JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

For a suitably chosen real constant a, let a
function, f:R - { - a } → R be defined by
f(x) = (a - x)/(a + x). Further suppose that for any real number x ≠ - a and f(x) ≠ - a,
(fof)(x) = x. Then f( - 1/2 ) is equal to :

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

Answer (d) 3

Solution

Maths sir · 4 steps

Put x = 0 into f(f(x)) = x: that one substitution fixes a, then evaluate at -1/2.

  1. Rule out zero. If a = 0, then f(x) = -x/x = -1 for every allowed x. A constant function cannot satisfy f(f(x)) = x, so a ≠ 0.
  2. Substitute zero. Since a ≠ 0, f(0) = a/a = 1. Then f(f(0)) = 0 means f(1) = 0, that is (a - 1)/(a + 1) = 0, so a = 1.

    f(x) = (1 - x)/(1 + x), f(f(x)) = ((1 + x) - (1 - x))/((1 + x) + (1 - x)) = 2x/2 = x

  3. Check and evaluate. The check above shows that a = 1 works for every allowed x. Now f(-1/2) = (1 + 1/2)/(1 - 1/2) = (3/2)/(1/2) = 3.
  4. So. f(-1/2) = 3, which is option (d).

The idea to keep. When a functional equation must hold for every x, plug in a convenient value like x = 0 to find the unknown constant. Then verify that the constant works in general.

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

Study Functions (Advanced)

Asked about 70 times in JEE Main since 2020. This question is from 2020.