यदि सभी x ∈ R के लिए x2 + 2ax + 10 - 3a > 0 हो तो

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  1. - 5 < a < 2
  2. a < -5
  3. a > 5
  4. 2 < a < 5

Answer (Detailed Solution Below)

Option 1 : - 5 < a < 2
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NIMCET 2020 Official Paper
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Detailed Solution

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संकल्पना:

ax2 + bx + c > 0 सभी x ∈ R के लिए यदि: a > 0 और b2 - 4ac < 0

गणना:

समीकरण x2 + 2ax + 10 - 3a > 0 सभी x ∈ R के लिए

⇒ (2a)2 - [4 × (10 - 3a)] < 0 

⇒ 4a2 - [40 - 12a] < 0

⇒ 4a2 + 12a - 40 < 0

⇒ a2 + 3a - 10 < 0

⇒ (a + 5)(a - 2) < 0 

∴  -5 < a < 2

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