Match List I with List II
| List I | List II | ||
| A. | Range of |x| | I. | \((-5, \infty)\) |
| B. | Range of \(9x^2 + 6x - 5\) for all \(x \ge 0\) | II. | \([0, \infty)\) |
| C. | Domain of \(\displaystyle\frac{1}{\sqrt{x+5}}\) | III. | {(1,1), (2,2), (3,3)} |
| D. | Smallest equivalence relation on Set {1,2,3} | IV. | \([-5, \infty)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-IV, C-I, D-III
(A) Range of |x|
This matches (II) \([0, \infty)\).
|x| is never negative, and it takes every value \(k \ge 0\) (at x = k).
(B) Range of \(9x^2 + 6x - 5\) for all \(x \ge 0\)
This matches (IV) \([-5, \infty)\).
\(f'(x) = 18x + 6 \gt 0\) for \(x \ge 0\)
So f is increasing on \([0, \infty)\); its least value is f(0) = −5 and it grows without bound.
(C) Domain of \(\frac{1}{\sqrt{x+5}}\)
This matches (I) \((-5, \infty)\).
The root must be defined and non-zero, since it is in the denominator:
\(x + 5 \gt 0,\quad x \gt -5\)
(D) Smallest equivalence relation on Set {1,2,3}
This matches (III) {(1,1), (2,2), (3,3)}.
Every equivalence relation is reflexive, so it must contain (1, 1), (2, 2), (3, 3). This set alone is already symmetric and transitive, so it is the smallest equivalence relation (the identity relation).
So the matching is A-II, B-IV, C-I, D-III.