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In the given questions, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer.I. 2x2 – 32x + 128 = 0II. 2y2 – 32y + 96 = 01. x > y2. x ≥ y3. x < y4. x ≤ y5. x = y or the relationship cannot be established. |
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Answer» Correct Answer - Option 5 : x = y or the relationship cannot be established. I. 2x2 – 32x + 128 = 0 ⇒ 2x2 – 16x – 16x + 128 = 0 ⇒ 2x(x – 8) – 16(x – 8) = 0 ⇒ (x – 8)(2x – 16) = 0 ⇒ x = 8 II. 2y2 – 32y + 96 = 0 ⇒ 2y2 – 24y – 8y + 96 = 0 ⇒ 2y(y – 12) – 8(y – 12) = 0 ⇒ (y – 12)(2y – 8) = 0 ⇒ y = 12 or 4 Comparison between x and y (via Tabulation):
∴ Relationship cannot be established between x and y. |
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