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Inequality questions

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Manager
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Inequality questions [#permalink] New post 11 Feb 2008, 04:52
Hi Guys,

I am studying topic Inequalities and have been facing few problems. I will be glad if anyone can tell me what fundamental mistake am I making.

Q1. What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

Correct ans is (d)
I am getting (a)

what am I doing wrong ?

Q2. Solve x^2 + 8x + 7 < 0

OA is -7< x < -1

When I try to solve I do -
(x+1) (x+7) < 0
x + 1 < 0
x < -1
And
x + 7 < 0
x < -7.

something wrong??

Similarly -

Q3 Solution of inequality -x^2 + 5x - 6 < 0

ans???

Please help me out here...

Thanks,

Cheers,
JG
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 06:31
greatchap wrote:
Hi Guys,

I am studying topic Inequalities and have been facing few problems. I will be glad if anyone can tell me what fundamental mistake am I making.

Q1. What is the solution set of inequality 4 / (x-2) < 2 ?

(a) x > 4
(b) x < 4
(c) x < 4 union x > 4
(d) x < 2 union x > 4
(e) infinity

Correct ans is (d)
I am getting (a)

what am I doing wrong ?

Q2. Solve x^2 + 8x + 7 < 0

OA is -7< x < -1

When I try to solve I do -
(x+1) (x+7) < 0
x + 1 < 0
x < -1
And
x + 7 < 0
x < -7.

something wrong??

Similarly -

Q3 Solution of inequality -x^2 + 5x - 6 < 0

ans???

Please help me out here...

Thanks,

Cheers,
JG


here is what's important: solution of (x-a)*(x-b) > 0 (where a < b) is {x < a} U {x > b} , consequently solution of < 0 is a < x < b.

let's look at the first problem 2/(x-2) < 1 //multiply both sides by [(x-2)^2]/2 - positive number - inequality sign doesn't change

2*(x-2) < x^2-4*x + 4 -> x^2-6*x + 8 > 0 -> a = 2, b = 4
solution: { x < 2} U {x > 4} -> D

Second problem:

x^2 + 8*x + 7 < 0 -> a = -7, b = -1 -> -7 < x < -1

Third problem:
-x^2 + 5*x - 6 < 0
x^2 - 5*x + 6 > 0
a = 2, b = 3 -> {x < 2} U {x > 3}
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 07:20
Thanks for the solutions maratikus.

I understood the second and the third problem.

In the first one (4/x-2 < 2) what is keeping me confused is why cant we straightaway cross multiply and get answer. ?
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 07:44
greatchap wrote:
Thanks for the solutions maratikus.

I understood the second and the third problem.

In the first one (4/x-2 < 2) what is keeping me confused is why cant we straightaway cross multiply and get answer. ?


When you solve a problem, you can replace it with equivalent problems and solve those. However, if you multiply both sides by x-2 the new problem is not going to be equivalent to the initial one because there are going to be two situations:

4 < 2*(x-2) if x-2 > 0
4 > 2*(x-2) if x-2 < 0

because when you multiply both sides by the same positive number, inequality sign doesn't change, and the sign changes when you multiply both sides by the same negative number (for example if (-2) < (-1) and you multiply both sides by (-1) it would be incorrect to say: (-1)*(-2) < (-1)*(-1) or 2 < 1)

In order to avoid this problem, I mutliply both sides by a positive number (x-2)^2 and keep the inequality sign unchanged. I hope that helps.
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 07:52
greatchap wrote:
Thanks for the solutions maratikus.

I understood the second and the third problem.

In the first one (4/x-2 < 2) what is keeping me confused is why cant we straightaway cross multiply and get answer. ?


because x is a variable and we have an inequality here. u cant cross multiply unles u know the sign of the variable. a negative flips the inequality sign
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 08:55
maratikus wrote:
greatchap wrote:
Thanks for the solutions maratikus.
4 < 2*(x-2) if x-2 > 0
4 > 2*(x-2) if x-2 < 0

because when you multiply both sides by the same positive number, inequality sign doesn't change, and the sign changes when you multiply


here is my question: when you solve both of those inequalities, you get that either x is less than 4, or that x is greater than 4. how do you get to the final answer of x<2 and x>4 ?
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 09:31
I still have a confusion so am writing again...

I came across a ques:

Q-Z Find the value of x in (3x^2 + x) / (x^2 + 2) < 3.

A- here we did a normal cross multiply -
3x^2 + x < 3(x^2 + 2)
3x^2 + x < 3x^2 + 6
x < 6. answer (correct)

so whats the difference in that ques & (solution set of inequality 4/x-2 < 2 ) ??

cant we do -> 4 < 2(x-2)
4 < 2x - 4
8 < 2x
x > 4. (wrong)

If we can't then, why cross multiplication in Q-Z. moreover with what no u multiplied in 4/x-2 < 2. why not x-2 to both sides. ??
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Re: Inequality questions [#permalink] New post 11 Feb 2008, 09:50
i think the difference here is that in Q-Z, you are dealing with x^2, and so those terms will always be positive no matter what.
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Re: Inequality questions [#permalink] New post 13 Feb 2008, 08:51
thanks for the answer guys... :)
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Re: Inequality questions [#permalink] New post 13 Feb 2008, 10:23
Expert's post
\frac{4}{(x-2)} < 2

my logic:

1. it is true for all negative (x-2) or for x<2
2. the function f=\frac{4}{(x-2)} is \infty at x close to 2 at the greater side.
3. When x increases from 2, the function f always decreases and equal to 2 at x=4. Therefore, f<2 at x>4
4. Combine two conditions: x<2 & x>4
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Re: Inequality questions [#permalink] New post 13 Feb 2008, 19:57
walker wrote:
\frac{4}{(x-2)} < 2

my logic:

1. it is true for all negative (x-2) or for x<2


how can you show this algebraically ?

im getting 4 >x....
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Re: Inequality questions [#permalink] New post 13 Feb 2008, 22:50
Expert's post
pmenon wrote:
walker wrote:
\frac{4}{(x-2)} < 2

my logic:

1. it is true for all negative (x-2) or for x<2


how can you show this algebraically ?

im getting 4 >x....


1. the inequality is true when 4/(x-2) is negative.
2. 4/(x-2) is negative when (x-2) is negative
3. (x-2) is negative when x<2 (x-2<0 --> x<2)
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Re: Inequality questions   [#permalink] 13 Feb 2008, 22:50
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