Algebra - Inequalities & Modulus - Previous Year CAT/MBA Questions
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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
Text 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
Text 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:
If x > 5 and y < −1, then which of the following statements is true?
- (a)
(x + 4y) > 1
- (b)
x > − 4y
- (c)
−4x < 5y
- (d)
None of these
Answer: Option D
Text Explanation :
Assume some values for x and y and substitute in the options and check.
For example, (x, y) = (6, −10), (7, −20) etc. None of the options are satisfied by these values.
Hence, option (d).
Workspace:
x and y are real numbers satisfying the conditions 2 < x < 3 and –8 < y < –7. Which of the following expressions will have the least value?
- (a)
x2y
- (b)
xy2
- (c)
5xy
- (d)
None of these
Answer: Option D
Text Explanation :
Option 2 can easily be eliminated as it will give positive value, while both option 1 and 3 will be negative.
∵ x2 will fall in the range 4 < x2 < 9.
∵ 5x will fall between 10 < 5x < 15.
∴ 5x > x2
Also, 5x < 6x < 7x < ... and –8 < y < –7
∴ 5xy > 6xy > 7xy > ...
Hence, option (d).
Workspace:
m is the smallest positive integer such that for any integer n > m, the quantity n3 – 7n2 + 11n – 5 is positive. What is the value of m?
- (a)
4
- (b)
5
- (c)
8
- (d)
None of these
Answer: Option B
Text Explanation :
n3 − 7n2 + 11n − 5
For n = 1, the expression reduces to zero
∴ (n − 1) is a factor.
n3 − 7n2 + 11n − 5 = n3 − n2 − 6n2 + 6n + 5n − 5
= n2(n − 1) − 6n(n − 1) + 5(n − 1)
= (n − 1)(n2 − 6n + 5)
= (n − 1)(n − 1)(n − 5)
= (n − 1)2(n − 5)
Since, (n − 1)2 is always positive.
∴ The expression is positive only when n > 5.
∴ n = 6 and m = 5
Hence, option (b).
Alternatively,
Check using options. If we take m = 4 then n = 5, then we do not get a positive value of the given equation. Therefore, check the next higher value.
n = 6 and m = 5 gives the positive value.
Hence, option (b).
Workspace:
If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)?
- (a)
4
- (b)
1
- (c)
16
- (d)
18
Answer: Option C
Text Explanation :
This is possible when, a = b = c = d = 1
∵ a, b, c, d are positive real numbers
∴ (1 + a)(1 + b)(1 + c)(1 + d) = 2 × 2 × 2 × 2 = 16
Hence, option (c).
Workspace:
Let x, y be two positive numbers such that x + y = 1. Then, the minimum value of is ______.
- (a)
12
- (b)
20
- (c)
12.5
- (d)
13.3
Answer: Option C
Text Explanation :
The sum of two numbers, when their product is known, is minimum when they are equal.
The product of two numbers, when their sum is known, is maximum when they are equal.
= (x2 + y2) +
This expression is minimum when (x2 + y2) is minimum and x2y2 is maximum.
We have x + y = 1
Squaring,
x2 + 2xy + y2 = 1
∴ x2 + y2 = 1 – 2xy
x2 + y2 is minimum when xy is maximum.
xy is maximum when x = y
∴ For minimum value, both x and y have to be equal.
∴ x = y =
= (2.5)2 + (2.5)2
= 12.5
Hence, option (c).
Workspace:
If x > 2 and y > – 1, Then which of the following statements is necessarily true?
- (a)
xy > –2
- (b)
–x < 2y
- (c)
xy < –2
- (d)
–x > 2y
Answer: Option B
Text Explanation :
y > –1
∴ –2y < 2 < x
∴ –x < 2y
Hence, option (b).
Workspace:
If x2 + y2 = 0.1 and |x – y| = 0.2, then |x| + |y| is equal to
- (a)
0.3
- (b)
0.4
- (c)
0.2
- (d)
0.6
Answer: Option B
Text Explanation :
x2 + y2 = 0.1 …(i)
|x – y| = 0.2
Squaring both the sides, we get,
x2 + y2 – 2xy = 0.04 …(ii)
From (i) and (ii),
2xy = 0.1 – 0.04 = 0.06
Thus, x and y both are positive or both are negative.
∴ 2xy = 2|x||y|= 0.06
(|x| + |y|)2 = x2 + y2 + 2|x||y| = 0.1 + 0.06 = 0.16
|x| + |y| > 0
∴ |x| + |y| = 0.4
Hence, option (b).
Workspace:
If |r − 6| = 11 and |2q − 12| = 8,what is the minimum possible value of ?
- (a)
- (b)
- (c)
- (d)
None of these
Answer: Option D
Text Explanation :
|r − 6| = 11 ⇒ r − 6 = 11, r = 17
or –(r – 6) = 11, r = –5
|2q – 12| = 8 ⇒ 2q –12 = 8 , q = 10
or 2q – 12 = –8 , q =2
Hence, minimum value of
Hence, option (d).
Workspace:
Which of the following values of x do not satisfy the inequality (x2 – 3x + 2 > 0) at all?
- (a)
1 ≤ x ≤ 2
- (b)
–1 ≥ x ≥ –2
- (c)
0 ≤ x ≤ 2
- (d)
0 ≥ x ≥ –2
Answer: Option A
Text Explanation :
If we simplify the expression x2 – 3x + 2 > 0, we get (x – 1)(x – 2) > 0. For this product to be greater than zero, either both the factors should be greater than zero or both of them should be less than zero. Therefore, (x – 1) > 0 and (x – 2) > 0 or (x – 1) < 0 and (x – 2) < 0.
Hence, x > 1 and x > 2 or x < 1 and x < 2. If we were to club the ranges, we would get either x > 2 or x < 1. So for any value of x equal to or between 1 and 2, the above equation does not follow.
Workspace:
What is the value of m which satisfies 3m2 – 21m + 30 < 0?
- (a)
m < 2 or m > 5
- (b)
m > 2
- (c)
2 < m < 5
- (d)
Both (a) and (c)
Answer: Option C
Text Explanation :
3m2 – 21m + 30 < 0
∴ m2 – 7m + 10 < 0 ⇒ (m – 5)(m – 2) < 0.
⇒ (m – 5) < 0 and (m – 2) > 0
∴ 2 < m < 5.
Hence, option (c).
Workspace:
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