The lines p(p2 + 1)x − y + q = 0 and (p2 + 1)2x + (p2 + 1)y + 2q = 0 are perpendicular to a common line for 

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  1. exactly one value of p 
  2. exactly two values of p
  3. more than two values of p
  4. no value of p 

Answer (Detailed Solution Below)

Option 1 : exactly one value of p 
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Calculation

Given:

The lines p(p² + 1)x - y + q = 0 and (p² + 1)x + (p² + 1)y + 2q = 0 are perpendicular to a common line.

Then these lines must be parallel to each other.

∴ m₁ = m₂ ⇒ -\(\frac{p(p^2+1)}{-1}\) = -\(\frac{(p^2+1)^2}{p^2+1}\)

⇒ (p² + 1)(p + 1) = 0

⇒ p = -1

Hence option 1 is correct

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