Consider the function f : [1/2, 1] → R defined by f(x) = 4√2x³ - 3√2x - 1. Consider the statements (

Q: Wait, why do we reject the negative root here? Or why don't we need to worry about other roots?

Great question! We don't 'reject' other roots in the general sense; we are restricted to the interval $[\frac{1}{2}, 1]$. Because we proved that $f'(x) \ge 0$ on this specific interval, the function is strictly increasing. A strictly increasing function can cross the $x$-axis at most once. Since we found one root in this interval, there cannot be another one hiding there.

Q: Why can't we just use energy conservation or some other physics trick? This looks like a kinematics problem.

I love the intuition! While this isn't a physics problem, the 'energy' of the function—its behavior—is what we are analyzing. In JEE Advanced, you will often find that math problems are disguised as physics or vice versa. Here, the 'trick' is recognizing the algebraic pattern $4x^3 - 3x$, which is the signature of the triple-angle identity for cosine. Always look for patterns before brute-forcing!

Q: How do I know for sure that $\cos(\frac{\pi}{12})$ is in the interval $[\frac{1}{2}, 1]$?

Excellent attention to detail. We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(0) = 1$. Since the cosine function is strictly decreasing in the first quadrant, and we know that $0 < \frac{\pi}{12} < \frac{\pi}{3}$, it follows logically that $\cos(\frac{\pi}{3}) < \cos(\frac{\pi}{12}) < \cos(0)$. Thus, $\frac{1}{2} < \cos(\frac{\pi}{12}) < 1$. It fits perfectly!

Question

Consider the function $f : [\frac{1}{2}, 1] 	o \mathbb{R}$ defined by $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$. Consider the statements
(I) The curve $y = f(x)$ intersects the $x$-axis exactly at one point
(II) The curve $y = f(x)$ intersects the $x$-axis at $x = \cos \frac{\pi}{12}$
Then

Options:

(A) Only (II) is correct

(B) Both (I) and (II) are incorrect

(C) Only (I) is correct

(D) Both (I) and (II) are correct

Sigma Percentile · Accepted Answer

### The Cubic Mystery: Unlocking the Roots

Welcome, fellow traveler on the JEE journey. Today, we are going to dissect a cubic function that might look intimidating at first glance, but hides a beautiful, elegant secret.

We are looking at the function $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$ defined on the interval $x \in [\frac{1}{2}, 1]$. Our mission is to verify two statements: one about the number of roots, and one about the specific location of that root.

### Phase 1: The Monotonicity Hunt

Before we start hunting for roots, we need to understand the 'personality' of this curve. To find out if it is rising or falling, we calculate the derivative, $f'(x)$.

Differentiating $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$ with respect to $x$, we get:

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

Now, let's make this expression talk to us. We can factor out $3\sqrt{2}$ to get:

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

Look closely at our interval, $x \in [\frac{1}{2}, 1]$. If $x \ge \frac{1}{2}$, then $x^2 \ge \frac{1}{4}$, which means $4x^2 \ge 1$. This implies that $4x^2 - 1 \ge 0$.

Because the derivative is non-negative throughout our interval, the function $f(x)$ is strictly increasing. It is a one-way street—it only goes up! This tells us that if the curve crosses the $x$-axis, it can only do so exactly once.

### Phase 2: The IVT Bridge

Now that we know the function is strictly increasing, we need to confirm it actually crosses the $x$-axis. We use the Intermediate Value Theorem (IVT).

At the lower bound, $x = \frac{1}{2}$, we have:

$$f\left(\frac{1}{2}ight) = 4\sqrt{2}\left(\frac{1}{8}ight) - 3\sqrt{2}\left(\frac{1}{2}ight) - 1 = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} - 1 = -\sqrt{2} - 1$$

This is clearly negative. At the upper bound, $x = 1$, we have:

$$f(1) = 4\sqrt{2} - 3\sqrt{2} - 1 = \sqrt{2} - 1$$

Since $\sqrt{2} \approx 1.414$, this value is positive. Because the function is continuous and changes sign from negative to positive, it must cross the $x$-axis. Combined with our monotonicity proof, we have confirmed Statement (I): there is exactly one root.

### Phase 3: The Trigonometric Revelation

Now for the grand finale: Statement (II). We need to see if the root is at $x = \cos\left(\frac{\pi}{12}ight)$. Let's set $f(x) = 0$:

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

Rearranging this, we get $\sqrt{2}(4x^3 - 3x) = 1$, or:

$$4x^3 - 3x = \frac{1}{\sqrt{2}}$$

Does the expression $4x^3 - 3x$ look familiar? It is the exact structure of the triple-angle identity for cosine: $\cos(3	heta) = 4\cos^3	heta - 3\cos	heta$.

If we substitute $x = \cos	heta$, our equation becomes:

$$\cos(3	heta) = \frac{1}{\sqrt{2}}$$

We know that $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$, so $3	heta = \frac{\pi}{4}$, which gives us $	heta = \frac{\pi}{12}$. Thus, **$x = \cos\left(\frac{\pi}{12}ight)$ is indeed a root**.

Since $\frac{\pi}{12}$ is in the first quadrant and corresponds to a value between $\frac{1}{2}$ and $1$, it lies perfectly within our interval. Both statements are true. You have just mastered the art of combining calculus, the IVT, and trigonometric identities to solve a complex cubic.

Select Answer:

Visualized Solution

Analyzing $f (x)$ on $[\frac{1}{2}, 1]$

Finding the Derivative $f^{'} (x)$

Factoring $f^{'} (x)$

Checking Monotonicity

Setup for $f (\frac{1}{2})$

Computing $f (\frac{1}{2})$

Setup for $f (1)$

Computing $f (1)$

Existence of Exactly One Root

Finding the Exact Root

Trigonometric Substitution

Solving for $θ$

Verifying the Interval

Final Conclusion

The Sigma Insight: Monotonicity

The Cubic Mystery

Phase 1

Phase 2

Phase 3

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Visualized Solution

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