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Q.

If (1−p)1+2x+4x2+8x3+16x4+32x5=1−p6⋅p≠1 then a value of px is

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a

12

b

2

c

14

d

4

answer is B.

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Detailed Solution

(1−2x)1+2x+4x2+8x3+16x4+32x51-2x=1−p6(1−p)1−(2x)61−2x=1−p61−p⇒p=2x⇒px=2
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