IPMAT/Algebra•6mos
missmatched

Algebra Doubt

What if we put 0? It is gonna be not defined.

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Comments (8)

S
scraggy_46
242/360 (IPMAT 2025)
•6mos

We are given:

x / |x| < x

First note: x = 0 is not allowed because x / |x| becomes 0/0, which is undefined. So consider only x ≠ 0.

Now split into cases:

Case 1: x > 0

Then |x| = x

So, x / |x| = 1

Inequality becomes:

1 < x

So, x > 1

Case 2: x < 0

Then |x| = −x

So, x / |x| = −1

Inequality becomes:

−1 < x

But since x < 0 already, we get:

−1 < x < 0

So the complete solution set is:

(−1, 0) ∪ (1, ∞)

Now check which option is true for every value in this set.

For all x in (−1, 0) and (1, ∞), we have:

x > −1

Hence, the correct answer is:

x > −1

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R
roxanne_67
•6mos

yea,but it means x also satisfies the interval (0,1). Which isn't true.

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S
scraggy_46
242/360 (IPMAT 2025)
•6mos

no, it does not include (0, 1) at all

that’s the key point

from the solving:

  • when x > 0 → you get x > 1, not (0,1)
  • when x < 0 → you get (-1, 0)

so the final set is only:

(-1, 0) ∪ (1, ∞)

now coming to your doubt

the question is not asking what values satisfy the inequality

it is asking: what is true for every value in the solution set

so we check options:

x > 1 → fails because (-1,0) is included

|x| < 1 → fails because (1, ∞) is included

|x|² > 1 → fails because (-1,0) is included

but x > -1 → works for both intervals

  • all values in (-1,0) are > -1
  • all values in (1,∞) are also > -1

so even though the solution is split, x > -1 is the only condition that holds for the entire solution set

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R
roxanne_67
•6mos

It feels illegal to do this. but ok I understand

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X
xyzzzzzzz
•6mos

ig we will first cancel out the variable num and den

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R
roxanne_67
•6mos

shouldn't the question be, which of the following must not be true.

...??

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M
missmatched
OP•6mos

I thought the same

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R
roxanne_67
•6mos

Yea

1