Algebra - Inequalities & Modulus - Previous Year CAT/MBA Questions
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The largest real value of a for which the equation |x + a| + |x - 1| = 2 has an infinite number of solutions for x is
- (a)
-1
- (b)
2
- (c)
0
- (d)
1
Answer: Option D
Explanation :
Given, |x + a| + |x - 1| = 2
⇒ Sum of distance of x from 1 and -a is 2.
This is possible when a = -3 or 1.
Hence, option (d).
Workspace:
Let 0 ≤ a ≤ x ≤ 100 and f(x) = |x - a| + |x - 100| + |x - a - 50|. Then the maximum value of f(x) becomes 100 when a is equal to
- (a)
0
- (b)
25
- (c)
50
- (d)
100
Answer: Option C
Explanation :
Since a ≤ x ≤ 100
⇒ f(x) = x - a + -(x – 100) + |x - a - 50|
⇒ f(x) = x - a - x + 100 + |x - a - 50|
⇒ f(x) = 100 – a + |x - a - 50| ≤ 100
Option (a): a = 0
⇒ f(x) = 100 – 0 + |x - 0 - 50|
⇒ f(x) = 100 + |x - 50|
Here, when x = 0, f(x) > 100.
∴ This option is rejected.
Option (b): a = 25
⇒ f(x) = 100 – 25 + |x - 25 - 50|
⇒ f(x) = 75 + |x - 75|
Here, when x = 25, f(x) > 100.
∴ This option is rejected.
Option (c): a = 50
⇒ f(x) = 100 – 50 + |x - 50 - 50|
⇒ f(x) = 50 + |x - 100|
For any value of x ≥ 50, f(x) ≤ 100.
∴ This option is correct.
Option (d): a = 100
⇒ f(x) = 100 – 100 + |x - 100 - 50|
⇒ f(x) = |x - 150|
The only value x can take in this case is 100.
⇒ f(x) = 50
Here, the maximum value of f(x) is not 100.
∴ This option is rejected.
Hence, option (c).
Workspace:
The minimum possibe value of , for x < 3, is
- (a)
-2
- (b)
- (c)
2
- (d)
-
Answer: Option C
Explanation :
We have
=
= +
= (3 - x) +
Here, since x < 3, 3 – x > 0
Also, we know that sum of a positive number and its reciprocal is always greater than or equal to 2.
⇒ (3 - x) + ≥ 2
Hence, option (c).
Workspace:
If c = + for some non-zero real numbers x and y, then c cannot take the value
- (a)
-60
- (b)
-70
- (c)
60
- (d)
-50
Answer: Option D
Explanation :
Given, c = +
Let = a
⇒ c = 16a + 49/a
Now, for two positive numbers, AM ≥ GM
If a > 0
⇒ ≥
⇒ ≥ 28
⇒ c ≥ 56
∴ option (c) is rejected.
Now, for two negative numbers, AM ≤ - GM
If a < 0
⇒ ≤ -
⇒ ≤ - 28
⇒ c ≤ - 56
∴ option (a) and (b) are rejected.
∴ c cannot be equal to -50
Hence, option (d).
Workspace:
The number of integers n that satisfy the inequalities |n - 60| < |n - 100| < |n - 20| is
- (a)
18
- (b)
21
- (c)
20
- (d)
19
Answer: Option D
Explanation :
Here the critical points are 20, 60 and 100.
Case 1: n ≥ 100
⇒ n - 60 < n - 100
⇒ -60 < -100
This can never be true.
This case is rejected.
Case 2: 60 ≤ n < 100
⇒ n - 60 < - n + 100
⇒ 2n < 160
⇒ n < 80
Also,
- n + 100 < n – 20
⇒ 2n > 120
⇒ n > 60
∴ Possible integral values of n are 61, 62, 63, …, 79 i.e., 19 values
Case 3: n < 60
⇒ - n + 100 < - n + 20
⇒ 100 < 20
This can never be true.
This case is rejected.
∴ n can take 19 integral values.
Hence, option (d).
Workspace:
For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if
- (a)
6 < x < 11
- (b)
9 < x < 14
- (c)
10 < x < 15
- (d)
7 < x < 12
Answer: Option D
Explanation :
Case 1: 3x ≥ 40 ⇒ x ≥ 13.33
⇒ 3x – 20 + 3x – 40 = 20
⇒ 6x = 80
⇒ x = 13.33
Case 2: 20 ≤ 3x < 40 ⇒ 6.67 ≤ x < 13.33
⇒ 3x – 20 - 3x + 40 = 20
⇒ 20 = 20
This is always true.
Case 3: 3x ≤ 20
⇒ - 3x + 20 - 3x + 40 = 20
⇒ -6x = -40
⇒ x = 6.67
Combining solution from all the three cases, we get
6.67 ≤ x ≤ 13.33
Only option (d) is a subset of this range.
Hence, option (d).
Workspace:
If 3x + 2|y| + y = 7 and x + |x| + 3y = 1, then x + 2y is
- (a)
8/3
- (b)
1
- (c)
0
- (d)
-4/3
Answer: Option C
Explanation :
Case 1: x, y > 0
⇒ 3x + 3y = 7 and 2x + 3y = 1
Solving these two equations we get,
x = 6 and y = -11/3
This is rejected as y should be positive.
Case 2: x > 0, y < 0
⇒ 3x - y = 7 and 2x + 3y = 1
Solving these two equations we get,
x = 2 and y = -1
This is accepted.
∴ x + 2y = 2 + 2 × -1 = 0
Case 3: x < 0, y > 0
⇒ 3x + 3y = 7 and 3y = 1
Solving these two equations we get,
x = 2 and y = 1/3
This is rejected as x should be negative.
Case 4: x, y < 0
⇒ 3x - 3y = 7 and 3y = 1
Solving these two equations we get,
x = 8/3 and y = 1/3
This is rejected as and y should be negative.
Hence, option (c).
Workspace:
The number of distinct pairs of integers (m, n) satisfying |1 + mn| < |m + n| < 5 is
Answer: 12
Explanation :
|1 + mn| < |m + n| < 5
Squaring all the expressions
(1 + mn)2 < (m + n)2 < 25
⇒ 1 + m2n2 + 2mn < m2 + n2 + 2mn
⇒ m2 + n2 - m2n2 – 1 > 0
⇒ (m2 – 1) - n2(m2 – 1) > 0
⇒ (m2 – 1)(1 – n2) > 0
⇒ (m2 – 1)(n2 – 1) < 0
Case 1: Either m2 > 1 and n2 < 1
⇒ n = 0 and since |1 + mn| < |m + n| < 5
⇒ 1 < |m| < 5
⇒ m = ±2, ±3 or ±4
∴ 6 possible pairs of (m, n)
Case 2: Either n2 > 1 and m2 < 1
⇒ m = 0 and since |1 + mn| < |m + n| < 5
⇒ 1 < |n| < 5
⇒ n = ±2, ±3 or ±4
∴ 6 possible pairs of (m, n)
∴ Total 6 + 6 = 12 possible pairs of (m, n)
Hence, 12.
Workspace:
In how many ways can a pair of integers (x , a) be chosen such that x2 − 2|x| + |a - 2| = 0?
- (a)
4
- (b)
5
- (c)
6
- (d)
7
Answer: Option D
Explanation :
Given, x2 − 2|x| + |a - 2| = 0
This can be written as
|x|2 − 2|x| + 1 – 1 + |a - 2| = 0
⇒ (|x| - 1)2 + |a - 2| - 1 = 0
Now, ((|x| - 1)2 ≥ 0 and |a - 2| - 1 will be an integer since a is an integer.
∴ |a - 2| - 1 can take only take non-positive values i.e., 0 or -1.
Case 1: |a - 2| - 1 = 0 and (|x| - 1) = 0
⇒ |a – 2| = 1 and |x| = 1
⇒ a = 1 or 3 and x = ± 1
∴ 4 possible combinations of (x, a)
Case 2: |a - 2| - 1 = -1 and (|x| - 1) = ±1
|a - 2| = 0 and |x| = 0 or 2
⇒ a = 2 and x = 0 or ± 2
∴ 3 possible combinations of (x, a)
∴ Total 7 possible combination of (x, a) are there.
Hence, option (d).
Workspace:
Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N?
Answer: 6
Explanation :
Given, 2 < x < 10
x can take any of the values from {3, 4, 5, 6, 7, 8, 9}
Also, 14 < y < 23
y can take any of the values from {15, 16, 17, 18, 19, 20, 21, 22}
The highest value N (i.e., x + y) can take = 9 + 22 = 31 (when x = 9; y = 22)
The lowest value N (i.e., x + y) can take = 3 + 15 = 18 (when x = 3; y = 15)
But, N = x + y > 25. Hence the different values of x + y are {31, 30, 29, 28, 27, 26}.
Hence, x + y, and thereby N can take 6 distinct values.
Hence, 6.
Workspace:
The number of the real roots of the equation 2cos(x(x + 1)) = 2x + 2–x is
- (a)
0
- (b)
2
- (c)
infinite
- (d)
1
Answer: Option D
Explanation :
−2 ≤ 2cos(x(x + 1)) ≤ 2
∴ −2 ≤ 2x + 2–x ≤ 2
Let 2x be a, so 2–x is 1/a.
So, −2 ≤ a + (1/a) ≤ 2
∴ −2 ≤ (a2 + 1)/a ≤ 2
∴ −2a ≤ (a2 + 1) ≤ 2a
∴ (a2 + 1 + 2a) ≥ 0 ⇒ (a + 1)2 ≥ 0, so a ∈ R.
Also, a2 + 1 − 2a ≤ 0 ⇒ (a − 1)2 ≤ 0, so a = 1.
Hence a = 1.
So, 2x = 1.
∴ x = 0.
So, there is only one real root.
Hence, option (d).
Workspace:
If x is a real number, then is a real number if and only if
- (a)
-3 ≤ x ≤ 3
- (b)
1 ≤ x ≤ 2
- (c)
1 ≤ x ≤ 3
- (d)
-1 ≤ x ≤ 3
Answer: Option C
Explanation :
It is given that, is a real number
Therefore, ≥ 0
⇒ ≥ 1
⇒ 4x – x2 ≥ 3
⇒ x2 - 4x + 3 ≤ 0
⇒ (x - 1)(x - 3) ≤ 0
⇒ x ∈ [1, 3]
Hence, option (c).
Workspace:
The smallest integer n for which 4n > 1719 holds, is closest to
- (a)
33
- (b)
37
- (c)
39
- (d)
35
Answer: Option C
Explanation :
We have, 4n > 1719
Taking log on both sides to the base 4,
n × log4 4 > 19 × log4 17
Now, 17 > 42.
∴ log4 17 > 2.
⇒ n × log4 4 > 19 × log4 17 > 19 × 2
Also, log4 4 = 1
∴ n > 38.
Hence, option (c).
Workspace:
If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
- (a)
16
- (b)
48
- (c)
–32
- (d)
64
Answer: Option D
Explanation :
Given: a2 + b2 = 97.
AM ≥ GM
⇒ ≥
⇒ 48.5 ≥ |a × b|
⇒ 48.5 ≥ a × b ≥ -48.5
Out of the given options, 64 definitely lies outside this range.
Hence, option (d).
Workspace:
If a and b are integers such that 2x2 − ax + 2 > 0 and x2 − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is
Answer: 36
Explanation :
We have to maximize the value of 2a - 6b. Therefore let us look for the largest possible value of a and the smallest possible value of b.
If 2x2 – ax + 2 > 0 for all value of x, the graph of 2x2 – ax + 2 is above the x-axis or the roots of the quadratic equations 2x2 – ax + 2 = 0 are imaginary or its discriminant is less than 0.
∴ We have a2 – 16 < 0 or a2 < 16 or –4 < a < 4.
⇒ The largest possible value of a is 3.
If x2 – bx + 8 ≥ 0, the discriminant of the equation is less than or equal to zero.
⇒ b2 - 32 ≤ 0
⇒ -4√2 ≤ b ≤ 4√2
⇒ -5.64 ≤ b ≤ 5.64
∴ The smallest possible value of b is –5.
Therefore the maximum possible value of 2a - 6b = 2(3) – 6(–5) = 36.
Hence, 36.
Workspace:
For how many integers n, will the inequality (n – 5) (n – 10) – 3(n – 2) ≤ 0 be satisfied?
Answer: 11
Explanation :
(n – 5) (n – 10) – 3(n – 2) ≤ 0
⇒ n2 – 15n + 50 - 3n + 6 ≤ 0
⇒ n2 – 18n + 56 ≤ 0
⇒ (n – 4) (n – 14) ≤ 0
As n is an integer, n can be 4, 5, 6 ……14, i.e. it can have 11 values.
Hence, 11.
Workspace:
If a, b, c, and d are integers such that a + b + c + d = 30, then the minimum possible value of (a - b)2 + (a - c)2 + (a - d)2 is
Answer: 2
Explanation :
a + b + c + d = 30, a, b, c, d are integers.
(a – b)2 + (a – c)2 + (a – d)2would have its maximum value when each bracket has the least possible value.
Let (a, b, c, d) = (8, 8, 7, 7)
The given expression would be 2. It cannot have a smaller value.
Hence, 2.
Workspace:
If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is
Answer: 200
Explanation :
Let ‘x’ and ‘y’ be the dimensions of the rectangle
Let us suppose 2x + y = 400 … (1)
Area = xy.
For xy to be maximum, 2xy should also be maximum.
Now the product 2xy will be maximum when 2x = y. [AM ≥ GM]
So y + y = 400 [From (1)]
⇒ 2y = 400 or y = 200
Substituting value of y in (1) we get,
2x = 200 or x = 100
In a rectangle, length is greater than breadth, so we take y as the length.
Hence area of the park is maximum when length is 200 ft.
Hence, 200.
Workspace:
If f(x) = x3 – 4x + p, and f(0) and f(1) are of opposite signs, then which of the following is necessarily true?
- (a)
–1 < p < 2
- (b)
0 < p < 3
- (c)
–2 < p < 1
- (d)
–3 < p < 0
Answer: Option B
Explanation :
f(x) = x3 – 4x + p
∴ f(0) = p and f(1) = p – 3
p – 3 < p
As p and p – 3 are of opposite signs, p – 3 < 0 and p > 0.
∴ p < 3 and p > 0
∴ 0 < p < 3
Hence, option (b).
Workspace:
Let f(x) = ax2 – b|x|, where a and b are constants. Then at x = 0, f(x) is
- (a)
maximized whenever a > 0, b > 0
- (b)
maximized whenever a > 0, b < 0
- (c)
minimized whenever a > 0, b > 0
- (d)
minimized whenever a > 0, b < 0
Answer: Option D
Explanation :
f(x) = ax2 – b|x|
x2 and |x| both are positive. Let x ≠ 0.
At x = 0, f(0) = 0
Consider the following cases:
1. a > 0, b > 0
∴ f(x) > 0, when ax2 > b|x|
∴ f(x) < 0, when ax2 < b|x|
f(x) is neither maximised or minimised when x = 0.
2. a > 0, b < 0
∴ f(x) = ax2 + |b||x| > 0
Thus f(x) is greater than 0 when x ≠ 0.
∴ f(x) is minimised at x = 0 whenever a > 0, b < 0.
Hence, option (d).
Workspace:
Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?
- (a)
The minimum possible value of a2 + b2 + c2 + d2 is 4m2 − 2m + 1
- (b)
The minimum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1
- (c)
The maximum possible value of a2 + b2 + c2 + d2 is 4m2 − 2m + 1
- (d)
The maximum possible value of a2 + b2 + c2 + d2 is 4m2 + 2m + 1
Answer: Option B
Explanation :
a + b + c + d = 4m + 1
a2 + b2 = (a + b)2 – 2ab
a2 + b2 is minimum when 2ab is maximum.
The product of two numbers is maximum when the numbers are equal.
∴ a² + b² is minimum when a = b
Similarly, c² + d² is minimum when c = d
∴ a² + b² + c² + d² is minimum when a = b and c = d
∴ (a² + b² + c² + d²)min = 2(a² + c²)
But, a² + c² is minimum when a = c
∴ a² + b² + c² + d² is minimum when a = b = c = d
When a = b = c = d, a + b + c + d is a multiple of 4.
But, a + b + c + d = 4m + 1
So, one out of a, b, c, d must be one greater than the other three.
∴ a = b = c = m and d = m + 1
∴ a² + b² + c² + d² = m² + m² + m² + (m + 1)² = 4m² + 2m + 1
Hence, option (b).
Workspace:
Given that −1 ≤ v ≤ 1, −2 ≤ u ≤ −0.5 and −2 ≤ z ≤ −0.5 and w = vz/u, then which of the following is necessarily true?
- (a)
−0.5 ≤ w ≤ 2
- (b)
−4 ≤ w ≤ 4
- (c)
−4 ≤ w ≤ 2
- (d)
−2 ≤ w ≤ −0.5
Answer: Option B
Explanation :
w = vz/u
From the given range of values for u, v and z, we have,
Maximum possible value of w is 4 when v is 1, z is −2 and u is −0.5.
Also minimum value of w is −4 when v is −1, z is −2 and u is −0.5.
Hence, option (b).
Workspace:
If x, y, z are distinct positive real numbers, then would be
- (a)
greater than 4
- (b)
greater than 5
- (c)
greater than 6
- (d)
None of these
Answer: Option C
Explanation :
The given expression may be represented as
We know that, A.M. ≥ G.M.
∴ The given expression will have a minimum value of 6.
But x, y and z are distinct, so the value will always be greater than 6.
Hence, option (c).
Workspace:
If |b| ≥ 1 and x = –|a|b, then which one of the following is necessarily true?
- (a)
a – xb < 0
- (b)
a – xb ≥ 0
- (c)
a – xb > 0
- (d)
a – xb ≤ 0
Answer: Option B
Explanation :
Consider the case when b is negative:
i.e. say b = –k, where k ≥ 1
Then, x = –|a|b = –|a| × (–k) = |a|k
∴ xb = –|a|k2
∴ a – xb = a + |a|k2
Now,
If a > 0, then a – xb = a + |a|k2 > 0 since all the terms will be positive
If a < 0 (say a = –2), then a – xb = –2 + 2k2 ≥ 0, since 2k2 ≥ 2 as k ≥ 1
However, if a = 0, then a – xb = 0 + 0 = 0
Hence, when b is negative, a – xb ≥ 0
Now, consider the case when b is positive:
i.e. say b = +k, where k ≥ 1
Then, x = –|a|b = –|a| × (k) = –|a|k
∴ xb = –|a|k2
This is the same value of xb as we got in the previous case. Hence, the same conclusions will hold.
∴ For all cases, a − xb ≥ 0
Hence, option (b).
Workspace:
A real number x satisfying for every positive integer n, is best described by:
- (a)
1 < x < 4
- (b)
1 < x ≤ 3
- (c)
0 < x ≤ 4
- (d)
1 ≤ x ≤ 3
Answer: Option C
Explanation :
wher n is a positive integer
If n = 1, then 0 < x ≤ 4
As the value of n increases from 1 to infinity, the range of x approaches (1 < x ≤ 3).
So, the lower limit of x increases from almost 0 (to almost 1), while the higher limit decreases from 4 (to 3).
Thus, the lowest possible limit of x will be > 0 and the highest possible limit will be ≤ 4.
∴ For all range of n, 0 < x ≤ 4
Hence, option (c).
Workspace:
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