JEE Main 2020 · 2 September · morning · Single correct · +4 / −1
If R = (x, y) : x, y ∈ Z, x² + 3y² ≤ 8 is a relation on the set of integers Z, then the domain of R –1 is :
- (a) {0, 1}
- (b) {–2, –1, 1, 2}
- (c) {–1, 0, 1}
- (d) {–2, –1, 0, 1, 2}
Answer and solution
Answer (c) {–1, 0, 1}
Solution
Maths sir · 3 steps
The domain of R⁻¹ is the range of R, so collect every y that can appear in a pair.
- Flip the pairs. R⁻¹ has every pair of R turned around, (y, x). So its first entries are the second entries of R, and the domain of R⁻¹ equals the range of R.
- Bound y. Since x² ≥ 0, the condition gives 3y² ≤ 8 - x² ≤ 8, so y² ≤ 8/3. For an integer that means y² ≤ 2, so y can only be -1, 0 or 1.
y Condition on x Integer x allowed 0 x² ≤ 8 -2, -1, 0, 1, 2 ± 1 x² ≤ 5 -2, -1, 0, 1, 2 ± 2 x² ≤ -4 none - So. Each of y = -1, 0, 1 pairs with at least one x, and no other y works. The domain of R⁻¹ is {-1, 0, 1}, option (c).
The idea to keep. Inverse swaps the roles: domain of R⁻¹ is the range of R. Find which coordinate the question really wants, then bound that variable alone.
Answer from the NTA final answer key. Solution by PrepWiser.