## Further Mathematics | Study Mode

Question 1
A
$$\frac{-x}{1 - x}, x \neq 1$$
B
$$\frac{1}{1 - x}, x \neq 1$$
C
$$\frac{-1}{1 - x}, x \neq 1$$
D
$$\frac{x}{1 - x}, x \neq 1$$
Explanation.

$$x * y = x + y - xy$$

Let $$x^{-1}$$ be the inverse of x, so that

$$x * x^{-1} = x + x^{-1} - x(x^{-1}) = 0$$

$$x + x^{-1} - x(x^{-1}) = 0 \implies x(x^{-1}) - x^{-1} = x$$

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

= $$\frac{x}{-(1 - x)} = \frac{-x}{1 - x}, x \neq 1$$

Question 2
A
200
B
90
C
60
D
Explanation.

$$\sin \theta = \tan \theta \implies \frac{\sin \theta}{1} = \frac{\sin \theta}{\cos \theta}$$

Equating, we have

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

= $$0$$

Question 3
A
-6
B
-1.2
C
0.83
D
1.2
Explanation.

$$a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1$$

$$\implies a^{\frac{5}{6} + \frac{-1}{n}} = a^{0}$$

Equating bases, we have

$$\frac{5}{6} - \frac{1}{n} = 0$$

$$\frac{5n - 6}{6n} = 0$$

$$5n - 6 = 0 \implies 5n = 6$$

$$n = \frac{6}{5} = 1.20$$

Question 4
A
3 log 2
B
4 log 2
C
-3 log 2
D
-4 log 2
Explanation.

$$\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}$$

= $$\log 2^{-3} + \log 2^{-1}$$

= $$-3 \log 2 - 1 \log 2 = -4 \log 2$$

Question 5
A
$$\sin^{2} x$$
B
$$\sin x^{2}$$
C
$$(\sin x)x^{2}$$
D
$$x \sin x$$
Explanation.

$$f(x) = x^{2}, g(x) = \sin x$$

$$g \circ f = g(x^{2}) = \sin x^{2}$$

Question 6
A
$$-15a^{4}b^{2}$$
B
$$15a^{4}b^{2}$$
C
$$-15a^{3}b^{3}$$
D
$$15a^{3}b^{3}$$
Explanation.

$$(a - b)^{6} = ^{6}C_{0}(a)^{6}(-b)^{0} + ^{6}C_{1}(a)^{5}(-b)^{1} + ^{6}C_{2}(a)^{4}(-b)^{2} + ...$$

Third term = $$^{6}C_{2}(a)^{4}(-b)^{2} = \frac{6!}{(6-2)! 2!}(a^4)(b^2)$$

= $$15a^{4}b^{2}$$

Question 7
A
1 and -1
B
-1 and 2
C
1 and 2
D
0 and -1
Explanation.

$$\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}$$

Squaring both sides, we have

$$(\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}$$

$$x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1$$

$$2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0$$

$$(2\sqrt{x(x + 1)})^{2}= 0^{2} \implies 4(x(x + 1)) = 0$$

$$\therefore x(x + 1) = 0$$

$$x = \text{0 or -1}$$

Question 8
A
$$\frac{24}{5}$$
B
$$\frac{8}{5}$$
C
$$\frac{5}{8}$$
D
$$\frac{5}{24}$$
Explanation.

$$2x^{2} - 6x + 5 = 0 \implies a = 2, b = -6, c = 5$$

$$\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{2} = 3$$

$$\alpha\beta = \frac{c}{a} = \frac{5}{2}$$

$$\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^{2} + \alpha^{2}}{\alpha\beta}$$

$$\frac{(\alpha + \beta)^{2} - 2\alpha\beta}{\alpha\beta} = \frac{3^{2} - 2(\frac{5}{2})}{\frac{5}{2}}$$

= $$\frac{4}{\frac{5}{2}} = \frac{8}{5}$$

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

Since f(3) = 0, then (x - 3) is a factor of f(x).

Dividing f(x) by (x - 3), we get $$2x^{2} + 3x - 2$$.

$$2x^{2} + 3x - 2 = 2x^{2} - x + 4x - 2$$

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

Therefore, $$f(x) = (x - 3)(x + 2)(2x -1)$$

Question 10
A
y + 5x + 3 = 0
B
2y - 5x - 9 = 0
C
5y + 2x - 8 = 0
D
5y - 2x - 12 = 0
Explanation.

$$Line:2y + 5x - 6 = 0$$

$$2y = 6 - 5x \implies y = 3 - \frac{5x}{2}$$

Gradient = $$\frac{-5}{2}$$

For the line perpendicular to the given line, Gradient = $$\frac{-1}{\frac{-5}{2}} = \frac{2}{5}$$

The midpoint of P(4, 3) and Q(-6, 1) = $$(\frac{-6 + 4}{2}, \frac{3 + 1}{2})$$

= (-1, 2).

Therefore, the line = $$\frac{y - 2}{x + 1} = \frac{2}{5}$$

$$2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0$$

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