A) \[\{-1,-4\}\]
B) \[\{1,4\}\]
C) \[\{-4,4\}\]
D) None of these
Correct Answer: D
Solution :
Case 1: \[\operatorname{x} \ge 0\] \[\therefore the \,equation \,becomes \,{{x}^{2}}+5x+4=0\] or \[\operatorname{x}=-1,\,\,-4\,\,but\, x\ge 0\] \[\therefore \] both values, non-admissible: Case 2: \[\operatorname{x} \le 0\] The eqn becomes \[{{\operatorname{x}}^{2}}- 5x + 4 = 0 \,or \,\,x = 1,\,\,\,4\] both values are non-admissible, \[\therefore \] No real roots. Alternatively, since \[{{\operatorname{x}}^{2}} \ge 0; \,\,\left| \,x\, \right| \ge 0\] \[\therefore \,{{x}^{2}} + \left| \,x\, \right| + 4 > 0 \,for \,all \,x \in R\] \[\therefore \,{{x}^{2}} + \left| \,x\, \right| + 4 \ne 0 \,for \,all \,x \in R\]
You need to login to perform this action.
You will be redirected in
3 sec
