If in the expansion of `(x^(3)-(2)/(sqrt(x)))^(n)` a term like `x^(2)` exists and `'n'` is a double digit number, then least value of `'n'` is (a) 10 (b) 11 (c) 12 (d) 13

Options

A. `10`
B. `11`
C. `12`
D. `13`

Explanation

To solve the problem, we need to find the least value of \( n \) such that a term like \( x^2 \) exists in the expansion of \( (x^3 - \frac{2}{\sqrt{x}})^n \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (x^3 - \frac{2}{\sqrt{x}})^n \) can be expressed as:
\[ T_{r+1} = \binom{n}{r} (x^3)^{n-r} \left(-\frac{2}{\sqrt{x}}\right)^r \]

2. **Simplify the General Term**:
Simplifying the expression:
\[ T_{r+1} = \binom{n}{r} (x^{3(n-r)}) \left(-2\right)^r (x^{-1/2})^r \]
\[ = \binom{n}{r} (-2)^r x^{3(n-r) - \frac{r}{2}} \]

3. **Combine the Exponents**:
The exponent of \( x \) in the term is:
\[ 3(n - r) - \frac{r}{2} = 3n - 3r - \frac{r}{2} = 3n - \frac{7r}{2} \]

4. **Set the Exponent Equal to 2**:
We want this exponent to equal 2:
\[ 3n - \frac{7r}{2} = 2 \]

5. **Clear the Fraction**:
Multiply through by 2 to eliminate the fraction:
\[ 6n - 7r = 4 \]

6. **Rearrange for \( r \)**:
Rearranging gives:
\[ 7r = 6n - 4 \quad \Rightarrow \quad r = \frac{6n - 4}{7} \]

7. **Check for Integer Values of \( r \)**:
For \( r \) to be an integer, \( 6n - 4 \) must be divisible by 7.

8. **Find the Least Double Digit \( n \)**:
We will check the smallest double-digit values of \( n \) starting from 10:
- For \( n = 10 \):
\[ r = \frac{6(10) - 4}{7} = \frac{60 - 4}{7} = \frac{56}{7} = 8 \quad (\text{integer}) \]
- For \( n = 11 \):
\[ r = \frac{6(11) - 4}{7} = \frac{66 - 4}{7} = \frac{62}{7} \quad (\text{not an integer}) \]
- For \( n = 12 \):
\[ r = \frac{6(12) - 4}{7} = \frac{72 - 4}{7} = \frac{68}{7} \quad (\text{not an integer}) \]
- For \( n = 13 \):
\[ r = \frac{6(13) - 4}{7} = \frac{78 - 4}{7} = \frac{74}{7} \quad (\text{not an integer}) \]

9. **Conclusion**:
The least value of \( n \) for which \( r \) is an integer is \( n = 10 \).

### Final Answer:
The least value of \( n \) is \( \boxed{10} \).