JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

Let ƒ : R → R be such that for all x ∈ R
(2¹⁺ˣ + 2 1–x ), ƒ(x) and (3ˣ + 3 –x ) are in A.P.,
then the minimum value of ƒ(x) is

  1. (a) 2
  2. (b) 0
  3. (c) 3
  4. (d) 4
Answer and solution

Answer (c) 3

Solution

Maths sir · 4 steps

Write f(x) as the average of the two outer terms, then bound each bracket with AM–GM.

  1. Middle term of AP. In an AP the middle term is the average of its neighbours. So f(x) is half the sum of the two outer terms.

    f(x) = (2¹⁺ˣ + 2¹⁻ˣ + 3ˣ + 3⁻ˣ)/2 = (2ˣ + 2⁻ˣ) + 1/2(3ˣ + 3⁻ˣ)

  2. Apply AM–GM. For any a > 0, aˣ + a⁻ˣ ≥ 2√(aˣ · a⁻ˣ) = 2, with equality at x = 0. So 2ˣ + 2⁻ˣ ≥ 2 and 1/2(3ˣ + 3⁻ˣ) ≥ 1.
  3. Check equality. Both bounds become equalities at the same point, x = 0. So the sum really reaches 2 + 1 = 3 there: f(0) = 2 + 1 = 3.
  4. So. The minimum value of f(x) is 3, option (c).

The idea to keep. Whenever you see aˣ + a⁻ˣ, think AM–GM: it is always at least 2, at x = 0. Add the bounds only after checking that they hold with equality at the same x.

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

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Asked about 250 times in JEE Main since 2020. This question is from 2020.