QJEEM19Ma9fa5: JEE Main 2019 Mathematics - Class 11 JEEM19Ma9fa5 Solution: The term independent of x in the expansion of is equal to :

Question

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The term independent of x in the expansion of

\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

is equal to :

Accepted Answer

Given expression =

\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

=

{1 \over {60}}{\left( {2{x^3} - {3 \over {{x^2}}}} \right)^6} - {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

So its general term is

T_{r + 1} =

{1 \over {60}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r} - {{{x^8}} \over {81}}{}^6{C_r}{\left( {2{x^2}} \right)^{6 - r}}{\left( { - {3 \over {{x^2}}}} \right)^r}

=

{1 \over {60}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{12 - 4r}} - {1 \over {81}}{}^6{C_r}{\left( 2 \right)^{6 - r}}{\left( { - 3} \right)^r}{x^{20 - 4r}}

.....(i)

For this term to be independent of x, put r = 3 in 1^st part and r = 5 in 2^nd part.

So from (i) the term independent of
x =

{1 \over {60}} \times {2^3} \times {\left( { - 3} \right)^3} \times {}^6{C_3} + \left( { - {1 \over {81}}} \right)(2){( - 3)^5} \times {}^6{C_5}

= -72 + 36 = -36

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