JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 18)
If the term independent of x in the expansion of
$${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$$ is k, then 18 k is equal to :
$${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$$ is k, then 18 k is equal to :
5
9
7
11
Explanation
General term,
$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}$$
$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}$$
For independent of x
18 $$ - $$ 3r = 0 $$ \Rightarrow $$ r = 6
$$ \therefore $$ $${T_7} = {}^9{C_6}{\left( {{3 \over 2}} \right)^3}{\left( { - {1 \over 3}} \right)^6} = {{21} \over {54}} = k$$
$$ \therefore $$ $$18k = {{21} \over {54}} \times 18 = 7$$
$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}$$
$${T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}$$
For independent of x
18 $$ - $$ 3r = 0 $$ \Rightarrow $$ r = 6
$$ \therefore $$ $${T_7} = {}^9{C_6}{\left( {{3 \over 2}} \right)^3}{\left( { - {1 \over 3}} \right)^6} = {{21} \over {54}} = k$$
$$ \therefore $$ $$18k = {{21} \over {54}} \times 18 = 7$$
Comments (0)
