SSC CGL 2024 Tier-I · 18 September 2024, Shift 2 · Quantitative Aptitude · Question 4 of 99

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

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SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

If x + 1⁄x = 2, then find the value of (x²⁷²³ + 1⁄x³³⁵⁶).

(a)−1
(b)1
(c)0
(d)2

Answer

Why

Correct — D. The condition pins x down completely, so the two huge exponents never have to be handled.

x + 1⁄x = 2
Multiply through by x: x² + 1 = 2x
Rearrange: x² − 2x + 1 = 0
Factorise: (x − 1)² = 0, so x = 1

Every power of 1 is 1:
x²⁷²³ = 1 and 1⁄x³³⁵⁶ = 1
Sum = 1 + 1 = 2 → option (d).

Why the others are wrong

(a)
−1 — −1 needs one of the two terms to be negative. x = 1 is the only root the condition allows, so both terms are exactly +1.
(b)
1 — 1 is the value of each term separately, not of their sum. The bracket asks for x²⁷²³ plus 1⁄x³³⁵⁶, which is 1 + 1.
(c)
0 — 0 would need the two terms to be equal and opposite. With x = 1 both are +1, so they add rather than cancel.

Concept

x + 1⁄x = 2 is the one value of that sum which leaves no freedom. Clearing the fraction gives a perfect square, (x − 1)² = 0, with a repeated root at x = 1.

For a positive real x, AM–GM gives x + 1⁄x ≥ 2, with equality only at x = 1. Two is the floor of the expression, and touching the floor fixes x.

That is why the exponents are decoration. Once x = 1, x raised to anything is 1, whether the power is 2 or 2,723.

The neighbouring condition x + 1⁄x = 1 behaves nothing like this: there x satisfies x³ = −1, the powers cycle with period 6, and the exponents matter enormously. Misreading the 2 as a 1 changes the whole method.

Key facts

x + 1⁄x = 2 clears to (x − 1)² = 0, whose only root is x = 1.
With x = 1, x raised to any power equals 1, so the exponents 2723 and 3356 change nothing.
For positive real x, AM–GM gives x + 1⁄x ≥ 2, with equality only at x = 1.

Study next

x + 1⁄x = 1 and the period-6 cycle it puts on the powers of xThe recurrence xⁿ + 1⁄xⁿ = (x + 1⁄x)(xⁿ⁻¹ + 1⁄xⁿ⁻¹) − (xⁿ⁻² + 1⁄xⁿ⁻²)Reading x² − 2x + 1 as a perfect square instead of running the quadratic formula

Common traps

Reducing 2723 and 3356 modulo 4 or 6, a habit carried over from the x + 1⁄x = 1 form
Answering 1, the value of one term, instead of the sum of the two
Assuming the unequal exponents must make the two terms differ

SSC leaves the stem text blank and puts the whole condition into an image, then picks exponents large enough to look unmanageable.

The identical trick runs at 19 Sep 2024, 09:00, Quant Q.8, where sin x + cosec x = 2 forces sin x = 1 and the seventh powers collapse the same way. The neighbouring form x² + 1 = x, which is x + 1⁄x = 1, appears at 25 Sep 2024, 16:00, Quant Q.21.

Related PYQs

No directly related past PYQ was found.