JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 7)
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 :
$$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$ is equal to :
36
- 108
- 36
- 72
Explanation
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
Tr + 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 1st part and r = 5 in 2nd 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
= $${1 \over {60}}{\left( {2{x^3} - {3 \over {{x^2}}}} \right)^6} - {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$$
So its general term is
Tr + 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 1st part and r = 5 in 2nd 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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