Hi Shlok02,Your algebra is perfect. You correctly cracked |x+3| > 2 into its two ranges:
x > -1 OR
x < -5. The only thing that's flipped is the
direction you're checking.
Here's the key idea. Statement II is satisfied by a
big set of values - everything with x >
-1and everything with x <
-5. The word "OR" means you only need to land in
one of those ranges for |x+3| > 2 to hold.
Now look at what the question actually hands you: x <
-5. Every value you could possibly have (
-5.1,
-6,
-100, ...) sits
inside one of statement II's ranges - the x <
-5 piece. So for
every allowed value of x, statement II is true. That's exactly what "must be true" means.
The x >
-1 range isn't a problem. It just means II happens to be true for some extra values too - values our x never takes. Extra coverage never hurts you. You don't need our range to match
both of II's pieces; you only need our range to fall
within II's true set. And it does.
So the test is one-directional: Is (given range) a subset of (statement's true range)? Here, x <
-5 is fully inside x <
-5 OR x >
-1. Yes - must be true.
A cleaner-to-feel version:Given: x <
-5. Must x <
0 be true?
x <
0 is true for tons of values you'll never have (like
-2,
-0.5), but every single x <
-5 value is also below
0. So yes - it
must be true. The extra values x <
0 allows are irrelevant; what matters is that your whole given range lives inside it.
Same logic drives statement II. Your ranges were right - just check containment, not matching.
Answer: DShlok02
I had a query regarding Statement 2
We get 2 conditions- x>-1 and x<-5
since the second range satisfies the inequality given, but the first one does not, how can we say that (2) MUST BE TRUE??