Test Level 2: Inequalities - 1 - CAT MCQ

x^2/3 + x^1/3 - 2 ≤ 0
Or (x^1/3 - 1)(x^1/3 + 2) ≤ 0
Or -8 ≤ x ≤ 1

f(0) = p and f(1) = p - 3
Since f(0) and f(1) have opposite signs, so if p < 0, then p - 3 > 0, i.e. p > 3, which is not possible.
Or, p > 0 and p - 3 < 0
⇒ p < 3
⇒ 0 < p < 3
Hence, answer option 2 is correct.

As (a - b)² is never negative.
a² + b² - 2ab is greater than or equal to zero.
Therefore, we can say that a² + b² ≥ 2ab.

Since AM ≥ GM,

x + y + z ≥ 6
To get minimum value, take x + y + z = 6.

⇒ 2^x + 2^y + 2^z ≥ 12
Minimum value = 12

(x, y) pairs satisfying the condition are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2) and (4, 1).
So, total 8 pairs are there.

|x| ≤ 2 ⇒ minimum value of x = - 2.
|y + 3| ≤ 5 ⇒ minimum value of y = - 8.
So, x + y = - 2 - 8 = - 10.

Given that

For any positive n,

Again, from inequality (ii), 

So, from inequalities (i) , (iii) and (iv), we get 0 < x ≤ 4

|x - 4| < 2
-2 < x - 4 < 2
2 < x < 6
x = (2, 4)  (4, 6)

The common part is x ≥ 0, y ≥ 0, 2x + 3y ≤ 12.

Total = 19

So, we can't find the value of N 

So, the answer can't be determined.