Question 208 of 347MediumMCQ
In the expansion of (xcosθ+xsinθ1)16, if l1 is the least value of the term independent of x when 8π≤θ≤4π, and l2 is the least value of the term independent of x when 16π≤θ≤8π, then the ratio l2:l1 is equal to:
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Official solution
Tr + 1 = 16Cr\[{\left( {{x \over {\cos \theta }}} \right)^{16 - r}}{\left( {{1 \over {x\sin \theta }}} \right)^r}\]
= 16Cr\[{\left( x \right)^{16 - 2r}} \times {1 \over {{{\left( {\cos \theta } \right)}^{16 - r}}{{\left( {\sin \theta } \right)}^r}}}\]
term is independent of x when
\[ \therefore \] 16 -- 2r = 0
\[ \Rightarrow \] r = 8
T9 = 16C8 \[ \times \] \[{1 \over {{{\cos }^8}\theta {{\sin }^8}\theta }}\]
= 16C8 \[ \times \] \[{{{2^8}} \over {{{\left( {\sin 2\theta } \right)}^8}}}\]
If \[\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]\] then \[2\theta \in \left[ {{\pi \over 4},{\pi \over 2}} \right]\]
In the range \[\left[ {{\pi \over 4},{\pi \over 2}} \right]\], sin 2\[\theta \] is increasing.
And value of T9 is least when sin 2\[\theta \] is maximum.
And sin 2\[\theta \] is maximum in the range \[\left[ {{\pi \over 4},{\pi \over 2}} \right]\]
when 2\[\theta \] = \[{{\pi \over 2}}\]
\[ \therefore \] l1 = 16C8 \[ \times \] 28
AgainIf \[\theta \in \left[ {{\pi \over {16}},{\pi \over 8}} \right]\] then \[2\theta \in \left[ {{\pi \over 8},{\pi \over 4}} \right]\]
In the range \[\left[ {{\pi \over 8},{\pi \over 4}} \right]\], sin 2\[\theta \] is increasing.
And value of T9 is least when sin 2\[\theta \] is maximum.
And sin 2\[\theta \] is maximum in the range \[\left[ {{\pi \over 8},{\pi \over 4}} \right]\]
when 2\[\theta \] = \[{{\pi \over 4}}\]
\[ \therefore \] l2 = 16C8 \[ \times \] \[{{{2^8}} \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^8}}}\]
= 16C8 \[ \times \] 2824
\[ \therefore \] \[{{{l_2}} \over {{l_1}}}\] = \[{{{2^4}} \over 1}\] = \[{{16} \over 1}\]
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