Let the numbers $2, b, c$ be in an A.P and $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2}\end{array}\right]$. If $\operatorname{det}(A) \in[2,16]$ then $c$ lies in the interval

(A) $\left[3,2+2^{2/4}\right]$

(B) $\left(2+2^{3/4}, 4\right)$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Determinants

Let the numbers

2, b, c

be in an A.P and

A = 124 1 b b^{2} 1 c c^{2}

. If

det (A) \in [2, 16]

then

c

lies in the interval

Answer: (D)

Since

2, b, c

are in A.P., write

b = 2 + d

and

c = 2 + 2 d

. Reducing the determinant with the column operations

C_{2} \to C_{2} - C_{1}

and

C_{3} \to C_{3} - C_{1}

brings it to a Vandermonde-type form,

det (A) = (b - 2) (c - 2) (c - b) = d \cdot 2 d \cdot d = 2 d^{3}

. The condition

det (A) \in [2, 16]

becomes

2 d^{3} \in [2, 16]

, so

d^{3} \in [1, 8]

and

d \in [1, 2]

. Then

c = 2 + 2 d

ranges over

[2 + 2, 2 + 4] = [4, 6]

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option