Let alpha>0, beta>0 be such that $lpha^{3}+eta^{2}=4$


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 $10 k$, then $k$ is equal to :

  1. (1) 336

  2. (2) 352

  3. (3) 84

  4. (4) 176

Correct Option: 1


General term of

$\left(\alpha x^{\frac{1}{9}}+\beta x^{\frac{-1}{6}}\right)^{10}={ }^{10} C_{r}\left(\alpha x^{\frac{1}{9}}\right)^{10-r}\left(\beta x^{\frac{-1}{6}}\right)^{r}$

$={ }^{10} C_{r} \alpha^{10-r} \beta^{r}(x)^{\frac{10-r}{9}-\frac{r}{6}}$

Term independent of $x$ if $\frac{10-r}{9}-\frac{r}{6}=0 \Rightarrow r=4$.

$\therefore$ Term independent of $x={ }^{10} C_{4} \alpha^{6} \beta^{4}$

Since $\alpha^{3}+\beta^{2}=4$

Then, by AM-GM inequality

$\frac{\alpha^{3}+\beta^{2}}{2} \geq\left(\alpha^{3} b^{2}\right)^{\frac{1}{2}}$

$\Rightarrow(2)^{2} \geq \alpha^{3} \beta^{2} \Rightarrow \alpha^{6} \beta^{4} \leq 16$

$\because$ The maximum value of the term independent of

$x=10 k$

$\therefore 10 k={ }^{10} C_{4} \cdot 16 \Rightarrow k=336$

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