Equations And Inequalities Question 17
Question: For $ x\in R,\langle x \rangle $ is defined as follows: $ \langle x \rangle = \begin{matrix} x+1, \\ | x-4 |, \\ \end{matrix},\begin{matrix} 0\le x<2 \\ x\ge 2 \\ \end{matrix}\begin{matrix} {} \\ {} \\ \end{matrix} . $ Then the solution set of the equation $ {{\langle x \rangle }^{2}}+x=\langle x \rangle +x^{2} $ is
Options:
A) $ {-1,1} $
B) $ [2,\infty ) $
C) $ [0,2) $
D) $ {0,2} $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Case 1: Let $ 0\le x<2, $ then $ \langle x \rangle =(x+1) $ and the equation becomes $ {{(x+1)}^{2}}+x=(x+1)+x^{2} $
$ \Rightarrow 2x=0\Rightarrow x=0 $ Case 2: Let $ x\ge 2 $ then $ \langle x \rangle =| x-4 | $ and the equation becomes $ {{| x-4 |}^{2}}+x=| x-4 |+x^{2} $
$ \Rightarrow x^{2}-8x+16+x=| x-4 |+x^{2} $
$ \Rightarrow | x-4 |=16-7x $
$ \therefore x-4=\pm (16-7x), $ provided $ 16-7x\ge 0 $
$ \therefore x=\frac{5}{2}or2, $ But for $ x=\frac{5}{2},16-7x<0, $ hence rejected
$ \therefore x=2. $ The solution set is $ {0,2} $
BETA
