CAT — Modulus

Step 1: Split the expression based on the absolute value. Given, x² − |x + 9| + x > 0. The absolute value changes at x = −9. So, consider two cases.

Case 1: x ≥ −9. Here, |x + 9| = x + 9. Substituting: x² − (x + 9) + x > 0 → x² − 9 > 0 → (x − 3)(x + 3) > 0. This is positive when x < −3 or x > 3. Since this case requires x ≥ −9, the solution becomes: −9 ≤ x < −3 or x > 3, i.e., [−9, −3) ∪ (3, ∞).

Case 2: x < −9. Here, |x + 9| = −(x + 9). Substituting: x² − [−(x + 9)] + x > 0 → x² + x + 9 + x > 0 → x² + 2x + 9 > 0. Completing the square: (x + 1)² + 8 > 0. This expression is always positive for every real value of x. Since this case requires x < −9, all values x < −9 satisfy the inequality. So the solution is: (−∞, −9).

Step 3: Combine both cases. (−∞, −9) ∪ [−9, −3) ∪ (3, ∞). Since x = −9 satisfies the inequality, (−∞, −9) ∪ [−9, −3) = (−∞, −3). Therefore, the complete solution set is (−∞, −3) ∪ (3, ∞).

Given, 2|x|(x² + 1) = 5x²
Since the equation contains |x|, consider two cases.
Case 1: x ≥ 0
Then, |x| = x
So, 2x(x² + 1) = 5x² → 2x³ + 2x = 5x² → 2x³ − 5x² + 2x = 0
x(2x² − 5x + 2) = 0
Factorize, x(2x − 1)(x − 2) = 0
Possible solutions are, x = 0, 1/2, 2
Among these, the integer solutions are, x = 0, 2
Case 2: x < 0
Then, |x| = −x
So, −2x(x² + 1) = 5x² → 2x³ + 5x² + 2x = 0
x(2x² + 5x + 2) = 0
Factorize, x(2x + 1)(x + 2) = 0
Possible solutions are, x = 0, −1/2, −2
Since x < 0, discard x = 0.
The only integer solution is, x = −2
Therefore, the integer solutions are, x = −2, 0, 2
Hence, the number of integer solutions is 3
Answer: 3

Given:
max{x, 2} − min{x, 2} = |x + 2| − |x − 2|
Find the number of distinct real values of x.

Step 1: Simplify the left-hand side
For any two numbers,
max(a, b) − min(a, b) = |a − b|
Hence,
max{x, 2} − min{x, 2}
= |x − 2|
The equation becomes
|x − 2| = |x + 2| − |x − 2|
or
2|x − 2| = |x + 2|

Step 2: Consider different intervals
The critical points are
x = −2 and x = 2.
Case 1: x ≥ 2
|x − 2| = x − 2
|x + 2| = x + 2
So,
2(x − 2) = x + 2
x = 6
This is valid.
Case 2: −2 ≤ x < 2
|x − 2| = 2 − x
|x + 2| = x + 2
So,
2(2 − x) = x + 2
4 − 2x = x + 2
3x = 2
x = 2/3
This is valid.
Case 3: x < −2
|x − 2| = 2 − x
|x + 2| = −x − 2
So,
2(2 − x) = −x − 2
4 − 2x = −x − 2
x = 6
This does not satisfy x < −2.
Hence, no solution in this interval.

Step 3: Count the solutions
The distinct real solutions are
x = 2/3
and
x = 6
Hence, the number of distinct real values is
2

Answer: 2

Given:
|x + y| + |x − y| = 2
Find the number of distinct integer solutions (x, y).

Step 1: Use a standard identity
For any real numbers x and y,
|x + y| + |x − y| = 2 max(|x|, |y|)
Hence,
2 max(|x|, |y|) = 2
So,
max(|x|, |y|) = 1

Step 2: Find all integer pairs
Since
max(|x|, |y|) = 1,
each of x and y must belong to
{−1, 0, 1},
and at least one of them must have absolute value 1.
The valid pairs are
(-1, -1)
(-1, 0)
(-1, 1)
(0, -1)
(0, 1)
(1, -1)
(1, 0)
(1, 1)
There are
8
such pairs.

Final Answer
8

Step 1: Simplify for 0 ≤ x ≤ 5 Since |x − 5| = 5 − x in this region: |x − y| = 2 + (5 − x) = 7 − x Step 2: Split into two cases Case 1: x − y = 7 − x → y = 2x − 7 Case 2: x − y = −(7 − x) → y = 7 Bounding lines: y = 7 and y = 2x − 7 Step 3: Find the vertices At x = 0: y = 7 and y = −7 → (0, 7) and (0, −7) At x = 5: y = 7 and y = 3 → (5, 7) and (5, 3) The figure is a trapezium with parallel sides 14 and 4, distance = 5. Step 4: Find the area Area = (1/2) × (14 + 4) × 5 = (1/2) × 18 × 5 = 45

Given:
|x| + x + y = 15
x + |y| − y = 20
Find the value of
x − y.

Step 1: Analyze the first equation
We know,
● If x ≥ 0, then |x| = x.
● If x < 0, then |x| = −x.
Case 1: x ≥ 0
Then,
2x + y = 15
Case 2: x < 0
Then,
y = 15

Step 2: Analyze the second equation
We know,
● If y ≥ 0, then |y| = y.
● If y < 0, then |y| = −y.
Case 1: y ≥ 0
Then,
x = 20
Case 2: y < 0
Then,
x − 2y = 20

Step 3: Check possible cases
Case 1: x ≥ 0 and y ≥ 0
From the equations,
x = 20
2x + y = 15
40 + y = 15
y = −25
Contradiction since y ≥ 0.
Not possible.
Case 2: x ≥ 0 and y < 0
From the equations,
2x + y = 15
x − 2y = 20
Multiply the first equation by 2,
4x + 2y = 30
Add the second equation,
5x = 50
x = 10
Substituting,
20 + y = 15
y = −5
This satisfies the conditions.
Case 3: x < 0 and y ≥ 0
From the equations,
y = 15
x = 20
Contradiction since x < 0.
Not possible.
Case 4: x < 0 and y < 0
From the equations,
y = 15
Contradiction since y < 0.
Not possible.

Step 4: Find the required value
x − y
= 10 − (−5)
= 15

Answer:
B. 15