## Further Mathematics | Study Mode

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

$$log_{y}\frac{1}{8} = 3 \implies y^{3} = \frac{1}{8}$$ (Laws of logarithm)

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

Equating both sides, we have

$$y = \frac{1}{2}$$

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

$$a \Delta b$$ = $$\frac{a+b}{\sqrt{ab}}$$

$$-3\Delta -1$$ = $$\frac{-3 + -1}{\sqrt{-3\times -1}}$$

$$\frac{-4}{\sqrt{3}}$$, rationalising, we have

$$\frac{-4 \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} = \frac{-4\sqrt{3}}{3}$$

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

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

$$(1-\sqrt{3})^{2} = (1-\sqrt{3})(1-\sqrt{3})$$

$$1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}$$

$$\frac{1}{4-2\sqrt{3}}$$

After rationalising (multiplying the denominator and numerator with $$4+2\sqrt{3}$$, we have

$$\frac{4+2\sqrt{3}}{4} = 1 + \frac{1}{2}\sqrt{3}$$

Question 4
A
3, 4
B
$$\pm3$$
C
$$\pm5$$
D
$$\pm6$$
Explanation.

For equal roots, we have that $$b^{2} = 4ac$$, so, given a=1, b = -k and c = 9,

$$(-k)^{2} = 4\times1\times9 \implies k^{2} = 36$$

$$k = \sqrt{36} = \pm6$$

Question 5
A
$$x<2$$
B
$$x \leq 2$$
C
$$x = 2$$
D
$$x > -2$$
Explanation.

$$f : x \to \sqrt{4 -2x}$$ defined on the set of real numbers, R, which has range from $$(-\infty, \infty)$$ but because of the root sign, it is defined from $$[0, \infty)$$.

This is because the root of numbers only has real number values from 0 and upwards.

$$\sqrt{4-2x} \geq 0 \implies 4-2x \geq 0$$

$$-2x \geq -4; x \leq 2$$

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

Let $$f(x) = y$$, then we have

$$y = \frac{x+1}{2} \implies 2y = x+1; x = 2y-1$$

Let $$f^{1}(x) = x; x = 2y-1$$, replacing y with x,

$$f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5$$

Question 7
A
20
B
12
C
-10
D
-22
Explanation.

Taking the LCM of the right hand side of the equation, we have

$$\frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{6x+m}{2x^{2}+7x-15}$$

Comparing the numerators, we have

$$4(2x-3) - 2(x+5) = 6x+m$$

$$8x-12-2x-10 = 6x -22 = 6x + m$$

$$\implies m = -22$$

Question 8
A
-320
B
-240
C
240
D
320
Explanation.

$$^{6}C_{4}(1)^{6-4}(-2x)^{4}$$ = $$15\times1\times16x^{4} = 240x^{4}$$

The coefficient of $$x^{4}$$= 240

Question 9
A
(-2,4)
B
($$\frac{-2}{3}, \frac{4}{3}$$)
C
($$\frac{2}{3}, \frac{-4}{3}$$)
D
(2, -4)
Explanation.

The equation for a circle with centre coordinates (a, b) and radius r is

$$(x-a)^{2} + (y-b)^{2} = r^{2}$$

Expanding the above equation, we have

$$x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0$$ so that

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

Taking the original equation given, $$3x^{2} + 3y^{2} - 4x + 8y = 2$$ and making the coefficients of $$x^{2}$$ and $$y^{2}$$ = 1,

$$x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}$$, comparing, we have

$$2a = \frac{4}{3}; 2b = \frac{-8}{3}$$

$$\implies a = \frac{2}{3}; b = \frac{-4}{3}$$

Question 10
A
43.5
B
46
C
48.5
D
51
Explanation.

$$T_{n} = a + (n-1)d$$

$$d = T_{2} - T_{1} = T_{3} - T_{2} = -1.5 - (-4) = 2.5$$

$$T_{21} = -4 + (21 - 1) \times 2.5 = -4 + (20\times 2.5)$$

= $$-4 + 50 = 46$$

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