JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 4)
If the greatest value of the term independent of 'x' in the
expansion of $${\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}}$$ is $${{10!} \over {{{(5!)}^2}}}$$, then the value of 'a' is equal to :
expansion of $${\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}}$$ is $${{10!} \over {{{(5!)}^2}}}$$, then the value of 'a' is equal to :
$$-$$1
1
$$-$$2
2
Explanation
$${T_{r + 1}} = {}^{10}{C_r}{(x\sin \alpha )^{10 - r}}{\left( {{{a\cos \alpha } \over x}} \right)^r}$$
r = 0, 1, 2, ......., 10
Tr + 1 will be independent of x when 10 $$-$$ 2r = 0 $$\Rightarrow$$ r = 5
$${T_6} = {}^{10}{C_5}{(x\sin \alpha )^5} \times {\left( {{{a\cos \alpha } \over x}} \right)^5}$$
$$ = {}^{10}{C_5} \times {a^5} \times {1 \over {{2^5}}}{(\sin 2\alpha )^5}$$
will be greatest when sin2$$\alpha$$ = 1
$$ \Rightarrow {}^{10}{C_5}{{{a^5}} \over {{2^5}}} = {}^{10}{C_5} \Rightarrow a = 2$$
r = 0, 1, 2, ......., 10
Tr + 1 will be independent of x when 10 $$-$$ 2r = 0 $$\Rightarrow$$ r = 5
$${T_6} = {}^{10}{C_5}{(x\sin \alpha )^5} \times {\left( {{{a\cos \alpha } \over x}} \right)^5}$$
$$ = {}^{10}{C_5} \times {a^5} \times {1 \over {{2^5}}}{(\sin 2\alpha )^5}$$
will be greatest when sin2$$\alpha$$ = 1
$$ \Rightarrow {}^{10}{C_5}{{{a^5}} \over {{2^5}}} = {}^{10}{C_5} \Rightarrow a = 2$$
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