Sets, Relations and Functions Questions and Answers

Answer: B

The value is 2.

Simplify \(h\) with the rules of logarithms:

\[ \begin{aligned} h(x) &= \ln 9x + \ln 9x^{2} + \ln 9x^{3} \\ &= 3\ln 9 + 6\ln x \end{aligned} \]

With \(x = \frac{1}{3}e^{1/3}\), \(\ln x = -\ln 3 + \frac{1}{3}\):

\[ \begin{aligned} h &= 6\ln 3 + 6\left(-\ln 3 + \tfrac{1}{3}\right) \\ &= 6\ln 3 - 6\ln 3 + 2 = 2 \end{aligned} \]

Here \(3\ln 9 = 6\ln 3\), which is why the logarithms cancel completely.

Answer: E

The answer is 54.

\((A \cap B)' \cap (A \cup B)\) is the set of students who play at least one game but not both, so it is the union minus the intersection.

\[ \begin{aligned} n(A \cup B) &= 45 + 35 - 13 = 67 \\ n &= 67 - 13 = 54 \end{aligned} \]

Check: 45 − 13 = 32 play only cricket and 35 − 13 = 22 play only football, and 32 + 22 = 54. 67 is the whole union, which is not offered, and none of the other values matches either count.

Answer: A

The domain is [5,7].

h is defined where both square roots are defined. \(\sqrt{7 - x}\) needs \(x \le 7\), and \(\sqrt{x - 5}\) needs \(x \ge 5\).

\[ (-\infty, 7] \cap [5, \infty) = [5, 7] \]

\((-\infty, 7]\) and \([5, \infty)\) are the domains of f and g separately, not of their product. [-7,-5] has the signs reversed, and \((-\infty, \infty)\) ignores both conditions. Both end points are included because each root is 0 there.

Answer: E

n(R) = 5.

A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7}. Since every a in A is odd, a + b is odd only when b is even, and the only even prime is 2.

So R = {(1,2), (3,2), (5,2), (7,2), (9,2)}, which has 5 elements.

Pairing an odd a with an odd prime (3, 5 or 7) gives an even sum, so those pairs are excluded. Counting only the odd primes, or only some of the five odd numbers, leads to the smaller options.

Answer: C

The keyed answer is an even function.

Replace \(x\) by \(-x\): \(\tan(-x)\sin(-x) = (-\tan x)(-\sin x) = \tan x \sin x\). So \(f(-x) = f(x)\) for every \(x\) in the domain, and \(f\) is even. That also rules out the two odd options and the option calling it 1-1 and odd, because an odd function satisfies \(f(-x) = -f(x)\).

The question is loosely set. Because \(f(-x) = f(x)\), distinct points \(x\) and \(-x\) give the same value, so \(f\) is also not one-one, which makes the first option a true statement too, and it is the more informative of the two. The key names the shorter property, even, and that is the one taught here. The function is also not onto \((-\infty, \infty)\), since \(\tan x \sin x \geq 0\) on the domain, and \(\tan x\) is not defined at the end points \(\pm \frac{\pi}{2}\) of the closed interval.

Answer: A

\(f^{-1}(g(x)) = \sin^{-1}(2 - x^{2})\).

Invert f: if \(y = 3\sin x\) with \(x \in [-\frac{\pi}{2}, \frac{\pi}{2}]\), then \(x = \sin^{-1}\frac{y}{3}\). So \(f^{-1}(y) = \sin^{-1}\frac{y}{3}\).

\[ \begin{aligned} f^{-1}(g(x)) &= \sin^{-1}\left(\frac{6 - 3x^{2}}{3}\right) \\ &= \sin^{-1}(2 - x^{2}) \end{aligned} \]

The options with a factor 3 or \(\frac{1}{3}\) outside the inverse sine put the 3 in the wrong place: it divides the argument. \(\sin^{-1}(6 - 3x^{2})\) forgets the division altogether.

Answer: D

The relation is reflexive, symmetric and transitive.

\(R\) contains exactly the pairs \((a_k, a_k)\), so it is the identity relation on \(X\). It is reflexive because every \((a_k, a_k)\) is in \(R\). It is symmetric because the only pairs are of the form \((a, a)\), and reversing them gives the same pair. It is transitive because \((a, b)\) and \((b, c)\) in \(R\) force \(a = b = c\), so \((a, c) = (a, a)\) is in \(R\).

The identity relation is therefore an equivalence relation, and every option that denies one of the three properties is wrong.

Answer: E

The true statement is one-to-one function cannot be defined from \(X\) to \(Y\).

\(X\) has 6 elements and \(Y\) has 5. A one-to-one function must send the 6 elements of \(X\) to 6 different elements of \(Y\), which is impossible (pigeonhole principle).

The sets in the first and third options leave \(f\) without an image, so they are not functions from \(X\) at all. The second sends both \(b\) and \(f\) to 11, so it is not one-to-one. The fourth sends both \(c\) and \(f\) to 9, so it is not one-to-one either.

Answer: C

\(h(h(10)) = 10\).

Compute \(f(f(x))\) directly:

\[ \begin{aligned} f(f(x)) &= \frac{2\cdot\frac{2x+3}{x-2} + 3}{\frac{2x+3}{x-2} - 2} \\ &= \frac{4x + 6 + 3x - 6}{2x + 3 - 2x + 4} \\ &= \frac{7x}{7} = x \end{aligned} \]

So \(f\) is its own inverse and \(h(x) = x\) for every \(x \neq 2\). Hence \(h(10) = 10\) and \(h(h(10)) = 10\).

Check: \(f(10) = \frac{23}{8}\) and \(f(\frac{23}{8}) = \frac{46/8 + 3}{23/8 - 2} = \frac{70/8}{7/8} = 10\). The other values come from treating \(h\) as a power or product of \(f\) rather than a composition.

Answer: A

The inverse is \(f^{-1}(x) = -2 - \sqrt{x}, x \geq 0\).

Write \(f(x) = (x + 2)^{2}\). On \(x \leq -2\) we have \(x + 2 \leq 0\), so the function is one-to-one there and its range is \([0, \infty)\).

\[ \begin{aligned} y &= (x + 2)^{2} \\ x + 2 &= -\sqrt{y} \quad (x + 2 \leq 0) \\ x &= -2 - \sqrt{y} \end{aligned} \]

The negative square root is taken because \(x + 2\) is not positive. The domain of the inverse is the range of \(f\), namely \(x \geq 0\). The options restricting the domain to \(x \geq 1, 2, 4\) or 5 leave out values that \(f\) does take, and \(\sqrt{x - 1}\) does not undo \((x + 2)^{2}\).