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In the expansion of $${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$$, if $${\ell _1}$$ is the least value of the term independent of x when $${\pi \over 8} \le \theta \le {\pi \over 4}$$ and $${\ell _2}$$ is the least value of the term independent of x when $${\pi \over {16}} \le \theta \le {\pi \over 8}$$, then the ratio $${\ell _2}$$ : $${\ell _1}$$ is equal to :

JEE · Math · previous-year question

  1. A.8 : 1
  2. B.16 : 1correct
  3. C.1 : 8
  4. D.1 : 16

Answer

B. 16 : 1

Explanation

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$$\theta $$ = $${{\pi \over 2}}$$ $$ \therefore $$ $${l_1}$$ = 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$$\theta $$ = $${{\pi \over 4}}$$ $$ \therefore $$ $${l_2}$$ = 16C8 $$ \times $$ $${{{2^8}} \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^8}}}$$ = 16C8 $$ \times $$ $${{2^8}{2^4}}$$ $$ \therefore $$ $${{{l_2}} \over {{l_1}}}$$ = $${{{2^4}} \over 1}$$ = $${{16} \over 1}$$

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