Let α 0, β 0 be such that α³ + β² = 4. If the maximum value of the term independent of x in the

Q: Wait, why do we set the exponent to zero? I thought we were looking for a coefficient.

Great question! A term is 'independent of $x$' if it does not contain $x$. Mathematically, this is equivalent to saying the power of $x$ is $x^0$, which is just $1$. So, by setting the net exponent of $x$ to zero, we are forcing $x$ to vanish, leaving us with only the constant coefficient.

Q: Can we use AM-GM directly on $\alpha^6 \beta^4$?

Not directly! AM-GM is most powerful when the terms in the inequality match the terms in your constraint. Since your constraint involves $\alpha^3$ and $\beta^2$, it is much easier to apply AM-GM to those specific terms first, and then square the result to reach $\alpha^6 \beta^4$.

Q: Is it possible for $r$ to be a fraction? What if I got $r=4.5$?

In a binomial expansion $(a+b)^n$, $r$ must be an integer such that $0 \le r \le n$. If you calculate a non-integer $r$, it usually means there is no term independent of $x$ in that specific expansion. In this problem, the math works out perfectly to $r=4$, which is a valid index.

Question

Let $\alpha > 0, \beta > 0$ be such that $\alpha^3 + \beta^2 = 4$. If the maximum value of the term independent of $x$ in the binomial expansion of $(\alpha x^{1/9} + \beta x^{-1/6})^{10}$ is $10k$, then $k$ is equal to :

Options:

(A) 84

(B) 176

(C) 336

(D) 352

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the maximum value of the term independent of $x$ in the expansion of $(\alpha x^{1/9} + \beta x^{-1/6})^{10}$, subject to the constraint $\alpha^3 + \beta^2 = 4$.

The general term of a binomial expansion $(a+b)^n$ is given by the formula:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Substituting our specific values where $n=10$, $a = \alpha x^{1/9}$, and $b = \beta x^{-1/6}$, we obtain:

$$T_{r+1} = \binom{10}{r} (\alpha x^{1/9})^{10-r} (\beta x^{-1/6})^r$$

### The Quest for Independence

To isolate the term independent of $x$, we must group the powers of $x$ and set their sum to zero. The exponent of $x$ in the general term is:

$$\frac{10-r}{9} - \frac{r}{6} = 0$$

Multiplying the entire equation by the least common multiple, $18$, we simplify the expression:

$$2(10-r) - 3r = 0$$

$$20 - 5r = 0 \Rightarrow r = 4$$

Substituting $r=4$ back into our general term expression, the independent term $T_5$ is:

$$T_5 = \binom{10}{4} \alpha^6 \beta^4$$

### The AM-GM Masterclass

We now maximize the product $\alpha^6 \beta^4$ subject to $\alpha^3 + \beta^2 = 4$. We apply the **Arithmetic Mean-Geometric Mean (AM-GM) Inequality**, which states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean.

Applying this to the terms $\alpha^3$ and $\beta^2$:

$$\frac{\alpha^3 + \beta^2}{2} \ge \sqrt{\alpha^3 \beta^2}$$

Substituting the constraint $\alpha^3 + \beta^2 = 4$:

$$\frac{4}{2} \ge \sqrt{\alpha^3 \beta^2} \Rightarrow 2 \ge \sqrt{\alpha^3 \beta^2}$$

Squaring both sides yields $4 \ge \alpha^3 \beta^2$. To reach our target term $\alpha^6 \beta^4$, we square the inequality once more:

$$16 \ge \alpha^6 \beta^4$$

### Final Calculation

The maximum value of the independent term is the product of the binomial coefficient and the maximum value of the variable component:

$$	ext{Max Value} = \binom{10}{4} 	imes 16$$

Calculating the binomial coefficient:

$$\binom{10}{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$

Thus, the maximum value is:

$$210 	imes 16 = 3360$$

Given that this value is represented as $10k$, we find $10k = 3360$, which results in **$k = 336$**.

Animated Solution for Mathematics - Binomial Theorem: Let $α > 0, β > 0$ be such that $α^{3} + β^{2} = 4$ . If the maximum value of the term independent of $x$ in the binomial expansion of $(α x^{1/9} + β x^{- 1/6})^{10}$ is $10 k$ , then $k$ is equal to :

Select Answer:

Visualized Solution

The Binomial Expression

General Term Formula

Substituting the Values

Isolating the Exponent of $x$

Condition for Independence

Solving for $r$

The Independent Term

Applying AM-GM Inequality

Executing AM-GM

Maximizing the Product

Maximum Value of the Term

Finding $k$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Quest for Independence

The AM-GM Masterclass

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

The Binomial Expression

General Term Formula

Substituting the Values

Isolating the Exponent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Condition for Independence

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

The Independent Term

Applying AM-GM Inequality

Executing AM-GM

Maximizing the Product

Maximum Value of the Term

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

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Quest for Independence

The AM-GM Masterclass

Final Calculation

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