(A) 16: 1 (B) 8: 1 (C) 1: 8 (D) 1: 16

Q: Why do we set $16-2r=0$ to find the term independent of $x$?

Think of the binomial expansion as a battle between the powers of $x$ in the numerator and the denominator. The general term is $T_{r+1} = ^{16}C_r (x/\cos \theta)^{16-r} (1/(x \sin \theta))^r$. When you group the $x$ terms, you get $x^{16-r} \cdot x^{-r} = x^{16-2r}$. For the term to be 'independent of $x$', the variable $x$ must effectively vanish, meaning its exponent must be zero. Setting $16-2r=0$ is the mathematical way of forcing that cancellation to happen.

Q: Why does maximizing the denominator minimize the whole term?

This is a fundamental principle of inequalities. Consider the expression $T_9 = \frac{K}{(\sin 2\theta)^8}$, where $K$ is a constant. Since the numerator is fixed, the value of the fraction is inversely proportional to the denominator. To make the *entire* fraction as small as possible, you must make the denominator as large as possible. It is a seesaw effect: when the denominator goes up, the fraction goes down.

Q: Is $r=8$ always the term independent of $x$?

Not at all! It is only $r=8$ here because the power of the binomial is $16$. If the power were $20$, the condition $20-2r=0$ would give $r=10$. Always derive the condition $n-2r=0$ based on the specific power given in the problem. Never memorize the 'middle term' rule blindly, as it only applies when the power is even and the structure of $x$ is symmetric.

Question

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

Options:

(A) $16: 1$

(B) $8: 1$

(C) $1: 8$

(D) $1: 16$

Sigma Percentile · Accepted Answer

### Analyzing the Binomial Anatomy

Imagine you are standing before the expression $(\frac{x}{\cos 	heta} + \frac{1}{x \sin 	heta})^{16}$. The Binomial Theorem is your most loyal ally in finding the term independent of $x$.

We invoke the general term formula:
$$T_{r+1} = ^{16}C_r \left( \frac{x}{\cos 	heta} ight)^{16-r} \left( \frac{1}{x \sin 	heta} ight)^r$$

When we combine the powers of $x$, we obtain $x^{16-2r}$. For the term to be independent of $x$, the exponent must be zero, which yields $16-2r=0$, or $r=8$.

We have found our target: the ninth term, $T_9$.

### The Trigonometric Bridge

With $r=8$, the $x$ terms vanish, leaving us with:
$$T_9 = ^{16}C_8 \frac{1}{(\cos 	heta)^8 (\sin 	heta)^8}$$

We utilize the double-angle identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$, which implies $\sin 	heta \cos 	heta = \frac{\sin 2	heta}{2}$. Substituting this into our expression, we get:
$$T_9 = ^{16}C_8 \frac{1}{(\frac{\sin 2	heta}{2})^8}$$

Simplifying this, the $2^8$ term moves to the numerator:
$$T_9 = ^{16}C_8 \frac{2^8}{(\sin 2	heta)^8}$$

### The Minimization Dance

We are given two intervals for $	heta$. Since $T_9$ is inversely proportional to $(\sin 2	heta)^8$, minimizing $T_9$ is equivalent to maximizing $\sin 2	heta$.

For the first interval, $\frac{\pi}{8} \leq 	heta \leq \frac{\pi}{4}$, we multiply by $2$ to get $\frac{\pi}{4} \leq 2	heta \leq \frac{\pi}{2}$. In this range, $\sin 2	heta$ reaches its maximum of $1$ at $2	heta = \frac{\pi}{2}$.
Thus, the minimum value is:
$$l_1 = ^{16}C_8 \frac{2^8}{(1)^8} = ^{16}C_8 \cdot 2^8$$

For the second interval, $\frac{\pi}{16} \leq 	heta \leq \frac{\pi}{8}$, we multiply by $2$ to get $\frac{\pi}{8} \leq 2	heta \leq \frac{\pi}{4}$. The maximum value of $\sin 2	heta$ occurs at the right endpoint, $2	heta = \frac{\pi}{4}$, where $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Thus, the minimum value is:
$$l_2 = ^{16}C_8 \frac{2^8}{(1/\sqrt{2})^8} = ^{16}C_8 \cdot 2^8 \cdot 2^4 = ^{16}C_8 \cdot 2^{12}$$

### Final Calculation

We now compute the ratio $l_2 : l_1$:
$$\frac{l_2}{l_1} = \frac{^{16}C_8 \cdot 2^{12}}{^{16}C_8 \cdot 2^8}$$

The binomial coefficient $^{16}C_8$ cancels out completely. We are left with:
$$\frac{2^{12}}{2^8} = 2^4 = 16$$

The final ratio is **16:1**.

Select Answer:

Visualized Solution

General Term $T_{r + 1}$

Substituting $A$ and $B$

Separating the Powers of $x$

Condition for Term Independent of $x$

Simplifying $T_{9}$

Final Expression for $T_{9}$

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 Final Answer

The Sigma Insight: General Term and Middle Term

Analyzing the Binomial Anatomy

The Trigonometric Bridge

The Minimization Dance

Final Calculation

Similar Questions

If the greatest value of the term independent of $x$ in the expansion of $(x sin α + a \frac{c o s α}{x})^{10}$ is $\frac{10 !}{( 5 ! ) ^{2}}$ , then the value of $a$ is equal to:

Let the coefficients of third, fourth and fifth terms in the expansion of $(x + \frac{a}{x ^{2}})^{n}, x \neq = 0$ , be in the ratio $12 : 8 : 3$ . Then the term independent of $x$ in the expansion, is equal to ____.

The term independent of x in the expansion of $(\frac{1}{60} - \frac{x ^{8}}{81}) (2 x^{2} - \frac{3}{x ^{2}})^{6}$ is equal to :

If the coefficients of $x^{3}$ and $x^{4}$ in the expansion of $(1 + a x + b x^{2}) (1 - 2 x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to

Let $α > 0$ , be the smallest number such that the expansion of $(x^{2/3} + \frac{2}{x ^{3}})^{30}$ has a term $β x^{- α}, β \in N$ . Then $α$ is equal to ______.

The term independent of $x$ in the expansion of $(\frac{( x + 1 )}{( x ^{2/3} + 1 - x ^{1/3} )} - \frac{( x + 1 )}{( x - x ^{1/2} )})^{10}, x > 1$ is:

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $(42 + \frac{1}{4 3})^{n}$ , in the increasing powers of $\frac{1}{4 3}$ be $46 : 1$ . If the sixth term from the beginning is $\frac{α}{4 3}$ , then $α$ is equal to ______.

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(42 + \frac{1}{4 3})^{n}$ is $6 : 1$ , then the third term from the beginning is:

Let $α > 0, β > 0$ be such that $α^{3} + β^{2} = 4$ . If the maximum value of the term independent of $x$ in the binomial expansion of $(α x^{1/9} + β x^{- 1/6})^{10}$ is $10 k$ , then $k$ is equal to :

Select Answer:

Visualized Solution

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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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

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

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T9​

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T9​

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​