Mathematics Questions and Answers

Answer: A

\(\psi(\phi(x)) = 2^{x^{2}}\).

Composition means applying \(\phi\) first and then \(\psi\) to the result. Since \(\phi(x) = x^{2}\), substitute \(x^{2}\) for \(x\) in \(\psi(x) = 2^{x}\):

\[ \psi(\phi(x)) = \psi(x^{2}) = 2^{x^{2}} \]

\(x^{2}\) is \(\phi(x)\) alone. \(2^{2x}\) would be \(\psi(2x)\), which doubles \(x\) instead of squaring it. \(x^{2x}\) mixes the two functions without composing them. Note also that the order matters: \(\phi(\psi(x)) = (2^{x})^{2} = 2^{2x}\), which is a different function.

Answer: E

\(n(A) = \) 20.

The number of elements in a Cartesian product is the product of the sizes: \(n(A \times B) = n(A)\,n(B)\). Write \(a, b, c\) for the three sizes:

\[ \begin{aligned} ab &= 160,\quad bc = 80,\quad ca = 200 \\ (abc)^{2} &= 160 \times 80 \times 200 = 2560000 \\ abc &= 1600 \\ a &= \frac{abc}{bc} = \frac{1600}{80} = 20 \end{aligned} \]

Check: then \(b = 8\) and \(c = 10\), and \(20 \times 8 = 160\), \(8 \times 10 = 80\), \(10 \times 20 = 200\). The other values (10, 12, 16, 18) do not divide the three products consistently.

Answer: B

\(f^{-1}(5) = \) \(\{-2, 12\}\).

The inverse image of 5 is the set of all real \(x\) with \(f(x) = 5\):

\[ \begin{aligned} x^{2} - 10x - 19 &= 5 \\ x^{2} - 10x - 24 &= 0 \\ (x - 12)(x + 2) &= 0 \end{aligned} \]

So \(x = 12\) or \(x = -2\). The other sets have a sign wrong on one root (a common slip when factorising \(-24\)), and the empty set is wrong because the quadratic has real roots.

Answer: D

The domain of \(g \circ f\) is \(\left[\frac{1}{5}, 5\right]\).

\(g(f(x)) = \cos^{-1}(\log_{5}x)\) needs \(x \gt 0\) for the logarithm and \(-1 \leq \log_{5}x \leq 1\) for the inverse cosine:

\[ -1 \leq \log_{5}x \leq 1 \iff 5^{-1} \leq x \leq 5^{1} \]

So the domain is \(\frac{1}{5} \leq x \leq 5\). The interval \([-1, 5]\) mixes the range condition with the answer and includes negative \(x\), where \(\log_{5}x\) is undefined. (0,1] and the half-lines starting at -1 or 0 (printed with \(\alpha\), evidently for infinity) ignore the restriction that the logarithm must lie between -1 and 1.

Answer: C

The domain is [3,5].

Each square root needs a non-negative argument: \(x - 3 \geq 0\) gives \(x \geq 3\), and \(5 - x \geq 0\) gives \(x \leq 5\). Both hold together on \(3 \leq x \leq 5\). The coefficient 4 in front of the second root does not affect the domain.

[3,20] and [12,20] come from wrongly multiplying the 5 by 4; [1,2] and [2,4] include points below 3 where \(\sqrt{x - 3}\) is not real.

Answer: D

\(f(x) + f(1 - x) = \) 1.

Simplify \(f(1 - x)\) by multiplying top and bottom by \(3^{x}\):

\[ \begin{aligned} f(1 - x) &= \frac{3 \cdot 3^{-x}}{3 \cdot 3^{-x} + \sqrt{3}} = \frac{3}{3 + \sqrt{3}\,3^{x}} \\ &= \frac{\sqrt{3}}{\sqrt{3} + 3^{x}} \end{aligned} \]

Adding \(f(x) = \frac{3^{x}}{3^{x} + \sqrt{3}}\), the numerators sum to the common denominator, so the total is 1. Check at \(x = \frac{1}{2}\): \(f(\frac{1}{2}) = \frac{1}{2}\), and twice that is 1. The surd options mistake the constant \(\sqrt{3}\) for the answer.

Answer: B

\(n(B') = \) 13.

Find \(n(B)\) from the union formula, then subtract it from \(n(U)\):

\[ \begin{aligned} n(B) &= n(A \cup B) - n(A) + n(A \cap B) \\ &= 20 - 14 + 6 = 12 \\ n(B') &= n(U) - n(B) = 25 - 12 = 13 \end{aligned} \]

12 is \(n(B)\) itself, the usual slip of stopping one step early; 14 is \(n(A)\), and 15 or 16 come from adding \(n(A \cap B)\) with the wrong sign.

Check: \(n(A \cup B) = 14 + 12 - 6 = 20\), as given.

Answer: E

The range is \(\left[\frac{1}{8}, \frac{1}{6}\right]\).

Reading the denominator as \(7 - \cos x\) (the \(x\) is missing in the paper): since \(-1 \le \cos x \le 1\),

\[ \begin{aligned} 6 \le 7 - \cos x &\le 8 \\ \frac{1}{8} \le f(x) &\le \frac{1}{6} \end{aligned} \]

Both end values are attained (at \(\cos x = -1\) and \(\cos x = 1\)), so the interval is closed. The negative intervals are impossible because the denominator is always positive, and intervals near 1 would need a denominator near 1.

Answer: C

The domain is \(\left(-\infty, \frac{7}{11}\right]\).

A square root is real only when its argument is non-negative:

\[ 7 - 11x \ge 0 \;\Rightarrow\; x \le \frac{7}{11} \]

So every \(x\) up to and including \(7/11\) is allowed. \(\left[\frac{7}{11}, \infty\right)\) reverses the inequality (the sign flips only when dividing by a negative number, which is not done here), and \((-\infty, -1]\), \(\left[\frac{7}{11}, 1\right]\) and \([-1, 1]\) are only parts of the domain or lie outside it.

Answer: D

\(n(B) =\) 1.

A relation from \(A\) to \(B\) is any subset of \(A \times B\), so the number of relations is \(2^{n(A)n(B)}\):

\[ \begin{aligned} 2^{7\,n(B)} &= 128 = 2^{7} \\ 7\,n(B) \\ &= 7 \\ n(B) \\ &= 1 \end{aligned} \]

With \(n(B) = 0\) there would be only \(2^{0} = 1\) relation (the empty one), and \(n(B) = 2, 3, 4\) would give \(2^{14}, 2^{21}, 2^{28}\) relations.

In general, with \(m\) and \(n\) elements there are \(mn\) ordered pairs and so \(2^{mn}\) relations.