CGL 2026 Tier 1EzSSC Practice
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2 questions from all 53 shifts of SSC CGL Tier 1 2026 · Quant

AllAlgebraic Identities · 28Surds · 9Identities · 8Componendo and Dividendo · 4Sum of Cubes Identity · 2Reciprocal Identity Powers · 2Difference of Squares · 2Linear Equations · 2Surds and Square Roots · 2Indices · 2
More subtopics (5)
Reciprocal Powers · 2Componendo-Dividendo · 2Identities with x + 1/x · 2Perfect Square Trinomial · 2Cubic Identities · 2

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Q109 Oct 2026 01:30 PM (Shift-2) · Q64Reciprocal Identity Powersmoderate
If $x + \frac{1}{x} = \sqrt{17}$, then find the value of $\frac{x^{4}-1}{x^{2}}$.
Answer: A
(X^4 - 1)/X^2 = X^2 - 1/X^2 = (X - 1/X)(X + 1/X). Since (X - 1/X)^2 = (X + 1/X)^2 - 4 = 17 - 4 = 13, X - 1/X = +/- sqrt13. Product = +/- sqrt13 x sqrt17 = +/- sqrt221. This is option A. Option A.
यदि $x + \frac{1}{x} = \sqrt{17}$ है, तो $\frac{x^{4}-1}{x^{2}}$ का मान ज्ञात कीजिए।
उत्तर: A
(X⁴ − 1)/X² = X² − 1/X² = (X − 1/X)(X + 1/X)। (X − 1/X)² = (X + 1/X)² − 4 = 17 − 4 = 13, अतः X − 1/X = ±√13। गुणनफल = ±√13 × √17 = ±√221। अतः सही विकल्प A है। विकल्प A।
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Q209 Oct 2026 01:30 PM (Shift-2) · Q73Reciprocal Identity Powershard
If $x + \frac{1}{x} = 12$, then find the value of $x^{4} + \frac{1}{x^{4}} + x^{2} + \frac{1}{x^{2}}$.
Answer: B
x^2 + 1/x^2 = 144 - 2 = 142. Then x^4 + 1/x^4 = 142^2 - 2 = 20164 - 2 = 20162. Adding x^2 + 1/x^2 = 142 gives 20162 + 142 = 20304. This is option B. Option B.
यदि $x + \frac{1}{x} = 12$ है, तो $x^{4} + \frac{1}{x^{4}} + x^{2} + \frac{1}{x^{2}}$ का मान ज्ञात कीजिए।
उत्तर: B
x² + 1/x² = 144 − 2 = 142। तब x⁴ + 1/x⁴ = 142² − 2 = 20164 − 2 = 20162। इसमें x² + 1/x² = 142 जोड़ने पर 20162 + 142 = 20304। अतः सही विकल्प B है। विकल्प B।
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