Multiple choice

If the equation ${ a }{ n }{ x }^{ n } + { a }{ n-1 }{ x }^{ n-1 } + \dots + { a }{ 1 }x = 0$ has a positive root $x = \alpha$, then the equation $ n{ a }{ n }{ x }^{ n-1 } + \left( n-1 \right) { a }{ n-1 }{ x }^{ n-2 } + \dots + { a }{ 1 } = 0$ has a positive root, which is

  1. Smaller than $\alpha$
  2. Greater than $\alpha$
  3. Equal to $\alpha$
  4. Greater than or equal to $\alpha$
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A Correct answer
Explanation

Let f(x) = a_n*x^n + ... + a_1*x. Since f(alpha) = 0 and f(0) = 0, by Rolle's Theorem, there exists a root of f'(x) in (0, alpha). The derivative is f'(x) = n*a_n*x^(n-1) + ... + a_1. Thus, the root of the second equation is smaller than alpha.

AI explanation

This result is derived directly from Rolle's Theorem applied to the polynomial. Let the original polynomial be P(x), and the second equation represents its derivative, P'(x). Because P(x) has a positive root at x = alpha and also has a root at x = 0, P(x) is continuous on the interval (0, alpha); therefore, P'(x) must have at least one root strictly between 0 and alpha, making it smaller than alpha.

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