Here are a few good probs.Pls solve them.
1)What is the solution of the inequality
|x+3|>|2x-1|
2)Find the solution of |x2+3x|+x2-2 ≥ 0
3)Find the solution of 2x+2|x| ≥ 2√2
106Q1. lx+3l > l2x-1l
==> lx+3l -l2x-1l>0
Let f(x) = lx+3l - l2x-1l
= x+3-2x+1 , x≥1/2 and x+3+2x-1, 1/2≥x≥ -3 and -x-3+2x-1, x≤ -3
So f(x) = -x+4 , x≥1/2
= 3x+2, 1/2≥x≥ -3
= x-4, x≤ -3
You can draw the required graph and see where the function is +ve and -ve
So, you get the interval for which f(x)>0 is x=(-2/3, 4)
106Q2. lx2+3xl = x2+3x, x = R-(-3,0)
= -(x2+3x), x=[-3,0]
Let f(x) = lx2+3xl +x2-2
= x2+3x+x2-2 , x=R-(-3,0)
= -(x2+3x) +x2-2 , x=[-3,0]
So f(x) = 2x2+3x-2 x=(-∞,-3]U[0,∞)
= -3x-2 x=[-3,0]
Drawing the graph u can see that the function ≥0 when (-∞,-2/3]U[1/2,∞)
11Number 3 is a sitter, just take cases when x is positive and negative....