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Bunuel
Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?
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Hi
sinhap07,
Please find the algebraic approach for the question
I. \(x^5 <x^3\)\(x^3(x^2 - 1) < 0\) i.e. \(x^3 (x -1) (x +1) < 0.\)
Using the wavy line method we know that for \(0 < x < 1\) the
inequality holds true.II.\(x^4 + x^5 < x^3 + x^2\)\(x^5 - x^3 + x^4 - x^2 < 0\) i.e. \(x^2(x -1) (x + 1)^2 < 0\)
Using wavy line method we know that for \(0 < x < 1\)
the inequality holds trueIII.\(x^4 - x^5 < x^2 - x^3\)\(x^4 - x^2 - x^5 + x^3 < 0\) i.e. \(x^2 (x + 1)(x - 1) (1 -x) < 0\)
\(x^2 (x -1)^2(x + 1) > 0\)
Using wavy line method we know that for \(0 < x < 1\)
the inequality holds true Thus all the three inequalities are true for the range \(0 < x < 1\)
Hope it's clear
Regards
Harsh
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Thanks Harsh for your reply. Can you please throw some more light on the wave method? Of what I knw, signs usually alternate but what I see in your workings, signs don't always alternate. Can you explain it?
Second, why are we ignoring the negative portions say in statement 1 that makes it less than -1. It shows x values dont only pertain to 0 and 1 region. So how can stmt 1 be sufficient?
Hi
sinhap07,
You are right when you say signs usually alternate around zero points of the inequality but they
only alternate when the expression is sign dependent i.e. for odd powers. For example sign of \(x^3\) is dependent on the sign of \(x\) but sign of \(x^2\) is independent of sign of \(x\). Hence for expressions, signs of which are independent of the signs of their base variable, the wave would not alternate but rather bounce back to the same region it is currently in.
I would also suggest you to go through this post on
Wavy Line method which will help you understand it better.
For your second query, in st-I the range of \(x\) for which the inequality holds is \(0 < x < 1\) or \(x < -1\) as seen from the wavy line diagram. However as the question asks us if the inequality is true for the range \(0 < x < 1\), we are not concerned about the other possible values of x for which the inequality is true. Once the inequality satisfies the range \(0 < x < 1\), we have our answer.
Hope it's clear
Regards
Harsh
Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?
I. x3 < x
II. x2 < |x|
III. x4 – x5 > x3 – x2