If x + 1⁄9x = 3, then the value of 9x² + 1⁄9x² is:

Correct — B. The stem is a figure: x + 1⁄9x = 3, find 9x² + 1⁄9x².

Multiply the given equation by 3, so that the squared term becomes 9x²:
3x + 1⁄(3x) = 9

Square both sides:
(3x + 1⁄(3x))² = 9² = 81
9x² + 2 + 1⁄9x² = 81

Take the middle term across:
9x² + 1⁄9x² = 81 − 2 = 79 → option (b)

Why the others are wrong

• (a)
  81 — 81 is the square, not the answer. (3x + 1⁄(3x))² comes to 81, but that expansion still carries the middle term 2, which has to be removed before the expression the question asks for is left standing.
• (c)
  77 — 77 is 81 − 4 — the middle term subtracted twice. The expansion produces 2·3x·1⁄(3x) exactly once, so one subtraction of 2 is all the identity allows.
• (d)
  87 — Nothing in the expansion reaches 87. Squaring gives 81 and the middle term is 2, so the two values in reach are 79 by subtracting and 83 by wrongly adding.

The whole family rests on one expansion: if a + 1⁄a = k, then squaring gives a² + 2 + 1⁄a² = k², so a² + 1⁄a² = k² − 2. The cross term is always 2, because a × 1⁄a = 1.

The work in this item is spotting which quantity plays the part of a. The stem's x + 1⁄9x is not in that shape, but multiplying through by 3 turns it into 3x + 1⁄(3x).

Now a = 3x, and a² = 9x² with 1⁄a² = 1⁄9x² — exactly the expression asked for.

Key facts

• If a + 1⁄a = k then a² + 1⁄a² = k² − 2.
• If a − 1⁄a = k then a² + 1⁄a² = k² + 2.
• Multiplying x + 1⁄9x = 3 through by 3 gives 3x + 1⁄(3x) = 9.
• With a = 3x, the square a² is 9x² and its reciprocal is 1⁄9x².

Common traps

• Squaring the stem as printed instead of scaling it to 3x first.
• Stopping at 81 and forgetting to subtract the middle term.
• Reading 1⁄9x as (1⁄9)x rather than 1⁄(9x), which changes the identity entirely.

No directly related past PYQ was found.