Consider the function $f:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R}$ defined by $f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$. Consider the statements
(I) The curve $y=f(x)$ intersects the $x$-axis exactly at one point.
(II) The curve $y=f(x)$ intersects the $x$-axis at $x=\cos \frac{\pi}{12}$.
Then
Solution
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text { for }\left[\frac{1}{2}, 1\right] \\
& \mathrm{f}\left(\frac{1}{2}\right)<0
\end{aligned}$$</p>
<p>$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$ is correct.</p>
<p>$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$</p>
<p>Let $\cos \alpha=\mathrm{x}$,</p>
<p>$\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}$</p>
<p>$\mathrm{x}=\cos \frac{\pi}{12}$</p>
<p>(4) is correct.</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Increasing and Decreasing Functions
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