In the expansion of (x cos θ + 1/(x sin θ))¹⁶, if l₁ is the least value of the term independent of x

Q: Wait, if it is 'independent of $x$', why does it still have $\theta$ in it?

That is a fantastic question! In the context of binomial expansions, 'independent of $x$' simply means the variable $x$ has an exponent of zero, so $x^0 = 1$. It does not mean the entire term is a constant number. It is a constant *with respect to $x$*, but it remains a function of $\theta$. You are finding the value of the term for a given $\theta$.

Q: Why do we maximize the denominator to minimize the function?

Think of the fraction $\frac{1}{y}$. As $y$ gets bigger, the fraction gets smaller. Since our function $f(\theta)$ is proportional to $\frac{1}{(\sin 2\theta)^8}$, maximizing the denominator $\sin 2\theta$ is the only way to make the whole expression reach its minimum value.

Q: Do I need to check the endpoints for the interval $(\frac{\pi}{16}, \frac{\pi}{8})$?

Yes! Even though the interval is open, we look at the supremum (the limit as it approaches the boundary). Since the function $\sin 2\theta$ is monotonic in that range, the minimum value of our function $f(\theta)$ will be approached as $\theta$ approaches the upper bound of the interval.

Question

In the expansion of $\left(x \cos 	heta + \frac{1}{x \sin 	heta}ight)^{16}$, if $l_1$ is the least value of the term independent of $x$ when $\frac{\pi}{8} \le 	heta \le \frac{\pi}{4}$ and $l_2$ is the least value of the term independent of $x$ when $\frac{\pi}{16} < 	heta < \frac{\pi}{8}$, then the ratio $l_2 : l_1$ is equal to:

Options:

(A) 1 : 8

(B) 1 : 16

(C) 8 : 1

(D) 16 : 1

Sigma Percentile · Accepted Answer

### The Dance of Variables: Unlocking the Binomial Mystery

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are peeling back the layers of a beautiful mathematical structure.

We are looking at the expansion of $\left(x \cos 	heta + \frac{1}{x \sin 	heta}ight)^{16}$. At first glance, it looks like a chaotic mix of algebra and trigonometry. But I promise you, there is a hidden order here, and finding it is the key to unlocking the solution.

### Phase 1: The Binomial Hunt

Every binomial expansion is a story told in terms. We start with the general term formula, $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

Here, our $n$ is $16$, our first term $a$ is $x \cos 	heta$, and our second term $b$ is $\frac{1}{x \sin 	heta}$. Let us substitute these into our formula:

$$T_{r+1} = \binom{16}{r} (x \cos 	heta)^{16-r} \left(\frac{1}{x \sin 	heta}ight)^r$$

Now, we need to group the $x$ terms. We have $x^{16-r}$ from the first part and $x^{-r}$ from the second.

Using the laws of exponents, we combine them: $x^{16-r} \cdot x^{-r} = x^{16-2r}$. For the term to be independent of $x$, the exponent must be zero.

Thus, $16-2r = 0$, which gives us $r=8$. We have found our target! The term independent of $x$ is the ninth term, $T_9$.

### Phase 2: The Trigonometric Bridge

Now that we have $r=8$, let us look at the constant part of our term:

$$T_9 = \binom{16}{8} \frac{(\cos 	heta)^8}{(\sin 	heta)^8}$$

This is where the magic happens. We want to simplify this into something we can easily analyze. We know that $\sin 2	heta = 2 \sin 	heta \cos 	heta$.

If we multiply and divide by $2^8$, we can force this identity into our expression:

$$T_9 = \binom{16}{8} \frac{2^8}{(2 \sin 	heta \cos 	heta)^8} = \binom{16}{8} \frac{2^8}{(\sin 2	heta)^8}$$

Let us call this function $f(	heta) = \binom{16}{8} \frac{2^8}{(\sin 2	heta)^8}$. This is the function we need to minimize.

### Phase 3: The Interval Analysis

This is the heart of the problem. We have two intervals for $	heta$. To minimize $f(	heta)$, we must maximize the denominator, $(\sin 2	heta)^8$.

Since the power is positive, this is equivalent to maximizing $\sin 2	heta$.

In the first interval, $	heta \in [\frac{\pi}{8}, \frac{\pi}{4}]$, the angle $2	heta$ ranges from $\frac{\pi}{4}$ to $\frac{\pi}{2}$. In this range, $\sin 2	heta$ is increasing, reaching its maximum of $1$ at $2	heta = \frac{\pi}{2}$.

Thus, $l_1 = \binom{16}{8} \frac{2^8}{1^8} = \binom{16}{8} 2^8$.

In the second interval, $	heta \in (\frac{\pi}{16}, \frac{\pi}{8})$, the angle $2	heta$ ranges from $(\frac{\pi}{8}, \frac{\pi}{4})$. Here, $\sin 2	heta$ is also increasing, approaching its supremum of $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

Thus, $l_2 = \binom{16}{8} \frac{2^8}{(\frac{1}{\sqrt{2}})^8} = \binom{16}{8} \frac{2^8}{2^{-4}} = \binom{16}{8} 2^8 \cdot 2^4$.

### Phase 4: The Final Ratio

Finally, we calculate the ratio $l_2 : l_1$. Look at how the complexity melts away:

$$\frac{l_2}{l_1} = \frac{\binom{16}{8} 2^8 \cdot 2^4}{\binom{16}{8} 2^8} = 2^4 = 16$$

The ratio is **$16:1$**. We have navigated the binomial expansion, bridged it with trigonometry, and analyzed the behavior of the sine function across intervals.

You have done it! Keep this confidence, and remember: every complex problem is just a series of simple steps waiting to be connected.

Select Answer:

Visualized Solution

The Binomial Expression

The General Term $T_{r + 1}$

Isolating the Power of $x$

Term Independent of $x$

Simplifying the Constant Term

Defining the Function $f (θ)$

Analyzing Interval 1 for $l_{1}$

Calculating $l_{1}$

Analyzing Interval 2 for $l_{2}$

Calculating $l_{2}$

Finding the Ratio $l_{2} : l_{1}$

The Sigma Insight: General Term and Middle Term

The Dance of Variables

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

The Binomial Expression

The General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​

Isolating the Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Term Independent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Simplifying the Constant Term

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(θ)

Analyzing Interval 1 for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​

Analyzing Interval 2 for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​

Finding the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​:l1​