Solutions to Review 1

Question 1
(i)
y = f(x) = \frac{x^2 + 14x + 50}{3(x+7)}

3y(x+7) = x^2 + 14x + 50

x^2 + (14-3y)x + 50 - 21 y = 0

\text{discriminant} \ge 0

(14-3y)^2 - 4(1)(50-21y) \ge 0

196 - 84y + 9y^2 - 200 + 84y \ge 0

9y^2 - 4 \ge 0

(3y - 2)(3y + 2) \ge 0

y \le - \frac{2}{3} \text{~or~} y \ge \frac{2}{3}

(ii)
Using long division, we find that

y = \frac{x^2 + 14x + 50}{3(x+7)} = \frac{x}{3} + \frac{7}{3} + \frac{1}{3(x+7)}

So the asymptotes are y = \frac{x}{3} + \frac{7}{3} and x = -7

Curve C of 1(ii)

Question 2
(i)
x^2 - 9y^2 + 18y = 18

x^2 - 9(y^2 - 2y) = 18

x^2 - 9[(y-1)^2 - 1^2] = 18

x^2 - 9(y-1)^2 + 9 = 18

x^2 - 9(y-1)^2 = 9

\frac{x^2}{9} - (y-1)^2 = 1

This is a hyperbola with centre (0, 1), asymptotes are y = \pm \frac{x}{3} + 1, and vertices (3, 1) and (-3, 1).

y = \frac{1}{x^2} + 1 is a graph with asymptotes x = 0 and y=1.

Use GC to plot.

(ii)
\frac{x^2}{9} - (y-1)^2 = 1—(1)

y = \frac{1}{x^2} + 1 —(2)

Subst (2) to (1),

\frac{x^2}{9} - (\frac{1}{x^2} + 1 - 1)^2 = 1

\frac{x^2}{9} - (\frac{1}{x^2})^2 = 1

x^2 - \frac{9}{x^4} = 9

x^6 - 9 = 9x^4

x^6 - 9x^4 - 9 = 0

(iii)
From graph, we observe two intersections. Thus, two roots.

Question 3
(ai)
\sum_{r=1}^n (r+1)(3r-1)

= \sum_{r=1}^n (3r^2 + 2r -1)

= \sum_{r=1}^n 3r^2 + \sum_{r=1}^n 2r - \sum_{r=1}^n 1

= 3 \sum_{r=1}^n r^2 + 2 \sum_{r=1}^n r - \sum_{r=1}^n 1

= 3 \frac{n}{6}(n+1)(2n+1) + 2 \frac{n}{2}(1 + n) - n

= \frac{n}{2}(n+1)(2n+1) + n(1+n) - n

= \frac{n}{2}(n+1)(2n+1) + n^2

(aii)
2 \times 4 + 3 \times 10 + 4 \times 16 + ... + 21 \times 118

= 2 [2 \times 2 + 3 \times 5 + 4 \times 8 + ... + 21 \times 59]

= 2 [(1+1) \times (3 \cdot 1 - 1) + (2+1) \times (3 \cdot 2 -1) + (3+1) \times (3 \cdot 3 -1)  + ... + (20+1) \times (3 \cdot 20 -1) ]

= 2 \sum_{r=1}^{20} (r+1)(3r-1)

= 2 [\frac{n}{2}(n+1)(2n+1) + n^2 ]

= n(n+1)(2n+1) + n^2

= n(2n^2 + 3n + 1) + n^2

= 2n^3 + 4n^2 + n

(bi)
\frac{2}{(r-1)(r+1)} = \frac{A}{r-1} - \frac{B}{r+1}

2 = A(r+1) - B(r-1)

Let r = -1

2 = - B(-2) \Rightarrow B = 1

Let r = 1

2 = A(2) \Rightarrow A = 1

\therefore \frac{2}{(r-1)(r+1)} = \frac{1}{r-1} - \frac{1}{r+1}

(bii)
\sum_{r=2}^n \frac{1}{(r-1)(r+1)}

= \frac{1}{2} \sum_{r=2}^n \frac{2}{(r-1)(r+1)}

= \frac{1}{2} \sum_{r=2}^n (\frac{1}{r-1} - \frac{1}{r+1})

= \frac{1}{2} [ 1 - \frac{1}{3}

+ \frac{1}{2} - \frac{1}{4}

+ \frac{1}{3} - \frac{1}{5}

...

+ \frac{1}{n-3} - \frac{1}{n-1}

+ \frac{1}{n-2} - \frac{1}{n}

+ \frac{1}{n-1} - \frac{1}{n+1}]

= \frac{1}{2} [1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}]

= \frac{1}{2} (\frac{3}{2} - \frac{n+1+n}{n(n+1)})

= \frac{3}{4} - \frac{2n+1}{2n(n+1)}

(biii)
As n \to \infty, \frac{1}{n} \to 0 and \frac{1}{n+1} \to 0, the sum of series tends to \frac{3}{4}, a constant. Thus, series is convergent.

(biv)

\sum_{r=5}^{n+3} \frac{1}{(r-3)(r-1)}

Replace r by r + 2. Then we have

\sum_{r=3}^{n+1} \frac{1}{(r-1)(r+1)}

= \sum_{r=2}^{n+1} \frac{1}{(r-1)(r+1)} - \frac{1}{(2-1)(2+1)}

= \frac{3}{4} - \frac{2(n+1)+1}{2(n+1)[(n+1)+1]} - \frac{1}{3}

= \frac{5}{12} - \frac{2n+3}{2(n+1)(n+2)}

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