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QJEEM21Ma68e9: JEE Main 2021 Mathematics - Class 11 JEEM21Ma68e9 Solution: If the greatest value of the term independent of 'x' in the…

If the greatest value of the term independent of 'x' in the

expansion of (xsin⁡α+acos⁡αx)10{\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}} is 10!(5!)2{{10!} \over {{{(5!)}^2}}} , then the value of 'a' is equal to :

Multiple Choice Options

  • A 1
  • B −- 1
  • C −- 2
  • D 2

Step-by-Step Verified Solution

Tr+1=10Cr(xsin⁡α)10−r(acos⁡αx)r{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

T6=10C5(xsin⁡α)5×(acos⁡αx)5{T_6} = {}^{10}{C_5}{(x\sin \alpha )^5} \times {\left( {{{a\cos \alpha } \over x}} \right)^5}

=10C5×a5×125(sin⁡2α)5= {}^{10}{C_5} \times {a^5} \times {1 \over {{2^5}}}{(\sin 2\alpha )^5}

will be greatest when sin2 α\alpha = 1

⇒10C5a525=10C5⇒a=2\Rightarrow {}^{10}{C_5}{{{a^5}} \over {{2^5}}} = {}^{10}{C_5} \Rightarrow a = 2

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