Sets, Relations and Functions Questions and Answers

Answer: B

\(f(x)\) is \(\frac{11 - 4x}{8}\).

Replace x by \(1 - x\) to get a second equation, then solve the pair:

\[ \begin{aligned} f(x) + 3f(1 - x) &= x + 4 \\ f(1 - x) + 3f(x) &= 5 - x \end{aligned} \]

Multiply the second by 3 and subtract the first: \(9f(x) - f(x) = 15 - 3x - x - 4\), so \(8f(x) = 11 - 4x\).

Check at x = 0: f(0) = 11/8 and f(1) = 7/8, and \(\frac{11}{8} + 3 \times \frac{7}{8} = 4\), as required. \(\frac{19 - 4x}{8}\) and the options over 9 fail this check; for example \(\frac{19}{8} + 3 \times \frac{15}{8}\) is far from 4.

Answer: B

The values are \(m = 6\) and \(n = 4\).

A set with \(k\) elements has \(2^{k}\) subsets, so the condition is a difference of two powers of 2:

\[ \begin{aligned} 2^{m} - 2^{n} &= 48 \\ 2^{n}(2^{m-n} - 1) &= 16 \times 3 \end{aligned} \]

The bracket is odd, so \(2^{n} = 16\) and \(2^{m-n} - 1 = 3\), giving \(n = 4\) and \(m - n = 2\), so \(m = 6\). Check: \(64 - 16 = 48\).

The other pairs fail: 6,3 gives \(64 - 8 = 56\); 5,6 and 2,6 make A the smaller set, so the difference is negative; 7,1 gives \(128 - 2 = 126\).

Answer: A

The ratio equals \(f(a^{2})\).

Substitute into \(f(x) = \frac{x}{x-1}\):

\[ \begin{aligned} f(a) &= \frac{a}{a-1}, \quad f(a+1) = \frac{a+1}{a} \\ \frac{f(a)}{f(a+1)} &= \frac{a}{a-1} \cdot \frac{a}{a+1} \\ &= \frac{a^{2}}{a^{2} - 1} \end{aligned} \]

That is exactly \(f\) evaluated at \(a^{2}\), since \(f(a^{2}) = \frac{a^{2}}{a^{2}-1}\).

The other choices do not match: \(f(-a) = \frac{a}{a+1}\), \(f(-a^{2}) = \frac{a^{2}}{a^{2}+1}\), \(f(1/a) = \frac{1}{1-a}\) and \(f(1/a^{2}) = \frac{1}{1-a^{2}}\).

Answer: D

The sum is 18.

A function with \(f(x+y) = f(x) + f(y)\) is additive. Putting \(x = y = 0\) gives \(f(0) = 2f(0)\), so \(f(0) = 0\). Repeated use gives \(f(n) = n f(1)\) for whole numbers, so with \(f(1) = 3\):

\[ \begin{aligned} f(2) &= f(1) + f(1) = 6 \\ f(3) &= f(2) + f(1) = 9 \\ \text{sum} &= 0 + 3 + 6 + 9 = 18 \end{aligned} \]

Once \(f(0) = 0\) and \(f(n) = 3n\) are fixed, the sum is determined, so 12, 14, 16 and 22 cannot occur. The common slip is to assume \(f(0) = f(1)\), which gives 21, not an option; 12 is \(f(1) + f(3)\) alone.

Answer: E

The domain is \([-4, -1] \cup [1, 4]\).

Both square roots need non-negative arguments:

\[ \begin{aligned} |x| - 1 \ge 0 &\Rightarrow |x| \ge 1 \\ 4 - |x| \ge 0 &\Rightarrow |x| \le 4 \end{aligned} \]

So \(1 \le |x| \le 4\), which on the number line is \(-4 \le x \le -1\) or \(1 \le x \le 4\). All four endpoints are included, because a square root of zero is defined.

The options with round brackets wrongly exclude an endpoint such as \(x = 1\) (where \(f(1) = 0 + \sqrt{3}\)) or \(x = 4\). \([-4, -1]\) alone forgets the positive half.

Answer: E

\(n(A') = 50\).

First find \(n(A)\) from the union formula, then subtract it from the universal set.

\[ \begin{aligned} n(A \cup B) &= n(A) + n(B) - n(A \cap B) \\ 99 &= n(A) + 61 - 28 \\ n(A) &= 66 \\ n(A') &= 116 - 66 = 50 \end{aligned} \]

Check: \(66 + 61 - 28 = 99\), as given. The common slip is to subtract \(n(A \cup B)\) or \(n(B)\) from 116 directly; \(116 - 61 = 55\) and \(116 - 99 = 17\) are not \(n(A')\), because \(A'\) contains the part of \(B\) outside \(A\) as well as everything outside both sets.

Answer: B

The largest domain is \([1, 3]\).

\(f(g(x)) = \sin^{-1}(x - 2)\), and \(\sin^{-1}\) is defined only for arguments between \(-1\) and \(1\).

\[ \begin{aligned} -1 &\le x - 2 \le 1 \\ 1 &\le x \le 3 \end{aligned} \]

The interval \([-1, 3]\) or \([-3, 3]\) would let \(x - 2\) fall below \(-1\), and \([2, 5]\) lets it exceed 1, so \(f \circ g\) would be undefined at some points. \([0, 2]\) works but is not the largest such interval.

Answer: C

The range is \(\left[\frac{5}{6}, 5\right]\).

The expression \(4\sin x + 3\cos x\) can be written as \(5\sin(x + \phi)\), so it takes every value from \(-5\) to \(5\), since \(\sqrt{4^{2} + 3^{2}} = 5\).

\[ \begin{aligned} 7 - 5 &\le 7 + 4\sin x + 3\cos x \le 7 + 5 \\ 2 &\le \text{denominator} \le 12 \\ \tfrac{10}{12} &\le f(x) \le \tfrac{10}{2} \end{aligned} \]

So \(f\) runs from \(\frac{5}{6}\), when the denominator is largest, to 5, when it is smallest. Every other option has a different lower bound, which would need a different maximum value of the denominator than 12.

Answer: A

The value is 30.

Compute each part.

\[ \begin{aligned} g(2x) &= e^{2x} + 5 \\ f(g(x)) &= (e^{x} + 5)^{2} - 10(e^{x} + 5) \\ &= e^{2x} + 10e^{x} + 25 - 10e^{x} - 50 \\ &= e^{2x} - 25 \end{aligned} \]

\[ g(2x) - (f \circ g)(x) = (e^{2x} + 5) - (e^{2x} - 25) = 30 \]

The \(e^{2x}\) terms cancel, so the result is the same constant for every \(x\). Getting \(-30\) means the subtraction was done the other way round, and 20 or \(-20\) come from mishandling the constant 5.

Answer: C

\(f^{-1}(x) = \tan^{-1}\left(\frac{1 - x}{x}\right)\).

Put \(y = \frac{1}{1 + \tan x}\) and solve for \(x\):

\[ \begin{aligned} 1 + \tan x &= \frac{1}{y} \\ \tan x &= \frac{1 - y}{y} \\ x &= \tan^{-1}\left(\frac{1 - y}{y}\right) \end{aligned} \]

Renaming \(y\) as \(x\) gives the inverse. For \(0 \lt x \lt \frac{\pi}{2}\), \(\tan x \gt 0\) so \(0 \lt y \lt 1\) and \(\tan^{-1}\) returns the right angle. The options \(\frac{1 - \tan x}{\tan x}\) and \(\frac{\tan x}{1 - \tan x}\) are not inverses at all, only rearrangements of \(\tan x\).