Question

Let the number \(2, b, c\) be in an A.P. and \(A = \begin{bmatrix}1 & 1 & 1\\ 2 & b & c\\ 4 & b^2 & c^2\end{bmatrix}\). If \(\det(A) \in [2, 16]\), then \(c\) lies in the interval :

• A. \([2, 3)\)
• B. \([4, 6]\)
• C. \((2 + 2^{3/4}, 4)\)
• D. \([3, 2 + 2^{3/4}]\)

Hint:

The determinant is of Vandermonde type: \((b - 2)(c - 2)(c - b)\). Put \(b = 2 + d\), \(c = 2 + 2d\).

Answer 1

The Correct Option is B

\(\det A\) is a Vandermonde determinant in \(2, b, c\): \(\det A = (b - 2)(c - 2)(c - b)\).
Since \(2, b, c\) are in A.P. with common difference \(d\): \(b = 2 + d\), \(c = 2 + 2d\).
\(\det A = d\cdot 2d\cdot d = 2d^3\).
\(2 \le 2d^3 \le 16 \Rightarrow 1 \le d^3 \le 8 \Rightarrow 1 \le d \le 2\).
\(c = 2 + 2d \in [4, 6]\).
Answer: (2) \([4, 6]\)