Modulus doubt

just hit and trial acc to interval

Yes, you can solve this by checking options one by one, but you must pay attention to the wording of the question. It asks which statement must be true for every value that satisfies the inequality. So the correct option has to hold for all solutions, not just some of them.

First, understand the expression x/|x|. This expression only depends on the sign of x. It equals 1 when x is positive and −1 when x is negative. It is not defined at x = 0, so zero cannot be a solution.

Now solve the inequality x/|x| < x by cases.

If x > 0, then x/|x| = 1.

So the inequality becomes 1 < x, which gives x > 1.

If x < 0, then x/|x| = −1.

So the inequality becomes −1 < x, which gives −1 < x < 0.

Combining both cases, the full solution set is:

−1 < x < 0, or x > 1.

So every solution is greater than −1, and x ≠ 0.

Now check the options using values that satisfy the inequality, such as x = −0.5 and x = 2.

Option: x > 1

Fails, because −0.5 satisfies the inequality but is not greater than 1.

Option: x > −1

Works, because both −0.5 and 2 are greater than −1, and in fact every solution is greater than −1.

Option: |x| < 1

Fails, because 2 satisfies the inequality but |2| is not less than 1.

Option: |x| > 1

Fails, because −0.5 satisfies the inequality but |−0.5| is not greater than 1.

Therefore, the only statement that must be true for every solution is:

x > −1.