Sets, Relations and Functions Questions and Answers

Answer: C

The domain is \((5, \infty)\).

Two conditions apply. The logarithm needs \(x - 5 \gt 0\), so \(x \gt 5\). The denominator must not be zero: \(x^{2} + 3x - 4 = (x + 4)(x - 1)\), so \(x \neq -4, 1\).

Both excluded values already lie outside \(x \gt 5\), so the domain is simply \((5, \infty)\).

\(\mathbb{R} \setminus \{-4, 1\}\) ignores the logarithm, \((1, \infty)\) includes values where \(x - 5 \leq 0\), and \((10, \infty)\) leaves out valid points such as \(x = 6\).

Answer: B

P has 2 elements.

\(A \setminus B = \{d\}\) means that, of the elements of A, only d is missing from B. So B must contain a, b and c, and must not contain d. The only element of X still undecided is e, which B may or may not contain:

\[ P = \{\{a, b, c\},\ \{a, b, c, e\}\} \]

So P has 2 elements. Counting all subsets of X (32), or treating e as forced in or out, leads to the wrong totals such as 1, 4 or 8.

Answer: E

R becomes reflexive on adding (2,2), (3,3), (4,4), (5,5), (6,6) and (7,7).

A relation on X is reflexive when \((x, x) \in R\) for every \(x \in X\). R already contains (1,1) but none of (2,2) to (7,7), so all six must be added.

The other options each fail somewhere: one swaps (2,2) for (2,7), one replaces (4,4) with (4,3), one has (7,6) instead of (7,7), and one leaves out (4,4) and (6,6). Pairs such as (2,7) or (4,3) do not help, since only pairs of the form \((x, x)\) count.

Answer: C

The value is 100.

The function has the form \(f(x) = \frac{ax + b}{cx - a}\) with \(a = 2025\), \(b = 2026\), \(c = 2027\). Such a function is its own inverse. Writing \(f(x) = y\):

\[ \begin{aligned} y(cx - a) &= ax + b \\ x(cy - a) &= ay + b \\ x &= \frac{ay + b}{cy - a} = f(y) \end{aligned} \]

So \(f(f(x)) = x\): applying \(f\) twice returns the starting value. 1000 is even, so \(f^{1000}\) is 500 double applications, and \(f^{1000}(100) = 100\). The large options come from trying to compute \(f(100)\) numerically.

Answer: E

The domain of the inverse is \([2, \infty)\).

The domain of \(f^{-1}\) is the range of \(f\). Complete the square:

\[ f(x) = x^{2} + 2x + 3 = (x + 1)^{2} + 2 \]

For \(x \leq -1\), \((x + 1)^{2}\) takes every value from 0 (at \(x = -1\)) upward, so \(f(x)\) takes every value from 2 upward. On this domain \(f\) is one-one (decreasing), so the inverse exists, with domain \([2, \infty)\). The other intervals do not start at the minimum value 2.

Answer: D

The range of R is the set \(\{1, 2, 3\}\), printed in the key's option as \(\{1, 2, 3,\}\).

Find the natural-number pairs with \(x + 3y = 10\), that is \(x = 10 - 3y \ge 1\):

• \(y = 1\): \(x = 7\)
• \(y = 2\): \(x = 4\)
• \(y = 3\): \(x = 1\)

\(y = 4\) would give \(x = -2\), which is not a natural number. So \(R = \{(7, 1), (4, 2), (1, 3)\}\) and its range, the set of second elements, is \(\{1, 2, 3\}\).

\(\{1, 2, 3, 4\}\) and \(\{2, 3, 4\}\) wrongly include 4, and \(\{1, 3, 4\}\) and \(\{1, 3\}\) leave out 2. The trailing comma in the keyed option is a printing slip; the set it lists is \(\{1, 2, 3\}\).

Answer: A

The domain is \(\mathbb{R}\), all real numbers.

A rational function is undefined only where its denominator is zero. The denominator \(x^{2} + x + 1\) has discriminant \(1 - 4 = -3 \lt 0\), so it has no real roots; in fact

\[ \begin{aligned} x^{2} + x + 1 &= \left(x + \tfrac{1}{2}\right)^{2} + \tfrac{3}{4} \\ &\gt 0 \end{aligned} \]

for every real \(x\). Since the denominator never vanishes, no real number is excluded, and the options removing 2, 1, 0 or \(\pm 1\) are wrong; none of those values makes the denominator zero.

Answer: C

There are 42 such subsets.

\(A \times B\) has \(3 \times 2 = 6\) elements, so it has \(2^{6} = 64\) subsets in all. Remove those with fewer than three elements:

\[ \begin{aligned} \text{fewer than 3} &= \binom{6}{0} + \binom{6}{1} + \binom{6}{2} \\ &= 1 + 6 + 15 = 22 \\ \text{at least 3} &= 64 - 22 = 42 \end{aligned} \]

64 is the total number of subsets, and 52 and 60 are too large: they leave in some of the 22 subsets with fewer than three elements.

Answer: E

The set has 8 elements.

Since \(3y^{2} \le 35\), \(y^{2}\) can only be 0, 1, 4 or 9. Check each:

• \(y^{2} = 0\): \(2x^{2} = 35\), not possible in integers.
• \(y^{2} = 1\): \(2x^{2} = 32\), \(x^{2} = 16\), \(x = \pm 4\).
• \(y^{2} = 4\): \(2x^{2} = 23\), not possible.
• \(y^{2} = 9\): \(2x^{2} = 8\), \(x^{2} = 4\), \(x = \pm 2\).

So \(x = \pm 4, y = \pm 1\) gives 4 pairs and \(x = \pm 2, y = \pm 3\) gives 4 more: 8 in all. Counting only positive values gives 2, and counting signs for one variable only gives 4.

Answer: B

The value is \(\lambda = -1\).

A function is its own inverse when \(f(f(x)) = x\) for every x.

\[ \begin{aligned} f(f(x)) &= \lambda(\lambda x + 1) + 1 \\ &= \lambda^{2}x + \lambda + 1 \end{aligned} \]

For this to equal x we need \(\lambda^{2} = 1\) and \(\lambda + 1 = 0\), so \(\lambda = -1\), giving \(f(x) = 1 - x\), which is indeed its own inverse. \(\lambda = 1\) satisfies the first condition but gives \(f(f(x)) = x + 2\), and the fractional values fail both.