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Question

The number of real roots of the equation x2−12|x|+20=0 is P. Then the values of a for which the equation ∣∣|x−2|+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<−4x2−12|x|+20=0⇒|x|2−12|x|+20=0⇒(|x|−10)(|x|−2)=0⇒|x|=10,|x|=2⇒x=±10,x=±2⇒P=4 Now, ∣∣|x−2|+a∣∣=4 Case (i) |x−2|+a=4 ⇒|x−2|=4−a It has 2 distinct solutions, if 4−a>0⇒a<4 …(1) Case (ii) |x−2|+a=−4 ⇒|x−2|=−4−a It has 2 distinct solutions, if −4−a>0⇒a<−4 …(2) So, equation ∣∣|x−2|+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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