## Further Mathematics | Study Mode

Question 1
A
B
C
-4
D
-5
Explanation.

$$\frac{1}{5^{-y}} = 25(5^{4-2y})$$

$$\implies 5^{y} = (5^{2})(5^{4-2y})$$

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

Comparing bases, we have

$$y = 6 - 2y$$

$$3y = 6 \implies y = 2$$

Question 2
A
$$\sin^{2} \theta$$
B
$$\sec^{2} \theta$$
C
$$\tan^{2} \theta$$
D
$$\cos^{2} \theta$$
Explanation.

$$(1 + \sin\theta)(1 - \sin\theta) = 1 - \sin \theta + \sin \theta - \sin^{2} \theta$$

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

Recall, $$\cos^{2} \theta + \sin^{2} \theta = 1$$

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

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

When you have two lines, $$y_{1}, y_{2}$$, perpendicular to each other, the product of their slopes = -1.

$$3x + 4y + 6 = 0 \implies 4y = -6 - 3x$$

$$\therefore y = \frac{-6}{4} - \frac{3}{4}x$$

$$\frac{\mathrm d y}{\mathrm d x} = \frac{-3}{4}$$

Also, $$4x - by + 3 = 0 \implies by = 4x + 3$$

$$y = \frac{4}{b}x + \frac{3}{b}$$

$$\frac{\mathrm d y}{\mathrm d x} = \frac{4}{b}$$

$$\frac{-3}{4} \times \frac{4}{b} = -1 \implies \frac{4}{b} = \frac{4}{3}$$

$$b = 3$$

Question 4
A
B
C
D
Explanation.

$$(x * y) = \frac{x+y}{2}$$

$$(3 * b) = \frac{3+b}{2}$$

$$x \circ y = \frac{x^{2}}{y}$$

$$(\frac{3+b}{2}) \circ 48 = \frac{(\frac{3+b}{2})^{2}}{48} = \frac{1}{3}$$

$$\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}$$

$$(3 + b)^{2} = \frac{48 \times 4}{3} = 64$$

$$b^{2} + 6b + 9 = 64 \implies b^{2} + 6b +9- 64 = 0$$

$$b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0$$

$$b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (} b + 11) = 0$$

Since b > 0, b - 5 = 0

b = 5.

Question 5
A
-67
B
-61
C
61
D
67
Explanation.

$$f(x) = 3x^{3} + 8x^{2} + 6x + k$$

$$f(2) = 3(2^{3}) + 8(2^{2}) + 6(2) + k = 1$$

$$\implies 24 + 32 + 12 + k = 1$$

$$68 + k = 1 \therefore k = 1 - 68 = -67$$

Question 6
A
$$\frac{5}{4}$$
B
$$\frac{3}{5}$$
C
$$1$$
D
$$\frac{2}{3}$$
Explanation.

$$8^{x} (\frac{1}{4})^{y} = 1$$

$$(2^{3})^{x} (2^{-2})^{y} = 2^{0}$$

$$2^{3x - (-2y)} = 2^{0}$$

$$\implies 3x + 2y = 0 .... (1)$$

$$\log_{2}(x - 2y) = 1$$

$$x - 2y = 2^{1} = 2 ..... (2)$$

Solving equations 1 and 2,

$$x = \frac{1}{2}, y = \frac{-3}{4}$$

$$(x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}$$

Question 7
A
$$7 + \sqrt{2}$$
B
$$7 + 7\sqrt{2}$$
C
$$1 - 7\sqrt{2}$$
D
$$1 + \sqrt{2}$$
Explanation.

$$\frac{1 + \sqrt{8}}{3 - \sqrt{2}}$$

Rationalizing by multiplying through with $$3 + \sqrt{2}$$,

$$(\frac{1 + \sqrt{8}}{3 - \sqrt{2}})(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}) = \frac{3 + \sqrt{2} + 3\sqrt{8} + 4}{9 - 2}$$

= $$\frac{3 + \sqrt{2} + 3\sqrt{4 \times 2} + 4}{7}$$

= $$\frac{7 + 7\sqrt{2}}{7} = 1 + \sqrt{2}$$

Question 8
A
64.245
B
61.255
C
60.255
D
60.245
Explanation.

$$(1.98)^{6} = (1 + 0.98)^{6} = 1 + 6(0.98) + 15(0.98)^{2} + 20(0.98)^{3} + 15(0.98)^{4} + 6(0.98)^{5} + (0.98)^{6}$$

$$\approxeq 1 + 5.88 + 14.406 + 18.823 + 13.836 + 5.424 + 0.886$$

= $$60.255$$

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

To get the third factor, take the product of the other 2 factors and then divide the main equation by their product.

Question 10
A
-3 and 5
B
5 and -5
C
3 and -3
D
-5 and 3
Explanation.

Given an exponential sequence, say $$a, b, c,...$$, as consecutive terms, then $$\sqrt{a \times c} = b$$.

$$\therefore 2, (k+1), 8 \implies \sqrt{2 \times 8} = k + 1$$

$$k + 1 = \pm{4} \implies k = \text{-5 or 3}$$

Question Map