The maximum value of the term independent of t in the expansion of (tx(1)/(5) + ((1 - x)(1)/(10))/(t)) ^ 10 where x ∈ (0, 1) is:

Correct answer: C. 2 . 10 ! 3 3 (5 !) ^ 2 T _ r + 1 = Cr 10 (t x ^ (1 / 5) ) ^ 10 - r [((1 - x)(1)/(10))/(t)] ^ r = Cr 10 (t) ^ 10 - 2r (x)((10 - r))/(5)(1 - x)(r)/(10) For the term independent of t , ⇒ 10 - 2r = 0 ⇒ r = 5 T _ 6 = f(x) = C5 10 (x 1 - x ); for maximum f ^ ' (x) = 0 ⇒ x = (2)/(3) and f ^ ' ' ((2)/(3)) so f (x) _ max . = C5 10 ((2)/(3)) · 1 3 = 2 . 10 ! 3 3 (5 !) ^ 2

JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2021 · 26 February · Morning Shift · Mathematics · Binomial Theorem

The maximum value of the term independent of t in the expansion of...

The maximum value of the term independent of

t

in the expansion of

{\left(tx\dfrac{1}{5} + \dfrac{\left(1 - x\right)\dfrac{1}{10}}{t}\right)}^{10}

where

x \in \left(0, 1\right)

is:

Options

A\dfrac{10 !}{\sqrt{3}{ $\left(5 !\right)}^{2}}$
B\dfrac{10 !}{3{ $\left(5 !\right)}^{2}}$
C\dfrac{2 . 10 !}{3\sqrt{3}{ $\left(5 !\right)}^{2}}$
D\dfrac{2 . 10 !}{3{ $\left(5 !\right)}^{2}}$

Correct answer

C. \dfrac{2 . 10 !}{3\sqrt{3}{ $\left(5 !\right)}^{2}}$

Step-by-step solution

${T}_{r + 1} = Cr 10 {\left(t{x}^{\left(1 / 5\right)}\right)}^{10 - r}{\left[\dfrac{\left(1 - x\right)\dfrac{1}{10}}{t}\right]}^{r}$

$= Cr 10 {\left(t\right)}^{10 - 2r}\left(x\right)\dfrac{\left(10 - r\right)}{5}\left(1 - x\right)\dfrac{r}{10}$

For the term independent of $t$ ,

$⇒ 10 - 2r = 0$

$⇒ r = 5$

${T}_{6} = f\left(x\right) = C5 10 \left(x\sqrt{1 - x}\right);$ for maximum

${f}^{'}\left(x\right) = 0 ⇒ x = \dfrac{2}{3} \text{ and } {f}^{' '}\left(\dfrac{2}{3}\right) < 0$

so $f{\left(x\right)}_{max .} = C5 10 \left(\dfrac{2}{3}\right) \cdot \dfrac{1}{\sqrt{3}}$

$= \dfrac{2 . 10 !}{3\sqrt{3}{\left(5 !\right)}^{2}}$

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