pappueshwar wrote:Is |a^2 -b^2|<6?
1) |a+b|<2
2) |a-b|<3
please xplain how to solve? i am heavily getting confused when solving modulus problems
Step 1: Can the main statement be proven independently of parts (1) and (2)?
No, since there are no restrictions on a and b. ie. a = b = 0 works and a = 3 | b = 0 does NOT.
Step 2: Let's look at parts (1) and (2)
Always keep your eye out for common themes. In this case, a^2 - b^2 = (a-b)(a+b)
|(a-b)(a+b)| < 6
(1) |a+b| < 2
If you picture a number line, |a+b| < 2 allows for two possibilities:
i.) |a|+|b| < 2
ii.) The distance between a and 0 is within 2 of the distance between b and 0 and the two have opposite signs, assuming |a|+|b| => 2
Since the second possiblity allows for (a-b) to be largely positive (if a is largely positive and b is largely negative for example), |(a-b)(a+b)| can certainly be much greater than 6, but since the first possibility allows for (a-b) to be very small (a and b are both very close to 0), |(a-b)(a+b)| can be less than 6. INSUFFICIENT
(2) |a-b| < 3
Again, picturing a number line, |a-b| represents the distance between a and b. Since a and b can be moved in unison to any area of the number line, clearly this allows for (a+b) to be as large as we want (or as negatively small)...or of course it can be 0 if we want. INSUFFICIENT
(1) and (2)
Since (a-b)(a+b) is within an absolute value sign, we can ignore negatives and simply try to make a+b and a-b each as big as possible. We already know that setting a = b = 0 works for all constraints, so let's try to violate the main statement.
|a+b| < 2, which can be rewritten as
-2 < a+b < 2
So, a+b cannot exceed 2.
|a-b| < 3, which can be rewritten as
-3 < a-b < 3
So a-b cannot exceed 3.
Therefore, |(a-b)(a+b)| < 6 is ALWAYS true.
Answer is C.
