Inequalities CAT Previous Year Questions with Answer PDF

Let the total amount be equal to T.
T = n × 352
“However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330”

So, the maximum value that n can take is 16.

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

Possible integers = -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
If we see the second expression that is provided, i.e 
3 + 3^n+1, it can be implied that n should be at least -1 for this expression to be an integer.
So, n = -1, 0, 1, 2, 3, 4, 5.
Hence, there are a total of 7 values.

Let us break this up into 2 inequations [ Let us assume x as m and y as n ]
| 1 + mn | < | m + n |
| m + n | < 5
Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.
Let us try out with the first quadrant and extend the results to the other quadrants.
We will also consider the +X and +Y axes along with the quadrant.
So, the first inequality becomes,
1 + mn < m + n
1 + mn - m - n < 0
1 - m + mn - n < 0
(1-m) + n(m-1) < 0
(1-m)(1-n) < 0
(m - 1)(n - 1) < 0
Let us try to plot the graph.
If we consider only mn < 0, then we get 

But, we have (m - 1)(n - 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.
So, we have,

But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.

So, if we look for only integer values, we get
(0,2), (0,3),.......
(0,-2), (0, -3),......
(2,0), (3,0), ......
(-2,0), (-3,0), .......
Now, let us consider the other inequation as well, in which |x + y| < 5
Since one of the values is always zero, the modulus of the other value is less than or equal to 4.
Hence, we get 
(0,2), (0,3), (0,4)
(0,-2), (0, -3), (0, -4)
(2,0), (3,0), (4,0)
(-2,0), (-3,0), (-4,0)
Hence, a total of 12 values.

Given, n³ – 11n² + 32n – 28 > 0
⇒ (n - 2)(n2 – 9n + 14) > 0
⇒ (n - 2)(n - 7)(n - 2) > 0
⇒ For n < 2, (n – 2)(n – 7)(n – 2) is negative.
⇒ For 2 < n < 7, (n – 2)(n – 7)(n – 2) is negative.
⇒ For n > 7, (n – 2)(n – 7)(n – 2) is positive.
When n = 8, (n – 2)(n – 7)(n – 2) = 36, which is greater than 0.
The least integral value of n which satisfies the inequation is 8.

When any quadratic equation is greater than 0 then D(Discriminant) will be less than 0.
Calculation:
⇒ 2x² – ax + 2 > 0; Where x belongs to R
As, 2x² – ax + 2 > 0
∴ D < 0
⇒ b2 – 4ac < 0
⇒ a² – 4 × 2 × 2 < 0
⇒ a² < 16 -4 < a < 4
Now, x² − bx + 8 ≥ 0; Where x belongs to R
∴ D ≤ 0 ⇒ b² – 4ac ≤ 0
⇒ b² – 4 × 1 × 8 ≤ 0
⇒ b² ≤ 32 -4√2 ≤ b ≤ 4√2
As b is an integer, -5 ≤ b ≤ 5
We have been asked about the maximum possible value of 2a – 6b
Therefore, We will take a = 3 and b = -5 to get the maximum value.
Hence, maximum possible value of 2a – 6b = 2(3) – 6(–5) = 36

⇒ (n - 5) (n - 10) - 3 × (n - 2) ≤ 0
⇒ n² - 15n + 50 - 3n + 6 ≤ 0
⇒ n² - 18n + 56 ≤ 0
⇒ (n - 14) (n - 4) ≤ 0
⇒ 4 ≤ n ≤ 14, n = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
⇒ Total integer is 11
∴ The required result will be 11.