Coded Inequalities -2

Question 13: For how many integer values does the following inequality hold good? (x + 2) (x + 4) (x + 6)........(x + 100) < 0? 

A. 25
B. 50
C. 49
D. 47

Answer. 25

Explanation.

(x + 2) (x + 4) (x + 6) ........(x + 100) < 0
Now, the above expression will be zero for x = -2, -4, -6, - 8.....-100.
For x > - 2, all the terms will be positive and so, the product will be positive.
For x < - 100, all the terms will be negative, and since there are 50 terms (even number), the product will be positive.
Now, if x = - 99, the term x + 100 would be positive, everything else would be negative, so the expression would have 49 negative terms and one positive term. So the product would be negative.
Overall, the expression will be negative if there are exactly 49 negative terms, or exactly 47 negative terms, or exactly 45 terms.... Or so on, up to exactly one negative term.
Exactly 49 negative terms= x = - 99
Exactly 47 negative terms= x= - 95
Exactly 45 negative terms= x = - 91
...............
Exactly 1 negative term=. x = - 3
So, x can take values {-3, -7, -11, -15, -19.... -99}. We need to compute the number of terms in this list.
In other words, how many terms are there in the list {3, 7, 11, ....99}. Now, these terms are separated by 4, so we can write each term as a multiple of 4 + 'some constant'.
Or 
3 = 0 * 4 + 3
7 = 1 * 4 + 3
11 = 2 * 4 + 3
......................
99=24 * 4 + 3
We go from 0 * 4 + 3 to 24 * 4 + 3, a total of 25 terms. 
Hence the answer is "25"
Choice A is the correct answer.

Question 14: If a, b, c are integers such that - 50 < a, b, c < 50 and a + b + c = 30, what is the maximum possible value of abc? 

Answer. 4410

Explanation.
To maximise abc, we prefer a positive product with large magnitudes.
If all three are non-negative, an even distribution maximises the product under a fixed sum, so 10, 10, 10 gives a product of 1000.
Consider two negatives and one positive; the product becomes positive and can be larger if the positive number is as large as possible in magnitude and the two negatives have large magnitudes (within given bounds).
The variables must satisfy -49 ≤ each ≤ 49 (since -50 < value < 50).
Take the positive value as 49; then the other two must sum to -19. The best choice is -9 and -10 (close in magnitude and within bounds).
Product = 49 × (-9) × (-10) = 49 × 90 = 4410.
Hence, the answer is "4410"
Choice C is the correct answer.

Question 15: Solve x² - |x + 3| + x > 0? 

A. x ∊ (-∞,-1] ∪ [√3, 3)
B. x ∊ (-∞,-3] ∪ [√3, ∞)
C. x ∊ (-4,-3) ∪ (4, ∞)
D. x ∊ (-8,-3] ∪ [2, ∞)

Answer. x ∊ (-∞,-√3) ∪ (√3, ∞)

Explanation.

x² - |x + 3| + x > 0
If x + 3 > 0 ⇒ x > -3
Then equation is in the form x² - x - 3+ x > 0 i.e., x² -3 > 0
x² > 3 ⇒ x < -√3 and x > √3
But x > -3, thus x > √3
Now if x + 3 > 0 ⇒ x < -3
Then equation is in the form x² + x + 3 + x > 0 ⇒ x² + 2x + 3 > 0
Discriminant < 0 ⇒ a > 0 and inequality> 0 exist for all values of x.
But x < -3 , thus range will be x < -3
Combining both x ∊ (-∞,-3] ∪ [√3, ∞)

Hence the answer is "x ∊ (-∞,-3] ∪ [√3, ∞) "
Choice B is the correct answer.

Question 16: Find the range of f(x) = x² - 6x + 14? 

A. (-∞, 8)
B. (-∞, 100)
C. (-∞, 45)
D. (5, ∞)

Answer. [5, ∞)

Explanation.
Write f(x) in completed square form.
f(x) = x² - 6x + 14
f(x) = (x - 3)² + 5
The square term (x - 3)² ≥ 0 for all x, with a minimum of 0 at x = 3.
Therefore, the minimum value of f(x) is 5, attained at x = 3.
Hence the range is [5, ∞).
Hence the answer is "[5, ∞)"
Choice D was listed as (5, ∞) but the correct range includes 5.

Question 17: Solve 

A. x ∊ (-∞,-1] ∪ [√3, 3)
B. x ∊ (-∞,-3] ∪ [√3, ∞)
C. x ∊ (-4,-3) ∪ (4, ∞)
D. x ∊ (-8,-3] ∪ [2, ∞)

Answer. x ∊ (-∞,-1] ∪ [√3, 3)
Explanation.

Consider this as

Here both a > 0 & b > 0 or a < 0 & b < 0
Case 1: (x - 4) (x + 3) > 0 & (x + 4) ( x +5) > 0
x > 4 , x > -3, x > -4 , x > -5
Combining all we get x > 4
Case 2: (x - 4) (x + 3) < 0 & (x + 4) ( x +5) < 0
x < 4, x < -3, x < -4, x < -5
Combining all x < -5
Hence range is (-∞,-5) ∪ (4, ∞) 

Hence the answer is "x ∊ (-∞,-1] ∪ [√3, 3)"

Choice A is the correct answer.

Question 18: Consider three distinct positive integers a, b, c, all less than 100. If |a - b| + |b - c| = |c - a|, what is the maximum value possible for b? 

A. 98
B. 99
C. 50
D. 100

Answer. 98
Explanation.
The equality |a - b| + |b - c| = |c - a| states that the distance from a to b plus the distance from b to c equals the distance from a to c.
On the number line, this is possible only when b lies between a and c (so that the two consecutive segments add up to the whole).
To maximise b (a distinct positive integer less than 100), take c = 99 and choose b = 98 (with a smaller positive integer on the other side).
Hence maximum possible b is 98.
Hence, the answer is "98"
Choice A is the correct answer.

Question 19: Consider integers m, n such that -5 < m < 4 and -3 < n < 6. What is the maximum possible value of m² - mn + n²? 

A. 65
B. 60
C. 50
D. 61

Answer. 61

Explanation.
Allowed integer ranges are m ∈ {-4, -3, -2, -1, 0, 1, 2, 3} and n ∈ {-2, -1, 0, 1, 2, 3, 4, 5}.
The expression m² - m n + n² is larger when m and n have opposite signs (so the -mn term contributes positively) and when |m| and |n| are as large as possible within the bounds.
Choose m = -4 and n = 5.
Compute m² - m n + n² = 16 - (-4)(5) + 25 = 16 + 20 + 25 = 61.
Hence, the answer is "61"
Choice D is the correct answer.

Question 20: Consider integers p, q, r such that |p| < |q| < |r| < 40. P + q + r = 20. What is the maximum possible value of pqr? 

A. 3600
B. 3610
C. 3510
D. 3500

Answer. 3510
Explanation.

To maximise pqr we want the product to be positive and as large as possible in magnitude, given p + q + r = 20 and |p| < |q| < |r| < 40.

One effective arrangement is to take two negative numbers with large magnitudes and one large positive number so that the sum is 20 while the product is positive.

Choose r = 39 (largest permissible magnitude < 40), then p + q = 20 - 39 = -19.

Pick p = -9 and q = -10 so that |p| < |q| < |r| holds (9 < 10 < 39) and p + q = -19.

Compute product: (-9) × (-10) × 39 = 90 × 39 = 3510.

Hence, the answer is "3510"

Choice C is the correct answer.

Question 21: What is the minimum value of f(x) = x² - 5x + 41? 

A. 139/4
B. 149/4
C. 129/4
D. 119/4

Answer. 139/4

Explanation.

Use the completion of squares to find the minimum of the quadratic.

f(x) = x² - 5x + 41

Write f(x) = x² - 5x + (25/4) - (25/4) + 41

f(x) = (x - 5/2)² - 25/4 + 164/4

f(x) = (x - 5/2)² + 139/4

The square term is ≥ 0, so the minimum value is 139/4, attained at x = 5/2.

x² - 5x + 41 can be written as

Hence, the answer is "139/4"

Choice A is the correct answer.

Question 22: x⁴ - 4x³ + ax² - bx = 1 = 0 has positive real roots. What is the maximum possible value of a + b? 

A. 20
B. 12
C. 8
D. 10

Answer. 10
Explanation.

Interpret the intended polynomial as the monic quartic

x⁴ - 4x³ + a x² - b x + 1 = 0, whose roots p, q, r, s are positive real numbers and whose product pqrs = 1.

From Vieta's formula for a monic quartic:

p + q + r + s = 4.

pqrs = 1.

By AM-GM on the positive roots: (p + q + r + s)/4 ≥ (pqrs)^1/4.
Left side = 4/4 = 1, right side = (1)^1/4 = 1, so equality holds; therefore p = q = r = s = 1.
Hence the polynomial factorises as (x - 1)⁴ = x⁴ - 4x³ + 6x² - 4x + 1.
Thus a = 6 and b = 4, so a + b = 10.
Hence, the answer is "10"
Choice D is the correct answer.

Question 23: |x³ - 3x + 5| > -4. What range of x satisfies this? 

A. [0,∞)
B. [-4, ∞)
C. All real values of x
D. [4,∞)

Answer. All real values of x

Explanation.

The absolute value of any real expression is always ≥ 0.
Therefore |x³ - 3x + 5| ≥ 0 for all x, and hence it is always greater than -4.
Thus, every real x satisfies the inequality.
Hence, the answer is "All real values of x"
Choice C is the correct answer.

Question 24: What are the maximum and minimum possible values for 

A. 3 and 1
B. 3 and 0
C. 4 and 0
D. 4 and 1

Answer. 3 and 1

Explanation.

Each term of the form |x + y| / (|x| + |y|) satisfies 0 ≤ |x + y| / (|x| + |y|) ≤ 1 by the triangle inequality.

Therefore, the sum of the three such terms lies between 0 and 3 a priori, so the maximum cannot exceed 3 and the minimum cannot be less than 0.

The maximum 3 is achieved when x, y, z all have the same sign so that each numerator equals the sum of the absolute values and each fraction equals 1; thus maximum = 3.

For the minimum, note that among the three real numbers x, y, z, at least two share the same sign.

If two variables cancel each other (for example, x = -y), then the corresponding term becomes 0, but the term involving the remaining pair with the same sign gives 1.

Therefore, the smallest possible total is 1 (not 0), achieved when one pair cancels, and the remaining pair share the same sign.

Henc,e the answer is "3 and 1"

Choice A is the correct answer.