Every Book Back multiple-choice question from Relations and Functions (10th Standard Mathematics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
If \(n(A \\times B) = 6\) and \(A = \\{1, 3\\}\) then \(n(B)\) is
- A. 1
- B. 2
- C. 3Correct
- D. 6
Explanation. Since \(n(A \\times B) = n(A) \\times n(B) = 6\) and \(A = \\{1, 3\\}\) implies \(n(A) = 2\), we have \(2 \\times n(B) = 6\). Solving this gives \(n(B) = 3\).
Q2
If \(A = \\{a, b, p\\}\), \(B = \\{2, 3\\}\), \(C = \\{p, q, r, s\\}\) then \(n((A \\cup C) \\times B)\) is
- A. 8
- B. 20
- C. 12Correct
- D. 16
Explanation. We find \(A \\cup C = \\{a, b, p, q, r, s\\}\) so \(n(A \\cup C) = 6\). Given \(n(B) = 2\), the cardinality of the Cartesian product is \(n((A \\cup C) \\times B) = 6 \\times 2 = 12\).
Q3
If \(A = \\{1, 2\\}\), \(B = \\{1, 2, 3, 4\\}\), \(C = \\{5, 6\\}\) and \(D = \\{5, 6, 7, 8\\}\) then state which of the following statement is true.
- A. \((A \\times C) \\subset (B \\times D)\)Correct
- B. \((B \\times D) \\subset (A \\times C)\)
- C. \((A \\times B) \\subset (A \\times D)\)
- D. \((D \\times A) \\subset (B \\times A)\)
Explanation. Since \(A = \\{1, 2\\} \\subset B = \\{1, 2, 3, 4\\}\) and \(C = \\{5, 6\\} \\subset D = \\{5, 6, 7, 8\\}\), it directly follows that the Cartesian product \(A \\times C\) is a subset of \(B \\times D\).
Q4
If there are 1024 relations from a set \(A = \\{1, 2, 3, 4, 5\\}\) to a set \(B\), then the number of elements in \(B\) is
- A. 3
- B. 2Correct
- C. 4
- D. 8
Explanation. The total number of relations from \(A\) to \(B\) is given by \(2^{n(A) \\times n(B)} = 1024 = 2^{10}\). Since \(n(A) = 5\), we have \(5 \\times n(B) = 10\), which gives \(n(B) = 2\).
Q5
The range of the relation \(R = \\{(x, x^2) \\mid x \\text{ is a prime number less than } 13\\}\) is
- A. \(\\{2, 3, 5, 7\\}\)
- B. \(\\{2, 3, 5, 7, 11\\}\)
- C. \(\\{4, 9, 25, 49, 121\\}\)Correct
- D. \(\\{1, 4, 9, 25, 49, 121\\}\)
Explanation. The prime numbers less than 13 are 2, 3, 5, 7, and 11. The range consists of the squares of these primes: \(2^2=4\), \(3^2=9\), \(5^2=25\), \(7^2=49\), and \(11^2=121\).
Q6
If the ordered pairs \((a+2, 4)\) and \((5, 2a+b)\) are equal then \((a, b)\) is
- A. (2, -2)
- B. (5, 1)
- C. (2, 3)
- D. (3, -2)Correct
Explanation. Equating the corresponding coordinates, we get \(a + 2 = 5 \\implies a = 3\). Then, \(2a + b = 4 \\implies 2(3) + b = 4 \\implies b = -2\). Thus, \((a, b) = (3, -2)\).
Q7
Let \(n(A) = m\) and \(n(B) = n\) then the total number of non-empty relations that can be defined from \(A\) to \(B\) is
- A. \(mn\)
- B. \(n^m\)
- C. \(2^{mn} - 1\)Correct
- D. \(2^{mn}\)
Explanation. The total number of relations from \(A\) to \(B\) is \(2^{n(A) \\times n(B)} = 2^{mn}\). To find the number of non-empty relations, we subtract the empty relation, yielding \(2^{mn} - 1\).
Q8
If \(\\{(a, 8), (6, b)\\}\) represents an identity function, then the value of \(a\) and \(b\) are respectively
- A. (8, 6)Correct
- B. (8, 8)
- C. (6, 8)
- D. (6, 6)
Explanation. An identity function maps each element to itself. For \(\\{(a, 8), (6, b)\\}\) to be an identity function, we must have \(a = 8\) and \(b = 6\). Thus, \((a, b) = (8, 6)\).
Q9
Let \(A = \\{1, 2, 3, 4\\}\) and \(B = \\{4, 8, 9, 10\\}\). A function \(f: A \\to B\) given by \(f = \\{(1, 4), (2, 8), (3, 9), (4, 10)\\}\) is a
- A. Many-one function
- B. Identity function
- C. One-to-one functionCorrect
- D. Into function
Explanation. Distinct elements in the domain \(A\) map to distinct elements in the codomain \(B\). Thus, the function is one-to-one.
Q10
If \(f(x) = 2x^2\) and \(g(x) = \\frac{1}{3x}\), then \(f \\circ g\) is
- A. \(\\frac{3}{2x^2}\)
- B. \(\\frac{2}{3x^2}\)
- C. \(\\frac{2}{9x^2}\)Correct
- D. \(\\frac{1}{6x^2}\)
Explanation. Substituting \(g(x)\) into \(f(x)\), we get \((f \\circ g)(x) = f(g(x)) = f(\\frac{1}{3x}) = 2(\\frac{1}{3x})^2 = 2(\\frac{1}{9x^2}) = \\frac{2}{9x^2}\).
Q11
If \(f: A \\to B\) is a bijective function and if \(n(B) = 7\), then \(n(A)\) is equal to
- A. 7Correct
- B. 49
- C. 1
- D. 14
Explanation. A bijective function is both one-to-one and onto, which requires the domain and codomain to have the same cardinality. Since \(n(B) = 7\), we must have \(n(A) = 7\).
Q12
Let \(f\) and \(g\) be two functions given by \(f = \\{(0, 1), (2, 0), (3, -4), (4, 2), (5, 7)\\}\) and \(g = \\{(0, 2), (1, 0), (2, 4), (-4, 2), (7, 0)\\}\), then the range of \(f \\circ g\) is
- A. \(\\{0, 2, 3, 4, 5\\}\)
- B. \(\\{-4, 1, 0, 2, 7\\}\)
- C. \(\\{1, 2, 3, 4, 5\\}\)
- D. \(\\{0, 1, 2\\}\)Correct
Explanation. The domain of \(g\) is \(\\{0, 1, 2, -4, 7\\}\). Finding \((f \\circ g)(x)\) for these inputs gives \(f(g(0))=0\), \(f(g(1))=1\), \(f(g(2))=2\), \(f(g(-4))=0\), and \(f(g(7))=1\). Thus, the range is \(\\{0, 1, 2\\}\).
Q13
Let \(f(x) = \\sqrt{1+x^2}\) then
- A. \(f(xy) = f(x) \\cdot f(y)\)
- B. \(f(xy) \\ge f(x) \\cdot f(y)\)
- C. \(f(xy) \\le f(x) \\cdot f(y)\)Correct
- D. None of these
Explanation. We have \(f(xy) = \\sqrt{1 + x^2y^2}\) and \(f(x)f(y) = \\sqrt{1 + x^2 + y^2 + x^2y^2}\). Since \(1 + x^2y^2 \\le 1 + x^2 + y^2 + x^2y^2\), we obtain \(f(xy) \\le f(x) \\cdot f(y)\).
Q14
If \(g = \\{(1, 1), (2, 3), (3, 5), (4, 7)\\}\) is a function given by \(g(x) = \\alpha x + \\beta\) then the values of \(\\alpha\) and \(\\beta\) are
- A. (-1, 2)
- B. (2, -1)Correct
- C. (-1, -2)
- D. (1, 2)
Explanation. Using \(g(1)=1 \\implies \\alpha+\\beta=1\) and \(g(2)=3 \\implies 2\\alpha+\\beta=3\), subtracting the first from the second gives \(\\alpha=2\). Substituting back yields \(\\beta=-1\). Thus, \((\\alpha, \\beta) = (2, -1)\).
Q15
Let \(f(x) = (x+1)^3 - (x-1)^3\) represents a function which is
- A. linear
- B. cubic
- C. reciprocal
- D. quadraticCorrect
Explanation. Expanding the expression gives \(f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) = 6x^2 + 2\), which is a quadratic function of degree 2.
