Equations And Inequalities Question 23
Question: The solution set of $ {{(x)}^{2}}+{{(x+1)}^{2}}=25, $ where (x) is the least integer greater than or equal to x, is
Options:
A) $ (2,4) $
B) $ (-5,-4]\cup (2,3] $
C) $ [-4,-3)\cup [3,4) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] $ {{(x)}^{2}}+{{(x+1)}^{2}}=25\Rightarrow {{(x)}^{2}}+{{{(x)+1}}^{2}}=25 $
$ \Rightarrow 2{{(x)}^{2}}+2(x)-24=0 $
$ \Rightarrow {{(x)}^{2}}+(x)-12=0\Rightarrow (x)=-4or3 $ Now $ (x)=-4\Rightarrow -5<x\le -4 $ and $ (x)=3\Rightarrow 2<x\le 3 $
$ \therefore $ Solution set is $ (-5,-,4]\cup (2,3] $
BETA
