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On the interval $[0,1]$, the function $x^{25}(1-x)^{75}$ takes its maximum value at the point
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Verified Answer
The correct answer is:
$\frac{1}{4}$
Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{25}(1-\mathrm{x})^{75}, \mathrm{x} \in[0,1]$
$\begin{aligned}
\Rightarrow f^{\prime}(x) &=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74} \\
&=25 x^{24}(1-x)^{74}\{(1-x)-3 x\} \\
&=25 x^{24}(1-x)^{74}(1-4 x)
\end{aligned}$
We can see that $f^{\prime}(x)$ is positive for $\mathrm{x} < \frac{1}{4}$ and $f^{\prime}(x)$ is negative for $x>\frac{1}{4}$.
Hence, $f(x)$ attains maximum at $\mathrm{x}=\frac{1}{4}$.
$\begin{aligned}
\Rightarrow f^{\prime}(x) &=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74} \\
&=25 x^{24}(1-x)^{74}\{(1-x)-3 x\} \\
&=25 x^{24}(1-x)^{74}(1-4 x)
\end{aligned}$
We can see that $f^{\prime}(x)$ is positive for $\mathrm{x} < \frac{1}{4}$ and $f^{\prime}(x)$ is negative for $x>\frac{1}{4}$.
Hence, $f(x)$ attains maximum at $\mathrm{x}=\frac{1}{4}$.
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