The range of values of the term independent of x in the
expansion of `(x sin^(-1) alpha + (cos^(-1)alpha)/(x))^(10), a in [-1,1]`is

Options

A. `[-(""^(10)C_(5)pi^(10))/(2^(5)),(""^(10)C_(5)pi^(10))/(2^(20))]`
B. `[-(""^(10)C_(5)pi^(2))/(2^(20)),(""^(10)C_(5)pi^(2))/(2^(5))]`
C. `[1, 2]`
D. `(1, 2)`

Answer

To find the range of values of the term independent of \( x \) in the expansion of

\[ \left( x \sin^{-1} \alpha + \frac{\cos^{-1} \alpha}{x} \right)^{10} \]

where \( \alpha \in [-1, 1] \), we follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_k \) in the binomial expansion of \( (a + b)^n \) is given by:

\[ T_k = \binom{n}{k} a^{n-k} b^k \]

In our case, let \( a = x \sin^{-1} \alpha \) and \( b = \frac{\cos^{-1} \alpha}{x} \). Thus, we have:

\[ T_k = \binom{10}{k} (x \sin^{-1} \alpha)^{10-k} \left(\frac{\cos^{-1} \alpha}{x}\right)^k \]

### Step 2: Simplify the general term
Simplifying \( T_k \):

\[ T_k = \binom{10}{k} (x^{10-k} (\sin^{-1} \alpha)^{10-k}) \left(\frac{(\cos^{-1} \alpha)^k}{x^k}\right) \]

This simplifies to:

\[ T_k = \binom{10}{k} (\sin^{-1} \alpha)^{10-k} (\cos^{-1} \alpha)^k x^{10-2k} \]

### Step 3: Find the term independent of \( x \)
For the term to be independent of \( x \), we need the exponent of \( x \) to be zero:

\[ 10 - 2k = 0 \implies k = 5 \]

### Step 4: Substitute \( k = 5 \) into the general term
Now we substitute \( k = 5 \) into the expression for \( T_k \):

\[ T_5 = \binom{10}{5} (\sin^{-1} \alpha)^{5} (\cos^{-1} \alpha)^{5} \]

### Step 5: Calculate the binomial coefficient
The binomial coefficient \( \binom{10}{5} \) is:

\[ \binom{10}{5} = 252 \]

Thus,

\[ T_5 = 252 (\sin^{-1} \alpha)^{5} (\cos^{-1} \alpha)^{5} \]

### Step 6: Determine the range of \( \sin^{-1} \alpha \) and \( \cos^{-1} \alpha \)
The values of \( \sin^{-1} \alpha \) and \( \cos^{-1} \alpha \) for \( \alpha \in [-1, 1] \) are:

- \( \sin^{-1} \alpha \) ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \)
- \( \cos^{-1} \alpha \) ranges from \( 0 \) to \( \pi \)

### Step 7: Find the maximum and minimum values of \( T_5 \)
To find the maximum and minimum values of \( T_5 \), we can evaluate it at the endpoints of the range of \( \alpha \):

- When \( \alpha = -1 \):
\[ \sin^{-1}(-1) = -\frac{\pi}{2}, \quad \cos^{-1}(-1) = \pi \]
\[ T_5 = 252 \left(-\frac{\pi}{2}\right)^{5} (\pi)^{5} = 252 \left(-\frac{\pi^5}{32}\right) = -\frac{252 \pi^{10}}{32} \]

- When \( \alpha = 1 \):
\[ \sin^{-1}(1) = \frac{\pi}{2}, \quad \cos^{-1}(1) = 0 \]
\[ T_5 = 252 \left(\frac{\pi}{2}\right)^{5} (0)^{5} = 0 \]

- When \( \alpha = 0 \):
\[ \sin^{-1}(0) = 0, \quad \cos^{-1}(0) = \frac{\pi}{2} \]
\[ T_5 = 252 (0)^{5} \left(\frac{\pi}{2}\right)^{5} = 0 \]

### Conclusion
The maximum value occurs at \( \alpha = -1 \) and the minimum value occurs at \( \alpha = 1 \) and \( \alpha = 0 \). Therefore, the range of the term independent of \( x \) is:

\[ \left[-\frac{252 \pi^{10}}{32}, 0\right] \]