IOE entrance Mathematics · Chapter 1
Set, Logic and Functions
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35 questions in 2 syllabus topics · 14 are 2-mark questions.
1.1 Sets, real numbers and logic
15 questions
1. If n(A) = 20, n(B) = 15 and n(A ∩ B) = 8, then n(A ∪ B) is
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Use the inclusion–exclusion formula for two sets.
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Answer: D. 27
n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 20 + 15 − 8 = 27.
2. In a class of 60 students, 35 like Mathematics, 30 like Physics and 10 like neither subject. How many students like both subjects?
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First find how many like at least one subject.
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Answer: C. 15
Students liking at least one = 60 − 10 = 50. Then 50 = 35 + 30 − n(both), so n(both) = 15.
3. The number of proper subsets of the set {1, 2, 3, 4, 5} is
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Count all subsets first, then remove the set itself.
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Answer: C. 31
A set with n elements has 2ⁿ subsets; 2⁵ = 32. Excluding the set itself, the number of proper subsets is 31.
4. For any two sets A and B, A − (A ∩ B) is equal to
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Think of a Venn diagram: what part of A is left?
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Answer: B. A − B
Removing the common part A ∩ B from A leaves the elements of A not in B, which is A − B.
5. Which of the following numbers is irrational?
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Simplify each expression completely.
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Answer: D. (√3 + 1)²
(√3 + 1)² = 4 + 2√3, which is irrational. The others simplify to 2, 2 and 4 respectively.
6. The set {x ∈ ℝ : −2 < x ≤ 5} in interval notation is
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Strict inequality uses a round bracket; ≤ uses a square bracket.
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Answer: B. (−2, 5]
−2 is excluded (round bracket) and 5 is included (square bracket), giving (−2, 5].
7. The solution set of |x − 3| < 2 is
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|y| < a means −a < y < a.
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Answer: A. (1, 5)
|x − 3| < 2 means −2 < x − 3 < 2, i.e. 1 < x < 5.
8. The solution set of |2x − 1| ≥ 5 is
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|y| ≥ a splits into y ≥ a or y ≤ −a.
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Answer: A. (−∞, −2] ∪ [3, ∞)
|2x − 1| ≥ 5 gives 2x − 1 ≥ 5 (x ≥ 3) or 2x − 1 ≤ −5 (x ≤ −2). So x ∈ (−∞, −2] ∪ [3, ∞).
9. The sum of all real roots of the equation x² − 5|x| + 6 = 0 is
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Treat |x| as the unknown in a quadratic.
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Answer: D. 0
Since x² = |x|², the equation is |x|² − 5|x| + 6 = 0, so |x| = 2 or 3, giving x = ±2, ±3. Their sum is 0.
10. The negation of the statement "All students passed the examination" is
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Negating 'all' gives 'at least one … not'.
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Answer: B. Some students did not pass the examination
The negation of "for all x, P(x)" is "there exists x for which P(x) is false", i.e. some students did not pass.
11. The conditional statement p → q is false only when
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A promise is broken only if the condition holds but the result doesn't.
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Answer: D. p is true and q is false
An implication fails only when the hypothesis is true and the conclusion is false.
12. The contrapositive of the statement p → q is
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Negate both parts and interchange them.
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Answer: A. ~q → ~p
The contrapositive negates and swaps both parts: ~q → ~p, and it is logically equivalent to p → q.
13. The statement ~(p ∧ ~q) is logically equivalent to
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Apply De Morgan's law and the double negation law.
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Answer: A. p → q
By De Morgan's law, ~(p ∧ ~q) ≡ ~p ∨ ~(~q) ≡ ~p ∨ q, and ~p ∨ q ≡ p → q.
14. Which of the following statements is a tautology?
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Check each statement for both truth values of p.
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Answer: D. p ∨ ~p
p ∨ ~p is true whether p is true or false (law of excluded middle). p ∧ ~p is always false (a contradiction).
15. If p is true and q is false, which of the following statements is true?
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Substitute the truth values step by step.
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Answer: B. (p ∨ q) → (p ∧ ~q)
p ∨ q = T and p ∧ ~q = T ∧ T = T, so T → T is true. p → q = F, p ↔ q = F and ~p ∨ q = F ∨ F = F.
1.2 Functions: types, inverse and composite
20 questions
16. The domain of f(x) = √(4 − x²) is
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The expression under a square root must be non-negative.
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Answer: D. [−2, 2]
We need 4 − x² ≥ 0, i.e. x² ≤ 4, so −2 ≤ x ≤ 2.
17. The domain of f(x) = log(x − 3) is
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The argument of a logarithm must be strictly positive.
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Answer: A. (3, ∞)
Logarithm is defined only for positive arguments: x − 3 > 0, so x > 3.
18. The domain of f(x) = 1/√(x² − 5x + 6) is
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Square root in the denominator: the radicand must be strictly positive.
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Answer: A. (−∞, 2) ∪ (3, ∞)
We need x² − 5x + 6 > 0 (strictly, since it is under a root in the denominator): (x − 2)(x − 3) > 0, so x < 2 or x > 3.
19. The range of f(x) = 3 sin x + 1 is
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Start from −1 ≤ sin x ≤ 1.
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Answer: C. [−2, 4]
−1 ≤ sin x ≤ 1 gives −3 ≤ 3 sin x ≤ 3, so −2 ≤ 3 sin x + 1 ≤ 4.
20. The range of f(x) = x²/(1 + x²), x ∈ ℝ, is
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Rewrite f(x) as 1 − 1/(1 + x²).
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Answer: A. [0, 1)
f(x) = 1 − 1/(1 + x²). Since 1 + x² ≥ 1, 0 < 1/(1 + x²) ≤ 1, so 0 ≤ f(x) < 1. f(0) = 0 but f(x) never equals 1.
21. Which of the following functions from ℝ to ℝ is one-one (injective)?
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Check whether two different inputs can give the same output.
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Answer: A. f(x) = 2x + 3
2x₁ + 3 = 2x₂ + 3 forces x₁ = x₂. The others repeat values, e.g. (−1)² = 1², |−1| = |1|, cos 0 = cos 2π.
22. The function f: ℝ → ℝ defined by f(x) = x² is
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Test f(−a) vs f(a), and look for a pre-image of −1.
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Answer: B. neither injective nor surjective
f(−1) = f(1), so it is not one-one; negative numbers have no pre-image, so it is not onto ℝ.
23. The number of one-one (injective) functions from a set with 3 elements to a set with 5 elements is
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Distinct elements must go to distinct images.
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Answer: B. 60
The first element has 5 choices, the second 4, the third 3: 5 × 4 × 3 = 60.
24. The function f: ℝ → [−1, 1] defined by f(x) = sin x is
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Compare the codomain with the range, then look for repeated values.
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Answer: B. surjective but not injective
Every value in [−1, 1] is taken by sin x, so it is onto; but sin 0 = sin π = 0, so it is not one-one.
25. If f(x) = 2x + 3 and g(x) = x², then (f∘g)(x) is
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f∘g means apply g first, then f.
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Answer: B. 2x² + 3
(f∘g)(x) = f(g(x)) = f(x²) = 2x² + 3.
26. If f(x) = (x + 1)/(x − 1), x ≠ 1, then f(f(x)) equals
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Substitute f(x) into f and simplify the compound fraction.
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Answer: A. x
f(f(x)) = [(x + 1)/(x − 1) + 1] / [(x + 1)/(x − 1) − 1] = [2x/(x − 1)] / [2/(x − 1)] = x.
27. The inverse of f(x) = (3x − 2)/5 is
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Write y = f(x) and solve for x.
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Answer: C. f⁻¹(x) = (5x + 2)/3
Put y = (3x − 2)/5, so 5y = 3x − 2 and x = (5y + 2)/3. Hence f⁻¹(x) = (5x + 2)/3.
28. If f(x) = (2x + 3)/(x − 2), x ≠ 2, then f⁻¹(x) is
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Set y = f(x), collect the x terms and solve.
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Answer: B. (2x + 3)/(x − 2)
y(x − 2) = 2x + 3 gives x(y − 2) = 2y + 3, so x = (2y + 3)/(y − 2). Thus f⁻¹(x) = (2x + 3)/(x − 2), i.e. f is its own inverse.
29. Which of the following is an even function?
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A function is even if f(−x) = f(x).
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Answer: C. f(x) = x sin x
(−x)·sin(−x) = (−x)(−sin x) = x sin x, so it is even. x³ + x and x cos x are odd; eˣ is neither.
30. The fundamental period of f(x) = sin 3x is
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Period of sin(kx) is 2π/|k|.
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Answer: C. 2π/3
sin(kx) has period 2π/k, so sin 3x has period 2π/3.
31. The fundamental period of f(x) = |sin x| + |cos x| is
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Try shifting x by π/2 and see how sin and cos swap.
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Answer: D. π/2
f(x + π/2) = |cos x| + |−sin x| = f(x). π/4 is not a period since f(0) = 1 but f(π/4) = √2. So the period is π/2.
32. For a > 1, the graph of y = logₐ x always passes through the point
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What is the logarithm of 1 in any base?
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Answer: B. (1, 0)
logₐ 1 = 0 for every valid base a, so the point (1, 0) lies on the graph.
33. If f(x) = log[(1 + x)/(1 − x)], −1 < x < 1, then f(2x/(1 + x²)) equals
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Simplify the argument; numerator and denominator become perfect squares.
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Answer: A. 2f(x)
(1 + 2x/(1 + x²)) / (1 − 2x/(1 + x²)) = (1 + x)²/(1 − x)². So f(2x/(1 + x²)) = log[(1 + x)/(1 − x)]² = 2f(x).
34. If f(x) = 3x − 1 and g(x) = x² + 2, then (g∘f)(1) − (f∘g)(1) is
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Evaluate the inner function first in each composition.
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Answer: D. −2
(g∘f)(1) = g(f(1)) = g(2) = 6; (f∘g)(1) = f(g(1)) = f(3) = 8. Difference = 6 − 8 = −2.
35. The range of f(x) = 3ˣ + 3⁻ˣ, x ∈ ℝ, is
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Use AM ≥ GM on the two positive terms.
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Answer: C. [2, ∞)
By AM ≥ GM, 3ˣ + 3⁻ˣ ≥ 2√(3ˣ·3⁻ˣ) = 2, with equality at x = 0; the function grows without bound. Range = [2, ∞).