2015 A-level H2 Mathematics (9740) Paper 2 Question 3 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)
f(x) = \frac{1}{1-x^2}, x \in \mathbb{R}, x>1

f'(c) = \frac{2x}{1-x^2} > 0 for all x \in \mathbb{R}, x>1

Since f(x) is increasing function, it has no turning points, f is 1-1, its inverse will exist.

(a)
(ii)
Let f(x) = y

x = \pm \sqrt{1- \frac{1}{y}} Reject -\sqrt{1- \frac{1}{y}} since x>1

\Rightarrow f^{-1}(x) = \sqrt{1- \frac{1}{x}}

D_{f^{-1}} = \{x \in \mathbb{R} | x ~ \textless ~ 0 \}

(b)
Let g(x) = y = \frac{2+x}{1-x^2}

We need b^2 - 4ac \ge 0 for all real values of g.

(1)^2 - 4(y)(2-y) \ge 0

1 - 8y + 4y^2 \ge 0

y = \frac{8 \pm \sqrt{48}}{8} = 1 \pm \frac{\sqrt{3}}{2}

y \ge 1 + \frac{\sqrt{3}}{2} \mathrm{~or~} y \le 1 - \frac{\sqrt{3}}{2}

R_g = (-\infty, 1- \frac{\sqrt{3}}{2} ] \cup [1 +  \frac{\sqrt{3}}{2}, \infty )

2011 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
(a)
P(faulty)
= P(made by A and faulty) + P(made by B and faulty)
= 0.6(0.05) + 0.4(0.07)
= 0.058

(b)
P(made by A | faulty)
= \frac{\mathrm{P(made~by~A~and~faulty)}}{\mathrm{P(faulty)}}
= \frac{0.6(0.05)}{0.058}
= \frac{15}{29}

(ii)
(a)
P(exactly one of them is faulty)
= 0.058 \times (1 - 0.058) \times 2!
= 0.109272 (exact)

(b)
P(both were made by A | exactly one is faulty)
= \frac{\mathrm{P(both~were~made~by~A~and~exactly~one~is~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}
= \frac{\mathrm{P(one~is~made~by~A~and~faulty,~the~other~is~made~by~A~and~not~faulty)}}{\mathrm{P(exactly~one~is~faulty)}}
= \frac{0.6(0.05) \times 0.6(0.95) \times 2!}{0.109272}
= \frac{1425}{4553}

KS Comments:

Question can be easily solved by drawing a tree diagram. Do take note that we only wrong off NON exact answers to 3sf, so for (iia), we keep the full exact answer.

2010 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let P(n) be the statement: \sum_{r=1}^n r(r+2) = \frac{1}{6}n(n+1)(2n+7), n \in \mathbb{Z}^+

When n = 1, ~\text{LHS} = 1(3)=3, \text{~and~RHS} = \frac{1}{6}(2)(9) = 3 = \text{LHS}

\therefore, P(n) is true.

Assume that P(k) is true for some k \in \mathbb{Z}^+, i.e. \sum_{r=1}^k r(r+2) = \frac{1}{6}k(k+1)(2k+7)

Want to prove that P(k+1) is true, i.e. \sum_{r=1}^{k+1} r(r+2) = \frac{1}{6}(k+1)(k+2)(2k+9)

LHS

= \sum_{r=1}^{k+1} r(r+2)

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

= \frac{1}{6}k(k+1)(2k+7) + (k+1)(k+3)

= \frac{1}{6}(k+1)[k(2k+7)+6(k+3)]

= \frac{1}{6}(k+1)(2k^2+13k+18)

= \frac{1}{6}(k+1)(k+2)(2k+9)

= \text{RHS}

Since P(1) is true and P(k) \text{~is~true} \Rightarrow P(k+1) is true, hence by Mathematical Induction, P(n) is true for all n \in \mathbb{Z}^+

(ii)
(a)

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

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

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

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

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

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

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

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

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

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

(b)
As n \rightarrow \infty, \frac{1}{2(n+1)} \rightarrow 0, \frac{1}{2(n+2)} \rightarrow 0

Hence, \sum_{r=1}^{\infty} \frac{1}{r(r+2)} is a convergent series, and the value of the sum to infinity is \frac{3}{4}

2010 A-level H2 Mathematics (9740) Paper 2 Question 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
x^2 -6x + 34 = 0

x = \frac{6 \pm \sqrt{36-136}}{2}

x = 3 \pm 5i

(ii)
Since the coefficients are all real, another root of the equation is -2-i

[x-(-2+i)][x-(-2-i)]

= x^2 + 4x +5

By comparing coefficients,

x^4 + 4x^3 + x^2 + ax + b = (x^2+4x+5)(x^2-4)

\therefore, a=-16, b = -20

The other three roots are -2 - i, 2, -2