In the expansion of (x/(cos θ) + 1/(x sin θ))¹⁶, if ₁ is the least value of the term independent of x when (π)/8 ≤ θ ≤ (π)/4 and ₂ is the least value of the term independent of x when (π)/16 ≤ θ ≤ (π)/8, then the ratio ₂: ₁ is equal to:

Binomial Theorem and Its Simple Applications: JEE Main 2020 Mathematics question

JEE Main 2020 · 9 Jan, Evening shift · Q2926

JEE Main 20209 Jan, Evening shiftMathematicsHardBinomial Theorem and Its Simple Applications131× since 2002

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 Main 2020 · 9 Jan, Evening shift · Q2926

JEE Main 20209 Jan, Evening shiftMathematicsHardBinomial Theorem and Its Simple Applications131× since 2002

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 :