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Let $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}$
$+x^{3}+x+19 .$ Then, $f(x)=0$ has
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$+x^{3}+x+19 .$ Then, $f(x)=0$ has
Solution:
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Verified Answer
The correct answer is:
not more than one real root
We have, $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19$
$$
\begin{array}{r}
\Rightarrow f^{\prime}(x)=13 x^{12}+11 x^{10}+9 x^{4} \\
+7 x^{6}+5 x^{4}+3 x^{2}+1
\end{array}
$$
$f^{\prime}(x)$ has no real root.
$\therefore f(x)=0$ has not more than one real root.
$$
\begin{array}{r}
\Rightarrow f^{\prime}(x)=13 x^{12}+11 x^{10}+9 x^{4} \\
+7 x^{6}+5 x^{4}+3 x^{2}+1
\end{array}
$$
$f^{\prime}(x)$ has no real root.
$\therefore f(x)=0$ has not more than one real root.
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