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Algebra - Inequalities & Modulus - Previous Year IPM/BBA Questions

The best way to prepare for Algebra - Inequalities & Modulus is by going through the previous year Algebra - Inequalities & Modulus questions for IPMAT - Indore. Here we bring you all previous year Algebra - Inequalities & Modulus IPMAT - Indore questions along with detailed solutions.

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It would be best if you clear your concepts before you practice previous year Algebra - Inequalities & Modulus questions for IPMAT - Indore.

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1. IPMAT (I) 2023 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

The set of all real values of x satisfying the inequality $\frac{x^{2} (x + 1)}{(x - 1) {(2 x + 1)}^{2}} > 0$ .

(a)
(-1, -1/2) ∪ (0, ∞)
(b)
(-1, -1/2) ∪ (1, ∞)
(c)
(-1, 0) ∪ (1, ∞)
(d)
(-∞, -1) ∪ (-1/2, 0) ∪ (1, ∞)

2. IPMAT (I) 2022 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

The sum of the squares of all the roots of the equation x² + |x + 4| + |x − 4| − 35 = 0 is

(a)
74
(b)
175
(c)
50
(d)
148

3. IPMAT (I) 2020 QASA | Algebra - Inequalities & Modulus IPMAT - Indore Question

The minimum value of f(x) = |3 - x| + |2 + x| + |5 - x| is equal to __________.

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7 8 9 4 5 6 1 2 3 0 . -

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4. IPMAT (I) 2020 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

Consider the following statements:

(i) When 0 < x < 1, then $\frac{1}{1 + x}$ < 1 - x + x²

(ii) When 0 < x < 1, then $\frac{1}{1 + x}$ > 1 - x + x²

(iii) When -1 < x < 0, then $\frac{1}{1 + x}$ < 1 - x + x²

(iv) When -1 < x < 0, then $\frac{1}{1 + x}$ > 1 - x + x²

Then the correct statements are

(a)
(i) and (ii)
(b)
(ii) and (iv)
(c)
(i) and (iv)
(d)
(ii) and (iii)

Answer: Option C

Text Explanation :

When 0 < x < 1, consider $\frac{1}{1 + x}$ and 1 - x + x²
Let us take x = 0.5
∴ $\frac{1}{1 + x}$ = 0.67 and 1 - x + x² = 1 - 0.5 + 0.25 = 0.75
⇒ $\frac{1}{1 + x}$ < 1 - x + x²
∴ (i) is correct.

When 0 < x < 1, consider $\frac{1}{1 + x}$ and 1 - x + x²
Let us take x = -0.5
∴ $\frac{1}{1 + x}$ = 2 and 1 - x + x² = 1 + 0.5 + 0.25 = 1.75
⇒ $\frac{1}{1 + x}$ > 1 - x + x²
∴ (iv) is correct.

Hence, option (d).

5. IPMAT (I) 2019 QASA | Algebra - Inequalities & Modulus IPMAT - Indore Question

If |x| < 100 and |y| < 100, then the number of integer solutions of (x, y) satisfying the equation 4x + 7y = 3 is

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6. IPMAT (I) 2019 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

If x ∈ (a, b) satisfies the inequality $\frac{x - 3}{x^{2} + 3 x + 2}$ ≥ 1, then the largest possible value of b - a is

(a)
3
(b)
1
(c)
2
(d)
No real values of x satisfies the inequality

7. IPMAT (I) 2019 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

The set of values of x which satisfy the inequality $0 . 7^{2 x^{2} - 3 x + 4}$ < 0.343 is

(a)
(1/2, 1)
(b)
(1/2, ∞)
(c)
(-∞, 1/2)
(d)
(-∞, 1/2) ∪ (1, ∞)

8. IPMAT (I) 2019 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

For a > b > c > 0, the minimum value of the function f(x) = |x - a| + |x - b| + |x - c| is

(a)
2a- b - c
(b)
a + b - 2c
(c)
a + b + c
(d)
a - c

9. IPMAT (I) 2019 QA MCQ | Algebra - Inequalities & Modulus IPMAT - Indore Question

The inequality log_af(x) < log_ag(x) implies that

(a)
f(x) > g(x) > 0 for 0 < a < 1 and g(x) > f(x) > 0 for a > 1
(b)
g(x) > f(x) > 0 for 0 < a < 1 and f(x) > g(x) > 0 for a > 1
(c)
f(x) > g(x) > 0 for a > 0
(d)
g(x) > f(x) > 0 for a > 0

Algebra - Inequalities & Modulus - Previous Year IPM/BBA Questions

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