If `|(1,1,1),(2,b,c),(4,b^(2),c^(2))|` and `|A| = in [2, 16]. 2, b, c` and in A.P. the range of c is

Options

A. `[3,2+2^(3//4)]`
B. `(2+2^(3//4), 4)`
C. `[4,6]`
D. `[2,3]`

Answer

To solve the problem, we need to find the range of \( c \) given the determinant \( |(1,1,1),(2,b,c),(4,b^2,c^2)| \) and the conditions that \( |A| \) lies in the interval [2, 16] and that \( 2, b, c \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We start with the determinant:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{vmatrix} \]

2. **Apply Row Operations**:
We can simplify the determinant using row operations. Subtract the first row from the second and third rows:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2-1 & b-1 & c-1 \\ 4-1 & b^2-1 & c^2-1 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & b-1 & c-1 \\ 3 & b^2-1 & c^2-1 \end{vmatrix} \]

3. **Calculate the Determinant**:
Now we can calculate the determinant using cofactor expansion along the first row:
\[ D = 1 \cdot \begin{vmatrix} b-1 & c-1 \\ b^2-1 & c^2-1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & c-1 \\ 3 & c^2-1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & b-1 \\ 3 & b^2-1 \end{vmatrix} \]

The first determinant simplifies to:
\[ (b-1)(c^2-1) - (c-1)(b^2-1) = (b-1)(c^2-1) - (c-1)(b^2-1) \]

The second determinant simplifies to:
\[ 1(c^2-1) - 3(c-1) = c^2 - 1 - 3c + 3 = c^2 - 3c + 2 \]

The third determinant simplifies to:
\[ 1(b^2-1) - 3(b-1) = b^2 - 1 - 3b + 3 = b^2 - 3b + 2 \]

4. **Combine the Determinants**:
Thus, we have:
\[ D = (b-1)(c^2-1) - (c-1)(b^2-1) - (c^2 - 3c + 2) + (b^2 - 3b + 2) \]

5. **Substituting A.P. Condition**:
Since \( 2, b, c \) are in A.P., we have:
\[ b = 2 + d, \quad c = 2 + 2d \]
for some \( d \).

6. **Expressing the Determinant in Terms of \( d \)**:
Substitute \( b \) and \( c \) into \( D \) and simplify:
\[ D = 2d^3 \]

7. **Finding the Range of \( d \)**:
Given \( |D| \in [2, 16] \):
\[ 2d^3 \in [2, 16] \implies d^3 \in [1, 8] \implies d \in [1, 2] \]

8. **Finding the Range of \( c \)**:
Since \( c = 2 + 2d \):
\[ c \in [2 + 2 \cdot 1, 2 + 2 \cdot 2] = [4, 6] \]

### Final Answer:
The range of \( c \) is \( [4, 6] \).