On the interval [0, 1] the function x²⁵ (1 - x)⁷⁵ takes its maximum value at the point (A) 0 (B) 1/4

Q: Why can't I just expand the brackets using the Binomial Theorem?

Expanding $(1-x)^{75}$ would result in 76 individual terms! It is a massive trap that will lead to algebraic exhaustion and a high probability of error. In JEE Advanced, we prioritize elegance over brute force. Factoring is the intended path.

Q: How do I know $x=1/4$ is a maximum and not a minimum?

Great question. We know that at $x=0$ and $x=1$, the function value is $0$. Since the function is positive for all $x$ between $0$ and $1$, the value at $x=1/4$ must be the absolute maximum. You can also verify this by checking the sign change of the derivative around $x=1/4$.

Q: Is there a general rule for functions of the form $x^m(1-x)^n$?

Yes! For a function $f(x) = x^m(1-x)^n$ on $[0, 1]$, the maximum always occurs at $x = \frac{m}{m+n}$. In our case, $m=25$ and $n=75$, so $x = \frac{25}{25+75} = \frac{25}{100} = \frac{1}{4}$. This is a powerful shortcut for your competitive exams.

Question

On the interval $[0, 1]$ the function $x^{25} (1 - x)^{75}$ takes its maximum value at the point

Options:

(A) $0$

(B) $\frac{1}{4}$

(C) $\frac{1}{2}$

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the peak of the function defined by:

$$f(x) = x^{25}(1-x)^{75}$$

This function is defined on the interval $[0, 1]$. Our goal is to locate the value of $x$ where the function reaches its maximum.

### The Trap of Expansion

Many students fall into the trap of trying to expand $(1-x)^{75}$ using the Binomial Theorem. Please, never do that!

You would be staring at 76 terms, and you would lose your way in the algebra. The JEE Advanced examiners love to see if you can spot the path of least resistance.

Instead, we embrace the **Product Rule**. We define $u = x^{25}$ and $v = (1-x)^{75}$. The derivative is given by:

$$\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$$

### The Power of Factoring

Let us differentiate. The derivative of $u$ is $25x^{24}$, and the derivative of $v$ is $75(1-x)^{74} \cdot (-1)$.

When we assemble this, we get:

$$f'(x) = x^{25} \cdot [-75(1-x)^{74}] + (1-x)^{75} \cdot [25x^{24}]$$

Now, look at the magic of factoring. We pull out the common terms: $25x^{24}(1-x)^{74}$.

What remains inside the bracket? From the first term, we are left with $-3x$. From the second term, we are left with $(1-x)$.

This simplifies to $(1-4x)$. So, our derivative becomes:

$$f'(x) = 25x^{24}(1-x)^{74}(1-4x)$$

### The Final Victory

Setting the derivative to zero gives us $x = \frac{1}{4}$. This is the beauty of calculus—it turns a complex-looking expression into a simple linear equation.

We ignore the roots at $x=0$ and $x=1$ because they are the boundaries where the function is zero. This leaves us with our maximum at:

**$$x = \frac{1}{4}$$**

Remember, in JEE Advanced, the path of least resistance is usually the path of most elegance. Keep practicing, and keep looking for those common factors!

Animated Solution for Mathematics - Differentiation: On the interval $[0, 1]$ the function $x^{25} (1 - x)^{75}$ takes its maximum value at the point

Select Answer:

Visualized Solution

Defining the Function $f (x)$

The Maximum Value Strategy

Applying the Product Rule

Differentiating the Terms

Constructing $f^{'} (x)$

Factoring the Expression

Simplifying the Bracket

Solving for Critical Points

Identifying the Maximum

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Trap of Expansion

The Power of Factoring

The Final Victory

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } On the interval [0,1] the function x25(1−x)75 takes its maximum value at the point