), f(1), and f at critical points. Compare to identify minimum: smallest value is the minimum.

What does (80)^80(n)^-81 mean?

This equals (80)^80 / n^81. Set equal to minimum value and solve for n.

How to solve exponential equations?

Take logs if needed, or recognize patterns. For (80/n)^80 = n, try n = 80 or similar values.

Why is x = (80/81)^(1/25) the critical point?

From 2025x^25 = 2000, get x^25 = 2000/2025 = 80/81, so x = (80/81)^(1/25).

Let f(x) = x^2025 - x^2000, x ∈ [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 |

Let f(x) = x^2025 – x^2000, x ∈ [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 | JEE Main Mathematics

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Math Calculus

Q MCQ Optimization

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

✅ Correct Answer

-81

✓

Solution Steps

Find the derivative

$f(x) = x^{2025} – x^{2000}$

$f'(x) = 2025x^{2024} – 2000x^{1999}$

Apply power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$

Factor the derivative

$f'(x) = x^{1999}(2025x^{25} – 2000)$

Factor out common term $x^{1999}$

Find critical points

Set $f'(x) = 0$:

$x^{1999}(2025x^{25} – 2000) = 0$

Either $x = 0$ or $2025x^{25} = 2000$

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

$x = \left(\frac{80}{81}\right)^{1/25}$

Evaluate at critical points and endpoints

At $x = 0$: $f(0) = 0$

At $x = 1$: $f(1) = 1 – 1 = 0$

At $x = \left(\frac{80}{81}\right)^{1/25}$:

Let $x^{25} = \frac{80}{81}$

Calculate f at critical point

$x^{2000} = (x^{25})^{80} = \left(\frac{80}{81}\right)^{80}$

$x^{2025} = x^{2000} \cdot x^{25}$

$= \left(\frac{80}{81}\right)^{80} \cdot \frac{80}{81} = \left(\frac{80}{81}\right)^{81}$

Compute the minimum

$f(x) = x^{2025} – x^{2000}$

$= \left(\frac{80}{81}\right)^{81} – \left(\frac{80}{81}\right)^{80}$

$= \left(\frac{80}{81}\right)^{80}\left[\frac{80}{81} – 1\right]$

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

Compare with given form

Minimum: $-\frac{1}{81}(80)^{80}(81)^{-80}$

$= -(80)^{80}(81)^{-81}$

Given form: $(80)^{80}(n)^{-81}$

Therefore: $n = -81$

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Key Insight

The critical point gives negative minimum. The coefficient n = -81 accounts for the negative sign in the form.

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Calculus Theory

1. Finding Extrema on Closed Intervals

To find maximum and minimum of continuous function on $[a,b]$: (1) Find critical points where $f'(x) = 0$ in $(a,b)$. (2) Evaluate $f$ at each critical point and endpoints. (3) Compare all values. Largest is maximum, smallest is minimum. This systematic approach ensures no extrema are missed on closed intervals.

2. Power Rule for Derivatives

The power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$ applies to any real $n$. For polynomials, apply term-by-term. For $f(x) = x^{2025} – x^{2000}$, get $f'(x) = 2025x^{2024} – 2000x^{1999}$. Always look for common factors in derivatives to simplify and find critical points efficiently.

3. Critical Points and Factoring

When $f'(x) = 0$ is a product, use zero product property: if $A \cdot B = 0$, then $A = 0$ or $B = 0$. For $x^{1999}(2025x^{25} – 2000) = 0$, get $x = 0$ or $x^{25} = \frac{80}{81}$. Solving $x^n = a$ gives $x = a^{1/n}$. Verify critical points lie in domain $[0,1]$.

4. Evaluating at Critical Points

For $x = \left(\frac{80}{81}\right)^{1/25}$ where $x^{25} = \frac{80}{81}$, use exponent rules: $x^{2000} = (x^{25})^{80}$ and $x^{2025} = x^{2000} \cdot x^{25}$. Factor carefully: $f(x) = \left(\frac{80}{81}\right)^{80}\left[\frac{80}{81} – 1\right] = -\frac{1}{81}(80/81)^{80}$. Algebraic precision prevents errors in final answer.

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FAQs

How to find extrema on [a,b]?

⌄

Find critical points where f'(x) = 0. Evaluate f at all critical points and endpoints a, b. Compare: max is largest value, min is smallest.

What is the power rule?

⌄

d/dx(x^n) = nx^(n-1). For f(x) = x^2025 – x^2000, get f'(x) = 2025x^2024 – 2000x^1999.

How to factor derivatives?

⌄

Look for common factors. In 2025x^2024 – 2000x^1999, factor x^1999. Get x^1999(2025x^25 – 2000).

What is zero product property?

⌄

If AB = 0, then A = 0 or B = 0. For x^1999(2025x^25 – 2000) = 0: either x = 0 or 2025x^25 = 2000.

How to solve x^25 = 80/81?

⌄

Take 25th root: x = (80/81)^(1/25). This is the critical point in interval (0,1).

How to compute x^2000?

⌄

If x^25 = 80/81, then x^2000 = (x^25)^80 = (80/81)^80 using exponent rules.

Why is f(x) negative?

⌄

At x = (80/81)^(1/25), f(x) = (80/81)^80[(80/81) – 1] = -(1/81)(80/81)^80 < 0 since 80/81 < 1.

How to identify minimum?

⌄

Compare: f(0) = 0, f(1) = 0, f(critical) < 0. Minimum is at critical point, which is smallest value.

Why n = -81?

⌄

Minimum = -(80)^80(81)^(-81). Given form (80)^80(n)^(-81), so n = -81 accounts for negative sign.

How to verify answer?

⌄

Substitute n = -81 into form: (80)^80(-81)^(-81) = -(80)^80(81)^(-81), which matches minimum value.

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