Sets, Relations and Functions Questions and Answers

Answer: A

The domain is \([1, 7]\).

The square root needs a non-negative argument:

\[ \begin{aligned} 8x - x^{2} - 7 &\ge 0 \\ x^{2} - 8x + 7 &\le 0 \\ (x - 1)(x - 7) &\le 0 \;\Rightarrow\; 1 \le x \le 7 \end{aligned} \]

Raising the root to the power \(\frac{3}{2}\) keeps it real, since the root is already non-negative. \([3, 7]\) and \([1, 4]\) are only parts of the domain, \([-7, -1]\) has the signs reversed, and \([-3, 3]\) is unrelated.

Answer: D

The relation is not symmetric.

A relation is symmetric when every pair (a, b) comes with its reverse (b, a). R contains (2, 3) but not (3, 2), and (2, 4) but not (4, 2), so it fails the test.

The other properties fail too. R is not reflexive, since (1, 1), (2, 2) and (3, 3) are missing. It is not transitive: (2, 3) and (3, 1) are in R but (2, 1) is not. It is not a function at all, because 2 is paired with both 3 and 4, so it can be neither one-one nor onto.

Answer: A

The number of common elements is 121.

An ordered pair (x, y) lies in both \(A \times B\) and \(B \times A\) exactly when x is in A and in B, and y is in B and in A. So

\[ \begin{aligned} &(A \times B) \cap (B \times A) \\ &= (A \cap B) \times (A \cap B) \end{aligned} \]

With \(n(A \cap B) = 11\), this set has \(11 \times 11 = 121\) elements. 22 adds the two counts instead of multiplying them, and 11 forgets that both coordinates range over the common elements.

Answer: D

The domain is \((-\infty, -2] \cup [1, \infty)\).

The square root is real only when the expression under it is non-negative.

\[ x^{2} + x - 2 = (x + 2)(x - 1) \ge 0 \]

A product of two factors is non-negative when both have the same sign or one is zero, which happens for \(x \le -2\) or \(x \ge 1\). Both end points belong to the domain, since \(\sqrt{0} = 0\) is defined. The options with round brackets at -2 or 1 wrongly leave out those points, and \((-\infty, 1) \cup [0, \infty)\) is the whole real line.

Answer: C

The relation is reflexive.

Every element of A is related to itself: (4, 4), (5, 5), (7, 7), (8, 8) and (9, 9) are all in R.

It is not symmetric, because (4, 5) is in R but (5, 4) is not. It is not transitive, because (4, 5) and (5, 7) are in R but (4, 7) is not. Without symmetry and transitivity it is not an equivalence relation, and it is not a function because 4 is paired with 4, 5 and 8.

Answer: D

The number of common elements is 4.

An ordered pair (x, y) lies in both \(A \times B\) and \(B \times A\) only if x and y each belong to both sets, so

\[ \begin{aligned} &(A \times B) \cap (B \times A) \\ &= (A \cap B) \times (A \cap B) \end{aligned} \]

Here \(A \cap B = \{3, 4\}\), which gives \(2 \times 2 = 4\) pairs: (3, 3), (3, 4), (4, 3), (4, 4).

2 counts only the common elements of the sets, not the pairs, and 8 or 12 count pairs that lie in only one of the products.

Answer: C

The intersection is {24}.

List the sets. A holds the positive even numbers below 36: 2, 4, ..., 34. B holds the multiples of 3 above 16: 18, 21, 24, 27, 30, 33, 36, ... C holds the multiples of 4 below 42: 4, 8, ..., 40.

\(A \cap B\) keeps the even multiples of 3 between 16 and 36, that is the multiples of 6: {18, 24, 30} (36 is not in A, which stops below 36). Of these only 24 is a multiple of 4, so \((A \cap B) \cap C = \{24\}\).

{12, 24} and {12, 24, 36} include 12, which is not greater than 16, and 36 is excluded from A, so {24, 36} fails too. The intersection is not empty, so φ is wrong.

Answer: D

The number of such subsets is 56.

A set of 8 elements has \(\binom{8}{r}\) subsets with exactly r elements. Subsets with 2 elements and subsets with 6 elements are disjoint groups, so add:

\[ \binom{8}{2} + \binom{8}{6} = 28 + 28 = 56 \]

The two counts are equal because choosing 2 elements to keep is the same as choosing 6 to leave out. 28 counts only one of the two groups. 216 does not come from a binomial coefficient of 8.

Answer: B

The image is {-5, 5}.

Double each element first, then take the greatest integer not exceeding the result:

\[ \begin{aligned} f(-2.3) &= [-4.6] = -5 \\ f(2.9) &= [5.8] = 5 \end{aligned} \]

The trap is the negative value: the greatest integer not exceeding -4.6 is -5, not -4, because -4 is larger than -4.6. That rules out {-4, 5} and {-4, 6}. {-5, 3} and {-3, 2} each get the value at 2.9 wrong, since 2 times 2.9 is 5.8, whose integer part is 5.

Answer: E

The coefficient of \(x^{3}\) is \(2ab^{2}\).

Substitute \(f(x)\) into itself:

\[ \begin{aligned} f(f(x)) &= a(ax + bx^{2}) + b(ax + bx^{2})^{2} \\ &= a^{2}x + abx^{2} + b(a^{2}x^{2} + 2abx^{3} + b^{2}x^{4}) \end{aligned} \]

The only \(x^{3}\) term comes from the middle term of the square, \(b \times 2abx^{3} = 2ab^{2}x^{3}\). \(ab^{2}\) forgets the factor 2 in \((p + q)^{2} = p^{2} + 2pq + q^{2}\); \(ab\) is the coefficient of one of the \(x^{2}\) terms, and \(a^{3}\) does not occur.