Understanding the Binomial Expansion The problem asks for the greatest value of the term independent of 'x' in the expansion of ${\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}}$.

The general term, $T_{r+1}$, in the binomial expansion of $(A+B)^n$ is given by the formula: $T_{r+1} = \binom{n}{r} A^{n-r} B^r$ where $n$ is the exponent of the binomial, and $r$ is the index of the term (starting from $r=0$).

In this specific problem, we identify the components as: $A = x\sin \alpha$ $B = \frac{a\cos \alpha}{x}$ $n = 10$

Substituting these into the general term formula, we get: $T_{r+1} = \binom{10}{r} (x\sin \alpha)^{10-r} \left(\frac{a\cos \alpha}{x}\right)^r$ To simplify, we distribute the exponents and separate the terms involving $x$: $T_{r+1} = \binom{10}{r} x^{10-r} (\sin \alpha)^{10-r} a^r (\cos \alpha)^r x^{-r}$ Now, we combine the terms involving 'x' by adding their exponents: $T_{r+1} = \binom{10}{r} a^r (\sin \alpha)^{10-r} (\cos \alpha)^r x^{10-r-r}$ $T_{r+1} = \binom{10}{r} a^r (\sin \alpha)^{10-r} (\cos \alpha)^r x^{10-2r}$ Here, $r$ can be any integer from $0$ to $10$.

Finding the Term Independent of 'x' A term is considered "independent of x" if the variable 'x' does not appear in it. Mathematically, this means the exponent of 'x' in that term must be zero.

From our general term $T_{r+1}$, the exponent of $x$ is $10-2r$. We set this exponent to $0$: $10 - 2r = 0$ Solving this linear equation for $r$: $2r = 10$ $r = 5$ This tells us that the term independent of 'x' is the $(5+1)^{th}$ term, which is the $6^{th}$ term ($T_6$) in the expansion.

Calculating and Simplifying the Term Now, we substitute $r=5$ back into the expression for $T_{r+1}$ (specifically, the part without 'x'): $T_6 = \binom{10}{5} a^5 (\sin \alpha)^{10-5} (\cos \alpha)^5$ $T_6 = \binom{10}{5} a^5 \sin^5 \alpha \cos^5 \alpha$ To further simplify this expression, we utilize the trigonometric identity for the sine of a double angle: $\sin 2\alpha = 2\sin\alpha \cos\alpha$. From this, we can express $\sin\alpha \cos\alpha$ as $\frac{\sin 2\alpha}{2}$. Therefore, $(\sin\alpha \cos\alpha)^5 = \left(\frac{\sin 2\alpha}{2}\right)^5 = \frac{(\sin 2\alpha)^5}{2^5} = \frac{(\sin 2\alpha)^5}{32}$. Substituting this back into the expression for $T_6$: $T_6 = \binom{10}{5} a^5 \frac{(\sin 2\alpha)^5}{32}$

Determining the Greatest Value and Solving for 'a' The problem states that the greatest value of this term independent of 'x' is $\frac{10!}{(5!)^2}$. We recognize that $\frac{10!}{(5!)^2}$ is the definition of the binomial coefficient $\binom{10}{5}$. So, the given greatest value of $T_6$ is $\binom{10}{5}$.

The expression for $T_6$ is $\binom{10}{5} \frac{a^5}{32} (\sin 2\alpha)^5$. We know that the range of $\sin 2\alpha$ is $[-1, 1]$. Consequently, the range of $(\sin 2\alpha)^5$ is also $[-1, 1]$.

To find the greatest value of $T_6$, we need to ensure the entire term is maximized. This depends on the sign of $a^5$:

• If $a > 0$, then $a^5 > 0$. For $T_6$ to be its greatest (most positive) value, $(\sin 2\alpha)^5$ must be as large as possible and positive, which means $(\sin 2\alpha)^5 = 1$. This occurs when $\sin 2\alpha = 1$. In this scenario, the greatest value of $T_6$ is: $T_6 = \binom{10}{5} \frac{a^5}{32} (1) = \binom{10}{5} \frac{a^5}{32}$ Equating this to the given greatest value: $\binom{10}{5} \frac{a^5}{32} = \binom{10}{5}$ Dividing both sides by $\binom{10}{5}$ (which is non-zero): $\frac{a^5}{32} = 1$ $a^5 = 32$ Since $32 = 2^5$, we have: $a^5 = 2^5$ $a = 2$

• If $a < 0$, then $a^5 < 0$. Let $a = -k$ where $k$ is a positive real number. Then $a^5 = (-k)^5 = -k^5$. $T_6 = \binom{10}{5} \frac{-k^5}{32} (\sin 2\alpha)^5$ For $T_6$ to be its greatest (most positive) value, we need the product $\frac{-k^5}{32} (\sin 2\alpha)^5$ to be positive and maximal. Since $-k^5$ is negative, $(\sin 2\alpha)^5$ must be negative and maximal in magnitude, which means $(\sin 2\alpha)^5 = -1$. This occurs when $\sin 2\alpha = -1$. In this scenario, the greatest value of $T_6$ is: $T_6 = \binom{10}{5} \frac{-k^5}{32} (-1) = \binom{10}{5} \frac{k^5}{32}$ Equating this to the given greatest value: $\binom{10}{5} \frac{k^5}{32} = \binom{10}{5}$ $\frac{k^5}{32} = 1$ $k^5 = 32$ $k^5 = 2^5$ $k = 2$ Since $a = -k$, we get $a = -2$.

Both $a=2$ and $a=-2$ lead to the same greatest value for the term independent of 'x', which is $\binom{10}{5}$. Given the options, $a=2$ and $a=-2$ are both plausible solutions. The provided correct answer (A) is $a=-1$, which is inconsistent with the derived result. The original solution followed the path that leads to $a=2$.

Tips and Common Mistakes

• General Term Formula: Always start by correctly identifying $A$, $B$, and $n$ for the general term formula $T_{r+1} = \binom{n}{r} A^{n-r} B^r$. A common mistake is misidentifying $A$ or $B$, especially their signs or powers of $x$.
• Term Independent of 'x': To find the term independent of 'x', ensure the exponent of 'x' in the general term simplifies to zero. Be careful with terms like $\frac{1}{x^k}$ which means $x^{-k}$.
• Trigonometric Identities: The identity $\sin 2\alpha = 2\sin\alpha \cos\alpha$ (or its rearrangements) is frequently useful in such problems to simplify terms involving products of sine and cosine.
• Greatest Value: When asked for the "greatest value", you must consider the range of trigonometric functions (like $\sin \theta$ or $\cos \theta$) and how they interact with coefficients. If a part of the expression (like $a^5$) can be positive or negative, choose the value of the trigonometric function (e.g., $\sin 2\alpha$) that makes the entire term maximum positive. This often leads to taking the absolute value of variables, e.g., $|a|^5 = 32 \Rightarrow |a|=2$.
• Combinatorial Notation: Remember that $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. It's important to recognize common forms like $\frac{10!}{(5!)^2} = \frac{10!}{5!5!} = \binom{10}{5}$.

Summary To solve this problem, we first found the general term of the binomial expansion. Then, we identified the value of $r$ that makes the term independent of 'x' by setting the exponent of 'x' to zero. After substituting $r$ back into the general term and simplifying it using the double angle identity for sine, we arrived at an expression for the term independent of 'x' that depends on $a$ and $\sin 2\alpha$. By equating the maximum possible value of this term (considering the range of $\sin 2\alpha$) to the given greatest value, we solved for $a$. Our analysis indicates that $a=2$ or $a=-2$ satisfies the condition that the greatest value of the term independent of 'x' is $\binom{10}{5}$.