A curve $G$ has equation $y = ax + b + \dfrac{c}{(x+1)^2}$, where $a$, $b$ and $c$ are positive integers. Let $R$ be the region bounded by the curve $G$, the axes and the line $x = 3$.

1. Show that the value of $y$ is strictly positive for $x > 0$

Given that $a$, $b$, and $c$ are positive integers, $a > 0$, $b > 0$, and $c > 0$.
For $x > 0$, we have $ax > 0$ and $(x+1)^2 > 1 > 0$, which means $\frac{c}{(x+1)^2} > 0$.
Since all individual terms are strictly positive, their sum must also be strictly positive:
$y = ax + b + \frac{c}{(x+1)^2} > 0$
Since $y > 0$ for all $x > 0$, the graph of $G$ lies entirely above the $x$-axis in this domain.

1. Set up the definite integral for the area of region $R$.

$\int_0^3 \left( ax + b + c(x+1)^{-2} \right) \, \mathrm{d}x = 42$

2. Integrate the function with respect to $x$.

$\left[ \frac{a}{2}x^2 + bx - c(x+1)^{-1} \right]_0^3 = 42$

3. Substitute the limits and simplify to prove the equation.

$\left( \frac{a}{2}(3)^2 + b(3) - \frac{c}{3+1} \right) - \left( \frac{a}{2}(0)^2 + b(0) - \frac{c}{0+1} \right) = 42$
$\left( \frac{9}{2}a + 3b - \frac{1}{4}c \right) - \left( -c \right) = 42$
$\frac{9}{2}a + 3b - \frac{1}{4}c + c = 42$
$\frac{9}{2}a + 3b + \frac{3}{4}c = 42 \quad \text{(shown)}$

1. Substitute the coordinates of the turning point $(0, 5)$ into the curve's equation.

$y = a(0) + b + \frac{c}{(0+1)^2} = 5$
$b + c = 5$

2. Find the first derivative and apply the stationary point condition at $x = 0$.

Rewrite $y = ax + b + c(x+1)^{-2}$.
$\frac{dy}{dx} = a - 2c(x+1)^{-3}$
Since there is a turning point at $x = 0$, $\frac{dy}{dx} = 0$:
$a - \frac{2c}{(0+1)^3} = 0$
$a - 2c = 0 \implies a = 2c$

3. Solve the system of equations using the result from part (ii) to find $a, b$, and $c$, then state the equation of $G$.

From part (ii), we are given:
$\frac{9}{2}a + 3b + \frac{3}{4}c = 42$
Multiply by $4$ to clear denominators:
$18a + 12b + 3c = 168 \implies 6a + 4b + c = 56$
Substitute $a = 2c$ and $b = 5 - c$ into the equation:
$6(2c) + 4(5 - c) + c = 56$
$12c + 20 - 4c + c = 56$
$9c = 36 \implies c = 4$
Therefore, $a = 2(4) = 8$ and $b = 5 - 4 = 1$.
The equation of the curve $G$ is:
$y = 8x + 1 + \frac{4}{(x+1)^2}$

1. Determine the values of $a$, $b$, and $c$ to obtain the full equation of curve $G$ (building on parts ii and iii).

From part (iii), the curve has a minimum turning point at $(0,5)$. Given $y = ax + b + c(x+1)^{-2}$:
$\frac{dy}{dx} = a - 2c(x+1)^{-3}$
At $x=0$, $y=5$ and $\frac{dy}{dx}=0$:
$b + c = 5 \implies b = 5 - c$
$a - 2c = 0 \implies a = 2c$
Substitute these into the equation given in part (ii), $\frac{9}{2}a + 3b + \frac{3}{4}c = 42$:
$\frac{9}{2}(2c) + 3(5-c) + \frac{3}{4}c = 42$
$9c + 15 - 3c + 0.75c = 42$
$6.75c = 27 \implies c = 4$
Using $c=4$, we find $a = 8$ and $b = 1$.
The equation of $G$ is $y = 8x + 1 + \frac{4}{(x+1)^2}$.

2. Find the equations of the vertical and oblique asymptotes.

Observe the term $\frac{4}{(x+1)^2}$ in the equation $y = 8x + 1 + \frac{4}{(x+1)^2}$.
As $x \to -1$, $\frac{4}{(x+1)^2} \to \infty$, meaning the curve has a vertical asymptote at $x = -1$.
As $x \to \pm\infty$, $\frac{4}{(x+1)^2} \to 0$, meaning the curve approaches the line $y = 8x + 1$.
Thus, the oblique asymptote is $y = 8x + 1$.

3. Identify the turning point and axial intercepts.

The minimum turning point is given as $(0, 5)$.
Since this point has $x=0$, it also serves as the $y$-intercept.
To find the $x$-intercept, set $y = 0$:
$8x + 1 + \frac{4}{(x+1)^2} = 0$
Using a graphing calculator to solve for $x$:
$x \approx -1.59 \text{ (to 3 s.f.)}$
The $x$-intercept is $(-1.59, 0)$.

4. Sketch the curve, correctly positioning its branches relative to the asymptotes.

Draw the coordinate axes and plot the asymptotes $x = -1$ and $y = 8x + 1$ with dashed lines.
Plot the minimum point / $y$-intercept at $(0, 5)$ and the $x$-intercept at $(-1.59, 0)$.
Since $\frac{4}{(x+1)^2} > 0$ for all valid $x$, the entire curve must be drawn strictly above the oblique asymptote $y = 8x + 1$.

• Right branch ($x > -1$): Comes down from the vertical asymptote $x = -1$, reaches a local minimum at $(0,5)$, and steadily rises, approaching $y = 8x + 1$.
• Left branch ($x < -1$): Approaches $y = 8x + 1$ from above as $x \to -\infty$, smoothly rises to pass through $(-1.59, 0)$, and approaches the vertical asymptote $x = -1$ towards $+\infty$.

1. Determine the constants $a$, $b$, and $c$ to find the equation of curve $G$, which was required in part (iii).

$y' = a - \frac{2c}{(x+1)^3}$
At the minimum turning point $(0, 5)$, $x=0$, $y=5$ and $y'=0$:
$b + c = 5 \quad \text{and} \quad a - 2c = 0 \implies a = 2c$
Substitute into the area result from (ii):
$\frac{9}{2}(2c) + 3(5-c) + \frac{3}{4}c = 42$
$9c + 15 - 3c + 0.75c = 42 \implies 6.75c = 27$
$c = 4, \quad a = 8, \quad b = 1$
Equation of curve $G$: $y = 8x + 1 + \frac{4}{(x+1)^2}$

2. Set up the definite integral for the volume of revolution of region $S$ using the washer method.

Region $S$ is bounded above by $G$ and below by the line $y = 8x + 1$.
The volume $V$ generated is:
$V = \pi \int_0^3 \left[ \left( 8x + 1 + \frac{4}{(x+1)^2} \right)^2 - (8x+1)^2 \right] dx$

3. Expand and simplify the integrand.

$V = \pi \int_0^3 \left[ (8x+1)^2 + 2(8x+1)\frac{4}{(x+1)^2} + \frac{16}{(x+1)^4} - (8x+1)^2 \right] dx$
$V = \pi \int_0^3 \left[ \frac{64x+8}{(x+1)^2} + 16(x+1)^{-4} \right] dx$
Rewrite the first fraction using $64x+8 = 64(x+1)-56$ to split it:
$V = \pi \int_0^3 \left[ \frac{64}{x+1} - 56(x+1)^{-2} + 16(x+1)^{-4} \right] dx$

4. Integrate and evaluate to find the volume.

$V = \pi \left[ 64 \ln|x+1| + \frac{56}{x+1} - \frac{16}{3(x+1)^3} \right]_0^3$
$V = \pi \left[ \left( 64 \ln 4 + \frac{56}{4} - \frac{16}{192} \right) - \left( 64 \ln 1 + 56 - \frac{16}{3} \right) \right]$
$V = \pi \left[ 64 \ln 4 + 14 - \frac{1}{12} - 56 + \frac{16}{3} \right]$
$V = \pi \left( 128 \ln 2 - 36.75 \right) \approx 163.28$
$V \approx 163 \text{ units}^3 \text{ (to 3 s.f.)}$