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 verify the statements concerning the function \( f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1 \) on the interval \([\frac{1}{2}, 1]\), each statement requires individual analysis.

Analysis of Statement (I):

Statement (I) asserts that the graph of \( y = f(x) \) crosses the x-axis at precisely one point. This necessitates confirming a unique solution to \( f(x) = 0 \).

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

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

To locate intersection points, we solve for \( x \), testing potential roots within \([\frac{1}{2}, 1]\). Statement (II) proposes a specific value for testing:

Statement (II) proposes:

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

The value of \( \cos\frac{\pi}{12} \) is:

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

Observation:

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

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

Algebraic simplification of this expression confirms that \( f\left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) = 0 \), establishing it as an x-intercept.

Given that \( f(x) \) is a cubic polynomial, and the confirmation of a root at \( x = \cos\frac{\pi}{12} \), polynomial properties imply that there can be at most one x-intercept within the specified domain.

Conclusion:

• Both Statement (I) and Statement (II) are validated. The curve has a single x-intercept at \( x = \cos\frac{\pi}{12} \).

Consequently, the accurate conclusion is that Both (I) and (II) are correct.