JEE Main 2024 — 29 January, Shift 1 — Question 13

Given: $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1,\quad x \in \left[\frac{1}{2}, 1\right]$

Let us find the roots of $f(x) = 0$

$f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$

Try $x = \cos\left(\frac{\pi}{12}\right)$

$\cos\left(\frac{\pi}{12}\right) = \cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$

$\text{Let } x = \frac{\sqrt{6} + \sqrt{2}}{4}$

$f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$

$f(x) = \sqrt{2} \left(4x^3 - 3x\right) - 1$

Now, recall the identity: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$

$\text{So } f(x) = \sqrt{2} \cdot \cos(3\theta) - 1, \text{ where } \theta = \frac{\pi}{12}$

$\Rightarrow 3\theta = \frac{\pi}{4}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

$f(x) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} - 1 = 1 - 1 = 0$

$\Rightarrow x = \cos\left(\frac{\pi}{12}\right)$ is a root of $f(x)$

$\text{Let } f(x) = \sqrt{2}(4x^3 - 3x) - 1$

$\text{Let } g(x) = 4x^3 - 3x$

$g'(x) = 12x^2 - 3 > 0 \text{ for } x \in \left[\frac{1}{2}, 1\right]$

$\Rightarrow$ $g(x)$ is strictly increasing on $\left[\frac{1}{2}, 1\right]$

$\Rightarrow f(x)$ is strictly increasing$

Therefore, $f(x) = 0$ has exactly one root in $\left[\frac{1}{2}, 1\right]$

$\Rightarrow$ Both statements (I) and (II) are correct