If the greatest value of the term independent of x in the expansion of (x sin α + a cos α / x)¹⁰ is

Q: Why do we set the exponent of $x$ to zero? Isn't there a way to just guess the term?

Great question! 'Independent of $x$' literally means $x^0$. Since $x^0 = 1$, the variable $x$ vanishes. Guessing is risky; setting the exponent to zero is the rigorous, foolproof way to ensure you've found the correct term. It is the mathematical equivalent of finding the 'sweet spot' where the variable influence cancels out completely.

Q: Why did we use the double angle identity $\sin 2\alpha = 2 \sin \alpha \cos \alpha$?

Because it collapses two separate trigonometric functions into one. It transforms the expression into a single function of $\alpha$, making it trivial to find the maximum value. Always look for ways to consolidate variables! In JEE problems, if you see $\sin \alpha \cos \alpha$, your brain should immediately scream 'double angle identity'.

Q: Is it always guaranteed that the factorials will cancel out like that?

In well-constructed JEE problems, yes! The problem setter often designs the constants to simplify beautifully. If you find yourself stuck with a massive, unmanageable factorial expression, pause and check your expansion—you might have missed a simplification step. The beauty of these problems is that they are designed to resolve elegantly.

Question

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

Options:

(A) $-1$

(B) 1

(C) $-2$

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are examining the expression $(x \sin \alpha + \frac{a \cos \alpha}{x})^{10}$. Our objective is to identify the term independent of $x$ and determine the value of $a$ based on the provided constraints.

In the world of JEE Advanced, complexity is often a mask for elegance. We are looking for the specific part of this expansion where the variable $x$ effectively vanishes, leaving behind only constants and trigonometric values.

### The DNA of the Expansion

Every binomial expansion has a heartbeat, and that heartbeat is the General Term formula:
$$T_{r+1} = \binom{n}{r} A^{n-r} B^r$$

Here, our $n$ is $10$, our $A$ is $x \sin \alpha$, and our $B$ is $\frac{a \cos \alpha}{x}$. Substituting these into the formula, we obtain:
$$T_{r+1} = \binom{10}{r} (x \sin \alpha)^{10-r} \left( \frac{a \cos \alpha}{x} \right)^r$$

Do not rush this step. Keep your $x$ terms separate from your constants to avoid algebraic errors.

### The Hunt for Independence

Now, let us isolate the $x$ variables. From the first part, we have $x^{10-r}$, and from the second part, we have $x^{-r}$.

Using the laws of exponents, we combine these:
$$x^{10-r} \cdot x^{-r} = x^{10-2r}$$

For a term to be independent of $x$, the power of $x$ must be zero, as $x^0 = 1$. Setting the exponent to zero, we get $10 - 2r = 0$, which yields $r = 5$. We are looking for the 6th term, $T_6$.

### The Trigonometric Bridge

With $r=5$, our term becomes:
$$T_6 = \binom{10}{5} (\sin \alpha)^5 (a \cos \alpha)^5$$

We can simplify this by pulling out the constant $a$:
$$T_6 = \binom{10}{5} a^5 (\sin \alpha \cos \alpha)^5$$

Recall the double-angle identity $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, which implies $\sin \alpha \cos \alpha = \frac{\sin 2\alpha}{2}$. Substituting this into our expression, we get:
$$T_6 = \binom{10}{5} \frac{a^5}{32} \sin^5 2\alpha$$

### The Grand Finale

The problem asks for the greatest value. Since $\sin 2\alpha$ oscillates between $-1$ and $1$, the maximum value of $\sin^5 2\alpha$ is $1^5 = 1$.

The maximum value of the term is therefore $\binom{10}{5} \frac{a^5}{32}$. We are given that this value equals $\frac{10!}{(5!)^2}$.

Note that $\binom{10}{5} = \frac{10!}{5!5!} = \frac{10!}{(5!)^2}$. Equating the two expressions:
$$\binom{10}{5} \frac{a^5}{32} = \binom{10}{5}$$

The binomial coefficients cancel out, leaving us with $\frac{a^5}{32} = 1$. This simplifies to $a^5 = 32$, which means **$a = 2$**.

Animated Solution for Mathematics - Binomial Theorem: 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:

Select Answer:

Visualized Solution

Identify the Binomial Expression

The General Term Formula

Substitute the Values

Separate the Powers of $x$

Condition for Independence

Substitute $r = 5$ back

Apply Trigonometric Identity

The Greatest Value Condition

Equate to Given Value

Simplify the Equation

Final Calculation for $a$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The DNA of the Expansion

The Hunt for Independence

The Trigonometric Bridge

The Grand Finale

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