JEE PYQ: Binomial Theorem Question 16
Question 16 - 2021 (26 Feb Shift 1)
The maximum value of the term independent of $t$ in the expansion of $\left(tx^{1/5} + \frac{(1-x)^{1/10}}{t}\right)^{10}$ where $x \in (0,1)$ is:
(1) $\frac{10!}{\sqrt{3}(5!)^2}$ (2) $\frac{2 \cdot 10!}{3(5!)^2}$ (3) $\frac{10!}{3(5!)^2}$ (4) $\frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$
Show Answer
Answer: (4)
Solution
Term independent of $t$: $r = 5$. $T_6 = {}^{10}C_5 x\sqrt{1-x}$. Maximum at $x = 2/3$.
BETA
