## Further Mathematics | Study Mode

Question 1
A
$${x : -5 < x < 5}$$
B
$${x : -5 \leq x \leq 5}$$
C
$${x : -5 \leq x < 5}$$
D
$${x : -5 < x \leq 5}$$
Explanation.

$$P = {x : -2 < x < 5}$$ and $$Q = {x: -5 < x < 2}$$

$$(P \cup Q) = {x : -5 < x < 5}$$

Question 2
A
$$7\sqrt{3} - \frac{17\sqrt{2}}{3}$$
B
$$7\sqrt{2} - \frac{17\sqrt{3}}{3}$$
C
$$-7\sqrt{2} + \frac{17\sqrt{3}}{3}$$
D
$$-7\sqrt{3} - \frac{17\sqrt{2}}{3}$$
Explanation.

Given $$\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}$$,

first, we rationalise by multiplying through with $$2\sqrt{3} - 3\sqrt{2}$$ (the inverse of the denominator).

$$(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}})(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}})$$

= $$\frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}$$

= $$\frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}$$

Question 3
A
$$\frac{-1}{2}$$
B
$$0$$
C
$$\frac{2}{3}$$
D
$$2$$
Explanation.

Given the formula for p * q as: $$p + q + 2pq$$ and its identity element is 0, such that if, say, t is the inverse of p, then

$$p * t = 0$$, then $$p + t + 2pt = 0 \therefore p + (1 + 2p)t = 0$$

$$t = \frac{-1}{1 + 2p}$$ is the formula for the inverse of p and is undefined on R when

$$1 + 2p) = 0$$ i.e when $$2p = -1; p = \frac{-1}{2}$$.

Question 4
A
$$p \vee q$$
B
$$p \vee \sim q$$
C
$$p \wedge \sim q$$
D
$$p \wedge q$$
Explanation.

No official Explanation yet!

Question 5
A
$$\frac{-3}{7}$$
B
$$0$$
C
$$\frac{1}{2}$$
D
$$4$$
Explanation.

$$f(x) = \frac{4}{x} - 1$$. Let y = f(x)

$$y = \frac{4 - x}{x} \implies xy + x = 4$$

$$x(y + 1) = 4 \therefore x = \frac{4}{y + 1}$$

$$f^{-1}(7) = \frac{4}{7 + 1} = \frac{1}{2}$$

Question 6
A
{-9, -1, 2,3, 4}
B
{-9, -2, 0, 1, 7}
C
{-5, -4, -3, -2}
D
{-9, -5, -1, 3, 7}
Explanation.

The elements of x are {-2, -1, 0, 1, 2}

$$y = 4x - 1$$ = 4(-2) - 1 = -9; 4(-1) - 1 = -5; 4(0) - 1 = -1; 4(1) - 1 = 3; 4(2) - 1 = 7.

The range of x is {-9, -5, -1, 3 7}.

Question 7
A
(x + 2)(x + 5)(y + 1)
B
(x + 2)(x - 5)(y + 1)
C
(x - 2)(x + 5)(y + 1)
D
(x - 2)(x - 5)(y + 1)
Explanation.

$$x^{2} + x^{2}y + 3x - 10y + 3xy -10$$

= $$x^{2} + x^{2}y + 3x + 3xy - 10y - 10 = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)$$

= $$(x^{2} + 3x - 10)(y + 1)$$

= $$(x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10$$

= $$x(x - 2) + 5(x - 2) = (x - 2)(x +5)$$

$$\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)$$.

Question 8
A
-6
B
-4
C
D
Explanation.

Given x = (-1, 5) for the equation $$x^{2} + kx - 5 = 0$$

$$x = -1 \implies x + 1 = 0$$; $$x = 5 \implies x - 5 = 0$$

$$(x + 1)(x - 5) = 0$$, expanding,

$$x^{2} - 5x + x - 5 = 0 \therefore x^{2} - 4x - 5 = 0$$

$$\therefore$$ k = -4.

Question 9
A
$$\frac{-7}{8}$$
B
$$\frac{-3}{8}$$
C
$$\frac{1}{8}$$
D
$$\frac{5}{8}$$
Explanation.

The remainder theorem states that if f(x) is divided by (x - a), the remainder is f(a).

$$f(x) = x^{3} - 2x + m$$ divided by (x - 1), so that a = 1.

Remainder = $$f(1) = 1^3 - 2(1) + m = -1 + m$$

$$f(x) = 2x^{3} + x - m$$ divided by (2x + 1), so that a = $$\frac{-1}{2}$$

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

$$\implies m - 1 = \frac{-3}{4} - m$$, collecting like terms,

$$2m = \frac{1}{4} \therefore m = \frac{1}{8}$$

Question 10
A
$$\frac{3}{2}$$ or $$1$$
B
$$\frac{3}{2}$$ or $$-1$$
C
$$\frac{-3}{2}$$ or $$-1$$
D
$$\frac{-3}{2}$$ or $$1$$
Explanation.

$$(2t - 3s)(t - s) = 0 \implies (2t - 3s) = \text{0 or} (t - s) = 0$$

$$2t - 3s = 0 \implies 2t = 3s \therefore \frac{t}{s} = \frac{3}{2}$$

$$t - s = 0 \implies t = s \therefore \frac{t}{s} = 1$$

$$\frac{t}{s} = \frac{3}{2} or 1$$

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