QJEEM17Ma7015: JEE Main 2017 Mathematics - Class 11 JEEM17Ma7015 Solution: The coefficient of x −5 in the binomial expansion of where x 0, …

Question

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The coefficient of x⁻⁵ in the binomial expansion of

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

where x

\ne

0, 1, 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 = -5

then

r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}

=

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

= 10

\therefore

T₁₁ is the term with x^-5.

\therefore

T₁₁ =

{}^{10}{C_{10}}

= 1

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