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

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