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17 Mar 2025, 08:10
Kudos
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This is actually simple. We know x>0, which means it is positive and we know x<1.
Which means we can multiply x on both sides of inequality (because multiplying with positive number on both sides of inequalities won't change sign of an inequality) using which we can get multiple inequalities:
1. x^2<x
2. x^3<x^2 (again multiplying the first inequality with x on both sides)
3. x^4<x^3 (again multiplying the first inequality with x on both sides)
4. x^5<x^4 (again multiplying the first inequality with x on both sides)
if x^3 > x^4 and x^4 > x^5 (from above inequalities), we can say that x^3>x^5 which checks the first option.
By the above logic we will get x^5<x^3 (I) and x^4<x^2(II). And since the inequalities signs are same, we can add and subtract the inequalities.
I+II gets us the second statement as true and II-I gets us the third statement as true.
I hope the above explanation helps. Please let me know if there is any flaw in the solution.
Which means we can multiply x on both sides of inequality (because multiplying with positive number on both sides of inequalities won't change sign of an inequality) using which we can get multiple inequalities:
1. x^2<x
2. x^3<x^2 (again multiplying the first inequality with x on both sides)
3. x^4<x^3 (again multiplying the first inequality with x on both sides)
4. x^5<x^4 (again multiplying the first inequality with x on both sides)
if x^3 > x^4 and x^4 > x^5 (from above inequalities), we can say that x^3>x^5 which checks the first option.
By the above logic we will get x^5<x^3 (I) and x^4<x^2(II). And since the inequalities signs are same, we can add and subtract the inequalities.
I+II gets us the second statement as true and II-I gets us the third statement as true.
I hope the above explanation helps. Please let me know if there is any flaw in the solution.
KshitijMittal
18 Mar 2026, 16:53
Kudos
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The big rule to remember here: when 0 < x < 1, higher powers of x get SMALLER. Think of x = 0.5: 0.5 > 0.25 > 0.125 > 0.0625... So the order is x > x2 > x3 > x4 > x5.
Statement I: x5 < x3
Since higher powers are smaller, x5 is definitely less than x3. TRUE.
Statement II: x4 + x5 < x3 + x2
Factor both sides: Left = x4(1 + x), Right = x2(1 + x).
Since (1 + x) is positive, we can divide both sides by it, leaving us with: x4 < x2.
Again, higher power = smaller number when 0 < x < 1. TRUE.
Statement III: x4 - x5 < x2 - x3
Factor both sides: Left = x4(1 - x), Right = x2(1 - x).
Since (1 - x) is positive (because x < 1), divide both sides by it, leaving: x4 < x2.
Same logic — TRUE.
Note that all 3 statements boil down to the same core fact: x4 < x2 (or equivalently x5 < x3), which is always true when 0 < x < 1.
The most common mistake is getting confused by the subtraction in Statement III and thinking it might flip the inequality. But factoring reveals it's the exact same comparison as Statement II.
The answer is E — all 3 must be true.
General principle: For 0 < x < 1, raising x to a higher power always makes it smaller. This means x^a < x^b whenever a > b.
Answer: E
Statement I: x5 < x3
Since higher powers are smaller, x5 is definitely less than x3. TRUE.
Statement II: x4 + x5 < x3 + x2
Factor both sides: Left = x4(1 + x), Right = x2(1 + x).
Since (1 + x) is positive, we can divide both sides by it, leaving us with: x4 < x2.
Again, higher power = smaller number when 0 < x < 1. TRUE.
Statement III: x4 - x5 < x2 - x3
Factor both sides: Left = x4(1 - x), Right = x2(1 - x).
Since (1 - x) is positive (because x < 1), divide both sides by it, leaving: x4 < x2.
Same logic — TRUE.
Note that all 3 statements boil down to the same core fact: x4 < x2 (or equivalently x5 < x3), which is always true when 0 < x < 1.
The most common mistake is getting confused by the subtraction in Statement III and thinking it might flip the inequality. But factoring reveals it's the exact same comparison as Statement II.
The answer is E — all 3 must be true.
General principle: For 0 < x < 1, raising x to a higher power always makes it smaller. This means x^a < x^b whenever a > b.
Answer: E
25 Apr 2026, 11:59
Kudos
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can i take factor variables and cancel on both sides as i know x will not be zero and negative , thus all 3 eq gives same eq when simplified , and x will be between 0 and 1 .
am i right ???
am i right ???
Bunuel
