A) \[\pm \,\,a\]
B) \[\pm \,\frac{1}{2}\,a\]
C) \[\pm \,\frac{3}{2}\,a\]
D) \[\pm \,2\,a\]
Correct Answer: B
Solution :
Given equation is \[(a+1){{x}^{2}}-(a+2)x+(a+3)=0\] Since, roots are equal in magnitude and opposite in sign. \[\therefore \] coefficient of x is zero ie, \[a+2=0\] \[\Rightarrow \] \[a=-2\] ?.(i) \[\therefore \] Equation is \[(-2+1){{x}^{2}}-(-2+2)x+(-2+3)=0\] \[\Rightarrow \] \[-{{x}^{2}}+1=0\] \[\Rightarrow \] \[x=\pm 1\] ?..(ii) Only option [b] ie, \[\pm \frac{1}{2}\] a satisfies Eqs. (i) and (ii).
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