If rth term in the expansion of `(2x^(2)-1/x)^(12)` is independent of x, then the value of r is
(i) 7
(ii) 8
(iii) 9
(iv) 10

Options

A. 7
B. 8
C. 9
D. 10

Answer

To find the value of \( r \) such that the \( r \)th term in the expansion of \( (2x^2 - \frac{1}{x})^{12} \) is independent of \( x \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The \( r \)th term (denoted as \( T_{r+1} \)) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 2x^2 \), \( b = -\frac{1}{x} \), and \( n = 12 \).

### Step 2: Write the expression for the \( r \)th term
Substituting the values into the formula, we have:
\[ T_{r+1} = \binom{12}{r} (2x^2)^{12-r} \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{12}{r} (2^{12-r} x^{2(12-r)}) \left(-1\right)^r \left(\frac{1}{x^r}\right) \]

### Step 3: Combine the powers of \( x \)
Now, we can combine the powers of \( x \):
\[ T_{r+1} = \binom{12}{r} (-1)^r 2^{12-r} x^{24 - 2r - r} = \binom{12}{r} (-1)^r 2^{12-r} x^{24 - 3r} \]

### Step 4: Set the exponent of \( x \) to zero
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 24 - 3r = 0 \]

### Step 5: Solve for \( r \)
Solving the equation:
\[ 24 = 3r \implies r = \frac{24}{3} = 8 \]

### Conclusion
Thus, the value of \( r \) is \( 8 \). Therefore, the answer is option (ii) 8.

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