The maximum value of term independent of ' t ' in the expansion of $\left(\mathrm{t} \mathrm{x}^{\frac{1}{5}}+\frac{(1-\mathrm{x})^{\frac{1}{10}}}{\mathrm{t}}\right)^{10}$ where $\mathrm{x} \in(0, 1)$ is :

(B) $\frac{2.10!}{3 \sqrt{3}(5!)^{2}}$

(A) $\frac{10!}{\sqrt{3}(5!)^{2}}$

(D) $\frac{2.10!}{3(5!)^{2}}$

Binomial Theorem — JEE Main · 2021 · 26 Feb 2021 · Shift 1 Question with Solution

The maximum value of term independent of ' t ' in the expansion of

(t x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{10}

where

x \in (0, 1)

is :

Answer: (B)

The general term is

T_{r + 1} = (r 10) (t x^{1/5})^{10 - r} (\frac{( 1 - x ) ^{1/10}}{t})^{r} = (r 10) t^{10 - 2 r} x^{(10 - r) /5} (1 - x)^{r /10}

.

For the term independent of

t

, we need

10 - 2 r = 0

, so

r = 5

.

The coefficient is

(5 10) x^{5/5} (1 - x)^{5/10} = (5 10) x^{1} (1 - x)^{1/2} = \frac{10 !}{( 5 ! ) ^{2}} x 1 - x

.

To maximize

f (x) = x 1 - x

(0, 1)

, take

f^{'} (x) = 1 - x + x \cdot \frac{- 1}{2 1 - x} = 0

.

This gives

1 - x = \frac{x}{2 1 - x}

, so

2 (1 - x) = x

, hence

x = \frac{2}{3}

.

At

x = \frac{2}{3}

f (\frac{2}{3}) = \frac{2}{3} \frac{1}{3} = \frac{2}{3 3}

.

Maximum value =

\frac{10 !}{( 5 ! ) ^{2}} \cdot \frac{2}{3 3} = \frac{2 \cdot 10 !}{3 3 ( 5 ! ) ^{2}}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option