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

Answer: Option B

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Text Explanation :

A = |x + 3| + |x - 2| - |2x - 8|

The critical points are -3, 2 and 4.

Case 1: x ≥ 4
∴ A = x + 3 + x - 2 - (2x - 8) 
⇒ A = 2x + 1 - 2x + 8
⇒ A = 9

Case 2: 2 ≤ x < 4
∴ A = x + 3 + x - 2 + (2x - 8) 
⇒ A = 2x + 1 + 2x - 8
⇒ A = 4x - 7
∴ A ∈ [1, 9)

Case 3: -3 ≤ x < 2
∴ A = x + 3 - (x - 2) + (2x - 8) 
⇒ A = x + 3 - x + 2 + 2x - 8
⇒ A = 2x - 3
∴ A ∈ [-9, 1)

Case 4: x < -3
∴ A = - (x + 3) - (x - 2) + (2x - 8) 
⇒ A = - x - 3 - x + 2 + 2x - 8
⇒ A = - 9

From the above cases, The maximum value of |A| = 9

Hence, option (b).

Answer: Option D

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Text Explanation :

Let |x − 5| = k.

Hence, the equation can be rewritten as: k² + 5k − 24 = 0

Solving the equation, we get k = 3, −8.

But, |x − 5| cannot be negative, so k ≠ −8.

∴ |x − 5| = 3

So, x − 5 = ±3. 

∴ x = 8 and x = 2

Required sum = 8 + 2 = 10.

Hence, option (d).

Answer: Option E

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Text Explanation :

Since 2 ≤ |x – 1| × |y + 3| ≤ 5, the value of |x – 1| × |y + 3| can be 2, 3, 4 or 5. Thus, we have the following cases.

Case 1: |x – 1| × |y + 3| = 2
⇒ |x – 1| = 2 and |y + 3| = 1
⇒ x = -1 and y = -4 or -2

Note: |x - 1| cannot be equal to 1 since x has to be a negative number.

∴ 2 possibilities.

Case 2: |x – 1| × |y + 3| = 3
⇒ |x – 1| = 3 and |y + 3| = 1
⇒ x = -2 and y = -4 or -2

∴ 2 possibilities.

Case 3: |x – 1| × |y + 3| = 4

⇒ |x – 1| = 4 and |y + 3| = 1
⇒ x = -3 and y = -4 or -2

⇒ |x – 1| = 2 and |y + 3| = 2
⇒ x = -1 and y = -1 or -5

∴ 4 possibilities.

Case 3: |x – 1| × |y + 3| = 5

⇒ |x – 1| = 5 and |y + 3| = 1
⇒ x = -4 and y = -4 or -2

∴ 2 possibilities.

∴ Total 2 + 2 + 4 + 2 = 10 possibilities.

Hence, option (e).

Answer: Option C

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Text Explanation :

abc – (a + b + c) = a(bc – 1) – b – c .

It is clear that the expression would be maximum when b and c are negative and their modulus are maximum possible.

∴ b = –10 , c = –9.

Now a has to be positive and maximum possible. It has to be 8.

Hence the expression becomes equal to 731.

Hence, option (c).

Answer: Option D

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Text Explanation :

The LHS is of the form: (x – 7) / (x + 9) (x – 4)

For x > 7, the expression will be positive.

Similarly, for any negative integer,

[(x – 7)/(x – 4)] will always be positive.

For the expression to be positive,

(x + 9) > 0 i.e. x > −9

Hence, the smallest integer that satisfies this inequality is x = −8

Hence, option (d)

Answer: Option C

Video Explanation

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Text Explanation :

Let a + b + c = X

We need to find the maximum value of:

$\frac{X+d}{X-d}=\frac{(X-d)+2d}{X-d}=1+\frac{2d}{X-d}$

$$\begin{gathered} \frac{}{} \\ \frac{2d}{X-d} \end{gathered}$$

$\frac{\mathrm{is\; maximum\; when\; d\; is\; maximum}}{}$ and X is minimum possible.

∴ d = 25 and X = 26
The highest possible value of the given expression = 1 + (2 × 25) = 51
Hence, option (c).