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. a≥0
Which of the following is the CORRECT combination ?
A
a→s,b→r,c→q,d→p
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B
a→p,b→r,c→q,d→s
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C
a→r,b→s,c→p,d→q
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D
a→s,b→p,c→q,d→r
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Solution
The correct option is Aa→s,b→r,c→q,d→p a.|x|2+a|x|+1=0
Clearly, if a≥0, then there are no real roots as all the terms in LHS are positive.
b. For a=−2, |x|2−2|x|+1=0 ⇒(|x|−1)2=0 ⇒|x|=1 ⇒x=±1
Therefore, for a=−2, the equation has two real roots.
c.|x|=−a±√a2−42
Obviously, the equation does not have three real roots for any value of a
d. If a<−2,
then a2−4>0 and |−a|>√a2−4 ⇒|x|=−a±√a2−42>0
Hence, the equation has four real roots if a<−2