## Further Mathematics | Study Mode

Question 1
A
$$\frac{1}{2}$$
B
C

$$2\sqrt{3}$$

D
Explanation.

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

= $$\frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$$

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

= $$\frac{6}{2} = 3$$

Question 2
A
$${x : x \in R, x = \frac{1}{2}}$$
B
$$x: x \in R, x\neq \frac{1}{3}$$
C
$$x : x \in R, x = \frac{1}{3}$$
D
$$x: x \in R$$
Explanation.

The domain of a function refers to the regions where the function is defined or has a value on a particular region.

$$\frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}$$ has a domain defined on all set of real numbers because the function is defined on the set of real numbers when the denominator $$\sqrt{9x^{2} + 1} \geq 0$$.

$$\sqrt{9x^{2} + 1} \geq 0 \implies 9x^{2} + 1 \geq 0$$ which because of the square sign has a value for all values of x, be it negative or positive.

Question 3
A
1, 3
B
-1, -3
C
1, -3
D
-1, 3
Explanation.

$$f(x) = 3x^{2} - 12x + 12$$ and $$f(x) = 3$$

$$\therefore f(x) = 3 = 3x^{2} - 12x + 12 \implies 3x^{2} - 12x + 12 - 3 = 0$$

$$3x^{2} - 12x + 9 = 0; 3x^{2} - 9x - 3x + 9 = 0$$

$$3x(x - 3) - 3(x - 3) = (3x - 3)(x - 3) = 0$$

3x - 3 = 0 or x - 3 = 0

$$x = 1 or 3$$

Question 4
A
B
C
D
Explanation.

$$(\sqrt{x} + 1) * (\sqrt{x} - 1) = 4 \implies \frac{\sqrt{x} + 1}{\sqrt{x} - 1} + \frac{\sqrt{x} - 1}{\sqrt{x} + 1} = 4$$

$$\frac{(\sqrt{x} + 1)(\sqrt{x} + 1) + (\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)}$$

= $$\frac{x + 2\sqrt{x} + 1 + x - 2\sqrt{x} + 1}{x - 1} \implies \frac{2x + 2}{x - 1} = 4$$

$$2x + 2 = 4x - 4 \therefore 4x - 2x = 2x = 2 + 4= 6$$

$$x = 3$$

Question 5
A
$$\frac{5\sqrt{10}}{4}$$
B
$$4\sqrt{10}$$
C
$$5\sqrt{10}$$
D
$$\frac{4\sqrt{10}}{5}$$
Explanation.

$$4x^{2} + 5kx + 10 = (2x + \sqrt{10})^{2}$$

Expanding the right hand side equation, we have

$$4x^{2} + 4x\sqrt{10} + 10$$

Comparing with the left hand side, we have

$$5k = 4\sqrt{10} \implies k = \frac{4}{5}\sqrt{10}$$

Question 6
A
-17
B
-7
C
D
13
Explanation.

$$f(x) = 3x^{3} - 2x^{2} + 7x + 5$$.

$$x - 1 = 0, x = 1$$

$$f(1) = 3(1)^{3} - 2(1)^{2} + 7(1) + 5 = 13$$

Question 7
A
$$Q \cap R = \varnothing$$
B
$$R \subset P$$
C
$$(R \cap P) \subset (R \cap U)$$
D
$$n(P' \cap R) = 2$$
Explanation.

All the statements are false except option C.

$$R \cap P = {3, 5, 7} and R \cap U = {2, 3, 5, 7, 11}$$

$$\therefore (R \cap P) \subset (R \cap U)$$

Question 8
A
$$a = b^{3} - 3$$
B
$$a = b^{3} - 9$$
C
$$a = 9b^{3}$$
D
$$a = \frac{b^{3}}{9}$$
Explanation.

$$\log_{3}a - 2 = 3\log_{3}b$$

Using the laws of logarithm, we know that $$2 = 2\log_{3}3 = \log_{3}3^{2}$$

$$\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}$$

= $$\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3} \implies \frac{a}{9} = b^{3}$$

$$\implies a = 9b^{3}$$

Question 9
A
$$2x^{2} - 9x + 15 = 0$$
B
$$2x^{2} - 9x + 13 = 0$$
C
$$2x^{2} - 9x - 13 = 0$$
D
$$2x^{2} - 9x - 15 = 0$$
Explanation.

Note: Given the sum of the roots and its product, we can get the equation using the formula:

$$x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0$$. This will be used later on in the course of our solution.

Given equation: $$2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6$$.

$$\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}$$

$$\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3$$

Given the roots of the new equation as $$(\alpha + 1)$$ and $$(\beta + 1)$$, their sum and product will be

$$(\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}$$

$$(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}$$

The new equation is given by: $$x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0$$

= $$x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0$$

Question 10
A
$$\frac{x}{2(x - 2)} - \frac{5}{(x - 2)^{2}}$$
B
$$\frac{5}{(x - 2)} + \frac{x}{2(x - 2)^{2}}$$
C
$$\frac{1}{2(x - 2)} + \frac{5x}{2(x- 2)^{2}}$$
D
$$\frac{-1}{2(x - 2)} + \frac{8x}{2(x - 2)^{2}}$$
Explanation.

$$\frac{3x - 1}{(x - 2)^{2}} = \frac{A}{(x - 2)} + \frac{Bx}{(x - 2)^{2}}$$

$$\frac{3x - 1}{(x - 2)^{2}} = \frac{A(x - 2) + Bx}{(x - 2)^{2}}$$

Comparing, we have

$$3x - 1 = Ax - 2A + Bx \implies -2A = -1; A + B = 3$$

$$\therefore A = \frac{1}{2}; B = \frac{5}{2}$$

= $$\frac{1}{2(x - 2)} + \frac{5x}{2(x - 2)^{2}}$$

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