JEE Main 2020 — Binomial Theorem Question with Solution

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

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

term is independent of x when

$$\therefore$$ 16 – 2r = 0

$$\Rightarrow$$ r = 8

T₉ = ¹⁶C₈ $$\times$$ $$\frac{1}{{\cos }^8\theta {\sin }^8\theta }$$

= ¹⁶C₈ $$\times$$ $$\frac{2^8}{{\left(\sin 2\theta \right)}^8}$$

If $$\theta \in \left[\frac{\pi }{8},\frac{\pi }{4}\right]$$ then $$2\theta \in \left[\frac{\pi }{4},\frac{\pi }{2}\right]$$

In the range $$\left[\frac{\pi }{4},\frac{\pi }{2}\right]$$, 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 $$\left[\frac{\pi }{4},\frac{\pi }{2}\right]$$
when 2$$\theta$$ = $$\frac{\pi }{2}$$

$$\therefore$$ $$l_1$$ = ¹⁶C₈ $$\times$$ 2⁸

AgainIf $$\theta \in \left[\frac{\pi }{16},\frac{\pi }{8}\right]$$ then $$2\theta \in \left[\frac{\pi }{8},\frac{\pi }{4}\right]$$

In the range $$\left[\frac{\pi }{8},\frac{\pi }{4}\right]$$, 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 $$\left[\frac{\pi }{8},\frac{\pi }{4}\right]$$
when 2$$\theta$$ = $$\frac{\pi }{4}$$

$$\therefore$$ $$l_2$$ = ¹⁶C₈ $$\times$$ $$\frac{2^8}{{\left(\frac{1}{\sqrt{2}}\right)}^8}$$

= ¹⁶C₈ $$\times$$ $$2^8{2}^4$$

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