Algebra - Inequalities & Modulus - Previous Year CAT/MBA Questions

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).

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).

Answer: Option C

Explanation :

We have $\frac{{\mathrm{x}}^2-6\mathrm{x}+10}{3-\mathrm{x}}$

= $\frac{{(\mathrm{x}-3)}^2+1}{3-\mathrm{x}}$

= $\frac{{(\mathrm{x}-3)}^2}{3-\mathrm{x}}$ + $\frac{1}{3-\mathrm{x}}$

= (3 - x) + $\frac{1}{3-\mathrm{x}}$

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) + $\frac{1}{3-\mathrm{x}}$ ≥ 2

Hence, option (c).

Answer: Option D

Explanation :

Given, c = $\frac{16x}{y}$ + $\frac{49y}{x}$ 

Let $\frac{\mathrm{x}}{\mathrm{y}}$ = a

⇒ c = 16a + 49/a

Now, for two positive numbers, AM ≥ GM
If a > 0
⇒ $\frac{16\mathrm{a}+{\displaystyle \frac{49}{\mathrm{a}}}}{2}$ ≥ $\sqrt{16\mathrm{a}\times \frac{49}{\mathrm{a}}}$

⇒ $\frac{\mathrm{c}}{2}$ ≥ 28

⇒ c ≥ 56

∴ option (c) is rejected.

Now, for two negative numbers, AM ≤ - GM
If a < 0
⇒ $\frac{16\mathrm{a}+{\displaystyle \frac{49}{\mathrm{a}}}}{2}$ ≤ - $\sqrt{16\mathrm{a}\times \frac{49}{\mathrm{a}}}$

⇒ $\frac{\mathrm{c}}{2}$ ≤ - 28

⇒ c ≤ - 56

∴ option (a) and (b) are rejected.

∴ c cannot be equal to -50

Hence, option (d).

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).

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).

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).

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.

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).

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.

Answer: Option D

Explanation :

−2 ≤ 2cos(x(x + 1)) ≤ 2 

∴ −2 ≤ 2^x + 2^–x ≤ 2 

Let 2^x 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, 2^x = 1. 

∴ x = 0.  

So, there is only one real root. 

Hence, option (d).

Answer: Option C

Explanation :

It is given that, $\sqrt{{\log }_e\left(\frac{4x-x^2}{3}\right)}$ is a real number

Therefore, ${\log }_e\left(\frac{4x-x^2}{3}\right)$ ≥ 0

⇒ $\frac{4x-x^2}{3}$ ≥ 1

⇒ 4x – x² ≥ 3

⇒ x² - 4x + 3 ≤ 0

⇒ (x - 1)(x - 3) ≤ 0

⇒ x ∈ [1, 3]

Hence, option (c).

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 > 4^2.

∴ log₄ 17 > 2.

⇒ n × log₄ 4 > 19 × log₄ 17 > 19 × 2

Also, log₄ 4 = 1

∴ n > 38.

Hence, option (c).

Answer: Option D

Explanation :

Given: a² + b² = 97.

AM ≥ GM

⇒ $\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2}$ ≥ $\sqrt{{\mathrm{a}}^2\times {\mathrm{b}}^2}$

⇒ 48.5 ≥ |a × b|

⇒ 48.5 ≥ a × b ≥ -48.5

Out of the given options, 64 definitely lies outside this range.

Hence, option (d).

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.

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.

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.

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.

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).

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).

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).

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).

Answer: Option C

Explanation :

The given expression may be represented as

$\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}$

We know that, A.M. ≥ G.M.

$\therefore \frac{\left({\displaystyle \frac{x}{y}}+{\displaystyle \frac{x}{z}}+{\displaystyle \frac{y}{x}}+{\displaystyle \frac{y}{z}}+{\displaystyle \frac{z}{x}}+{\displaystyle \frac{z}{y}}\right)}{6}\ge \sqrt[6]{\frac{x}{y}\times \frac{x}{z}\times \frac{y}{x}\times \frac{y}{z}\times \frac{z}{x}\times \frac{z}{y}}$

$\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\ge 6$

∴ 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).

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).

Answer: Option C

Explanation :

$\left(1-\frac{1}{n}\right)<x\le \left(3+\frac{1}{n}\right),$ 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).