JEE Main 2020 · 9 January · evening · Single correct · +4 / −1

Let a – 2b + c = 1.
If f(x)=| x + a, x + 2, x + 1; x + b, x + 3, x + 2; x + c, x + 4, x + 3 |, then:

  1. (a) ƒ(50) = 1
  2. (b) ƒ(–50) = –1
  3. (c) ƒ(50) = –501
  4. (d) ƒ(–50) = 501
Answer and solution

Answer (a) ƒ(50) = 1

Solution

Maths sir · 4 steps

The condition a - 2b + c = 1 tells you the row operation to use: R₁ → R₁ - 2R₂ + R₃.

  1. Choose the operation. Replace R₁ by R₁ - 2R₂ + R₃. The coefficient of R₁ is 1, so the value of the determinant does not change.
  2. New first row. Column 1 becomes (x+a) - 2(x+b) + (x+c) = a - 2b + c = 1. Column 2 becomes (x+2) - 2(x+3) + (x+4) = 0, and column 3 becomes (x+1) - 2(x+2) + (x+3) = 0. Every x cancels, so the first row is (1, 0, 0).

    f(x) = |1, 0, 0; x+b, x+3, x+2; x+c, x+4, x+3| = (x+3)² - (x+2)(x+4)

  3. Expand. (x² + 6x + 9) - (x² + 6x + 8) = 1. So f(x) = 1 for every value of x.
  4. So. f(50) = 1, which is option (a). Note that f(-50) = 1 as well.

The idea to keep. When a condition like a - 2b + c = 1 is given, the same coefficients are the row operation to use. Apply it and look for a row that turns into 1s and 0s.

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

Study Determinants

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