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Re: Which of the following inequalities have at least two soluti [#permalink]
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Expert Reply
based on which assumption do you think they are wrong ??

Because when you say such statement you have to explain and demonstrate your reasons. Is not enough saying "I think they are wrong"

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Re: Which of the following inequalities have at least two soluti [#permalink]
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Chill bro
The answers are A B C
Option A:(3/7)x < x
Take x=0.1
0.04<0.1
Take x=0.2
0.071<0.2
A is correct
Option B: x<\(x^2\)
Take x=-0.1
-0.1<0.01
Take x=-0.2
-0.2<0.04
B is correct
Option C: x-(1/2) < x+(2/3)
Take x=0.5
0<1.1666
Take x=0.6
0.1<1.266
C is correct
Means all the inequalities have at least two solutions in between-1 and 1
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Re: Which of the following inequalities have at least two soluti [#permalink]
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So Sir

If as I can see all the answer are correct, what should I need to change in the above question?

I do not understand ☹️

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Re: Which of the following inequalities have at least two soluti [#permalink]
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At starting I got confused and after your reply I solved and got the answer

Posted from my mobile device Image
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Re: Which of the following inequalities have at least two soluti [#permalink]
This is a tricky one...I evaluated 1/2 and -1/2 and only selected B and C.

Important lesson...read carefully. "at least two solutions" not "at least one positive and one negative."
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Re: Which of the following inequalities have at least two soluti [#permalink]
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A. implies- x>0 => solutions between -1 and 1 is possible

B. implies- x(x-1)>0 => the soulutions are x>1 or x<0 => given x<0, solutions between -1 and 1 is possible

C. the inequality is satisfied for all the values of x => solutions between -1 and 1 is possible

Ans: all options are valied
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Which of the following inequalities have at least two soluti [#permalink]
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This question can be solved by understanding number properties for the numbers between -1 and 1.

A. For the infinity of numbers greater than 0 and up to 1,
(3/7)x < x will be true so at least two solutions satisfy this condition. This is because a positive fraction of a positive number will always be less than the positive number.

B. For the infinity of numbers greater than -1 and less than 0,
x < x^2 will be true so at least two solutions satisfy this condition. This is because numbers greater than -1 and less than zero (negative proper fractions) are get larger as the power is raised but not as large as zero.

C. For the all real numbers including all the numbers between than zero -1 and 1, subtracting a positive number from it will make it smaller and adding a positive number will make it larger.

So the left hand side of the inequality x - 1/2 < x + 2/3 will always be less than the right hand side of the inequality, hence it satisfies the condition that there are at least two solutions to the inequality as well.
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