If the maximum value of the term independent of t in the expansion of $\left(t^{2}x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right)^{15}$, $x \geq 0$, is K, then 8K is equal to ____.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Binomial Theorem

If the maximum value of the term independent of t in the expansion of

(t^{2} x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{15}

x \geq 0

, is K, then 8K is equal to ____.

Answer: 6006

The general term in the binomial expansion is

T_{r + 1} = (r 15) (t^{2} x^{1/5})^{15 - r} (\frac{( 1 - x ) ^{1/10}}{t})^{r} = (r 15) t^{30 - 3 r} x^{(15 - r) /5} (1 - x)^{r /10}

. For the term independent of

t

, we need

30 - 3 r = 0

, so

r = 10

. This term is

(10 15) x^{5/5} (1 - x)^{10/10} = (10 15) x (1 - x) = 3003 \cdot x (1 - x)

. The maximum of

f (x) = x (1 - x)

for

x \in [0, 1]

occurs at

x = 1/2

, giving

f (1/2) = 1/4

. Thus,

K = 3003 \cdot \frac{1}{4} = \frac{3003}{4}

, and

8 K = 2 \cdot 3003 = 6006

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer