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Q.

The lines pp2+1x−y+q=0 and p2+12x+p2+1y+2q=0 are perpendicular to a common line for

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a

exactly one value of p

b

no value of p

c

exactly two values of p

d

more than two values of p

answer is A.

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Detailed Solution

Lines are parallel⇒pp2+1p2+12=-1p2+1 ⇒p=-1 ⇒p takes exactly one value.
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