Question

( operatorname{tat} quad f(x)=x^{2}+a x+b )
[
g(x)=x^{2}+a|x|+b
]
Equation ( g(x)=0 ) has only one solution tomerer it is seen that if ( alpha ) satisfies the equ then ( -alpha ) alsodoes as well .
[
begin{array}{l}
therefore alpha=-alpha
Rightarrow alpha=0
end{array}
]
The root of ( g(x)=0 ) is 0
[
0^{2}+a^{2}left|0^{0}right|^{2}+b=0
]
[
therefore b=0
]
( therefore g(x)=x^{2}+a|x| )
[
=|x|^{2}+a|x|=|x|(|x|+a)
]
If ( g(x)=0 ) either ( |x|=0 ) or ( |x|+a=0 ) Sunce o is the only nomible solution, ageqslanto lemence observe that if ( a=0 )
(2) ( f(x)=x^{2}, ) thich does not han
( therefore quad ) AM
り ( quad a>0, b=0 ) ( Rightarrow ) Option ( (A) )

# OBJECTIVE TYPE Quin 0.35 has four choice (A),B),(C), D) out of which ONLY ONE is correct. 01 le equation x + +b= has distinct real roots and x2+2x+b=0 has only one to root testach one of the following is trud? (4) 6=9,220 B) b=0,2<0 (C)b>0,250

Solution