Verified Solution • 2025–26 Session moderate

QJEEM24Ma97d7: JEE Main 2024 Mathematics - Class 12 JEEM24Ma97d7 Solution: Consider the function defined by . Consider the statements (I)…

Consider the function f:[12,1]→Rf:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R} defined by f(x)=42x3−32x−1f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1 . Consider the statements

(I) The curve y=f(x)y=f(x) intersects the xx -axis exactly at one point.

(II) The curve y=f(x)y=f(x) intersects the xx -axis at x=cos⁡π12x=\cos \frac{\pi}{12} .

Then

Multiple Choice Options

  • A Both (I) and (II) are correct.
  • B Only (I) is correct.
  • C Both (I) and (II) are incorrect.
  • D Only (II) is correct.

Step-by-Step Verified Solution

amp;f′(x)=122x2−32≥0 for [12,1]amp;f(12)lt;0\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)=2(4x3−3x)−1=0f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0

Let cos⁡α=x\cos \alpha=\mathrm{x} ,

cos⁡3α=cos⁡π4⇒α=π12\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}

x=cos⁡π12\mathrm{x}=\cos \frac{\pi}{12}

(4) is correct.

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