## Further Mathematics | Study Mode

Question 1
A
B
$$P \cap Q$$
C
$$P \cup Q$$
D
$$\emptyset$$
Explanation.

No official Explanation yet!

Question 2
A
$$-\tan \theta$$
B
$$-\cos \theta$$
C
$$\tan \theta$$
D
$$\cos \theta$$
Explanation.

$$\frac{\cos 2\theta - 1}{\sin 2\theta}$$

$$\cos (x + y) = \cos x \cos y - \sin x \sin y \implies \cos 2\theta = \cos^{2} \theta - \sin^{2} \theta$$

$$\cos^{2} \theta = 1 - \sin^{2} \theta \implies \cos 2\theta = 1 - 2\sin^{2} \theta$$

$$\sin 2\theta = 2\sin \theta \cos \theta$$

$$\therefore \frac{\cos 2\theta - 1}{\sin 2\theta} = \frac{1 - 2\sin^{2}\theta - 1}{2\sin \theta \cos \theta}$$

= $$\frac{-2 \sin^{2} \theta}{2\sin \theta \cos \theta} = \frac{- \sin \theta}{\cos \theta}$$

= $$-\tan \theta$$

Question 3
A
$$-1 \leq x \leq 3$$
B
$$x \geq 3$$ and $$x \leq -1$$
C
$$x \geq 3$$ or $$x < -1$$
D
$$-1 \leq x < 3$$
Explanation.

$$x^{2} - 2x \geq 3 \implies x^{2} - 2x - 3 \geq 0$$

$$x^{2} + x - 3x - 3 = (x + 1)(x - 3) \geq 0$$

$$x = -1 ; x = 3$$

Check: $$x = -1 : (-1)^{2} - 2(-1) = 1 + 2 \geq 3$$ (satisfied)

$$-1 < x < 3 : 0^{2} - 2(0) = 0 \geq 3$$ (not satisfied)

$$x < -1 : (-2)^{2} - 2(-2) = 4 + 4 = 8 \geq 3$$ (satisfied)

$$x = 3 : 3^{2} - 2(3) = 9 - 6 = 3 \geq 3$$ (satisfied)

$$x > 3 : 4^{2} - 2(4) = 16 - 8 = 8 \geq 3$$ (satisfied)

$$\therefore x^{2} - 2x \geq\text{3 is satisfied in the region x}\leq \text{-1 and x} \geq 3$$

Question 4
A
$$27\sqrt{2}$$
B
$$27\sqrt{6}$$
C
$$81\sqrt{2}$$
D
$$81\sqrt{6}$$
Explanation.

$$T_{n} = ar^{n - 1}$$ (Geometric progression)

$$a = \sqrt{6}, r = \frac{T_{2}}{T_{1}} = \frac{3\sqrt{2}}{\sqrt{6}}$$

$$r = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}$$

$$\therefore T_{8} = (\sqrt{6})(\sqrt{3})^{8 - 1}$$

= $$(\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}$$

Question 5
A
300
B
240
C
120
D
60
Explanation.

In order to do this, simply find the option in the range where only the cos is +ve. This occurs in the range $$270 \leq x \leq 360$$.

Check: $$\sin 300 = - \sin 60 = \frac{-\sqrt{3}}{2}$$

$$\cos 300 = \cos 60 = \frac{1}{2}$$

Question 6
A

-3

B

0

C

$$\frac{5}{6}$$

D

1

Explanation.

$$\log_{10} (\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})$$

$$\frac{1}{3} + \frac{1}{4} = \frac{7}{12}$$

= $$\log_{10} (\frac{7}{12} \times 2^{2} \times \frac{3}{7})$$

= $$\log_{10} 1 = 0$$

Question 7
A
8i + j
B
2i - j
C
-2i - 3j
D
-8i - j
Explanation.

$$\overrightarrow{SQ} = \overrightarrow{SR} + \overrightarrow{RQ}$$

$$\overrightarrow{RQ} = -\overrightarrow{QR} = - (3i + 2j) = -3i - 2j$$

$$\overrightarrow{SQ} = (-5i + 3j) - 3i - 2j = -8i + j$$

Question 8
A
-8
B
-4
C
D
Explanation.

If (x + 1) is a factor, then f(-1) = 0.

$$(-1)^{3} + p(-1)^{2} + (-1) + 6 = 0$$

$$-1 + p - 1 + 6 = 0 \implies p + 4 = 0$$

$$p = -4$$

Question 9
A
-8
B
-2
C
D
Explanation.

Given: $$f(x + 1) = x^{3} + 3x^{2} - 4x + 2$$.

$$f(2) = f(x + 1) \implies x + 1 = 2; x = 1$$

$$f(2) = 1^{3} + 3(1)^{2} - 4(1) + 2 = 1 + 3 - 4 + 2 = 2$$

Question 10
A
B
C
D
Explanation.

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

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

$$\implies x^{2} + y^{2} - 2ax - 2by = r^{2} - a^{2} - b^{2}$$

Comparing with the given equation: $$3x^{2} + 3y^{2} + 24x - 12y = 15$$

Making the coefficients of $$x^{2}$$ and $$y^{2}$$ = 1, we have

$$x^{2} + y^{2} + 8x - 4y = 5$$

$$2a = -8 \implies a = -4$$

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

$$r^{2} - a^{2} - b^{2} = 5 \implies r^{2} = 5 + (-4)^{2} + (2)^{2} = 5 + 16 + 4 = 25$$

$$\therefore r = 5$$

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