Question

Consider the function \(f :[\frac{1}{2},1]\)⇢R defined by  \(f(x)=4\sqrt2x^3-3\sqrt2x-1\).Consider the statements
(1)The curve y=f(x) intersect the x-axis exactly at one point
(2)The curve y=f(x) intersect the x-axis at \(x=cos\frac{\pi}{12}\)
Then 

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

Answer 1

The Correct Option is D

To determine the correctness of the statements related to the function \( f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1 \) defined over the interval \([\frac{1}{2}, 1]\), we need to consider each statement and analyze them individually.

Analysis of Statement (I): 

The statement claims that the curve \( y = f(x) \) intersects the x-axis exactly at one point. To check this, we need to determine if there is a unique solution to the equation \( f(x) = 0 \).

The equation \( f(x) = 0 \) becomes:

\( 4\sqrt{2}x^3 - 3\sqrt{2}x - 1 = 0 \)

To find the intersection points, we solve for \( x \) by substituting possible roots within the interval \([\frac{1}{2}, 1]\). Let's test at a specific point, say \( x = \cos\frac{\pi}{12} \), noted in Statement (II).

Statement (II) suggests:

\( x = \cos\frac{\pi}{12} \)

Calculating \( \cos\frac{\pi}{12} \):

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

Observation:

Plug this value into \( f(x) \):

\( f\left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) = 4\sqrt{2}\left( \frac{\sqrt{6} + \sqrt{2}}{4} \right)^3 - 3\sqrt{2}\left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) - 1 \)

The complex calculations will simplify this to confirm that \( f\left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) \) results in 0, verifying that this is indeed an x-intercept.

As \( f(x) \) is a cubic polynomial and given the analysis at \( x = \cos\frac{\pi}{12} \) confirms a root, and considering the nature of polynomials and complex conjugate roots or real roots, there can't be more than one x-intercept within the domain.

Conclusion:

• Both Statement (I) and Statement (II) are correct. The curve intersects the x-axis exactly at the root \( x = \cos\frac{\pi}{12} \).

Thus, the correct answer is that Both (I) and (II) are correct.

Answer 2

Step 1: Check the Derivative \(f'(x)\) for Monotonicity

\(f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} \geq 0\) for \(\left[\frac{1}{2}, 1\right]\)

Step 2: Evaluate \(f(x)\) at the Endpoints

\(f\left(\frac{1}{2}\right) < 0\)

\(f(1) > 0\)

Since \(f(x)\) changes sign from negative to positive, there must be exactly one root in \(\left[\frac{1}{2}, 1\right]\), confirming that statement (I) is correct.

Step 3: Check if \(x = \cos \frac{\pi}{12}\) is a Root

Rewrite \(f(x)\) in terms of \(\cos \alpha\):

\(f(x) = \sqrt{2}(4x^3 - 3x) - 1 = 0\)

Let \(\cos \alpha = x\), then \(\cos 3\alpha = x\) gives \(\alpha = \frac{\pi}{12}\), so:

\(x = \cos \frac{\pi}{12}\)

This confirms statement (II) is also correct.

So, the correct answer is: Both (I) and (II) are correct