Sets Relations And Functions Question 90
Question: If $ | x^{2}-2x+2 |-| 2x^{2}-5x+2 | $ = $ | x^{2}-3x | $ , then the set of values of x is
Options:
A) $ ( -\infty ,0 ]\cup [ 3,\infty ) $
B) $ [ 0,\frac{1}{2} ]\cup [ 2,3 ] $
C) $ ( -\infty ,0 ]\cup [ \frac{1}{2},2 ]\cup [ 3,\infty ) $
D) $ [ 0,2 ]\cup [ 3,\infty ) $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] we have,
$ | 2x^{2}-5x+2 |+| x^{2}-3x |=| x^{2}-2x+2 | $ = $ | 2x^{2}-5x+2 - (x^{2}-3x) | $
$ \therefore (2x^{2}-5x+2)(x^{2}-3x)\le 0 $
$ \Rightarrow (2x-1)(x-2)x(x-3)\le 0 $ From the sign scheme, we have $ x\in [ \frac{1}{2},2 ]\cup [3,\infty) $
BETA
