Test Level 1: Inequalities - 1 CAT MCQs & solutions - Free

The absolute value of any number N is always non-negative.
|N| = N, when N > 0 and |N| = -N, when N < 0.
Since N is a negative number, therefore |N| = N is not true.

|x² - 4| = 4 + x²
Case I: If x² - 4 ≥ 0
i.e. x ≤ -2 and x ≥ 2
Then, |x² - 4| = x² - 4
⇒ The given equation becomes x² - 4 = 4 + x².
⇒ -4 = 4; which is not possible.
Case II: If x² - 4 < 0
⇒ -2 < x < 2
Then, |x² - 4| = - (x² - 4)
Thus, the given equation becomes - (x² - 4) = 4 + x².
Or 2x² = 0
⇒ x = 0
Hence, the given equation has only one solution.

Take x = 2, then option (1), (2) will become true and hence those are not the answers.
Take x = – 2, option (4) will be true and hence not the answer.
But for any value of x, 2x > x.
So option (3) is not true at all. It is always false.

w - z > 0
x - z < 0; this means z - x > 0
y - z < 0; this means z - y > 0
The product of cubes of three positive numbers will also be positive.

x² - (a + b) x + ab < 0
(x - a) (x - b) < 0
a < x < b

3m² - 21m + 30 < 0
m² - 7m + 10 < 0
⇒ (m - 5)(m - 2) < 0
⇒ 2 < m < 5

x² - 5x + 6 ≤ 0
⇒ (x - 3)(x - 2) ≤ 0
⇒ x ∈ [2, 3] or 2 ≤ x ≤ 3
Therefore, option 4 is the correct answer.

|x| is always positive.
If x is positive, x – |x| = 0.
If x is negative, x – |x| < 0.

(i) 5x + 2 > 3x - 1 or 2x > -3

(ii) 3x + 1 > 7x - 4

At x = 0, inequality is not satisfied. Thus, option (c) is rejected. Also x = 0 is not a solution of the equation. Since, this is a continuous function, the solution cannot start from 0. Thus options (a) and (b) are not right. Further, we see that the given function is quadratic with real roots. Hence, option (d) is also rejected.