Multiple choice

lf $ \mathrm{f}(\mathrm{x})=0$ has a repeated root $ \alpha$, then another equation having $\alpha$ as root, is

  1. $\mathrm{f}(2\mathrm{x})=0$
  2. $\mathrm{f}(3\mathrm{x})=0$
  3. $\mathrm{f}^{'}(\mathrm{x})=0$
  4. $\mathrm{f}^{''}(\mathrm{x})=0$
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C Correct answer
Explanation

If f(x) has a repeated root alpha, then f(alpha) = 0 and f'(alpha) = 0. Therefore, alpha is also a root of f'(x) = 0.

AI explanation

If a polynomial equation f(x) = 0 has a repeated root alpha, then the function f(x) can be factored as (x - alpha)^2 * g(x) for some polynomial g(x). Differentiating this function using the product rule gives the derivative f'(x) = 2(x - alpha)g(x) + (x - alpha)^2g'(x). Factoring out (x - alpha) from the derivative yields f'(x) = (x - alpha)[2g(x) + (x - alpha)g'(x)], which proves that substituting x = alpha will make the derivative zero. Thus, the equation having alpha as a root is f'(x) = 0.

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