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Q.

The set of all the values of ‘P’ for which the equation  (log2x)24|log2x|+P=0 has four real solutions is

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a

(0,4)

b

(0,2)

c

(2,4)

d

(0,)

answer is C.

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Detailed Solution

Let   |log2x|=yy24y+P=0...(1)
Equation (1) should have 2 positive roots.
P>0 and D = 16 – 4P > 0
P(0,4)
 

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