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Please help

by vinni.k » Wed Feb 29, 2012 1:16 am
(1) y > 0
(2) x = y

Answer is C

My answer is B, but it is wrong.

My method:-

y^2 + 3 > y^2 (After plugging y for x)
Now, if I plug any value for y, whether it is positive or negative the y^2 + 3 > y^2 is true.

(-2)^2 + 3 > (-2)^2
4 + 3 > 4
7 > 4

Now putting positive value.
(2)^2 + 3 > (2)^2
4 + 3 > 4
7 > 4

I am not able to understand where is the problem. Even if i take fraction for y, then also i am concluding on the same terms.

Please help

Thanks & Regards
Vinni
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by kmittal82 » Wed Feb 29, 2012 2:57 am
I think this is where you are wrong:
y^2 + 3 > y^2

The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
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by neelgandham » Wed Feb 29, 2012 8:50 am
Is 1/x > y/((y^2)+3)
(1) y > 0
Doesn't speak of y. *&**
(2) x = y
Is 1/x > y/((y^2)+3)
Is 1/y > y/((y^2)+3)
Is 1/y - y/((y^2)+3) > 0
Is (y^2)+3-(y^2)/(y*((y^2)+3)) > 0
Is 3/(y*((y^2)+3)) > 0 ?
(y^2)+3 > 0 , but for y can be > 0 or < 0, So we cannot, for certain, answer the question
From 1 and 2
We know that y > 0
So 3/(y*((y^2)+3)) > 0
Hence sufficient to answer the question.
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by vinni.k » Thu Mar 01, 2012 8:16 am
kmittal82 wrote:I think this is where you are wrong:
y^2 + 3 > y^2

The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative number , then inequality sign changes. Here i am not multiplying a value by negative number. I am just plugging a negative value inside the square. I am not convinced by your explanation.

I liked the neelgandham's method, but I will really appreciate if someone can clear my doubt.

Thanks & Regards
Vinni
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by krusta80 » Thu Mar 01, 2012 8:29 am
vinni.k wrote:
kmittal82 wrote:I think this is where you are wrong:
y^2 + 3 > y^2

The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative number , then inequality sign changes. Here i am not multiplying a value by negative number. I am just plugging a negative value inside the square. I am not convinced by your explanation.

I liked the neelgandham's method, but I will really appreciate if someone can clear my doubt.

Thanks & Regards
Vinni
kmittal82 is referring to when you cross-multiplied. You ALWAYS need to consider the possible signs of the denominators when cross multiplying inequalities, because in essence you are multiplying both sides by the denominators when you cross multiply.

Now, let's look at the denominators...

y^2+3 will always be positive, but...

Since x can be negative or positive in part (2) alone, you need to consider both cases. So, in essence, x is the negative number that kmittal is referring to.
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by kmittal82 » Thu Mar 01, 2012 8:41 am
krusta80 wrote:
vinni.k wrote:
kmittal82 wrote:I think this is where you are wrong:
y^2 + 3 > y^2

The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative number , then inequality sign changes. Here i am not multiplying a value by negative number. I am just plugging a negative value inside the square. I am not convinced by your explanation.

I liked the neelgandham's method, but I will really appreciate if someone can clear my doubt.

Thanks & Regards
Vinni
kmittal82 is referring to when you cross-multiplied. You ALWAYS need to consider the possible signs of the denominators when cross multiplying inequalities, because in essence you are multiplying both sides by the denominators when you cross multiply.

Now, let's look at the denominators...

y^2+3 will always be positive, but...

Since x can be negative or positive in part (2) alone, you need to consider both cases. So, in essence, x is the negative number that kmittal is referring to.
thanks for clarifying on my behalf krutsa, couldn't have explained it better myself :)
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