Inequality questions

Question

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) 4
(b) x 4
(d) 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

maratikus · Answer

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 /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}

greatchap · Answer

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. ?

maratikus · 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.

bmwhype2 · 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

pmenon · Answer

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 ?

greatchap · Answer

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. ??

pmenon · Answer

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.

greatchap · Answer

thanks for the answer guys...

walker · Answer

\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

pmenon · Answer

how can you show this algebraically ?

im getting 4 >x....

walker · Answer

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 x<2)

Inequality questions

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