Ahmed9955 wrote:
Skywalker18 wrote:
Bunuel wrote:
Official Solution:
If \(x\) and \(y\) are positive integers, is \(x\) a prime number?
(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(2-y \le{0}\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).
So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.
(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.
Answer: D
1. \(|x - 2| \lt 2 - y\)
=> x-2 < 2-y
=> x+y<4 -- eqn 1
or -(x-2)< 2-y
=> x>y -- eqn 2
Combining equations 1 and 2, we get
x=2(prime) , y=1 (since x and y are +ve integers)
(2) \(x + y - 3 = |1-y|\)
=> x+y-3 = 1-y
=> x+2y=4 --
eqn 3or x+y-3 = y-1
=> x= 2 -
- eqn 4Hi
Bunueli also solved in the same way, but for St.2, i got stuck after obtaining eqn3 & eqn 4 as highlighted above.
-for eqn-4; we get x=2
-for eqn-3 ; as x & y are positive integers. The minimum value for y could be 1 and x has to be 2. in any other case the eqn- x+2y=4 will not satisfy.
Is my approach right for St 2?
Also,
When we obtain two values after considering x>=0 and x<0
- Do i have to combine and check for a unique value?
OR
-Do i have to obtain a unique value for each "x>=0 and x<0" case?
1. Yes, you can solve (2) the way you did.
2. When considering cases for modulus questions.
Say we have |x - 3| = 4.
When x <= 3, then we'd have -(x - 3) = 4 -->
x = -1. Since x = -1 is in the range we consider (x <= 3), then this is a valid root. So, you can check already at this stage.
When x > 3, then we'd have x - 3 = 4 -->
x = 7. Since x = 7 is in the range we consider (x > 3), then this is a valid root. So, you can check already at this stage.
So, |x - 3| = 4 has two roots x = -1 and x = 7.
Another example, x^2 - 4x + 6 = 3 - |x - 1|.
If \(x<1\), then \(|x - 1| = -(x-1)=1-x\), so in this case we'll have \(x^2 - 4x + 6 = 3-(1-x)\) --> \(x^2-5x+4=0\) --> \(x=1\) or \(x=4\) -->
discard both solutions since neither is in the range \(x<1\).If \(x\geq{1}\), then \(|x - 1| = x-1\), so in this case we'll have \(x^2 - 4x + 6 = 3-(x-1)\) --> \(x^2-3x+2=0\) --> \(x=1\) or \(x=2\). Since both \(x=1\) and \(x=2\) are in the range we consider \(x\geq{1}\), then both are valid.
So, x^2 - 4x + 6 = 3 - |x - 1| has two roots: \(x=1\) and \(x=2\).
Hope it helsp.
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