The correct option is B a<−4
x2−12|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