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Let $\mathrm{n}$ be a natural number and let 'a' be a real number. The number of zeros of $\mathrm{x}^{2 \mathrm{n}+1}-(2 \mathrm{n}+1) \mathrm{x}+\mathrm{a}=0$ in the interval $[-1,1]$ is $-$
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Verified Answer
The correct answer is:
At most one for every value of a
$\begin{array}{l}
f(x)=x^{2 n+1}-(2 n+1) x+a \\
f^{\prime}(x)=(2 n+1) x^{2 n}-(2 n+1)
\end{array}$
$=(2 n+1)\left(x^{2 n}-1\right) < 0$ when $x \in[-1,1]$
$\mathrm{f}(\mathrm{x})$ is strictly decreasing $\mathrm{in}[-1,1]$
$\mathrm{f}(\mathrm{x})$ cut $\mathrm{x}$ axis at most one point in given interval
f(x)=x^{2 n+1}-(2 n+1) x+a \\
f^{\prime}(x)=(2 n+1) x^{2 n}-(2 n+1)
\end{array}$
$=(2 n+1)\left(x^{2 n}-1\right) < 0$ when $x \in[-1,1]$
$\mathrm{f}(\mathrm{x})$ is strictly decreasing $\mathrm{in}[-1,1]$
$\mathrm{f}(\mathrm{x})$ cut $\mathrm{x}$ axis at most one point in given interval
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