JEE Main 2023 · 31 January · evening · Single correct · +4 / −1

Let f: ℝ-{2,6} → ℝ be real valued function
defined as f(x)=(x²+2 x+1)/(x²-8 x+12).
Then range of f is

  1. (a) (-∞,-21/4] ∪[1, ∞)
  2. (b) (-∞,-21/4) ∪(0, ∞)
  3. (c) (-∞,-21/4] ∪[0, ∞)
  4. (d) (-∞,-21/4] ∪[21/4, ∞)
Answer and solution

Answer (c) (-∞,-21/4] ∪[0, ∞)

Solution

Maths sir · 4 steps

Turn y = f(x) into a quadratic in x and ask which values of y give a real x.

  1. Set up quadratic. Put y = (x²+2x+1)/(x²-8x+12) and cross-multiply. Collecting terms gives (y-1)x² - (8y+2)x + (12y-1) = 0. A value y is in the range only if this equation has a real root x other than 2 and 6.
  2. Discriminant condition. For y ≠ 1 we need D ≥ 0. Here (8y+2)² = 64y²+32y+4 and 4(y-1)(12y-1) = 48y²-52y+4, so their difference is 16y²+84y.

    D = 16y² + 84y = 4y(4y+21) ≥ 0 ⇒ y ≤ -21/4 or y ≥ 0

  3. Check special values. At y=1 the equation becomes linear, -10x+11=0, so x = 1.1 works and 1 is in the range. Both endpoints are reached: f(-1) = 0 and f(16/5) = (441/25)/(-84/25) = -21/4. The root is never 2 or 6, because x=2 or x=6 would make the numerator 0, which it is not.
  4. So. The range is (-∞, -21/4] ∪ [0, ∞), which is option (c).

The idea to keep. For the range of a quadratic over quadratic, write it as a quadratic in x and use D ≥ 0. Then check the leading-coefficient-zero case and the excluded points separately.

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 2023.