JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 1)
The maximum value of the term independent of 't' in the expansion
of $${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$$ where x$$\in$$(0, 1) is :
of $${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$$ where x$$\in$$(0, 1) is :
$${{10!} \over {\sqrt 3 {{(5!)}^2}}}$$
$${{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}$$
$${{10!} \over {3{{(5!)}^2}}}$$
$${{2.10!} \over {3{{(5!)}^2}}}$$
Explanation
$${T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}$$
$$ = {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}$$
$$ \Rightarrow 10 - 2r = 0 \Rightarrow r = 5$$
$$ \therefore $$ $${T_6} = {}^{10}{C_5} \times x\sqrt {1 - x} $$
At maximum, $${{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0$$
$$ \Rightarrow $$ $$ 1 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3$$
$${T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }}$$ = $${{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}$$
$$ = {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}$$
$$ \Rightarrow 10 - 2r = 0 \Rightarrow r = 5$$
$$ \therefore $$ $${T_6} = {}^{10}{C_5} \times x\sqrt {1 - x} $$
At maximum, $${{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0$$
$$ \Rightarrow $$ $$ 1 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3$$
$${T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }}$$ = $${{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}$$
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