JEE Main 2024 — Functions Question with Solution

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

$\Rightarrow f^{\text{'}}\left(x\right)=12\sqrt{2}{x}^2-3\sqrt{2}$

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

$\Rightarrow f^{\text{'}}\left(x\right)=3\sqrt{2}\left(2x-1\right)\left(2x+1\right)$

So, the function is increasing in $x\in \left(\frac{1}{2},1\right)$.

Thus, the function will intersect the $x-$axis only at one point.

Hence, statement $\mathrm{I}$ is correct.

Now, putting $x=\frac{\pi }{12}$in the equation $\cos 3x=4{\cos }^3\frac{\pi }{12}-3\cos \frac{\pi }{12}$.

$\Rightarrow \cos \frac{\pi }{4}=4{\cos }^3\frac{\pi }{12}-3\cos \frac{\pi }{12}$

$\Rightarrow 4\sqrt{2}{\cos }^3\frac{\pi }{12}-3\sqrt{2}\cos \frac{\pi }{12}-1=0$

Now, comparing with $f\left(x\right)=4\sqrt{2}{x}^3-3\sqrt{2}x-1$ we get.

$\cos \frac{\pi }{12}$ is a solution of $f\left(x\right)$.

So, statement-$\mathrm{II}$ is also correct.

Hence, both the statements are correct.