Please solve this questions in English please

For $S_1: x^2 + y^2 + 2x + 2y + 1 = 0$, the center $C_1 = (-1, -1)$ and the radius $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1+1-1} = \sqrt{1} = 1$.

For $S_2: x^2 + y^2 - 4x + 2y + 1 = 0$, the center $C_2 = (2, -1)$ and the radius $r_2 = \sqrt{(2)^2 + (-1)^2 - 1} = \sqrt{4+1-1} = \sqrt{4} = 2$.

The sum of the radii is $r_1 + r_2 = 1 + 2 = 3$.
Since $C_1C_2 = r_1 + r_2 = 3$, the circles $S_1$ and $S_2$ touch externally.

Part 2: Obtain the equation of the common tangent T at the point of contact.

The equation of the common tangent T is $\boxed{x=0}$.

9. (ii) A father and a mother have 5 children. This family is to occupy a particular front-line bench in church on a special thanksgiving service. Given that this bench has a capacity of 7 persons, in how many ways can this family be seated on the bench

(a) if the parents must seat adjacent to each other.

(b) if the parents must seat adjacent to each other at one end of the bench.

The family has 2 parents and 5 children, totaling 7 people. The bench has 7 seats.

Part (a): If the parents must seat adjacent to each other.

This leaves 6 entities to arrange (the parent unit and 5 children). The number of ways to arrange these 6 entities is $6!$.

Within the parent unit, the father and mother can swap positions in $2!$ ways.

Total ways = $6! \times 2! = 720 \times 2 = 1440$.
The number of ways is $\boxed{1440}$.

Part (b): If the parents must seat adjacent to each other at one end of the bench.

There are 2 possible ends for the parents to sit (left end or right end).

The parents can be arranged in $2!$ ways (FM or MF) at the chosen end.

The remaining 5 children can be arranged in the remaining 5 seats in $5!$ ways.

Total ways = (Number of ends) $\times$ (Ways to arrange parents) $\times$ (Ways to arrange children)
Total ways = $2 \times 2! \times 5! = 2 \times 2 \times 120 = 480$.
The number of ways is $\boxed{480}$.

10. (i) Given that $f(x) = \frac{2}{x(x+1)(x+2)}$, express $f(x)$ in partial fractions. Hence show that $\int_{2}^{4} f(x)\,dx = \ln\left(\frac{27}{25}\right)$.

Part 1: Express $f(x)$ in partial fractions.

• Let $x=0$: $2 = A(1)(2) \Rightarrow 2A = 2 \Rightarrow A = 1$.
• Let $x=-1$: $2 = B(-1)(-1+2) \Rightarrow 2 = B(-1)(1) \Rightarrow -B = 2 \Rightarrow B = -2$.
• Let $x=-2$: $2 = C(-2)(-2+1) \Rightarrow 2 = C(-2)(-1) \Rightarrow 2C = 2 \Rightarrow C = 1$.

Part 2: Hence show that $\int_{2}^{4} f(x)\,dx = \ln\left(\frac{27}{25}\right)$.

Using logarithm properties, this can be written as:

Thus, $\boxed{\int_{2}^{4} f(x)\,dx = \ln\left(\frac{27}{25}\right)}$.

10. (ii) Sketch the graph of the curve whose equation is given by $y = \frac{2x-7}{x-4}$, showing clearly the points where the curve meets the coordinate axes and the behavior of the curve near its asymptotes.

• x-intercept (set $y=0$): $0 = \frac{2x-7}{x-4} \Rightarrow 2x-7=0 \Rightarrow x=\frac{7}{2}=3.5$. The x-intercept is $(3.5, 0)$.
• y-intercept (set $x=0$): $y = \frac{2(0)-7}{0-4} = \frac{-7}{-4} = \frac{7}{4}=1.75$. The y-intercept is $(0, 1.75)$.

• Vertical Asymptote: Set the denominator to zero: $x-4=0 \Rightarrow x=4$.
• Horizontal Asymptote: As $x \to \pm\infty$, $y = \lim_{x \to \pm\infty} \frac{2x-7}{x-4} = \lim_{x \to \pm\infty} \frac{2 - \frac{7}{x}}{1 - \frac{4}{x}} = \frac{2}{1} = 2$. So $y=2$.

• As $x \to 4^+$ (e.g., $x=4.1$), $y = \frac{2(4.1)-7}{4.1-4} = \frac{1.2}{0.1} = 12$. So $y \to +\infty$.
• As $x \to 4^-$ (e.g., $x=3.9$), $y = \frac{2(3.9)-7}{3.9-4} = \frac{0.8}{-0.1} = -8$. So $y \to -\infty$.

• As $x \to +\infty$, $y = \frac{2x-7}{x-4} = \frac{2(x-4)+8-7}{x-4} = 2 + \frac{1}{x-4}$. Since $x-4 > 0$, $y$ approaches 2 from above.
• As $x \to -\infty$, $y = 2 + \frac{1}{x-4}$. Since $x-4 < 0$, $y$ approaches 2 from below.

Sketch Description:
The graph of $y = \frac{2x-7}{x-4}$ is a hyperbola.

1. Draw the x and y axes.
2. Plot the x-intercept at $(3.5, 0)$ and the y-intercept at $(0, 1.75)$.
3. Draw a vertical dashed line at $x=4$ (vertical asymptote).
4. Draw a horizontal dashed line at $y=2$ (horizontal asymptote).
5. The curve has two branches:
   • For $x > 4$, the curve starts from positive infinity near the vertical asymptote ($x=4$) and approaches the horizontal asymptote ($y=2$) from above as $x$ increases.
   • For $x < 4$, the curve starts from negative infinity near the vertical asymptote ($x=4$), passes through the intercepts $(3.5, 0)$ and $(0, 1.75)$, and approaches the horizontal asymptote ($y=2$) from below as $x$ decreases.