QJEEM13Maa4be: JEE Main 2013 Mathematics - Class 11 JEEM13Maa4be Solution: The term independent of in expansion of is

Question

Verified Solution • 2025–26 Session moderate

The term independent of

x

in expansion of

{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}

is

Accepted Answer

{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}

=

{\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}

=

{\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}

=

{\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}

=

{\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}

=

{\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}

[Note:

For

{\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}

the

\left( {r + 1} \right)

^th term with power m of x is

r = {{n\alpha - m} \over {\alpha + \beta }}

]

Here

\alpha = {1 \over 3}

,

\beta = {1 \over 2}

and m = 0

then

r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}

=

{{10} \over 3} \times {6 \over 5}

= 4

\therefore

T₅ is the term independent of x.

\therefore

T₅ =

{}^{10}{C_4}

= 210

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