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Question

The number of real roots of the equation x212|x|+20=0 is P. Then the values of a for which the equation |x2|+a=P can have four distinct solutions, is

A
a4
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B
a<4
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C
a2
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D
a<2
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Solution

The correct option is B a<4
x212|x|+20=0|x|212|x|+20=0(|x|10)(|x|2)=0|x|=10,|x|=2x=±10,x=±2P=4

Now, |x2|+a=4

Case (i) |x2|+a=4
|x2|=4a
It has 2 distinct solutions, if
4a>0a<4 (1)

Case (ii) |x2|+a=4
|x2|=4a
It has 2 distinct solutions, if
4a>0a<4 (2)

So, equation |x2|+a=4 will have four distinct solutions, when (1) and (2) both satisfy simultaneously.
Hence, final range of a is (1)(2)
a<4

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