## Further Mathematics | Study Mode

Question 1
A
$${x : x \in R, x \neq 3}$$
B
$${x : x \in R, x \neq 1}$$
C
$${x : x \in R, x \neq 0}$$
D
$${x : x \in R, x\neq -3}$$
Explanation.

$$f(x) = \frac{x}{3 - x}$$

f(x) has a defined value except at x = 3 where the function is undefined.

Question 2
A
$$\frac{6 + \sqrt{2}}{4}$$
B
$$\frac{3 + \sqrt{6}}{4}$$
C
$$\frac{\sqrt{2} - \sqrt{6}}{4}$$
D
$$\frac{3 - \sqrt{6}}{4}$$
Explanation.

$$\cos (x + y) = \cos x \cos y - \sin x \sin y$$

$$\cos (60 + 45) = \cos 60 \cos 45 - \sin 60 \sin 45$$

= $$\frac{1}{2} \times \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}$$

= $$\frac{\sqrt{2} - \sqrt{6}}{4}$$

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

$$\frac{5}{\sqrt{2}} = \frac{5 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{5\sqrt{2}}{2}$$

$$\frac{\sqrt{8}}{8} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}$$

$$\frac{5}{\sqrt{2}} - \frac{\sqrt{8}}{8} = (\frac{5}{2} - \frac{1}{4})\sqrt{2}$$

= $$\frac{9}{4}\sqrt{2}$$

= $$2\frac{1}{4}\sqrt{2}$$

Question 4
A
$$-1$$
B
$$\frac{-1}{3}$$
C
$$\frac{-3}{7}$$
D
$$\frac{-5}{19}$$
Explanation.

$$16^{3x} = \frac{1}{4}(32^{x - 1})$$

$$(2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})$$

$$2^{12x} = 2^{-2 + 5x - 5}$$

$$12x = -7 + 5x$$

$$7x = -7 \implies x = -1$$

Question 5
A
-2
B
$$\frac{-1}{2}$$
C
$$\frac{1}{2}$$
D
Explanation.

$$\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\log_{5} 8^{\frac{1}{2}}}$$

= $$\frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8}$$

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

= 2

Question 6
A
$$\frac{560}{243}$$
B
$$\frac{841}{243}$$
C
$$\frac{1120}{243}$$
D
$$\frac{4481}{243}$$
Explanation.

$$^{10}C_{7 - 1} (2^{10 - 6}) (\frac{-1}{3})^{6}$$

$$\frac{10!}{(10 - 6)! 6!} \times 16 \times \frac{1}{243}$$

= $$210 \times 16 \times \frac{1}{729}$$

= $$\frac{1120}{243}$$

Question 7
A
$$x^{2} - 6x - 9 = 0$$
B
$$x^{2} - 6x + 6 = 0$$
C
$$x^{2} + 6x - 9 = 0$$
D
$$x^{2} + 6x + 6 = 0$$
Explanation.

$$(x - \alpha)(x - \beta) = 0$$

$$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$

$$(x^{2} - (3 - \sqrt{3})x - (3 + \sqrt{3})x + (9 + 3\sqrt{3} - 3\sqrt{3} - 3) = 0$$

$$x^{2} - 3x - x\sqrt{3} - 3x + x\sqrt{3} + 6 = 0$$

$$x^{2} - 6x + 6 = 0$$

Question 8
A
-12
B
-6
C
D
Explanation.

Using remainder theorem, since x - 3 is a factor, then

given $$2x^{2} - 2x + p$$, f(3) = 0

$$2(3^{2}) - 2(3) + p = 0 \implies 18 - 6 = -p$$

$$p = -12$$

Question 9
A
110, 250
B
110, 290
C
200, 250
D
250, 290
Explanation.

The valueof the sine of an angle is negative in the third and fourth quadrant. Hence options A and B are not the options.

$$\sin 250 = -\sin (250 - 180) = - \sin 70$$

$$\sin 290 = - \sin (360 - 290) = - \sin 70$$

Question 10
A
6 or 3
B
-18 or -9
C
-7 or 5
D
-5 or 7
Explanation.

An fractionis undefined when the denominator has value = 0.

$$\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}$$ is undefined when $$x^{2} + 2x - 35 = 0$$

$$x^2 + 7x - 5x - 35 = 0 \implies (x + 7)(x - 5) = 0$$

$$x = \text{-7 or 5}$$

Question Map