Consider the function f:[ 1 2,1]→ R defined by f(x)=4√2 x^3-3√2x-1. Consider the statements (I) The curve y=f(x) intersects the x-axis exactly at one…

Given: $f(x)=4\sqrt{2}{x}^{3}-3\sqrt{2}x-1$f(x)=42​x3−32​x−1

$$\Rightarrow {f}^{'}(x)=12\sqrt{2}{x}^{2}-3\sqrt{2}$$⇒f′(x)=122​x2−32​

$$\Rightarrow {f}^{'}(x)=3\sqrt{2}(4{x}^{2}-1)$$⇒f′(x)=32​(4x2−1)

$$\Rightarrow {f}^{'}(x)=3\sqrt{2}(2x-1)(2x+1)$$⇒f′(x)=32​(2x−1)(2x+1)

So, the function is increasing in $x\in (\frac{1}{2},1)$x∈(21​,1).

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

Hence, statement $I$I is correct.

Now, putting $x=\frac{\pi }{12}$x=12π​in the equation $\mathrm{cos}3x=4{\mathrm{cos}}^{3}\frac{\pi }{12}-3\mathrm{cos}\frac{\pi }{12}$cos3x=4cos312π​−3cos12π​.

$$\Rightarrow \mathrm{cos}\frac{\pi }{4}=4{\mathrm{cos}}^{3}\frac{\pi }{12}-3\mathrm{cos}\frac{\pi }{12}$$⇒cos4π​=4cos312π​−3cos12π​

$$\Rightarrow 4\sqrt{2}{\mathrm{cos}}^{3}\frac{\pi }{12}-3\sqrt{2}\mathrm{cos}\frac{\pi }{12}-1=0$$⇒42​cos312π​−32​cos12π​−1=0

Now, comparing with $f(x)=4\sqrt{2}{x}^{3}-3\sqrt{2}x-1$f(x)=42​x3−32​x−1 we get.

$\mathrm{cos}\frac{\pi }{12}$cos12π​ is a solution of $f(x)$f(x).

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

Hence, both the statements are correct.