The least value of a ∈ R for which 4αx² + 1/x ≥ 1, for all x 0, is (A) 1/64 (B) 1/32 (C) 1/27 (D)

Q: Why can't we just use the AM-GM inequality here? It looks like a sum of two terms.

You absolutely can! AM-GM is a beautiful shortcut. If you write $4\alpha x^2 + \frac{1}{2x} + \frac{1}{2x} \ge 3 \sqrt[3]{(4\alpha x^2) \cdot (\frac{1}{2x}) \cdot (\frac{1}{2x})}$, the $x$ terms cancel out perfectly. It is a fantastic way to verify your calculus result!

Q: Wait, why do we set the derivative to zero? How do we know it's a minimum and not a maximum?

Great intuition! The second derivative test confirms it: $f''(x) = 8\alpha + \frac{2}{x^3}$. Since $\alpha > 0$ and $x > 0$, the second derivative is always positive, meaning the curve is concave up everywhere. A zero slope in a concave-up curve is guaranteed to be a global minimum.

Q: What happens if $\alpha$ is negative? The problem says $x > 0$, but does $\alpha$ have to be positive?

If $\alpha$ were negative, the term $4\alpha x^2$ would eventually dominate as $x$ increases, pulling the function toward negative infinity. Since we need $f(x) \ge 1$ for all $x > 0$, $\alpha$ must be positive to ensure the function stays bounded from below.

Q: I got $x^3 = \frac{1}{8\alpha}$. How do I handle the cube root correctly?

Take a deep breath. The cube root is a one-to-one function, so there is only one real solution for $x$. Just write $x = (8\alpha)^{-1/3}$ and substitute it carefully. Don't rush the algebra—keep your exponents organized!

Q: Is it possible for the minimum to be exactly 1?

Yes! The problem asks for the *least* value of $\alpha$. If $\alpha = \frac{1}{27}$, the minimum value of the function is exactly 1. If $\alpha$ were any smaller, the 'valley' would dip below 1, violating the condition. So, $\alpha = \frac{1}{27}$ is the boundary case.

Question

The least value of $a \in R$ for which $4\alpha x^2 + \frac{1}{x} \ge 1$, for all $x > 0$, is

Options:

(A) $\frac{1}{64}$

(B) $\frac{1}{32}$

(C) $\frac{1}{27}$

(D) $\frac{1}{25}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a landscape defined by the function $f(x) = 4\alpha x^2 + \frac{1}{x}$. As you look toward the horizon where $x$ approaches zero, the term $\frac{1}{x}$ shoots up toward the heavens.

As you look toward the distant future where $x$ grows large, the term $4\alpha x^2$ dominates, also climbing toward infinity. Between these two extremes, there must be a lowest point—a valley where the slope of the land is perfectly flat.

Our mission is to ensure that this valley never dips below the elevation of $y=1$. If the lowest point is at least $1$, then the entire landscape is guaranteed to be safe.

### The Calculus Toolkit

To find this valley, we turn to the most powerful tool in our arsenal: the derivative. We define our function as $f(x) = 4\alpha x^2 + \frac{1}{x}$.

To find where the slope is zero, we differentiate with respect to $x$:

$$f'(x) = \frac{d}{dx}(4\alpha x^2) + \frac{d}{dx}(x^{-1}) = 8\alpha x - \frac{1}{x^2}$$

Setting this derivative to zero is the moment of truth. We are looking for the critical point $x_0$ where the function stops descending and begins its ascent:

$$8\alpha x - \frac{1}{x^2} = 0 \Rightarrow 8\alpha x = \frac{1}{x^2}$$

Since we are given $x > 0$, we can multiply both sides by $x^2$ without fear of losing information. This gives us $8\alpha x^3 = 1$, or simply $x^3 = \frac{1}{8\alpha}$.

Taking the cube root, we find our critical point:

$$x = \frac{1}{2\alpha^{1/3}}$$

### The Moment of Substitution

Now, we must see how high this valley sits. We substitute our critical $x$ back into the original function $f(x)$.

Let's break it down term by term:

$$f_{min} = 4\alpha \left(\frac{1}{2\alpha^{1/3}}\right)^2 + \frac{1}{\left(\frac{1}{2\alpha^{1/3}}\right)}$$

For the first term, we square the fraction: $4\alpha \cdot \frac{1}{4\alpha^{2/3}}$. The $4$s cancel, and $\frac{\alpha}{\alpha^{2/3}}$ simplifies beautifully to $\alpha^{1/3}$.

For the second term, the reciprocal of the fraction is simply $2\alpha^{1/3}$. Adding them together, we get:

$$f_{min} = \alpha^{1/3} + 2\alpha^{1/3} = 3\alpha^{1/3}$$

### The Final Constraint

We have arrived at the heart of the problem. We know the minimum value of our function is $3\alpha^{1/3}$.

The problem demands that $f(x) \ge 1$ for all $x$. Therefore, our minimum must satisfy:

$$3\alpha^{1/3} \ge 1 \Rightarrow \alpha^{1/3} \ge \frac{1}{3}$$

To isolate $\alpha$, we cube both sides. The inequality remains unchanged because the cube function is strictly increasing:

$$\alpha \ge \left(\frac{1}{3}\right)^3 \Rightarrow \alpha \ge \frac{1}{27}$$

And there it is! The **least value of $\alpha$** that keeps our function above the threshold of $1$ is **$\frac{1}{27}$**.

Animated Solution for Mathematics - Differentiation: The least value of $a \in R$ for which $4 α x^{2} + \frac{1}{x} \geq 1$ , for all $x > 0$ , is

Select Answer:

Visualized Solution

Defining the Function $f (x)$

The Condition for $f (x) \geq 1$

Setting up the Derivative

Calculating $f^{'} (x)$

Setting $f^{'} (x) = 0$

Finding the Critical $x$ -value

Substituting $x$ back into $f (x)$

Simplifying the Expression

Setting $f_{min} \geq 1$

Cubing Both Sides

Final Answer

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Calculus Toolkit

The Moment of Substitution

The Final Constraint

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least value of a∈R for which 4αx2+x1​≥1, for all x>0, is

Select Answer:

Visualized Solution

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)≥1

Setting up the Derivative

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Setting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)=0

Finding the Critical .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-value

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x back into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)