A variable takes value x with frequency...

MathematicsMeasures of Central Tendency (Mean, Median, Mode)JEE Advanced 1982Moderate

A variable takes value $x$ with frequency $^{n + x - 1} C_{x}$ , $x = 0, 1, 2, \dots, n$ . The mode of the variable is .........

Answer:

n

Visualized Solution (Hindi)

Understanding the Frequency Distribution

Variable $x$ takes values $0, 1, 2, \dots, n$ .
Frequency of $x$ is given by $f (x) = (x n + x - 1)$ .
Objective: Find the value of $x$ that maximizes $f (x)$ (the Mode).

Defining the Frequency Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x)

Let $f (x) = (x n + x - 1)$ .
To find the maximum, we analyze the trend of $f (x)$ as $x$ increases.
We will compare $f (x)$ with its preceding term $f (x - 1)$ .

Setting up the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \frac{f ( x )}{f ( x - 1 )}

Consider the ratio: $\frac{f ( x )}{f ( x - 1 )} = \frac{( x n + x - 1 )}{( x - 1 n + x - 2 )}$
Using $(r n) = \frac{n !}{r ! ( n - r )!}$ :
$\frac{f ( x )}{f ( x - 1 )} = \frac{( n + x - 1 )!}{x ! ( n - 1 )!} \cdot \frac{( x - 1 )! ( n - 1 )!}{( n + x - 2 )!}$

Simplifying the Ratio

Simplifying the factorials:
$\frac{( n + x - 1 )!}{( n + x - 2 )!} = n + x - 1$
$\frac{( x - 1 )!}{x !} = \frac{1}{x}$
Therefore, $\frac{f ( x )}{f ( x - 1 )} = \frac{n + x - 1}{x} = 1 + \frac{n - 1}{x}$

Analyzing the Growth of Frequency

Since $n \geq 1$ and $x \geq 1$ , we have $\frac{n - 1}{x} \geq 0$ .
Thus, $\frac{f ( x )}{f ( x - 1 )} \geq 1$ , which means $f (x) \geq f (x - 1)$ .
The frequency $f (x)$ is a non-decreasing function of $x$ .

Finding the Mode

The maximum value of $x$ in the given range is $n$ .
Since $f (x)$ is increasing, $f (n)$ is the maximum frequency.
Conclusion: The mode of the variable is $n$ .

Key Takeaway and Expansion

Key Takeaway: For an increasing frequency distribution, the mode is the maximum value of the variable.
Next Challenge: What would be the mode if the frequency was $f (x) = (x n)$ for $x = 0, 1, \dots, n$ ?
Hint: Check the symmetry of the Pascal's triangle.

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