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Question

Consider the equation x2+a|x|+1=0, where a is a parameter.
Column IColumn IIa. No real rootsp. a<2b. Two real rootsq. ϕc. Three real rootsr. a=2a. Four distinct real rootss. a0
Which of the following is the CORRECT combination ?

A
as,br,cq,dp
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B
ap,br,cq,ds
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C
ar,bs,cp,dq
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D
as,bp,cq,dr
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Solution

The correct option is A as,br,cq,dp
a. |x|2+a|x|+1=0
Clearly, if a0, then there are no real roots as all the terms in LHS are positive.

b. For a=2,
|x|22|x|+1=0
(|x|1)2=0
|x|=1
x=±1
Therefore, for a=2, the equation has two real roots.

c. |x|=a±a242
Obviously, the equation does not have three real roots for any value of a

d. If a<2,
then a24>0 and |a|>a24
|x|=a±a242>0
Hence, the equation has four real roots if a<2

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