If 9 @ 1 = 82 and 10 @ 1 = 101, what is 8 @ 2?

Correct — B. SSC's key marks 84. Both examples have 1 as the second number, so they do not fix a single rule; the note on the key records the doubt.

Rule the key fits: a @ b = (a + b)² − 2a.

9 @ 1 = 10² − 18 = 100 − 18 = 82 ✓
10 @ 1 = 11² − 20 = 121 − 20 = 101 ✓

8 @ 2 = (8 + 2)² − 2 × 8
= 100 − 16 = 84 → option (b).

Why the others are wrong

• (a)
  68 — 68 is 8² + 2², the rule a² + b². It fits both examples as well (81 + 1 and 100 + 1, since 1² = 1). SSC's key marks (b) instead; the note on the key records this.
• (c)
  98 — 98 is 10² − 2. The key's rule subtracts twice the first number from (a + b)²: 18 for 9 @ 1 and 20 for 10 @ 1, so 8 @ 2 subtracts 16, not 2.
• (d)
  88 — 88 is 10² − 12. Twice the first number is 16 for 8 @ 2, as it was 18 and 20 in the examples, so the key's rule gives 100 − 16 = 84.

An operator defined by examples is solved by finding one rule that fits every example, then applying it to the new pair.

Results just above squares are a good first clue: 82 is 9² + 1 and 101 is 10² + 1.

But with 1 as the second number in both examples, 1, 1² and 1³ are all 1. So a² + b, a² + b² and (a + b)² − 2a all fit, and they give different answers for 8 @ 2: 66, 68 and 84.

Key facts

• 9² + 1 = 82 and 10² + 1 = 101.
• (9 + 1)² − 2 × 9 = 82 and (10 + 1)² − 2 × 10 = 101.
• When b = 1, the rules a² + b, a² + b² and (a + b)² − 2a all equal a² + 1.
• For 8 @ 2 those three rules give 66, 68 and 84.

Common traps

• Taking the first rule that fits as settled. With b = 1 in both examples, b, b² and b³ cannot be told apart.
• Applying a² + b from a similar item: it gives 8 @ 2 = 66, which is not offered.

No directly related past PYQ was found.