For the expansion `(x sin p + x^(-1)cos p)^(10), (p in R)`,

Options

A. the greatest value of the term independent of x is `10!//2^(5)(5!)^(2)`
B. the least value of sum of coefficient is zero
C. the greatest value of sum of coefficient is 12
D. the last value of the term independent of x occurs when `p = (2n+1)'(pi)/(4), n in Z`

Explanation

To solve the problem of finding the independent term in the expansion of \((x \sin p + x^{-1} \cos p)^{10}\), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ (x \sin p + x^{-1} \cos p)^{10} \]
This can be rewritten as:
\[ \left(x \sin p + \frac{\cos p}{x}\right)^{10} \]

### Step 2: Identify \(a\), \(b\), and \(n\)
In the expression \(a + b\) where \(a = x \sin p\) and \(b = \frac{\cos p}{x}\), we identify:
- \(a = x \sin p\)
- \(b = \frac{\cos p}{x}\)
- \(n = 10\)

### Step 3: Use the Binomial Theorem
According to the Binomial Theorem, the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Substituting our values, we have:
\[ T_{r+1} = \binom{10}{r} (x \sin p)^{10-r} \left(\frac{\cos p}{x}\right)^r \]

### Step 4: Simplify the Term
Now, simplifying the term:
\[ T_{r+1} = \binom{10}{r} (x^{10-r} \sin^{10-r} p) \left(\frac{\cos^r p}{x^r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{10}{r} \sin^{10-r} p \cos^r p \cdot x^{10-2r} \]

### Step 5: Find the Independent Term
To find the independent term, we need the power of \(x\) to be zero:
\[ 10 - 2r = 0 \]
Solving for \(r\):
\[ 2r = 10 \implies r = 5 \]

### Step 6: Substitute \(r\) Back into the Term
Now, substituting \(r = 5\) back into our term:
\[ T_{6} = \binom{10}{5} \sin^{5} p \cos^{5} p \cdot x^{0} \]
Thus, the independent term is:
\[ \binom{10}{5} \sin^{5} p \cos^{5} p \]

### Step 7: Simplify the Result
Using the identity \(2 \sin p \cos p = \sin 2p\):
\[ \sin^5 p \cos^5 p = \left(\frac{1}{2} \sin 2p\right)^5 = \frac{1}{32} \sin^5 2p \]
Hence, the independent term becomes:
\[ \binom{10}{5} \cdot \frac{1}{32} \sin^5 2p \]

### Step 8: Calculate \(\binom{10}{5}\)
Calculating \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = 252 \]
Thus, the independent term is:
\[ \frac{252}{32} \sin^5 2p = \frac{63}{8} \sin^5 2p \]

### Final Result
The independent term in the expansion is:
\[ \frac{63}{8} \sin^5 2p \]

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