## Further Mathematics | Study Mode

Question 1
A
-1
B
C
D
Explanation.

$$a * a^{-1} = e = -1$$

$$\therefore a^{-1} = -1$$

Question 2
A
$$\frac{5\pi}{12}$$
B
$$\frac{3\pi}{4}$$
C
$$\frac{5\pi}{6}$$
D
$$\frac{7\pi}{6}$$
Explanation.

$$180 = \pi rads$$

$$1 = \frac{\pi}{180}$$

$$\therefore 75 = \frac{\pi}{180} \times 75$$

= $$\frac{5\pi}{12}$$

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

$$\log_{9} 3 = \log_{9} (9^{\frac{1}{2}}) = \frac{1}{2}\log_{9} 9 = \frac{1}{2}$$

$$\frac{1}{2} + 2x = 1 \implies 2x = \frac{1}{2}$$

$$x = \frac{1}{4}$$

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

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

$$\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}$$

= $$(0 \times \frac{1}{2}) - (1 \times \frac{\sqrt{3}}{2})$$

= $$0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}$$

Question 5
A
-51
B
-23
C
29
D
49
Explanation.

Using remainder theorem, the remainder when $$5x^{3} + 2x^{2} - 7x -5$$ is divided by (x - 2) = f(2)

$$f(2) = 5(2^{3}) + 2(2^{2}) - 7(2) -5 = 40 + 8 - 14 - 5$$

= 29

Question 6
A
$$-1\frac{1}{4}$$
B
$$-1$$
C
$$\frac{4}{5}$$
D
$$1$$
Explanation.

$$f(x) = \frac{3x + 1}{x^{2} - 1}$$

$$f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1$$

Question 7
A
$$\frac{-5}{6}$$
B
$$-\frac{4}{27}$$
C
$$0$$
D
$$\frac{2}{9}$$
Explanation.

$$\sqrt[3]{\frac{8}{27}} - (\frac{4}{9})^{\frac{-1}{2}}$$

$$\frac{2}{3} - (\frac{9}{4})^{\frac{1}{2}}$$

= $$\frac{2}{3} - \frac{3}{2}$$

= $$\frac{-5}{6}$$

Question 8
A
$$x < -1, x < -\frac{1}{3}$$
B
$$x > -1, x > -\frac{1}{3}$$
C
$$x > \frac{1}{3}, x < -1$$
D
$$x < \frac{1}{3}, x > -1$$
Explanation.

$$3x^{2} + 4x + 1 > 0$$

$$3x^{2} + 3x + x + 1 > 0$$

$$3x(x + 1) + 1(x + 1) > 0$$

$$(3x + 1)(x + 1) > 0$$

$$3x + 1 > 0 \implies 3x > -1$$

$$x > -\frac{1}{3}$$

$$x + 1 > 0 \implies x > -1$$

$$x > -1, x > -\frac{1}{3}$$

Question 9
A
B
$$\sqrt{3}$$
C
$$\sqrt{11}$$
D
$$\sqrt{6}$$
Explanation.

The equation of a circle is given as $$(x - a)^{2} + (y - b)^{2} = r^{2}$$

Expanding this, we have $$x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}$$

Comparing with the given equation, $$3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0$$ (making the coefficients of $$x^{2}$$ and $$y^{2}$$ = 1, we get that

$$-2a = 2\implies a = -1$$

$$2b = 4\implies b = 2$$

$$r^{2} - a^{2} - b^{2} = -2$$

$$\therefore r^{2} - (-1)^{2} - (2)^{2} = -2\implies r^{2} = -2+ 1 + 4 = 3$$

$$r = \sqrt{3}$$

Question 10
A
13
B
15
C
17
D
26
Explanation.

$$f(x) = p + qx$$

$$f(1) = p + q(1) \implies p + q = 7 .... (1)$$

$$f(5) = p + 5q = 19 .....(2)$$

Solving for p and q using simultaneous equation, p = 4, q = 3

$$f(3) = 4 + 3(3) = 13$$

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