## Further Mathematics | Study Mode

Question 1
A
$$14(2\sqrt{2} + 6\sqrt{5} - 4\sqrt{10})$$
B
$$\frac{1}{14}(2 - 3\sqrt{2} - 4\sqrt{5} - 6\sqrt{10})$$
C
$$\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 6\sqrt{10} - 2)$$
D
$$14(2 + 3\sqrt{2} - 6\sqrt{5} + 4\sqrt{10})$$
Explanation.

$$\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}} = (\frac{1 - 2\sqrt{5}}{2 + 3\sqrt{2}})(\frac{2 - 3\sqrt{2}}{2 - 3\sqrt{2}})$$

= $$\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{4 - 6\sqrt{2} + 6\sqrt{2} - 18}$$

= $$\frac{2 - 3\sqrt{2} - 4\sqrt{5} + 6\sqrt{10}}{-14}$$

= $$\frac{1}{14}(3\sqrt{2} + 4\sqrt{5} - 2 - 6\sqrt{10})$$ (dividing through with the minus sign)

Question 2
A
$$(\frac{2\pi}{3}, \frac{4\pi}{3})$$
B
$$(\frac{\pi}{6}, \frac{5\pi}{6})$$
C
$$(\frac{\pi}{5}, \frac{2\pi}{5})$$
D
$$(\frac{\pi}{3}, \frac{5\pi}{3})$$
Explanation.

$$2\cos x - 1 = 0 \implies 2\cos x = 1$$

$$\cos x = \frac{1}{2}$$

$$x = \cos^{-1} (\frac{1}{2})$$

= $$\frac{\pi}{3}$$ = $$\frac{5\pi}{3}$$

Question 3
A
(1, 2)
B
(1, -2)
C
(-1, 2)
D
(-1, -2)
Explanation.

$$4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}$$

$$\implies 2^{2 + x^{2}} = 2^{3x}$$

Comparing bases, we have

$$2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0$$

$$x^{2} - 2x - x + 2 = 0$$

$$x(x - 2) - 1(x - 2) = 0$$

$$(x - 1) = 0$$ or $$(x - 2) = 0$$

$$x = \text{1 or 2}$$

Question 4
A
$$3^{2}$$
B
$$3^{\frac{1}{2}}$$
C
$$3^{\frac{1}{3}}$$
D
$$2^{13}$$
Explanation.

$$\log_{3} x = \log_{9} 3 \implies \log_{3} x = \log_{9} 9^{\frac{1}{2}} = \frac{1}{2}\log_{9} 9$$

$$\log_{3} x = \frac{1}{2}$$

$$\therefore x = 3^{\frac{1}{2}}$$

Question 5
A
$$\frac{x^{7}}{8}$$
B
$$\frac{7x^{6}}{16}$$
C
$$\frac{7x^{5}}{4}$$
D
$$\frac{35x^{4}}{8}$$
Explanation.

$$(\frac{x}{2} - 1)^{8} = ^{8}C_{8}(\frac{x}{2})^{8}(-1)^{0} + ^{8}C_{7}(\frac{x}{2})^{7}(-1)^{1} + ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2} + ...$$

$$\text{The third term in the expansion =} ^{8}C_{6}(\frac{x}{2})^{6}(-1)^{2}$$

= $$\frac{8!}{6!2!}(\frac{x^{6}}{64})(1)$$

= $$28 \times \frac{x^{6}}{64} = \frac{7x^{6}}{16}$$

Question 6
A
25
B
C
D
Explanation.

$$f : x \to x^{2} ; g : x \to x + 3$$

$$g(2) = 2 + 3 = 5$$

$$f o g(2) = f(5) = 5^{2} = 25$$

Question 7
A
P = 4 and Q = 2
B
P = 2 and Q = 4
C
P = 4 and Q = -2
D
P = -2 and Q = 4
Explanation.

$$\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}$$

$$\frac{2x}{(x + 6)(x + 3)} = \frac{P(x + 3) + Q(x + 6)}{(x + 6)(x + 3)}$$

Comparing equations, we have

$$2x = Px + 3P + Qx + 6Q$$

$$\implies 3P + 6Q = 0 ... (1) ; P + Q = 2 .... (2)$$

From equation (1), $$3P = -6Q \implies P = -2Q$$

$$\therefore -2Q + Q = -Q = 2$$

$$Q = -2$$

$$P = -2Q = -2(-2) = 4$$

$$P = 4, Q = -2$$

Question 8
A
$$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$
B
$$\begin{pmatrix} 27 & 12 \\ 16 & -15 \end{pmatrix}$$
C
$$\begin{pmatrix} -20 & -6 \\ 12 & -8 \end{pmatrix}$$
D
$$\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}$$
Explanation.

$$P = \begin{pmatrix} -2& 1\\ 3& 4\end{pmatrix}; Q = \begin{pmatrix} 5& -3\\ 2& -1\end{pmatrix}$$

= $$PQ = \begin{pmatrix} -10+2 & 6-1 \\ 15+8 & -9-4 \end{pmatrix}$$

= $$\begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix}$$

$$QP = \begin{pmatrix} -10-9 & 5-12 \\ -4-3 & 2-4 \end{pmatrix}$$

= $$\begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}$$

$$PQ - QP = \begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix} - \begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}$$

= $$\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}$$

Question 9
A
3x + 1
B
x + 1
C
2x + 1
D
x + 2
Explanation.

Using the remainder theorem, if (x - a) is a factor of f(x), then f(a) = 0.

Check the options and get the answer.

Question 10
A
$$f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2$$
B
$$f^{-1} : x \to \frac{1-2x}{x+2}, x \neq -2$$
C
$$f^{-1} : x \to \frac{1-2x}{x-2}, x \neq 2$$
D
$$f^{-1} : x \to \frac{1+2x}{x+2}, x \neq -2$$
Explanation.

$$f(x) = \frac{2x - 1}{x + 2}$$

$$y = \frac{2x - 1}{x + 2}$$

$$x = \frac{2y - 1}{y + 2} \implies x(y + 2) = 2y - 1$$

$$xy - 2y = -1 - 2x \implies y = \frac{-1 - 2x}{x - 2}$$

$$f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2$$

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