Question

Let \(\alpha \gt 0\), \(\beta \gt 0\) be such that
\(\alpha^3 + \beta^2 = 4\). If the maximum value of the term independent of x in
the binomial expansion of \(\left(\alpha x^{\frac19} + \beta x^{-\frac16}\right)^{10}\)
is \(10k\),
then k is equal to :

• A. 176
• B. 336
• C. 352
• D. 84

Hint:

Find r for which the power of x vanishes, then maximise \(\alpha^6\beta^4\) subject to \(\alpha^3+\beta^2=4\).

Answer 1

The Correct Option is B

General term: \(T_{r+1} = \binom{10}{r}\left(\alpha x^{1/9}\right)^{10-r}\left(\beta x^{-1/6}\right)^{r}\), with power of x equal to \(\frac{10-r}{9} - \frac r6\).
Independent of x: \(2(10-r) = 3r \Rightarrow r = 4\).
Term \(= \binom{10}{4}\alpha^6\beta^4 = 210\,\alpha^6\beta^4 = 210\,(\alpha^3\beta^2)^2\).
By AM-GM, \(\alpha^3\beta^2 \le \left(\frac{\alpha^3+\beta^2}{2}\right)^2 = 4\), with equality at \(\alpha^3=\beta^2=2\).
Maximum value \(= 210\times 16 = 3360 = 10k \Rightarrow k = 336\).
Answer: (2) 336