The problem consists of three parts: (i) Evaluate the definite integral $\int_{2}^{3} (t^2 - 3) dt$. (ii) Given $g(x) = 5 - 2x$ for $x < 1$, sketch $g(x)$, state the range, and find the value of $a$ such that $g(a) = 19$. (iii) Express the repeating decimal $1.171717...$ in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $\frac{a}{b}$ is in its lowest terms.

1. Problem Description

The problem consists of three parts:

(i) Evaluate the definite integral $\int_{2}^{3} (t^2 - 3) dt$.

(ii) Given $g(x) = 5 - 2x$ for $x < 1$, sketch $g(x)$, state the range, and find the value of $a$ such that $g(a) = 19$.

(iii) Express the repeating decimal $1.171717...$ in the form $\frac{a}{b}$, where $a$ and $b$ are integers and $\frac{a}{b}$ is in its lowest terms.

2. Solution Steps

(i) Evaluate $\int_{2}^{3} (t^2 - 3) dt$.

First, find the antiderivative of $t^2 - 3$:

$\int (t^2 - 3) dt = \frac{t^3}{3} - 3t + C$.

Now, evaluate the definite integral:

$\int_{2}^{3} (t^2 - 3) dt = [\frac{t^3}{3} - 3t]_{2}^{3} = (\frac{3^3}{3} - 3(3)) - (\frac{2^3}{3} - 3(2)) = (9 - 9) - (\frac{8}{3} - 6) = 0 - (\frac{8}{3} - \frac{18}{3}) = 0 - (-\frac{10}{3}) = \frac{10}{3}$.

(ii) Given $g(x) = 5 - 2x$ for $x < 1$.

Sketching $g(x)$ is a straight line with slope $-2$ and y-intercept $5$. Since $x < 1$, the domain is $x < 1$.

When $x = 1$, $g(1) = 5 - 2(1) = 3$. Since $x < 1$, $g(x) > 3$. The range of $g(x)$ is $g(x) > 3$.

Now, find the value of $a$ such that $g(a) = 19$.

$g(a) = 5 - 2a = 19$

$-2a = 19 - 5 = 14$

$a = \frac{14}{-2} = -7$

Since $a = -7 < 1$, this is a valid solution.

(iii) Express $1.171717...$ in the form $\frac{a}{b}$.

Let $x = 1.171717...$

$100x = 117.171717...$

$100x - x = 117.171717... - 1.171717...$

$99x = 116$

$x = \frac{116}{99}$

Since $116 = 2^2 \cdot 29$ and $99 = 3^2 \cdot 11$, there are no common factors between 116 and

9

9. Thus, the fraction is in its lowest terms.

3. Final Answer

(i) $\frac{10}{3}$

(ii) $g(x) > 3$, $a = -7$

(iii) $\frac{116}{99}$