Let λ ∈ ℝ and let the equation E be |x|² - 2|x| + |λ - 3| = 0. Then the largest element in the set S = x+λ:x is an integer solution of E is ______

Solution

Maths sir · 3 steps

Fix the integer x first; the equation then tells you exactly which λ goes with it.

1. Solve for lambda. Rewrite E as |λ - 3| = 2|x| - |x|² = |x|(2 - |x|). The left side is never negative, so the right side must satisfy |x|(2-|x|) ≥ 0.
2. Possible integers. That needs 0 ≤ |x| ≤ 2, so the integer solutions can only be x = 0, ± 1, ± 2. For each one, |λ - 3| is fixed, and then λ = 3 ± |λ - 3|.

   x	|λ - 3| = 2|x| - x²	λ	Largest x + λ
   0	0	3	3
   ± 1	1	2 or 4	1 + 4 = 5
   ± 2	0	3	2 + 3 = 5

3. So. The largest value of x + λ is reached at x = 1, λ = 4 and at x = 2, λ = 3. Both give 5.

The idea to keep. When an equation links x and a parameter, solve for the parameter in terms of x. A modulus or square being non-negative then fences in the allowed values of x.

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