15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
On \(A = \{1, 2, 3\}\), consider \(R = \{(1,1), (2,2), (3,3), (1,2)\}\). Which property does \(R\) fail to have?
Why: (1, 2) is in R but (2, 1) is not.
Reflexive: \((1,1), (2,2), (3,3) \in R\). Transitive: the only chains \((1,1),(1,2)\) and \((1,2),(2,2)\) give \((1,2) \in R\). Not symmetric: \((1,2) \in R\) but \((2,1) \notin R\).
The number of symmetric relations on a set \(\{a, b\}\) with two elements is
Why: (a,a) and (b,b) are free; (a,b) and (b,a) must come together: 2 × 2 × 2.
Choices: include \((a,a)\) or not, include \((b,b)\) or not, include the pair \(\{(a,b), (b,a)\}\) or not: \(2^3 = 8\).
Which of the following functions from \(\mathbb{R}\) to \(\mathbb{R}\) is a bijection?
Why: 3x − 5 is one-one, and every y has the pre-image (y + 5)/3.
\(3x_1 - 5 = 3x_2 - 5 \Rightarrow x_1 = x_2\), and \(x = \dfrac{y + 5}{3}\) maps to any \(y\). The others fail: \(x^2\) and \(|x|\) give \(f(-1) = f(1)\), and \(\sin x\) is never \(2\).
Which of the following functions from \(\mathbb{R}\) to \(\mathbb{R}\) is neither one-one nor onto?
Why: x² + 1 takes the same value at ±1 and never takes values below 1.
\(x^2 + 1\): \(f(-1) = f(1) = 2\) (not one-one) and \(f(x) \ge 1\), so \(0\) has no pre-image (not onto). \(2x + 1\) and \(x^3\) are bijections; \(e^x\) is one-one but not onto.
The number of ordered pairs in the smallest equivalence relation on \(\{a, b, c\}\) that contains \((a, b)\) is
Why: It needs (a,a), (b,b), (c,c), (a,b) and, for symmetry, (b,a).
Reflexivity needs \((a,a), (b,b), (c,c)\); \((a,b)\) is given; symmetry adds \((b,a)\). This set of \(5\) pairs is already transitive.
On \(\mathbb{Z}\), \(a\,R\,b\) if and only if \(a - b\) is divisible by \(5\). The equivalence class \([2]\) is
Why: b ∈ [2] exactly when b − 2 is a multiple of 5.
\(b \in [2] \Leftrightarrow 5 \mid (b - 2) \Leftrightarrow b = 5k + 2\): \([2] = \{\ldots, -8, -3, 2, 7, 12, \ldots\}\).
Which of the following functions \(f : \mathbb{N} \to \mathbb{N}\) is one-one but not onto?
Why: 2m = 2n forces m = n, but odd numbers are never reached.
\(2n\): one-one, and \(3\) has no pre-image. \(f(n) = n\) is a bijection; \(|n - 3| + 1\) is onto but \(f(2) = f(4)\); the constant \(5\) is neither.
The number of one-one functions from \(\{1, 2\}\) to \(\{a, b, c, d, e\}\) is
Why: 1 has 5 choices, then 2 has 4 remaining.
\(5 \times 4 = 20\). (All functions: \(5^2 = 25\); one-one ones need different images.)
How many functions from \(A = \{1, 2, 3, 4\}\) onto \(B = \{a, b\}\) (every element of \(B\) is an image) are there?
Why: All 2⁴ = 16 functions except the 2 constant ones.
There are \(2^4 = 16\) functions; only the two constant functions miss a value. Onto functions: \(16 - 2 = 14\).
\(R = \{(x, y) : x + 2y = 8\}\) is a relation on \(\mathbb{N}\). The domain of \(R\) is
Why: y = 1, 2, 3 give x = 6, 4, 2 (y = 4 gives x = 0 ∉ ℕ).
\(x = 8 - 2y\) with \(x, y \in \mathbb{N}\): \(y = 1, 2, 3\) give \(x = 6, 4, 2\). Domain \(= \{2, 4, 6\}\).
On the set of all lines in a plane, the relation \(L_1\,R\,L_2\) if \(L_1\) is perpendicular to \(L_2\) is
Why: L₁ ⟂ L₂ gives L₂ ⟂ L₁, but no line is ⟂ itself, and L₁ ⟂ L₂ ⟂ L₃ makes L₁ ∥ L₃.
Symmetric: yes. Not reflexive: a line is not perpendicular to itself. Not transitive: if \(L_1 \perp L_2\) and \(L_2 \perp L_3\) then \(L_1 \parallel L_3\), not perpendicular.
The number of relations on \(A = \{1, 2, 3\}\) that contain both \((1, 2)\) and \((2, 1)\) is
Why: Two of the 9 pairs are fixed; the other 7 are free.
\(A \times A\) has \(9\) pairs; \((1,2)\) and \((2,1)\) must be in, and each of the other \(7\) may be in or out: \(2^7 = 128\).
For \(f : \mathbb{R} - \{-1\} \to \mathbb{R}\), \(f(x) = \dfrac{x}{x + 1}\), which real number is not in the range of \(f\)?
Why: x/(x + 1) = 1 would need x = x + 1; so f is not onto.
\(\dfrac{x}{x + 1} = y \Rightarrow x = \dfrac{y}{1 - y}\), defined for every \(y \ne 1\). For \(y = 1\): \(x = x + 1\), impossible. So \(1\) is not in the range (and \(f\) is not onto).
For \(f : \mathbb{R} \to \mathbb{R}\), \(f(x) = x^2 - 2\), the set of pre-images of \(7\) is
Why: x² − 2 = 7 gives x² = 9.
\(x^2 - 2 = 7 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3\). (\(47 = f(7)\) is the image of \(7\), not a pre-image.)
On \(\{1, 2, 3, 4\}\), \(R = \{(1,1), (2,2), (3,3), (4,4), (1,2), (2,1)\}\). The number of distinct equivalence classes of \(R\) is
Why: The classes are {1, 2}, {3} and {4}.
\(R\) is an equivalence relation with classes \([1] = [2] = \{1, 2\}\), \([3] = \{3\}\), \([4] = \{4\}\): three classes.
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A school allots its \(12\) new students, with roll numbers \(1\) to \(12\), to three houses by roll number: students whose roll numbers leave the same remainder on division by \(3\) go to the same house. On \(A = \{1, 2, \ldots, 12\}\) define \(a\,R\,b\) if and only if \(a - b\) is divisible by \(3\).
A weather app converts a temperature of \(x\) degrees Celsius into degrees Fahrenheit using \(f(x) = \dfrac{9x}{5} + 32\). Treat \(f\) as a function from \(\mathbb{R}\) to \(\mathbb{R}\).
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