JEE PYQ: Binomial Theorem Question 18
Question 18 - 2020 (02 Sep Shift 1)
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:
(1) 336 (2) 352 (3) 84 (4) 176
Show Answer
Answer: (1)
Solution
$T_5 = {}^{10}C_4 \alpha^6 \beta^4$. By AM-GM, $\alpha^6\beta^4 \leq 16$. Max $= 210 \times 16 = 3360 = 10k$, $k = 336$.
BETA
