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Step-by-Step Solution
Step 1: Write the General Term of the Binomial Expansion
Consider the binomial expansion of
$ \left(\frac{x}{\cos \theta} + \frac{1}{x \sin \theta}\right)^{16} $.
The general term (the $(r+1)$-th term) in the expansion can be written as:
$ T_{r+1} = \binom{16}{r} \left(\frac{x}{\cos \theta}\right)^{16 - r} \left(\frac{1}{x \sin \theta}\right)^{r}.$
Simplifying the expression in terms of powers of $x$ and trigonometric functions gives:
$ T_{r+1} = \binom{16}{r} \, x^{16-r} \, \frac{1}{(\cos \theta)^{16-r}} \, x^{-r} \, \frac{1}{(\sin \theta)^{r}}
= \binom{16}{r} \, x^{\,16 - 2r} \, \frac{1}{(\cos \theta)^{16-r} (\sin \theta)^{r}}.
$
Step 2: Find the Condition for the Term Independent of x
A term is independent of $x$ if the power of $x$ in that term is zero, i.e.
$16 - 2r = 0.$
Hence,
$ r = 8.$
Therefore, the term independent of $x$ occurs when $r=8$, which is the 9th term
$T_9$ (since $r+1 = 9$).
Step 3: Write the Independent Term Explicitly
Substituting $r=8$ into the general term, we get:
$ T_9 = \binom{16}{8} \frac{1}{(\cos \theta)^{16-8} \, (\sin \theta)^8}
= \binom{16}{8} \frac{1}{(\cos \theta)^8 (\sin \theta)^8}.
$
Notice that
$ \frac{1}{(\cos \theta)^8 (\sin \theta)^8}
= \frac{1}{(\cos^8 \theta)\,(\sin^8 \theta)}. $
We can rewrite $\cos^8\theta\,\sin^8\theta$ using $\sin(2\theta) = 2\sin\theta \cos\theta$:
$ \sin(2\theta)^8 = (2\sin\theta \cos\theta)^8 = 2^8 (\sin\theta \cos\theta)^8. $
Thus,
$ T_9 = \binom{16}{8} \cdot \frac{2^8}{(\sin(2\theta))^8}.
$
Step 4: Determine the Least Value of the Term for Given Ranges of θ
We want the minimum possible value of
$ T_9 = \binom{16}{8} \cdot \frac{2^8}{(\sin(2\theta))^8}, $
which depends on maximizing $\sin(2\theta)$ in the relevant interval of $\theta$.
The smaller $(\sin(2\theta))^8$ is, the larger $T_9$ becomes, so for the least value of $T_9$, we need the largest possible $\sin(2\theta)$.
Step 4A: Compute ℓ₁ for θ in [π/8, π/4]
When
$ \theta \in \left[\frac{\pi}{8},\, \frac{\pi}{4}\right], $
then
$ 2\theta \in \left[\frac{\pi}{4},\, \frac{\pi}{2}\right]. $
In the interval
$ \left[\frac{\pi}{4}, \frac{\pi}{2}\right], $
$\sin(2\theta)$ is increasing, so its maximum value occurs at
$ 2\theta = \frac{\pi}{2}, $
or
$ \theta = \frac{\pi}{4}. $
Hence, the least value of $T_9$ (independent term) occurs when
$ \theta = \frac{\pi}{4}, $
giving
$ \sin(2 \cdot \frac{\pi}{4}) = \sin\left(\frac{\pi}{2}\right) = 1.
$
Therefore,
$
\ell_1 = \binom{16}{8} \cdot \frac{2^8}{(1)^8} = \binom{16}{8} \cdot 2^8.
$
Step 4B: Compute ℓ₂ for θ in [π/16, π/8]
When
$ \theta \in \left[\frac{\pi}{16},\, \frac{\pi}{8}\right], $
then
$ 2\theta \in \left[\frac{\pi}{8},\, \frac{\pi}{4}\right]. $
In the interval
$ \left[\frac{\pi}{8},\, \frac{\pi}{4}\right], $
$\sin(2\theta)$ is also increasing, so its maximum value occurs at
$ 2\theta = \frac{\pi}{4}, $
or
$ \theta = \frac{\pi}{8}. $
Hence, the least value of $T_9$ (independent term) in this interval occurs when
$ \theta = \frac{\pi}{8}, $
giving
$ \sin\left(2 \cdot \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}.
$
Therefore,
$
\ell_2 = \binom{16}{8} \cdot \frac{2^8}{\left(\frac{1}{\sqrt{2}}\right)^8}
= \binom{16}{8} \cdot 2^8 \cdot 2^4
= \binom{16}{8} \cdot 2^{12}.
$
(Since $\left(\frac{1}{\sqrt{2}}\right)^8 = \frac{1}{2^4}$, multiplying by $2^8$ yields $2^{8+4}=2^{12}$.)
Step 5: Determine the Ratio ℓ₂ : ℓ₁
We have
$
\ell_1 = \binom{16}{8} \cdot 2^8
$
and
$
\ell_2 = \binom{16}{8} \cdot 2^{12}.
$
Therefore,
$
\frac{\ell_2}{\ell_1}
= \frac{\binom{16}{8} \cdot 2^{12}}{\binom{16}{8} \cdot 2^8}
= 2^{12-8}
= 2^4
= 16.
$
Thus, the ratio
$ \ell_2 : \ell_1 $
is
$ 16 : 1, $
which matches the given correct answer.
