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 ${\left(\alpha x^{\frac{1}{9}}+\beta x^{-\frac{1}{6}}\right)}^{10}$ is $10k$, then $k$ is equal to:

• A. $176$
• B. $336$
• C. $352$
• D. $84$

Answer 1

The correct option is B

$336$

Explanation for the correct option:

Step-1: Finding the term independent of $x$

For the term independent of $x$, we have

$$\begin{gathered} T_{r+1}={}^{10}{C}_r{\left(\alpha x^{\frac{1}{9}}\right)}^{\left(10-r\right)}{\left(\beta x^{-\frac{1}{6}}\right)}^r \\ ={}^{10}{C}_r{\alpha }^{10-r}{\beta }^r{\left(x\right)}^{\frac{\left(10-r\right)}{9}}{\left(x\right)}^{-\frac{r}{6}} \\ ={}^{10}{C}_r{\alpha }^{10-r}{\beta }^r{{\left(x\right)}^{\frac{\left(10-r\right)}{9}}}^{-\frac{r}{6}}........\left(i\right) \end{gathered}$$

Step-2: Finding the value of $r$

Since the term $T_{r+1}$ is independent of $x$, hence from $\left(i\right)$, we must get the power of $x$ to be zero i.e. $\frac{10-r}{9}-\frac{r}{6}=0$. So,

$$\begin{gathered} \frac{\left(10-r\right)}{9}-\frac{r}{6}=0 \\ \Rightarrow \frac{6\left(10-r\right)-9r}{54}=0 \\ \Rightarrow 6\left(10-r\right)-9r=0 \\ \Rightarrow 60-6r-9r=0 \\ \Rightarrow 15r=60 \\ \Rightarrow r=4 \\ \therefore T_5={}^{10}{C}_4{\alpha }^6{\beta }^4....\left(ii\right)[from(i)] \end{gathered}$$

Step-3: Finding the upper bound of $T_5$

We know that $AM\ge GM$

So for $\frac{{\alpha }^3}{2},\frac{{\alpha }^3}{2},\frac{{\beta }^2}{2},\frac{{\beta }^2}{2}$, we get:

$$\begin{gathered} \Rightarrow \begin{array}{l}\frac{\frac{{\alpha }^3}{2}+\frac{{\alpha }^3}{2}+\frac{{\beta }^2}{2}+\frac{{\beta }^2}{2}}{4}\ge \sqrt[4]{\frac{{\alpha }^3}{2}\times \frac{{\alpha }^3}{2}\times \frac{{\beta }^2}{2}\times \frac{{\beta }^2}{2}}\end{array} \\ \Rightarrow {\left(\frac{{\alpha }^3+{\beta }^2}{4}\right)}^4\ge \left(\frac{{\alpha }^6{\beta }^4}{2^4}\right) \\ \Rightarrow {\left(\frac{4}{4}\right)}^4\ge \left(\frac{{\alpha }^6{\beta }^4}{2^4}\right)\left[\because {\alpha }^3+{\beta }^2=4\right] \\ \Rightarrow {\alpha }^6{\beta }^4\le 2^4 \\ \Rightarrow {}^{10}{C}_4{\alpha }^6{\beta }^4\le {}^{10}{C}_4{2}^4 \\ \Rightarrow T_5\le {}^{10}{C}_4{2}^4[from(ii\left)\right] \\ \Rightarrow T_5\le \frac{10!}{4!\left(10-4\right)!}{2}^4 \\ \Rightarrow T_5\le \frac{\left(10\times 9\times 8\times 7\times 2^4\right)}{(4\times 3\times 2\times 1)} \end{gathered}$$

So, the maximum value of $T_5=10\times 3\times 7\times 16=3360$.

Step-4: Finding the value of $k$

Now, given that the maximum value is $10k$. Therefore, we must have:

$$\begin{gathered} T_5=10k \\ \Rightarrow 10k=3360 \\ \Rightarrow k=\frac{3360}{10} \\ \therefore k=336 \end{gathered}$$

Hence, the correct option is (B).