Verified Solution • 2025–26 Session moderate

QJEEM21Maa608: JEE Main 2021 Mathematics - Class 11 JEEM21Maa608 Solution: The maximum value of the term independent of 't' in the…

The maximum value of the term independent of 't' in the expansion
of (tx15+(1−x)110t)10{\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}} where x ∈\in (0, 1) is :

Multiple Choice Options

  • A 10!3(5!)2{{10!} \over {\sqrt 3 {{(5!)}^2}}}
  • B 2.10!33(5!)2{{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}
  • C 10!3(5!)2{{10!} \over {3{{(5!)}^2}}}
  • D 2.10!3(5!)2{{2.10!} \over {3{{(5!)}^2}}}

Step-by-Step Verified Solution

Tr+1=10Cr(tx1/5)10−r[(1−x)1/10t]r{T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}

=10Crt(10−2r)×x10−r5×(1−x)r10= {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}

⇒10−2r=0⇒r=5\Rightarrow 10 - 2r = 0 \Rightarrow r = 5

∴\therefore T6=10C5×x1−x{T_6} = {}^{10}{C_5} \times x\sqrt {1 - x}

At maximum, dT6dx=10C5[1−x−x21−x]=0{{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0

⇒\Rightarrow 1−x=x/2⇒3x=2⇒x=2/31 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3

T6∣max⁡=10!5!5!×233{T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }} = 2.10!33(5!)2{{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}

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