QJEEM20Ma715b: JEE Main 2020 Mathematics - Class 11 JEEM20Ma715b Solution: In the expansion of , if is the least value of the term…

T_{r + 1} = ¹⁶C_r ${\left( {{x \over {\cos \theta }}} \right)^{16 - r}}{\left( {{1 \over {x\sin \theta }}} \right)^r}$

= ¹⁶C_r ${\left( x \right)^{16 - 2r}} \times {1 \over {{{\left( {\cos \theta } \right)}^{16 - r}}{{\left( {\sin \theta } \right)}^r}}}$

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 $\times$ ${{2^8}{2^4}}$

$\therefore$ ${{{l_2}} \over {{l_1}}}$ = ${{{2^4}} \over 1}$ = ${{16} \over 1}$