JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 16)
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
Explanation
$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text { for }\left[\frac{1}{2}, 1\right] \\ & \mathrm{f}\left(\frac{1}{2}\right)<0 \end{aligned}$$
$$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$$ is correct.
$$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$$
Let $$\cos \alpha=\mathrm{x}$$,
$$\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}$$
$$\mathrm{x}=\cos \frac{\pi}{12}$$
(4) is correct.
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