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$f(x)$ is invertible polynomial function of degree $n\geq 3$ then $f''(x) = 0$ has exactly $n - 2$ distinct real roots if

A)$f′(x)=0$ has $n−1/2$ distinct real roots

B)$f′(x)=0$ has $n−1$ distinct real roots

C)all the roots of $f′(x)=0$ are distinct

D)none of these

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  • $\begingroup$ Because f(x) is invertible function so all the roots of equation f'(x) = 0 are also the roots of equation f"(x) = 0 ⇒ If f'(x) = 0 has n−12 roots then number of roots of f"(x) = 0 equal to n−12+(n−12−1) =n−2 $\endgroup$ – user660100 Apr 17 at 11:28
  • $\begingroup$ What's your question? $\endgroup$ – Matti P. Apr 17 at 11:33

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