Let α = Σ (k=1 to ∞) sin²ᵏ(π/6). Let g : [0, 1] → R be the function defined by g(x) = 2^(αx) + 2^(α(

Q: Wait, how did you get $\alpha = \frac{1}{3}$? I thought the sum formula was $\frac{a}{1-r}$, but I got confused with the powers.

That is a great question! Let's look at the series: $\sum_{k=1}^{\infty} (\frac{1}{2})^{2k}$. This is $\sum_{k=1}^{\infty} (\frac{1}{4})^k$. The first term ($k=1$) is $a = \frac{1}{4}$. The common ratio is $r = \frac{1}{4}$. Plugging these into $\frac{a}{1-r}$, we get $\frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$. You were likely just one step away from the answer!

Q: Why do we have to check the boundaries $x=0$ and $x=1$? Isn't the derivative enough to find the maximum?

In calculus, the derivative helps us find 'local' peaks and valleys (critical points). However, when we are restricted to a closed interval like $[0, 1]$, the function might be increasing or decreasing as it hits the wall of the interval. The absolute maximum or minimum often hides at the endpoints. Always check the boundaries when the domain is restricted!

Q: Why does the function have the same value at $x=0$ and $x=1$? Is that a coincidence?

It is not a coincidence at all! Notice the symmetry in the function $g(x) = 2^{\frac{x}{3}} + 2^{\frac{1-x}{3}}$. If you replace $x$ with $(1-x)$, the function remains unchanged: $g(1-x) = 2^{\frac{1-x}{3}} + 2^{\frac{1-(1-x)}{3}} = 2^{\frac{1-x}{3}} + 2^{\frac{x}{3}} = g(x)$. This symmetry guarantees that the function behaves identically at both ends of the interval.

Question

Let $\alpha = \sum_{k=1}^{\infty} \sin^{2k} \left( \frac{\pi}{6} ight)$. Let $g : [0, 1] 	o \mathbb{R}$ be the function defined by $g(x) = 2^{\alpha x} + 2^{\alpha(1-x)}$. Then, which of the following statements is/are TRUE?

Options:

(A) The minimum value of $g(x)$ is $2^{7/6}$

(B) The maximum value of $g(x)$ is $1 + 2^{1/3}$

(C) The function $g(x)$ attains its maximum at more than one point

(D) The function $g(x)$ attains its minimum at more than one point

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the expression $\alpha = \sum_{k=1}^{\infty} \sin^{2k} \left( \frac{\pi}{6} \right)$. Recognizing that $\sin(\frac{\pi}{6}) = \frac{1}{2}$, the term inside the summation simplifies to $(\frac{1}{2})^{2k} = (\frac{1}{4})^k$.

The series becomes $\sum_{k=1}^{\infty} (\frac{1}{4})^k$, which is an infinite geometric progression. Here, the first term $a = \frac{1}{4}$ and the common ratio $r = \frac{1}{4}$.

Using the sum formula for an infinite geometric series, $S = \frac{a}{1-r}$, we calculate:

$$\alpha = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$$

### The Anatomy of $g(x)$

With $\alpha = \frac{1}{3}$ determined, our function $g(x)$ on the interval $x \in [0, 1]$ is defined as:

$$g(x) = 2^{\frac{x}{3}} + 2^{\frac{1-x}{3}}$$

Observe the inherent symmetry: replacing $x$ with $1-x$ leaves the function unchanged. This indicates that the graph is symmetric about the vertical line $x = \frac{1}{2}$.

### The Calculus Journey

To locate the extrema, we compute the derivative $g'(x)$ using the chain rule, where the derivative of $a^u$ is $a^u \ln(a) \frac{du}{dx}$:

$$g'(x) = \frac{d}{dx} \left( 2^{\frac{x}{3}} \right) + \frac{d}{dx} \left( 2^{\frac{1-x}{3}} \right)$$

$$g'(x) = 2^{\frac{x}{3}} \ln(2) \cdot \frac{1}{3} + 2^{\frac{1-x}{3}} \ln(2) \cdot \left( -\frac{1}{3} \right)$$

Factoring out the constants, we obtain:

$$g'(x) = \frac{\ln 2}{3} \left( 2^{\frac{x}{3}} - 2^{\frac{1-x}{3}} \right)$$

Setting $g'(x) = 0$ to find critical points leads to $2^{\frac{x}{3}} = 2^{\frac{1-x}{3}}$. Equating the exponents, we find the critical point at $x = \frac{1}{2}$.

### Final Calculation

We evaluate the function at the critical point $x = \frac{1}{2}$:

$$g\left(\frac{1}{2}\right) = 2^{\frac{1}{6}} + 2^{\frac{1/2}{3}} = 2^{\frac{1}{6}} + 2^{\frac{1}{6}} = 2 \cdot 2^{\frac{1}{6}} = 2^{\frac{7}{6}}$$

Next, we evaluate the function at the boundaries $x=0$ and $x=1$:

$$g(0) = 2^0 + 2^{\frac{1}{3}} = 1 + 2^{\frac{1}{3}}$$

$$g(1) = 2^{\frac{1}{3}} + 2^0 = 1 + 2^{\frac{1}{3}}$$

Comparing the values, $2^{\frac{7}{6}} \approx 2.2449$ and $1 + 2^{\frac{1}{3}} \approx 2.2599$. We conclude that the **minimum occurs at $x = \frac{1}{2}$**, while the **maximum occurs at the boundaries $x=0$ and $x=1$**.

Animated Solution for Mathematics - Differentiation: Let $α = \sum_{k = 1}^{\infty} sin^{2 k} (\frac{π}{6})$ . Let $g : [0, 1] \to R$ be the function defined by $g (x) = 2^{α x} + 2^{α (1 - x)}$ . Then, which of the following statements is/are TRUE?

Select Answer:

Visualized Solution

Analyze $α$

Sum the Geometric Progression

Define the Function $g (x)$

Differentiate $g (x)$

Find Critical Points

Solve for $x$

Calculate Minimum Value

Evaluate Boundaries

Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Anatomy of $g (x)$

The Calculus Journey

Final Calculation

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Select Answer:

Visualized Solution

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Sum the Geometric Progression

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

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

Find Critical Points

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Calculate Minimum Value

Evaluate Boundaries

Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Anatomy of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

The Calculus Journey

Final Calculation

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