The problem consists of multiple parts. We need to solve the problems from question 7 to 13 from the provided image.

1. Problem Description

The problem consists of multiple parts. We need to solve the problems from question 7 to 13 from the provided image.

2. Solution Steps

7. If $f(x) = \begin{cases} x^2 & \text{when } x < 0 \\ x & \text{when } 0 \leq x \leq 1 \\ \frac{1}{x} & \text{when } x > 1 \end{cases}$, find (i) $f(\frac{1}{2})$, (ii) $f(-2)$, (iii) $f(\sqrt{3})$, and (iv) $f(-\sqrt{3})$.

(i) Since $0 \leq \frac{1}{2} \leq 1$, $f(\frac{1}{2}) = \frac{1}{2}$.

(ii) Since $-2 < 0$, $f(-2) = (-2)^2 = 4$.

(iii) Since $\sqrt{3} > 1$, $f(\sqrt{3}) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

(iv) Since $-\sqrt{3} < 0$, $f(-\sqrt{3}) = (-\sqrt{3})^2 = 3$.

8. If $f(x) = x^3 - \frac{1}{x^3}$, show that $f(x) + f(\frac{1}{x}) = 0$.

$f(\frac{1}{x}) = (\frac{1}{x})^3 - \frac{1}{(\frac{1}{x})^3} = \frac{1}{x^3} - x^3$.

Then $f(x) + f(\frac{1}{x}) = x^3 - \frac{1}{x^3} + \frac{1}{x^3} - x^3 = 0$.

9. If $f(x) = \frac{2x}{1+x^2}$, show that $f(\tan \theta) = \sin 2\theta$.

$f(\tan \theta) = \frac{2\tan \theta}{1+(\tan \theta)^2} = \frac{2\tan \theta}{1+\tan^2 \theta}$.

Using the identity $\sin 2\theta = \frac{2\tan \theta}{1+\tan^2 \theta}$, we get $f(\tan \theta) = \sin 2\theta$.

1

0. If $f(x) = \frac{1-x}{1+x}$, then show that (i) $f(\frac{1}{x}) = -f(x)$ (ii) $f(-\frac{1}{x}) = \frac{1}{f(x)}$.

(i) $f(\frac{1}{x}) = \frac{1-\frac{1}{x}}{1+\frac{1}{x}} = \frac{\frac{x-1}{x}}{\frac{x+1}{x}} = \frac{x-1}{x+1} = -\frac{1-x}{1+x} = -f(x)$.

(ii) $f(-\frac{1}{x}) = \frac{1-(-\frac{1}{x})}{1+(-\frac{1}{x})} = \frac{1+\frac{1}{x}}{1-\frac{1}{x}} = \frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1} = \frac{1}{\frac{x-1}{x+1}} = \frac{1}{f(x)}$.

1

1. If for non-zero $x$, $a f(x) + b f(\frac{1}{x}) = \frac{5}{x} - 5$, where $a \neq b$, then find $f(x)$.

Let the given equation be

$a f(x) + b f(\frac{1}{x}) = \frac{5}{x} - 5$ ...(1)

Replacing $x$ with $\frac{1}{x}$, we get

$a f(\frac{1}{x}) + b f(x) = 5x - 5$ ...(2)

Multiplying equation (1) by $a$, and equation (2) by $b$, we have

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

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

Subtracting the two equations, we get

$(a^2 - b^2) f(x) = a(\frac{5}{x} - 5) - b(5x - 5)$

$(a^2 - b^2) f(x) = \frac{5a}{x} - 5a - 5bx + 5b$

$f(x) = \frac{\frac{5a}{x} - 5a - 5bx + 5b}{a^2 - b^2}$

$f(x) = \frac{5a - 5ax - 5bx^2 + 5bx}{x(a^2 - b^2)}$

$f(x) = \frac{5(a - ax - bx^2 + bx)}{x(a^2 - b^2)}$

1

2. If $f(x) = \frac{9}{5}x+32$, find $f(-10)$.

$f(-10) = \frac{9}{5}(-10) + 32 = 9(-2) + 32 = -18 + 32 = 14$.

1

3. If $f(x) = x^3$, find the value of $\frac{f(5)-f(1)}{5-1}$.

$f(5) = 5^3 = 125$.

$f(1) = 1^3 = 1$.

$\frac{f(5)-f(1)}{5-1} = \frac{125-1}{5-1} = \frac{124}{4} = 31$.

3. Final Answer

7. (i) $f(\frac{1}{2}) = \frac{1}{2}$ (ii) $f(-2) = 4$ (iii) $f(\sqrt{3}) = \frac{\sqrt{3}}{3}$ (iv) $f(-\sqrt{3}) = 3$

8. $f(x) + f(\frac{1}{x}) = 0$

9. $f(\tan \theta) = \sin 2\theta$

1

0. (i) $f(\frac{1}{x}) = -f(x)$ (ii) $f(-\frac{1}{x}) = \frac{1}{f(x)}$

1

1. $f(x) = \frac{5(a - ax - bx^2 + bx)}{x(a^2 - b^2)}$

1

2. $f(-10) = 14$

1

3. $\frac{f(5)-f(1)}{5-1} = 31$