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### 1.1

We are asked to determine the integral:

\[
\int \left( \frac{12}{4\sqrt{x}} - \frac{4}{3x^3} \right) dx
\]

Simplify the terms first:

- \(\frac{12}{4\sqrt{x}} = 3x^{-\frac{1}{2}}\)
- \(\frac{4}{3x^3} = \frac{4}{3}x^{-3}\)

Now the integral becomes:

\[
\int \left( 3x^{-\frac{1}{2}} - \frac{4}{3}x^{-3} \right) dx
\]

We can integrate term by term:

1. For \( 3x^{-\frac{1}{2}} \), apply the power rule:
   \[
   \int 3x^{-\frac{1}{2}} dx = 3 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 6\sqrt{x}
   \]

2. For \( -\frac{4}{3}x^{-3} \), apply the power rule:
   \[
   \int -\frac{4}{3}x^{-3} dx = -\frac{4}{3} \cdot \frac{x^{-2}}{-2} = \frac{2}{3}x^{-2}
   \]

Thus, the full solution to the integral is:

\[
6\sqrt{x} + \frac{2}{3x^2} + C
\]

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### 1.2

The derivative of the function \( f(x) \) is given as:

\[
f'(x) = \frac{3x^3 + 5x^2 - x}{x}
\]

Simplify the expression:

\[
f'(x) = 3x^2 + 5x - 1
\]

To find \( f(x) \), integrate \( f'(x) \):

\[
f(x) = \int (3x^2 + 5x - 1) dx
\]

Integrating term by term:

\[
f(x) = x^3 + \frac{5}{2}x^2 - x + C
\]

Next, use the point \( (1, \frac{11}{2}) \) to find \( C \). Plug \( x = 1 \) and \( f(1) = \frac{11}{2} \) into the equation:

\[
f(1) = 1^3 + \frac{5}{2}(1)^2 - 1 + C = \frac{11}{2}
\]

This simplifies to:

\[
1 + \frac{5}{2} - 1 + C = \frac{11}{2}
\]
\[
C = \frac{11}{2} - \frac{5}{2} = 3
\]

Thus, the function \( f(x) \) is:

\[
f(x) = x^3 + \frac{5}{2}x^2 - x + 3
\]

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### 1.3

We are asked to determine \( k \) such that:

\[
\int_k^{12} (x - 6) dx = 16
\]

First, compute the indefinite integral:

\[
\int (x - 6) dx = \frac{x^2}{2} - 6x
\]

Now apply the limits from \( k \) to 12:

\[
\left[ \frac{x^2}{2} - 6x \right]_k^{12} = 16
\]

Substitute the upper limit \( x = 12 \):

\[
\frac{12^2}{2} - 6(12) = 72 - 72 = 0
\]

Thus, the equation becomes:

\[
0 - \left( \frac{k^2}{2} - 6k \right) = 16
\]

Simplify:

\[
-\frac{k^2}{2} + 6k = 16
\]

Multiply the entire equation by 2 to eliminate the fraction:

\[
-k^2 + 12k = 32
\]

Rearrange:

\[
k^2 - 12k + 32 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
k = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2}
\]
\[
k = \frac{12 \pm 4}{2}
\]

Thus, the two possible solutions for \( k \) are:

\[
k = \frac{12 + 4}{2} = 8 \quad \text{or} \quad k = \frac{12 - 4}{2} = 4
\]

Therefore, \( k = 4 \) or \( k = 8 \).

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Would you like more details on any of these steps or have any questions?

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Here are five related questions:
1. What is the process for simplifying complex integrals?
2. How do you apply the quadratic formula in different situations?
3. What are some methods for integrating rational functions?
4. How can you verify the solution to an indefinite integral?
5. What are some common mistakes to avoid when solving for constants in integrals?

**Tip:** Always simplify an expression before integrating or differentiating, as it makes calculations easier and less prone to errors.

1.1 Bepaal ∫(12 / 4√x - 4 / 3x^3) dx. 1.2 Die afgeleide van die funksie f word gegee: f'(x) = (3x^3 + 5x^2 - x) / x. Bepaal die funksie f indien (1 ; 11/2) 'n punt op die grafiek van f is. 1.3 Bepaal k indien ∫(x - 6) dx van k tot 12 = 16.

Solve integrals and derivatives with detailed steps. This problem covers integration techniques, finding functions from derivatives, and solving quadratic equations, suitable for Grades 11-12.

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