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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Matrices and Determinants

Let the numbers 2, bb, cc be in an AP and A=[1112bc4b2c2]A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix}

If det⁡(A)∈[2,16]\det(A) \in [2, 16], then cc lies in the interval
  1. A.
    [3,2+23/4][3, 2 + 2^{3/4}]
  2. B.
    (2+23/4,4)(2 + 2^{3/4}, 4)
  3. C.
    [4,6][4, 6]
  4. D.
    [2,3)[2, 3)
Show answer and solution

Correct option: C

Solution

Since 2, b, c are in AP, b = (2+c)/2. The determinant of A is a Vandermonde determinant: det(A) = (b-2)(c-2)(c-b).

With b = (2+c)/2, we get (b-2) = (c-2)/2 and (c-b) = (c-2)/2.

So det(A) = (c-2)/2 · (c-2) · (c-2)/2 = (c-2)³/4.

Now apply the condition: 2 ≤ (c-2)³/4 ≤ 16, which gives 8 ≤ (c-2)³ ≤ 64, so 2 ≤ c-2 ≤ 4, meaning c ∈ [4, 6].

[JEE Main 2019 · 8 Apr · Shift II]

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