If the middle term in the binomial expansion of `(1/x+xsinx)^(10)` is equal to `(63)/8,` find the value of `xdot`

Options

A. `2n pi + (pi)/(6)`
B. `n pi + (pi)/(6)`
C. `n pi + (-1)^(n) (pi)/(6)`
D. `n pi + (-1)^(n) (pi)/(3)`

Explanation

To solve the problem step by step, we need to find the middle term in the binomial expansion of \((\frac{1}{x} + x \sin x)^{10}\) and set it equal to \(\frac{63}{8}\).

### Step 1: Identify the middle term
In a binomial expansion of the form \((a + b)^n\), the middle term is given by the formula:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For \(n = 10\), the middle term will be the \(T_{6}\) term (since \(n/2 + 1 = 5 + 1 = 6\)).

### Step 2: Set up the expression for the middle term
Here, \(a = \frac{1}{x}\) and \(b = x \sin x\). Thus, we can express the middle term as:
\[ T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^{10-5} (x \sin x)^5 \]

### Step 3: Simplify the middle term
Substituting the values we have:
\[ T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^{5} (x \sin x)^5 \]
This simplifies to:
\[ T_6 = \binom{10}{5} \frac{(x \sin x)^5}{x^5} = \binom{10}{5} (\sin x)^5 \]

### Step 4: Calculate \(\binom{10}{5}\)
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Thus, we have:
\[ T_6 = 252 (\sin x)^5 \]

### Step 5: Set the middle term equal to \(\frac{63}{8}\)
We set the equation:
\[ 252 (\sin x)^5 = \frac{63}{8} \]

### Step 6: Solve for \((\sin x)^5\)
Dividing both sides by 252:
\[ (\sin x)^5 = \frac{63}{8 \times 252} \]
Calculating \(8 \times 252 = 2016\):
\[ (\sin x)^5 = \frac{63}{2016} \]
Simplifying \(\frac{63}{2016}\):
\[ (\sin x)^5 = \frac{1}{32} \]

### Step 7: Solve for \(\sin x\)
Taking the fifth root:
\[ \sin x = \left(\frac{1}{32}\right)^{\frac{1}{5}} = \frac{1}{2} \]

### Step 8: Find the value of \(x\)
The general solution for \(\sin x = \frac{1}{2}\) is:
\[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]

### Conclusion
Thus, the values of \(x\) can be expressed as:
\[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]