JEE Main 2021 · 20 July · morning · Single correct · +4 / −1

Let A = [aᵢⱼ] be a 3 × 3 matrix, where aᵢⱼ = { 1, ,, if i = j; - x, ,, if | i - j | = 1; 2x + 1, ,, otherwise.
Let a function f : R → R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:

  1. (a) - 20/27
  2. (b) 88/27
  3. (c) 20/27
  4. (d) - 88/27
Answer and solution

Answer (d) - 88/27

Solution

Maths sir · 5 steps

Write the matrix out, expand it into a cubic, then add the values at its two turning points.

  1. Build the matrix. The diagonal entries are 1. The entries next to the diagonal (|i-j| = 1) are -x. The two corner entries a₁₃ and a₃₁ are 2x+1.

    f(x) = |1, -x, 2x+1; -x, 1, -x; 2x+1, -x, 1|

  2. Expand. Expand along row 1 and write a = 2x+1. This gives f = (1-x²) + x(ax - x) + a(x² - a) = 1 - 2x² + 2ax² - a². Putting a = 2x+1 back in gives f(x) = 4x³ - 4x² - 4x.
  3. Turning points. f'(x) = 12x² - 8x - 4 = 4(3x+1)(x-1), so the turning points are at x = -1/3 and x = 1. A cubic has no overall maximum or minimum on ℝ, so the question means these local values.
  4. Evaluate. f(-1/3) = -4/27 - 12/27 + 36/27 = 20/27, which is the local maximum. f(1) = 4 - 4 - 4 = -4, which is the local minimum.
  5. So. The sum is 20/27 - 4 = (20 - 108)/27 = -88/27, which is option (d).

The idea to keep. When a determinant has x in its entries, expand it into a polynomial first. Then it becomes an ordinary calculus problem: set f' = 0 and evaluate f at those points.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 190 times in JEE Main since 2020. This question is from 2021.