Given, a = √5, what is (a + 1)² + (a−1)² ?

Correct — B.
Identity: (a + 1)² + (a − 1)² = 2(a² + 1), because the +2a and −2a cross terms cancel.

Square a: a² = (√5)² = 5
Add 1: a² + 1 = 6
Double: 2 × 6 = 12 → option (b)

Direct check: (√5 + 1)² = 6 + 2√5 and (√5 − 1)² = 6 − 2√5. The √5 parts cancel, leaving 12.

Why the others are wrong

• (a)
  13 — 13 is odd, but the sum is 2(a² + 1), twice a whole number when a² = 5. Reaching 13 would need a² = 5.5.
• (c)
  14 — 14 = 2(a² + 1) would need a² = 6, but (√5)² = 5. Squaring the root removes it and leaves 5.
• (d)
  15 — 15 is odd too, and 2(a² + 1) is even whenever a² is a whole number. Reaching 15 would need a² = 6.5.

Two identities meet here: (x + y)² = x² + 2xy + y² and (x − y)² = x² − 2xy + y².

Adding them cancels the ±2xy terms, so (x + y)² + (x − y)² = 2(x² + y²).

With x = a and y = 1 this is 2(a² + 1). Simplifying first and substituting last keeps the surd out of the arithmetic: only a² = 5 is ever needed.

Key facts

• (x + y)² + (x − y)² = 2(x² + y²).
• (x + y)² − (x − y)² = 4xy.
• (√5)² = 5, so (√5 + 1)² = 6 + 2√5.

Common traps

• Treating (a − 1)² as a² − 1, which drops the −2a term and turns +1 into −1. The full square is a² − 2a + 1.
• Subtracting the squares instead of adding them, which gives 4a = 4√5.

No directly related past PYQ was found.