If the equation `a_(n)x^(n)+a_(n-1)x^(n-1)+..+a_(1)x=0, a_(1)!=0, n ge2`, has a positive root `x=alpha` then the equation `na_(n)x^(n-1)+(n-1)a_(n-1)x^(n-2)+….+a_(1)=0` has a positive root which is

Options

A. greater than or equal to `alpha`
B. equal to `alpha`
C. greater than `alpha`
D. smaller than `alpha`

Explanation

To solve the problem step by step, we will analyze the given polynomial equation and its derivative.

### Step 1: Define the polynomial function
Let \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x \).

### Step 2: Identify the root
We know that \( f(\alpha) = 0 \) where \( \alpha \) is a positive root of the polynomial \( f(x) \).

### Step 3: Differentiate the polynomial
Now, we differentiate \( f(x) \) to find \( f'(x) \):
\[ f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_1 \]

### Step 4: Analyze the derivative at the root
Since \( f(\alpha) = 0 \), we need to evaluate \( f'(\alpha) \):
- By the properties of polynomials, if \( f(x) \) has a positive root, then \( f'(x) \) will have at least one root in the interval \( (0, \alpha) \).

### Step 5: Establish the existence of a positive root for the derivative
Since \( f'(x) \) is a polynomial of degree \( n-1 \) (which is at least 1 because \( n \geq 2 \)), it can have at most \( n-1 \) roots. Given that \( f(\alpha) = 0 \) and \( f'(x) \) is continuous, we can conclude that \( f'(x) \) must cross the x-axis at least once in the interval \( (0, \alpha) \).

### Step 6: Conclude about the positive root of the derivative
Thus, there exists a positive root \( \beta \) of the equation:
\[ n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_1 = 0 \]
where \( \beta < \alpha \).

### Final Conclusion
Therefore, we conclude that the equation \( n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_1 = 0 \) has a positive root \( \beta \) which is less than \( \alpha \).

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