Prove the following


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 $\left(\alpha x^{\frac{1}{4}}+\beta x^{-\frac{1}{6}}\right)^{10}$ is $10 k$,

then $\mathrm{k}$ is equal to :

  1. (1) 176

  2. 336

  3. 352

  4. 84

Correct Option: , 2


Let $t_{r}+1$ denotes

$r+1^{\text {th }}$ term of $\left(\alpha x^{\frac{1}{9}}+\beta x^{-\frac{1}{6}}\right)^{10}$

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

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

If $t_{r+1}$ is independent of $x$

$\frac{10-r}{9}-\frac{r}{6}=0 \quad \Rightarrow r=4$

maximum value of $\mathrm{t}_{5}$ is $10 \mathrm{~K}$ (given)

$\Rightarrow{ }^{10} \mathrm{C}_{4} \alpha^{6} \beta^{4}$ is maximum

By $\mathrm{AM} \geq \mathrm{GM}$ (for positive numbers)

$\frac{\frac{\alpha^{3}}{2}+\frac{\alpha^{3}}{2}+\frac{\beta^{2}}{2}+\frac{\beta^{2}}{2}}{4} \geq\left(\frac{\alpha^{6} \beta^{4}}{16}\right)^{\frac{1}{4}}$

$\Rightarrow \alpha^{6} \beta^{4} \leq 16$

$\Rightarrow K=336$

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