If x + 1/x = 2, then which one of the following is the value of x³² + 1/x³²?

Correct — D, 2. Multiply x + 1/x = 2 through by x to get x² − 2x + 1 = 0, which is (x − 1)² = 0, so x = 1 is the only value that satisfies the condition. Then x³² = 1 and 1/x³² = 1, and their sum is 2. The general route reaches the same place without solving: squaring x + 1/x = 2 gives x² + 1/x² = 4 − 2 = 2, squaring again gives x⁴ + 1/x⁴ = 2, and the value stays at 2 through every doubling to the thirty-second power.

• (a)−1 — −1 is the value of x + 1/x when x is a primitive cube root of unity, a different condition entirely. With x = 1 no expression of the form xⁿ + 1/xⁿ can be negative.
• (b)0 — 0 would need x³² = −1/x³², which is impossible for the real value x = 1.
• (c)1 — 1 is the value of x itself, and of x³². The question asks for x³² + 1/x³², which adds two such terms.

The family x + 1/x = k is the standard gateway to powers, because squaring gives x² + 1/x² = k² − 2 and the process repeats: each squaring reaches the next doubled power. The value k = 2 is the fixed point of that recursion, since 2² − 2 = 2, so the expression never moves. It is also the boundary case where the quadratic x² − kx + 1 = 0 has equal roots.

Nothing here requires 32 steps. Either solve the quadratic and see that x = 1, or notice that k = 2 reproduces itself under squaring. Candidates lose the mark by reading the exponent as a threat and reaching for logarithms or binomial expansion. The companion case worth memorising is x + 1/x = −2, which forces x = −1; then xⁿ + 1/xⁿ is 2 for even n and −2 for odd n.

• x + 1/x = 2 has the unique real solution x = 1, from (x − 1)² = 0.
• Squaring gives x² + 1/x² = (x + 1/x)² − 2, the recursion that reaches higher powers.
• k = 2 is a fixed point of that recursion: 2² − 2 = 2, so every doubled power keeps the value 2.
• x + 1/x = −2 forces x = −1, and then xⁿ + 1/xⁿ alternates between 2 and −2.
• For |k| < 2 the equation x² − kx + 1 = 0 has no real root, and the powers cycle instead of settling.

• Answering 1, which is the value of x³² rather than of the whole expression.
• Attempting a binomial expansion of the thirty-second power.
• Assuming a large exponent must produce a large answer.

An algebra identity item where the large exponent is the disguise and the condition collapses to a single value of x.

No directly related past PYQ was found.

• practice — not a real PYQ

  If x + 1/x = 3, then x² + 1/x² equals

  • (a)5
  • (b)7
  • (c)9
  • (d)11

  Answer(b) 7 — squaring gives x² + 1/x² = 3² − 2 = 7.

• practice — not a real PYQ

  If x + 1/x = −2, then x²⁵ + 1/x²⁵ equals

  • (a)−2
  • (b)−1
  • (c)0
  • (d)2

  Answer(a) −2 — the condition forces x = −1, and an odd power of −1 gives −1 + (−1) = −2.