F 5 maths Revision chapter

F.5 Additional Mathematics Revision Notes / P.1

Quadratic Equations, Quadratic Functions, Remainder Theorem and Factor Theorem Note: 1. Question will state that α and β are the roots of an equation, Ax2 + Bx + C = 0,

then α+β = -BA , αβ =

CA and discriminant, ∆ = B2 – 4AC

2. If the roots are real and distinct B2 – 4AC > 0 real and equal B2 – 4AC = 0 real B2 – 4AC > 0 imaginary B2 – 4AC < 0 3. For finding value of any expression, write down the sum and product of previous roots. α2 +β2 = (α+β)2 - 2αβ α3 +β3 = (α+β)[(α+β)2 - 3αβ] 4. Note whether a new quadratic equation have to be formed. Past Question Section A Questions:

1. Let x2 + 5x + 1x2 - x + 1 = r ...(*).

Express (*) in the form ax2 + bx + c = 0. Hence find the range of the values of r for real values of x. [89-P1-Q5] 2. α, β are the roots of the quadratic equation x2 – (k + 2)x + k = 0. (a) Find α + β and αβ in terms of k. (b) If (α + 1)(β + 2) = 4, show that α = -2k. Hence find the two values of k. [90-P1-Q4]

3. p, q and k are real numbers satisfying the following conditions: p + q + k = 2 pq + qk + kp = 1

(a) Express pq in terms of k. (b) Find a quadratic equation, with coefficients in terms of k, whose roots are p and q.

Hence, find the range of possible values of k. [91-P1-Q7]

4. α, β are the roots of the equation x2 + px + q = 0 and α+3, β+3 are the roots of the equation x2 + qx + p = 0. Find the value of p and q. [93-P1-Q3] [Hint] From the first equation, α+β = -p , αβ = q From the second equation, (α+3)+(β+3) = -q , (α+3)(β+3) = p From above, -p + 6 = -q , q - 3p + 9 = p

Solving the last two equations, p = 1 and q = -5 5. Let f(x) = x2 + (1-m)x + 2m-5, where m is a constant. Find the discriminant of the equation f(x) = 0.

Hence, find the range of values of m so that f(x) > 0 for all real values of x. [95-P1-Q1] [Solution] The discriminant for x2 + (1-m)x + 2m-5 = 0, ∆ = (1-m)2 – 4(1)(2m-5) = m2 – 10m + 21 If f(x) > 0 for all real values of x, the equation has no real roots, ie. ∆ < 0 m2 – 10m + 21 < 0 (m-3)(m-7) < 0

3 < m < 7 6. Given x2 – 6x + 11 ≡ (x + a)2 + b, where x is real.

(a) Find the values of a and b. Hence, write down the least value of x2 – 6x + 11.

(b) Using (a), or otherwise, write down the range of possible values of 1

x2 – 6x + 11 [96-P1-Q4]

[Solution] (a) Given x2 – 6x + 11 ≡ (x + a)2 + b ≡ x2 + 2ax + (a2 + b) Equating coefficients, a = -3 and b = 2 Hence, the least value of x2 – 6x + 11 is 2 (when x = 3). (b) From (a), ∞ > x2 – 6x + 11 > 2

So, 0 < 1

x2 – 6x + 11 < 12

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q5

P1-Q11

P1-Q4

P1-Q9

P1-Q7

P1-Q9

P1-Q6

P1-Q9

P1-Q3P1-Q8P1-Q10

P1-Q8

P1-Q1P1-Q10

P1-Q4P1-Q8P1-Q13

P1-Q8 P1-Q2P1-Q3P1-Q11

P1-Q4

P1-Q7

P1-Q12 Marks 5

16

6

16

7

16

6

16

6 16 16

16

5 16

5 7

16

7 44

16

5 7

16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q09 Q11 Q03 Q15 Q04 Marks 6 7 4 12 4


7. The graph of y = x2 – (k-2)x + k+1 intersects the x-axis at two distinct points (α,0) and (β,0), where k is real.

(a) Find the range of possible values of k. (b) Furthermore, if |α+β| < 5, find the range of possible values of k. [96-P1-Q8][Hint]

(a) Since the equation has 2 real and distinct roots, ∆ > 0 (k-2)2 – 4(1)(k+1) > 0 k2 – 8k > 0 k(k-8) > 0 Range of possible values of k is k < 0 or k > 8. (b) If |α+β| < 5, |k-2| < 5 (sum of roots = negative of coefficient of x)

ie. –5 < k-2 < 5 or, -3 < k < 7 Combining the result of (a), the range of possible values of k is -3 < k < 0.

8. Let α and β be the roots of the equation x2 + (k+2)x + 2(k-1) = 0, where k is real.

(a) Show that α and β are real and distinct. (b) If |α-β| > 3, find the range of possible values of k. [97-P1-Q8]

9. α, β are the roots of the quadratic equation x2 – 2x + 7 = 0. Find the quadratic equation whose roots are (α+2) and (β+2). [98-P1-Q2] [Solution] Given α + β = 2 and αβ = 7 (α+2) + (β+2) = 6 and (α+2)(β+2) = αβ + 2(α+β) + 4 = 15 The required equation is x2 – 6x + 15 = 0

10. The quadratic equations x2 – 6x + 2k = 0 and x2 – 5x + k = 0 have a common root α (ie. α is

a root of both equations). Show that α = k and hence find the value(s) of k. [98-P1-Q3] [Hint] Refer to Section B Questions, 95-P1-Q10 11. Let f(x) = 2x2 + 2(k-4)x + k, where k is real.

(a) Find the discriminant of the equation f(x) = 0. (b) If the graph of y = f(x) lies above the x-axis for all values of x, find the range of possible values of

k. [99-P1-Q4]

12. α and β are the roots of the quadratic equation x2 + (p – 2)x + p = 0 where p is real. (a) Express α+β and αβ in terms of p.

(b) If α and β are real such that α2 + β2 = 11, find the value(s) of p. [00-P1-Q7]

13. Let p = x2 + 2x + 8

x - 2 ...(*) where x is real.

By expressing (*) in the form ax2 + bx + c = 0, find the range of possible values of x2 + 2x + 8

x - 2 .

Hence find the range of possible values of |x2 + 2x + 8

x - 2 |. [01-Q09]

14. Let f(x) = x2 – 2x – 6 and g(x) = 2x + 6. The graphs of y = f(x)

and y = g(x) intersect at points A and B. C is the vertex of the graph of y = f(x).

(a) Find the coordinates of points A, B and C. (b) Write down the range of values of x such that f(x) < g(x). Hence write down the value(s) of k such that the equation

f(x) = k has only one real root in this range. [02-Q11] 15. α and β are the roots of the quadratic equation x2 – 5x + k = 0 such that | α - β | = 3. Find the

value of k. [03-Q03] 16. If kx2 + x + k > 0 for all real values of x, where k ≠ 0, find the range of possible value of k. [06-Q04]


Section B Questions 1. α, β are the roots of the quadratic equation x2 + (p+1)x + (p-1) = 0, where p is a real number. (a) Show that α, β are real and distinct. (b) Express (α-2)(β-2) in terms of p. [92-P1-Q9(a,b)] [Hint] (a) Consider the discriminant, ∆ = (p+1)2 – 4(1)(p-1)

= p2 –2p + 5 = (p-1)2 + 4 > 0

The roots of the equation are real and distinct. (b) (α-2)(β-2) = αβ - 2(α+β) + 4

= (p-1) + 2(p+1) + 4 = 3p + 5

2. Let f(x) = x[x2 + kx + (2k-3)], where k is a real number. It is given that the equation f(x) = 0 has one

real and two imaginary roots. (a) Find the real root of f(x) = 0. (b) Show that 2 < k < 6. (c) It is also given that the equation f’(x) = 0 has tow distinct real roots α and β.

(1) Express (α - β)2 in terms of k. Hence, or otherwise, show that k ≠ 3.

(2) Suppose |α - β| < 23 and k is an integer.

Using the results of (b) and (c)(1), find the value(s) of k. [94-P1-Q8]

3. Let f(x) = 12x2 + 2px – q and g(x) = 12x2 + 2qx – p where p, q are distinct real numbers. α, β are the roots of the equation f(x) = 0 and α, γ are the roots of the equation g(x) = 0.

(a) Using the fact that f(α) = g(α), find the value of α. Hence, show that p + q = 3.

(b) Express β and γ in terms of p.

(c) Suppose |β3 + γ3| < 724 .

(1) Find the range of possible values of p. (2) Furthermore, if p > q, write down the possible integral values of p and q. [95-P1-Q10]

[Hint] (a) Since α is a root of both equations, f(α) = g(α) = 0, ie. 12α2 + 2pα – q = 0 and 12α2 + 2qα – p = 0 Subtracting second from first, 2pα - 2qα = -(p – q)

α = -12

Substituting the value back into f(α) = 0, 12(-12 )2 + 2p(

-12 ) – q = 0

p + q = 3

(b) Hence, (-12 ) + β =

-p6 and (

-12 )γ =

-p12

ie. β = -p + 3

6 and γ = p6

(c) (1) Substituting in, |p2 – 3p + 3| < 7 Solving, -1 < p < 4

(2) If p > q and p + q = 3, the integral solutions are p=2,q=1 or p=3,q=0


Inequalities and Absolute Value Sign Note: 1. For the absolute value part being the subject of a quadratic equation, solve and reject negative

root of the equation. 2. In other problems, consider each possible cases. 3. For inequality, inequality sign has to be changed when multiplying or dividing negative values. Type 1: Fraction comparing with a value. Past Questions

1. Solve the inequality 2(x + 1)

x - 2 > 1 [94-P1-Q1]

[Solution]

2(x + 1)

x - 2 - 1 > 0

x + 4x - 2 > 0

x < -4 or x > 2

2. Solve the inequality 2x - 3x + 1 < 1 [96-P1-Q3]

3. Solve the inequality x

x - 1 > 2 [99-P1-Q2]

4. Solve 1x > 1 [00-P1-Q1]

5. Solve y

y - 2 < 2.

Hence, or otherwise, solve 2x

2x - 2 < 2. [01-Q11]

Type 2: An absolute sign comparing with a value.

Past Questions

1. Solve |x(x+5)| > 6 for real values of x. [92-P1-Q3]

[Solution] -6 > x(x+5) or x(x+5) > 6 (x+2)(x+3) < 0 or (x-1)(x+6) > 0 -3 < x < -2 or (-6 > x or x > 1)

Combining, the solution is x < -6 or -3 < x < -2 or x > 1. 2. Solve |-x2 + 2x + 3| > 5 for real values of x. [93-P1-Q5]

[Solution] -5 > -x2 + 2x + 3 or -x2 + 2x + 3 > 5

(x+2)(x-4) > 0 or x2 – 2x + 2 < 0 (no possible solution) The solution is -2 > x or x > 4.

3. Solve |3x – 4| < 2 and 1

2x - 1 < 1 [97-P1-Q5]

[Solution]

-2 < 3x – 4 < 2 and 1

2x - 1 – 1 < 0

2 < 3x < 6 and 2 - 2x2x - 1 < 0

23 < x < 2 and (x <

12 or x > 1)

Combining the two solutions, 1 < x < 2. 4. (a) Solve | 1 – x | = 2 [00-P1-Q5a]

5. Solve the following inequalities: (a) | x – 1 | > 2 (b) | | y | - 1 | > 2 [02-Q07] 6. Solve the following equations:

(a) | x - 3 | = 1 (b) | x - 1 | = | x2 - 4x + 3 | [04-Q08]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q3 P1-Q5 P1-Q5 P1-Q5(a)Marks 6 6 6 2

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q07 Q08 Q07Marks 5 6 5

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q1 P1-Q3 P1-Q2 P1-Q1 Marks 4 4 4 3

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q11 Marks 6


7. Solve |x - x2| = -4x [05-Q07]


Type 3: Quadratic Functions involving absolute value sign.

Past Questions 1. Solve (x – 2)2 – 5 | x – 2 | - 6 = 0. [89-P1-Q8] 2. Solve (x + 2)2 – 8 | x + 2 | + 15 > 0. [90-P1-Q6] 3. Solve | x – 2 | = | x2 – 4 | [91-P1-Q3]

4. Solve (x-3)2 - |x-3| - 12 = 0 [94-P1-Q7] [Solution] (|x-3| + 3)(|x-3| - 4) = 0 |x – 3| = -3 (rej.) or |x – 3| = 4 x – 3 = -4 or x –3 = 4 The solution is x = -1 or x = 7. 5. (a) Solve x2 – 6x – 16 > 0

(b) Using (a), or otherwise, solve (y+1)2 – 6|y+1| - 16 > 0 [98-P1-Q6] [Solution] (a) Factorizing (x+2)(x-8) > 0 The solution is x < -2 or x > 8 (b) Substituting x with |y+1| gives |y+1| < -2 (rej.) or |y+1| > 8 ie. –8 > y + 1 or y + 1 > 8 The solution is y < -9 or y > 7.

6. Solve |x – 3| = |x2 – 4x + 3| [99-P1-Q3] [Hint] Factorizing |x – 3| = |x – 1||x – 3| |x – 3| ( |x – 1| - 1) = 0

The solutions are x = 3, x = 0 or x = 2. 7. Solve the inequality x2 > | x | [03-Q05] 8. Find the range of values of k such that x2 - x - 1 > k(x - 2) for all values of x. [05-Q05] 9. Solve x |x| + 5x + 6 = 0 [06-Q06]

Others: Consider case by case.

Past Questions

1. α, β are the real roots of the equation x2 + (k-2)x – (k-1) = 0. If |α| = |β|, find k. [92-P1-Q6] [Solution] α+β = 2-k and αβ = 1-k Case 1 α=β, 2α = 2-k and α2 = 1-k 4(1-k) = (2-k)2

k2 = 0 k = 0 (α = β = 1) Case 2 α= -β 0 = 2-k and -αβ = 1-k k = 2 (α = -β = +1)

Hence, k = 0 or 2.

2. By considering the two cases x>0 and x<0, or otherwise, solve the inequality x - 5x > 4.

[95-P1-Q4] [Solution] Case 1 Consider x > 0, x2 – 5 > 4x (x + 1)(x – 5) > 0 x < -1 or x > 5 Since x > 0, the solution is x > 5. Case 2 Consider x < 0, x2 – 5 < 4x -1 < x < 5 Since x < 0, the solution is –1 < x < 0 Combining the two cases, the solution is –1 < x < 0 or x > 5.

3. (b) By considering the cases x < 1 and x > 1, or otherwise, solve | 1 – x | = x – 1. [00-P1-Q5b] 4. (a) Solve | x – 1 | = | x | - 1, where 0 < x < 1. (b) Solve | x – 1 | = | x | - 1. [07-Q11]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q8 P1-Q6 P1-Q3 P1-Q7 P1-Q6 P1-Q3Marks 6 6 5 7 6 5

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q05 Q05 Q04 Marks 4 4 4

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q6 P1-Q4 P1-Q5Marks 6 6 5

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q11 Marks 5


Mathematical Induction Type 1: Given terms on LHS and expression on RHS. Note: 1. For positive integer: begin from 1, then k+1 after k. For non-negative integer: begin from 0, then k+1 after k. For even positive integer: begin from 2, then 2(k+1) after 2k. For odd positive integer: begin from 1, then 2k+1 after 2k-1. 2. Step 1 Declare proposition. Step 2 Show that LHS = RHS for the initial condition. Step 3 Assume that LHS = RHS for n = k (or values mentioned above). Step 4 Adding an extra term to LHS, transfer to previous expression, factorize with fractions. Show that both sides are still agree. Step 5 Concluding the proposition matches. Past Questions:

1. Prove, by mathematical induction, that 1×2×3 + 2×3×4 +...+ n(n+1)(n+2) = n(n+1)(n+2)(n+3)

4

for all positive integers n. [89-P2-Q2] 2. Let Tn = n2 + n for any positive integer n.

Prove, by mathematical induction, that T1 + T2 +...+ Tn = 13 n(n+1)(n+2) for any positive integer n.

[90-P2-Q2]

3. (a) Prove, by mathematical induction, that 12 + 22 +…+ n2 = 16 n(n+1)(2n+1) for all positive

integers n. (b) Using the formula in (a), find the sum 1×2 + 2×3 +…+ n(n+1) [91-P2-Q7] 4. Prove, by Mathematical Induction, that 1×2+2×5+3×8+…+n(3n-1) = n2(n+1) for all positive

integers n. [92-P2-Q1]

5. Prove that 12×2+22×3+…+n2(n+1) = n(n+1)(n+2)(3n+1)

12 for any positive integer n. [93-P2-Q1]

[Solution]

Let P be the proposition that “12×2+22×3+…+n2(n+1) = n(n+1)(n+2)(3n+1)

12 ”.

When n=1, LHS = 12×2 = 2

RHS = 1(1+1)(1+2)(3×1+1)

12 = 2

∴ P is true for n = 1. Assume P is true for n = k, where k is an integer

ie. 12×2+22×3+…+k2(k+1) = k(k+1)(k+2)(3k+1)

12

then 12×2+22×3+…+k2(k+1) + (k+1)2(k+1+1) = k(k+1)(k+2)(3k+1)

12 + (k+1)2(k+1+1)

= (k+1)(k+2)

12 [ k(3k+1) + 12(k+1) ]

= (k+1)(k+2)

12 (3k2 + 13k + 12)

= (k+1)(k+2)

12 (k+3)(3k+4)

= (k+1)[(k+1)+1][(k+1)+2][3(k+1)+1]

12

If P is true for n = k, it is also true for n = k+1. By the Principle of Mathematical Induction, P is true for any positive integer n.

6. Prove that 12 +

322 +

523 +…+

2n-12n = 3 -

2n+32n for any positive integer n. [94-P2-Q5]

7. Let Tn = (n2+1)(n!) for any positive integer n.

Prove, by Mathematical Induction, that T1 + T2 +…+ Tn = n[(n+1)!] for any positive integer n. [Note: n! = n(n-1)(n-2)…3×2×1] [97-P2-Q7] [Hint] T1 + T2 +…+ Tk = k[(k+1)!] T1 + T2 +…+ Tk + Tk+1 = k[(k+1)!] + [(k+1)2+1][(k+1)!] = [(k+1)!][k + (k+1)2 + 1] = [(k+1)!](k+1)(k+2) = (k+1)[(k+1)+1]!

8. Prove, by Mathematical Induction, that 1×2+2×3+22×4+…+2n-1(n+1) = 2n(n) for all positive

integers n. [98-P2-Q3]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q2 P2-Q2 P2-Q7 P2-Q1 P2-Q1 P2-Q5 P2-Q7 P2-Q3 P2-Q12a P2-Q4 Marks 5 5 8 5 5 5 5 6 6 6

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q12 Q12 Q07 Q08 Q05Marks 8 8 5 5 5


9. Prove, by Mathematical Induction, that cos θ + cos 3θ + cos 5θ +…+ cos (2n-1)θ = sin 2nθ2.sin θ ,

where sin θ ≠ 0, for all positive integers n. [99-P2-Q12(a)]

[Hint]

cos θ +…+ cos (2k-1)θ = sin 2kθ2.sin θ

cos θ +…+ cos (2k-1)θ + cos [2(k+1)-1] θ = sin 2kθ2.sin θ + cos (2k+1) θ

= 1

2.sin θ [ sin 2kθ + 2.sin θ.cos (2k+1) θ ]

= 1

2.sin θ [ sin 2kθ + sin (2k+2) θ - sin 2kθ ]

= sin 2(k+1)θ

2.sin θ

10. Prove, by mathematical induction, that 12 – 22 + 32 – 42 +...+ (-1)n-1 n(n + 1)

2 for all positive integers n. [00-P2-Q4]

11. Prove, by mathematical induction, that

1×2+2×3+3×4+...+n(n+1) = 13 n(n+1)(n+2) for all positive integers n.

Hence evaluate 1×3+2×4+3×5+...+50×52. [01-Q12] 12. (a) Prove, by mathematical induction, that 2(2) + 3(22) + 4(23) +...+ (n + 1)(2n) = n(2n + 1) for all

positive integers n. (b) Show that 1(2) + 2(22) + 3(23) +...+ 98(298) = 97(299) + 2. [02-Q12] 13. Prove, by mathematical induction, that

12 +

222 +

323 +…+

n2n = 2 -

n+22n for all positive integers n. [03-Q07]

14. Prove, by mathematical induction, that

1×22×3 +

2×22

3×4 + 3×23

4×5 +...+ n×2n

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

n + 2 - 1 [05-Q08]

15. Let a ≠ 0 and a ≠ 1. Prove by mathematical induction that

1

a - 1 - 1a -

1a2 - ... -

1an =

1an(a - 1) for all positive integers n. [07-Q05]

Type 2: An expression on LHS is divisible by RHS. Note: 1. Beware of the condition of integers. 2. Step 1 Declare proposition Step 2 Show the divisibility for the initial condition Step 3 Assume that LHS is a multiple of RHS for n = k Step 4 Change the (k+1)-th term in the form of k-th term. Show that the expression is also a

multiple. Step 5 Concluding the proposition Past Questions

1. Prove, by Mathematical Induction, that (8n-1) is divisible by 7 for all positive integers n. [95-P2-Q6] 2. Prove, by Mathematical Induction, that for all positive integers n, (2n3+n) is divisible by 3. [96-P2-Q4]

[Solution] Let P be the proposition that “ (2n3 + n) is divisible by 3”. When n = 1, 2(1)3 + (1) = 3 which is divisible by 3. ∴ P is true for n = 1. Assume P is true for n = k, where k is an integer ie. 2k3 + k = 3N where N is a natural number then 2(k+1)3 + (k+1) = 2k3 + 6k2 + 6k + 2 + k + 1 = (2k3 + k) + (6k2 + 6k + 3) = 3N + 3(2k2 + 2k + 1) = 3(N + 2k2 + 2k + 1) Since k is a positive integer, (N + 2k2 + 2k + 1) is also a positive integer. ie. 2(k+1)3 + (k+1) is divisible by 3. Hence, P is true for n = k+1 if it is true for n = k. By the Principle of Mathematical Induction, P is true for all positive integers n.

3. Prove that 9n - 1 is divisible by 8 for all positive integers n. [04-Q07]

4. Prove that n3 - n + 3 is divisible by 3 for all positive integers n. [06-Q08]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q6 P2-Q4 Marks 6 6

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q07 Q08Marks 5 5


Binomial Theorem Type 1: Expanding binomial to power m and adding or multiplying to another binomial to power n.

Note: 1. Expand the binomial according to Binomial Theorem. 2. Equating coefficients of appropriate terms to evaluate the unknown. 3. When multiplying to another polynomial, consider the power of x clearly. Past Questions

1. Find the constant term in the expansion of (1 + x)10(1 - 2x )3. [89-P2-Q1]

2. In the expansion of (1 + 3x)2(1 + x)n, where n is a positive integer, the coefficient of x is 10.

(a) Find the value of n. (b) Find the coefficient of x2. [92-P2-Q2]

3. (a) Expand (1 – 2x)3 and (1 + 1x )5

(b) Find, in the expansion of (1 – 2x)3(1 + 1x )5

(1) the constant term, and (2) the coefficient of x. [94-P2-Q3]

4. Given (x2 + 1x )5 - (x2 -

1x )5 = ax7 + bx +

cx5 , find the values of a, b and c.

Hence evluate (2 + 12 )5 - (2 -

12 )5 [95-P2-Q4]

5. If the coefficient of x2 in the expansion of (1 – 2x + x2)n is 66, find the value of n and the

coefficient of x3. [07-Q12]

5. Expand (1 + x)n(1 – 2x)4 in ascending powers of x up to the term x2, where n is a positive integer. If the coefficient of x2 is 54, find the coefficient of x. [97-P2-Q8] [Solution] (1 + x)n(1 – 2x)4

= (1 + nx + n(n-1)

2 x2 +…)(1 – 8x + 24x2 +…)

= 1 + (n-8)x + [24 – 8n + n(n-1)

2 ]x2 +…

If the coefficient of x2 is 54, n(n-1) – 16n + 48 = 108 n2 – 17n – 60 = 0 (n + 3)(n – 20) = 0

Since n is a positive integer, n = 20, and the coefficient of x = 12

6. Find the coefficient of x2 in the expansion of (x - 2x )6 [98-P2-Q1]

7. (a) Expand (1 + 2x)n in ascending powers of x up to the term x3, where n is a positive integer.

(b) In the expansion of (x - 3x )2(1 + 2x)n, the constant term is 210. Find the value of n.

[99-P2-Q7] 8. Expand (1 + 2x)7(2 – x)2 in ascending powers of x up to the terms x2. [00-P2-Q2]

9. Find the constant term in the expansion of (2x3 + 1x )8. [01-Q04]

10. If n is a positive integer and the coefficient of x2 in the expansion of (1 + x)n + (1 + 2x)n is 75,

find the value(s) of n. [02-Q01]

11. Determine whether the expansion of (2x2 + 1x )9 consists of

(a) a constant term, (b) an x2 term.

In each part, find the term if it exists. [03-Q12] 12. (a) Expand (1 + 2x)6 in ascending powers of x up to the term x3.

(b) Find the constant term in the expansion of (1 - 1x +

1x2 )(1 + 2x)6. [04-Q02]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q1 P2-Q2 P2-Q3 P2-Q4 P2-Q8 P2-Q1 P2-Q7 P2-Q2Marks 5 5 5 6 7 4 6 5



Type 2: Extending to trinomial

Past Questions

1. Given (1 + 2x – 3x2)n = 1 + ax + bx2 + terms involving higher powers of x, where n is a positive

integer. (a) Express a and b in terms of n. (b) If b = 63, find the value of n. [90-P2-Q1]

2. Given that (1 + x + ax2)8 = 1 + 8x + k1x2 + k2x3 + terms involving higher powers of x. (a) Express k1 and k2 in terms of a. (b) If k1 = 4, find the value of a.

Hence, find the value of k2. [91-P2-Q1]

3. Given (1 + 4x + x2)n = 1 + ax + bx2 + other terms involving higher powers of x, where n is a positive integer. (a) Express a and b in terms of n. (b) If a = 20, find n and b. [93-P2-Q3][Solution]

(a) [1 + (4x + x2)]n

= 1 + n(4x + x2) + n(n-1)

2 (4x + x2)2 +…

= 1 + 4nx + [8n(n-1) + n]x2 +… Equating coefficients, a = 4n and b = 8n2 – 7n (b) If a = 20, n = 5 and b = 165

4. It is given that (1 + x + ax2)6 = 1 + 6x + k1x2 + k2x3 + terms involving higher powers of x. (a) Express k1 and k2 in terms of a. (b) If 6, k1 and k2 are in A.P., find the value of a. [96-P2-Q2][Solution] (a) [1 + (x + ax2)]6

= 1 + 6(x + ax2) + 15(x + ax2)2 + 20(x + ax2)3 +… = 1 + 6x + (6a + 15)x2 + (30a + 20)x3 +… Equating coefficients, k1 = 6a + 15 k2 = 30a + 20 (b) If 6, k1 and k2 are in A.P., 2k1 = 6 + k2 2(6a + 15) = 6 + (30a + 20)

a = 29

4. (a) Expand (1 + y)5

(b) Using (a), or otherwise, expand (1 + x + 2x2)5 in ascending powers of x up to the term x2. [05-Q02]

5. It is given that (1 - 2x + 3x2)n = 1 - 10x + kx2 + terms involving higher powers of x, where n is a positive integer and k is a constant. Find the values of n and k. [06-Q03]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q1 P2-Q1 P2-Q3 P2-Q2Marks 5 5 6 6

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q03 Marks 5


Trigonometric Functions

Note: 1. Remember or look up all trigonometric formulae. 2. Don’t mix up the general solutions of all trigonometric functions. Past Questions 1. Let y = 5 sin θ - 12 sin θ + 7. (a) Express y in the form r sin(θ - α) + p where r, α and p are constants and 0o < α < 90o. (b) Using the result in (a), find the least value of y. [89-P2-Q5] 2. Find the general solution of 2 cos 2θ + 5 sin θ - 3 = 0. [89-P2-Q6]

3. Find the general solution of the equation 2 sin x2 sin

3x2 = 1. [90-P2-Q5]

4. (a) If cos θ + 3 sin θ = r cos(θ - α), where r > 0 and 0o < α < 90o, find r and α. [90-P2-Q6]

5. Find the general solution of cos 4θ + cos 2θ = cos θ [91-P2-Q3] 6. By using the identity cos 3θ = 4.cos3θ - 3.cos θ, find the general solution of the equation sin 2θ(4.cos2θ - 3) – sin θ = 0 [92-P2-Q5] [Solution] sin 2θ(4.cos2θ - 3) – sin θ = 0 2.sin θ.cosθ(4.cos2θ - 3) – sin θ = 0 sin θ(2.cos 3θ - 1) = 0 sin θ = 0 or cos 3θ = 0.5

θ = nπ or 3θ = 2nπ + π3

θ = nπ or θ = (6n + 1)π

9 where n is an integer

7. By expressing 3 cos x – sin x in the form r.cos (x+α), find the general solution of the equation 3 cos x – sin x = 1, giving your answer in terms of π. [93-P2-Q2] [Solution] 3 cos x – sin x = 1

2.cos(x + π6 ) = 1

x + π6 = 2nπ +

π3

x = (12n + 1)π

6 or (4n - 1)π

2 where n is an integer

8. Find the general solution of the equation cos(x – 7o) = 2.cos(x + 7o), giving the answer correct to

the nearest 0.1 degree. [94-P2-Q2] [Solution]

cos x cos 7o + sin x sin 7o = 2.cos x cos 7o – 2.sin x sin 7o

cos x cos 7o = 3.sin x sin 7o

tan x = 13 cot 7o

x = (180n + 69.8)o where n is an integer 9. (a) Show that cos2A – cos2B = sin(A+B).sin(B-A)

(b) ABC is a triangle. (1) Using (a), show that cos2A – cos2B + sin2C = 2.cos A.sin B.sin C (2) If cos2A – cos2B – cos2C = -1, show that ABC is a right-angled triangle.

(c) Using (a), or otherwise, show that cos2x – sin2y = cos(x+y).cos(x-y) Hence, find the general solution of cos22θ - sin23θ + cos θ.sin 5θ = 0 [95-P2-Q9]

10. Find the general solution of the equation sin 5θ + sin 3θ = cos θ [96-P2-Q1] [Solution] sin 5θ + sin 3θ = cos θ 2 sin 4θ.cos θ = cos θ cos θ(2.sin 4θ - 1) = 0 cos θ = 0 or sin 4θ = 0.5

θ = (2n + 12 )π or 4θ = nπ + (-1)n(

π6 )

θ = 4nπ + 1

2 π or θ = nπ4 + (-1)n(

π24 ) where n is an integer

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q5

P2-Q6 P2-Q5 P2-Q6

P2-Q3 P2-Q8

P2-Q5 P2-Q12

P2-Q2 P2-Q2 P2-Q6P2-Q9

P2-Q1 P2-Q1P2-Q4

P2-Q7 P2-Q8 P2-Q7

Marks 55

5 5

5 16

6 16

5

4 616

5 45

6 7 8

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q03 Q06 Q08 Q10 Q05 Q04 Q02

Q09 Marks 4 5 5 5 5 4 3

6


11. Show that sin 3θsin θ +

cos 3θcos θ = 4.cos 2θ [97-P2-Q1]

[Solution]

sin 3θsin θ +

cos 3θcos θ =

sin 3θ.cos θ + cos θ.sin 3θsin θ.cos θ

= sin 4θ

sin θ.cos θ

= 4.sin 2θ.cos 2θ

sin 2θ

= 4.cos 2θ

12. By expressing 6.sin x + 8.cos x in the form r.sin(x+α), find the general solution of the equation 6.sin x + 8.cos x = 5, and give your answer correct to the nearest degree. [97-P2-Q4]

13. Show that sin(3x+π4 ).cos(3x-

π4 ) =

1 + sin 6x2

Hence, or otherwise, find the general solution of the equation sin(3x+π4 ).cos(3x-

π4 ) =

34

[98-P2-Q7] 14. (a) Show that cos 3θ = 4.cos3θ - 3.cos θ

(b) Find the general solution of the equation cos 6x + 4.cos 2x = 0 [99-P2-Q8]

15. (a) By expressing cos x - 3 sin x in the form r cos(x + θ), or otherwise, find the general solution of the equation cos x - 3 sin x = 2.

(b) Find the number of points of intersection of the curves y = cos x and y = 2 + 3 sin x for 0 < x < 9π [00-P2-Q7]

16. (a) If sin x + cos x = r cos(x + θ) for all values of x, where r > 0 and -π < θ < π, find the values of r

and θ. (b) Find the general solution of the equation sin x + cos x = 2

[01-Q06]

17. Given 0 < x < π2 . Show that

tan x - sin2xtan x + sin2x =

42 + sin 2x – 1.

Hence, or otherwise, find the least value of tan x - sin2xtan x + sin2x . [02-Q08]

18. Given two acute angles α and β.

(a) Show that sin α + sin βcos α + cos β = tan(

α + β2 )

(b) If 3sin α - 4cos α = 4cos β - 3sin β, find the value of tan(α + β). [03-Q10]

19. Find the general solution of the equation sin 3x + sin x = cos x. [04-Q05] 20. Find the general solution of the equation sin(θ + 30o) = cos θ. [05-Q04] 21. Prove the identity cos2x - cos2y ≡ - sin(x + y) sin (x - y) [06-Q02] 22. (a) Express cos θ - 3 sin θ in the form r cos(θ + α), where r > 0 and 0o < α < 90o

(b) Find the general solution of the equation cos 2x - 3 sin 2x = 1 [06-Q09] 23. Find the general solution of the equation cos x - 2 cos 2x + cos 3x = 0 [07-Q03]


Three-dimensional Problems Note: 1. Familiarize the usage of Sine Law and Cosine Law. 2. State clearly the name of angle between line (or plane) and a plane. Past Questions Section A Questions

1. In the figure below, VABCD is a right pyramid with a square base of sides of length 4 cm. ∠VAB = 60o. Find, correct to the nearest 0.1 degree, (a) the angle between the plane VAB and the base ABCD, (b) the angle between the planes VAB and VAD. [91-P2-Q6]

V C B D A 2. In the figure above, VABCD is a right pyramid with a square base of side

6 cm. VB = 9 cm. Find, correct to the nearest 0.1 degree, (a) the angle between edge VB and the base ABCD, (b) the angle between the planes VAB and VAD. [92-P2-Q7]

3. In the figure, VABC is a right pyramid whose base ABC is an equilateral

triangle. AB = 12 cm and VA = 24 cm. D is a point on VB such that AD is perpendicular to VB. Find, correct to 3 significant figures, (a) ∠VBA and AD,

(b) the angle between the faces VAB and VBC. [93-P2-Q7]

4. In the figure, VPQRST is a right pyramid whose base PQRST is a regular pentagon. PQ = 10 cm and ∠PVQ = 42o. U is a point on VQ such that PU is perpendicular to VQ. Find, correct to 3 significant figures, (a) PU and PR, (b) the angle between the faces VPQ and VQR. [95-P2-Q7]

5. In the Figure, OABC is a pyramid such that OA = 3, OB = 5, BC = 12, ∠AOC = 120o and ∠OAB = ∠OBC = 90o. (a) Find AC. (b) A student says that the angle between the

planes OBC and ABC can be represented by ∠OBA. Determine whether the student is correct or not. [04-Q11]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q

P2-Q12

P2-Q13 P2-Q6 P2-Q7

P2-Q12 P2-Q7

P2-Q12P2-Q7

P2-Q12

P2-Q12

P2-Q13

P2-Q11

P2-Q12 Marks

16

16 7

16 8

16 7

16 7

16

16

16

16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q15 Q18 Q11 Marks 12 12 6


F.4 Summer Vacation Assignment Phase 1 (Additional Mathematics)

Phase 2 (Additional Mathematics) (Mathematics) Odd-numbered problems of P.221-236 in Canotta 4B textbook (except M.C.)

NB: There will have a test in the first week for Mathematics.

HKCE 01 15 3D

16 Int.Vol.

18 Sketch

HKCE 00 Paper 1 10 Sketch

12 Quad.

13 Rate

Paper 2 8Red.Form

11 Int.Vol.

12 3D


12 Quad.

13 Diff.

Paper 2 113D

12 MI+Int.

13 Int.Vol.


11 Quad.

13 Rate

Paper 2 9Int.

12 Int.Vol.

13 3D


12 Diff.

13 Rate

Paper 2 10Int.

11 Red.Form

12 3D

HKCE 96 Paper 1 Paper 2 12

HKCE 01 1 Diff.

2 Ind.Int.

4 Binom.

6 Gen.Sol.

7 Slope

9 Quad.

11 Ineq.

12 MI

HKCE 00 Paper 1 1 Ineq

2 Diff

3 1st Pr

4 Slope

5 Abs

7 Quad

Paper 2 1Ind.Int

2 Binom

4 MI

7 Gen.Sol

HKCE 99 Paper 1 1 Diff

2 Ineq

3 Abs

4 Quad

6 Slope

8 Rate

Paper 2 1Def.Int

2 Ind.Int

6 Slope

7 Binom

8 Gen.Sol

HKCE 98 Paper 1 1 1st Pr

2 Quad

3 Quad

6 Abs

8 Slope

Paper 2 1Binom

3 MI

4 Slope

6 Def.Int

7 Gen.Sol


2 Diff

4 Rate

5 Abs

8 Quad

Paper 2 1Trig

2 Ind.Int

4 Gen.Sol

5 Slope

7 MI

8 Binom


2 1st Pr

3 Ineq

4 Quad

6 Slope

8 Quad

Paper 2 1Gen.Sol

2 Binom

4 MI

6 Slope


Differentiation Type 1: First Principle of Differentiation

Past Questions

1. Let f(x) = 1

1 + x

Find f’(x) from first principles. [91-P1-Q2]

2. (a) Simplify ( 2(x + ∆x) - 2x )( 2(x + ∆x) + 2x )

(b) Find ddx ( 2x ) from first principles. [93-P1-Q1]

[Solution] (a) ( 2(x + ∆x) - 2x )( 2(x + ∆x) + 2x )

= 2(x + ∆x) – 2x = 2∆x

(b) ddx ( 2x )

= 1

∆x0lim→∆x

( 2(x + ∆x) - 2x )

= 1

∆x0lim→∆x

( 2(x + ∆x) - 2x )(2(x + ∆x) + 2x 2(x + ∆x) + 2x

)

= 1

∆x0lim→∆x

(2∆x

2(x + ∆x) + 2x )

= (2

0lim→∆x 2(x + ∆x) + 2x

)

= 12x

3. Find ddx (x2) from first principles. [96-P1-Q2]

4. Find ddx ( x ) from first principles. [98-P1-Q1]

5. (a) Show that 1

x + ∆x -

1x =

-∆xx( x + ∆x)( x + x + ∆x)

(b) Find ddx (

1x ) from first principles.

[00-P1-Q3]

6. Find ddx (x3) from first principles. [03-Q02]

7. Find ddx (

1x ) from first principles. [05-Q03]

8. Find ddx (x2 + 1) from first principles. [07-Q04]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q2 P1-Q1 P1-Q2 P1-Q1 P1-Q3Marks 5 5 4 4 5

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q02 Q03 Q04Marks 4 4 4


Type 2: Differentiation and Differential Equation

Past Questions Section A Questions

1. Let y = x sin 5x.

Find dydx and

d2ydx2 .

Hence find d2ydx2 + 25y. [89-P1-Q1]

2. Let tan θ = 2 tan x and y = tan(θ - x) where 0 < x < π2 .

(a) Express y in terms of tan x.

(b) When dydx = 0, find the value of x. [89-P1-Q4]

3. Let f(x) = x2 + k sin 2x, where k is a constant. If f’(0) = 1, find the value of k. [90-P1-Q1]

4. Let y = tan(1x )

Show that x2dydx + (y2 + 1) = 0

Hence show that d2ydx2 +

2(x + y)x2

dydx = 0 [94-P1-Q4]

[Solution]

Given y = tan(1x )

dydx =

-1x2 sec2(

1x )

x2dydx + (y2 + 1) = x2

-1x2 sec2(

1x ) + [tan2(

1x ) + 1]

= 0

Differentiating x2dydx + (y2 + 1) = 0 with respect to x,

2xdydx + x2d

2ydx2 + 2y

dydx = 0

x2d2y

dx2 + 2(x+y) dydx = 0

d2ydx2 +

2(x + y)x2

dydx = 0

5. Let f(x) = sin3x

Find f’(x) and f”(x) [96-P1-Q1]

6. Let f(x) = 3 + x2 . Find f’(-1). [97-P1-Q1]

7. Find (a) ddx sin(x2 + 1)

(b) ddx [

sin(x2 + 1)x ] [99-P1-Q1]

8. Find (a) ddx sin2x

(b) ddx sin2(3x + 1) [00-P1-Q2]

9. Find ddx (

x2

2x + 1 ) [01-Q01]

10. Let x sin y = 2002. Find dydx . [02-Q03]

11. (a) Find ddx sin3(x2 + 1)

(b) Let xy + y2 = 2005. Find dydx . [05-Q09]

12. Find ddx [

sin(2x + 1)x ] [06-Q01]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q1

P1-Q4 P1-Q1 P1-Q5 P1-Q4 P1-Q1 P1-Q1

P1-Q2 P1-Q1

Marks 45

5 6 6 3 33

4

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q01 Q03 Q09 Q01 Marks 3 4 6 3


Type 3: Find equation of tangent or normal from a curve Past Questions 1. Find the coordinates of the two points on the curve y = x3 at which the tangents to the curve have a

slope of 34 .

Hence find the equations of the two tangents to the curve y = x3 which are parallel to the line 3x – 4y = 0. [89-P1-Q3]

2. Given the curve C : x2 + 4xy + 5y2 = 1, find dydx .

Hence find the equations of the two tangents to C which are parallel to the line y = -12 x.

[90-P1-Q7]

3. Let C be the curve y = 1x + x , where x ≠ 0. P(1, 2) and Q(

12 ,

52 ) are two points on C.

(a) Find equations of the tangent and normal to C at P. (b) Show that the tangent to C at Q passes through the point A(0, 4) [91-P1-Q6]

4. The curve (x-2)(y2+3) = -8 cuts the y-axis at two points. Find (a) the coordinates of the two points; (b) the slope of the tangent to the curve at each of the two points. [92-P1-Q5][Solution]

(a) Put x = 0, y = +1 ie. the points are (0, 1) and (0, -1)

(b) Differentiate with respect to x,

(y2 + 3) + (x – 2)(2ydydx ) = 0

dydx =

y2 + 32y(2 - x)

dydx |(0,1) = 1

dydx |(0, -1) = -1

5. Given the curve C : x2 – 2xy2 + y3 + 1 = 0

(a) Find dydx .

(b) Find the equation of the tangent to C at the point (2, -1) [93-P1-Q7] [Solution]

(a) 2x – 2y2 – 4xydydx + 3y2dy

dx = 0

dydx =

2x - 2y2

4xy - 3y2

(b) At (2, -1), dydx =

2(2) - 2(-1)2

4(2)(-1) - 3(-1)2

= -211

Equation of tangent is y + 1x - 2 =

-211

2x + 11y + 7 = 0 6. Given the curve C : x2 + y.cos x – y2 = 0

(a) Find dydx .

(b) P(π2 ,

-π2 ) is a point on the curve C. Find the equation of the tangent to the curve at P.

[94-P1-Q6] 7. P(4, 1) is a point on the curve y2 + y x = 3, where x>0.

(a) Find the value of dydx at P.

(b) Find the equation of the normal to the curve at P. [95-P1-Q6] [Solution]

(a) Differentiate the equation with respect to x,

2ydydx + x

dydx +

y2 x

= 0

At P(4, 1), 2(1) dydx + ( 4 )

dydx +

(1)2( 4)

= 0

dydx =

-116

(b) Slope of normal = 16

The equation of the normal is y - 1x - 4 = 16

16x – y – 63 = 0

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q3 P1-Q7 P1-Q6 P1-Q5 P1-Q7 P1-Q6 P1-Q6 P1-Q6 P1-Q2 P1-Q8 P1-Q6 P1-Q4 Marks 5 7 7 6 7 7 7 7 3 7 6 5

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q07 Q02 Q04 Q09 Q12 Marks 5 4 4 6 5


8. Find the equations of the two tangents to the curve C : y = 6

x + 1 which are parallel to the line

x + 6y + 10 = 0 [96-P1-Q6]

9. P(8, 1) is a point on the curve y2 + 3 x y – 3 = 0. Find the value of dydx at P. [97-P1-Q2]

10. P(0, 2) is a point on the curve x2 – xy + 3y2 = 12.


(b) Find the equation of the normal to the curve at P. [98-P1-Q8]

11. The point P(a, a) is on the curve 3x2 – xy – y2 – a2 = 0, where a is a non-zero constant.


(b) Find the equation of the tangent to the curve at P. [99-P1-Q6]

12. P (-1 , 2) is a point on the curve (x + 2)(y + 3) = 5. Find

(a) the value of dydx at P.

(b) the equation of the tangent to the curve at P. [00-P1-Q4] 13. P (2 , 0) is a point on the curve x – (1 + sin y)5 = 1. Find the equation of the tangent to the curve

at P. [01-Q07] 14. Find the equation of the tangent to the curve C : y = (x – 1)4 + 4 which is parallel to the line

y = 4x + 8. [02-Q02]

15. Given that 3x2 + 3y2 – 2xy = 12, find dydx when x = 2, y = 0. [03-Q04]

16. In the Figure, P (a , b) is a point on the curve C : y = x3. The

tangent to C at P passes through (0 , 2). (a) Show that b = 3a3 + 2. (b) Find the values of a and b. [04-Q09]

17. (a) Let x2 - xy + y2 = 7. Find dydx .

(b) Find the equation of the normal to the curve x2 - xy + y2 = 7 at the point (1 , 3) [06-Q12]

Type 4: Rate of Change Past Questions Section A Questions

1. A vessel is in the shape of a right circular cone and with semi-vertical angle 30o. Water is flowing out of the cone through its apex at a constant rate of π cm3/s.

(a) Let V cm3 be the volume of water in the vessel when the depth of water is h cm. Express V

in terms of h. (b) How fast is the water level falling when the depth of water is 4 cm? [92-P1-Q7] [Solution]

(a) Let r cm be the radius of water surface when the depth of water is h cm.

tan 30o = rh ie. r =

h3

therefore, V = 13 π(

h3 )2h

= πh3

9

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q

P1-Q12P1-Q7

P1-Q12

P1-Q9

P1-Q11 P1-Q4

P1-Q13 P1-Q8

P1-Q13 Marks

16 7

16

16

16 5

16 7

16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q19 Q09Marks 12 5


(b) dVdt =

π3 h2dh

dt

Put dVdt = -π, and at h = 4, (-π) =

π3 (4)2dh

dt

dhdt =

-316

ie. water is falling at a rate of 3

16 cm/s.

2. A man stands at a horizontal distance of 30 m from a sight-seeing elevator of a building. The

elevator is rising vertically with a uniform speed of 1.5 m/s. When the elevator is a height h m above the ground, its angle of elevation from the man is θ. Find the rate of change of θ with respect to time when the elevator is at a height 30 3 m above the ground. [Note: You may assume the sizes of the elevator and the man are negligible.]

[97-P1-Q4] 3. A ball is thrown vertically upwards from the roof of a building 40 m in height. After t s, the

height of the ball above the roof is h m, where h = 20t – 5t2. At this instant, the angle of elevation of the ball from a point O, which is at a horizontal distance of 55 m from the building, is θ.

h m

40m

55m

Oθ

• ball

Figure 1

(a) Find (1) tan θ in terms of t, (2) the value of θ when t = 3

(b) Find the rate of change of θ with respect to time when t = 3. [99-P1-Q8]

4. Two rods HA and HB, each of length 5 m, are hinged at H. The rods slide such that A, B, H are on the same vertical plane and A, B move in opposite directions on the horizontal floor, as shown in the figure. Let AB be x m and the distance of H from the floor be y m.

(a) Write down an equation connecting x and y. (b) When H is 3 m from the ground, its falling speed is 2 m/s. Find the rate of change of the

distance between A and B with respect to time at that moment. [07-Q09]


Type 5: Maxima and minima Past Questions

Section A Question 1. Let y = x + sin 2x, where 0 < x < π.

Find (a) dydx and

d2ydx2

(b) the maximum and minimum values of y. [91-P1-Q4] 2. Let f(x) = 2 sin x – x for 0 < x < π. Find the greatest and least values of f(x). [03-Q13]

Type 6: Sketch curve Note: 1. Find the x- and y-intercepts, turning points and extreme points (and asymptotes, if required). 2. See whether the question asks for the testing of turning points. Past Questions Section A Question 1. Using the information in the following table, sketch the graph of y = f(x), where f(x) is a

polynomial. [Solution] 2. Let f(x) be a polynomial. Figure 2 shows a sketch of the

curve y = f '(x), where -2 < x < 6. The curve cuts the x-axis at the origin and (a , 0), where 0 < a < 6. It is known that the areas of the shaded regions R1 and R2 as shown in Figure 2 are 3 and 1 respectively.

(a) Write down the x-coordinates of the maximum and minimum points of the curve y = f(x) for -2 < x < 6.

(b) It is known that f(-2) = 2 and f(0) = 1.

(i) By considering f∫a

0 '(x) dx, find the value of f(a).

(ii) In Figure 3, sketch the curve y = f(x) for -2 < x < 6.

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q

P1-Q10 P1-Q12

P1-Q11

P1-Q4 P1-Q11

P1-Q11

P1-Q9 P1-Q12 P1-Q12

P1-Q13

P1-Q13

P1-Q12

Marks 16 16

16

7 16

16

16

16 16

16

16

16

Year 2001 2002 2003 2004 2005 2006 2007P&Q 13 16 Q Q Marks 7 12

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q

P1-Q10

P1-Q11

P1-Q12

P1-Q11

P1-Q9P1-Q3

P1-Q9

P1-Q10

P1-Q10

P1-Q9

P1-Q10 Marks

16

16

16

16

16 5

16

16

16

16

16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q18 Q17 Q17 Q13 Q10Marks

12

12

12 7 5

x < 0 x = 0 0 < x < 1 x = 1 1 < x < 2 x = 2 x > 2 f(x) 1 2 1 f'(x) < 0 0 > 0 0 < 0 0 > 0

x

y = f(x) (1 , 2)

(0 , 1) (2 , 1)

y


3. Let f(x) be a function of x. The figure shows the graph of y = f’(x) which is a straight line with x- and y-intercepts 4 and 2 respectively.

(a) Find the slope of the tangent to the curve y = f(x) at x = 1. (b) Find the x-coordinate(s) of all the turning point(s) of the curve y = f(x). For each turning

points, determine whether it is a minimum point or a maximum point. [07-Q10]


Integration Type 1: Indefinite and Definite Integration Note: 1. Make clear whether the integration is definite or indefinite. 2. Use Trigonetric Identities for those problems involving trigonometric functions. Past Questions 1. (a) Find cos∫ 22x dx.

(b) using the result in (a), find sin∫ 22x dx. [89-P2-Q4]

2. Evaluate ∫ 20

π

(sin x + cos x)2dx [91-P2-Q2]

3. Find (sin x – cos x)∫ 2 dx [94-P2-Q1]

[Solution]

∫ (sin x – cos x)2 dx = (1 - 2sin x cos x) dx ∫ = x – sin2x + C where C is a constant

(or x + cos2x + C or x + 12 cos 2x + C where C is a constant)

4. Evaluate ∫ 20

π

cos2x dx [99-P2-Q1]

5. Find ∫ 2x + 1 dx [00-P2-Q1]

6. Find cos∫ 2θ dθ. [03-Q01]

7. Find (a) ∫ cos(3x + 1) dx

(b) ∫ (2 - x)2004 dx [04-Q01]

8. Find ∫ (2x - 3)7 dx [05-Q01]

9. Find ∫x4 + 1

x2 dx [07-Q01] Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q4 P2-Q2 P2-Q1 P2-Q1 P2-Q1Marks 5 5 4 3 4

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q01 Q01 Q01 Q01Marks 3 4 2 3


Type 2: Integration with substitution Note: 1. By means of substitution, change the problem into a simpler one. 2. For trigonometric integration, think about sinmx cosnx and tanmx secnx. Past Questions 1. Using the substitution u = 2x2 + 1, evaluate ∫

+

2

0 2

3

d12

8 xxx [89-P2-Q3]

2. Using the substitution u = sin2x, find ∫

+x

xxxx dcos4sin9

cossin22

[90-P2-Q3]

3. Find x∫ x - 1 dx

[Hint: Let u = x – 1 ] [97-P2-Q2] [Solution] Let u = x – 1 , du = dx

∫ x x - 1 dx = (u + 1)∫ u du

= (u∫ 3/2 + u1/2) du

= 25 u5/2 +

23 u3/2 + C where C is a constant

= 25 (x – 1)5/2 +

23 (x – 1)3/2 + C

4. Using the substitution u = sin θ, evaluate ∫ 20

π

cos5θ sin2θ dθ [98-P2-Q6]

5. Find x(x + 2)∫ 99dx [99-P2-Q2]

6. Find x

∫ 3x2 + 1 dx.

(Hint: Let u = 3x2 + 1.) [01-Q02]

7. Find ∫ 21

0

11 - x2 dx.

(Hint: Let x = sin θ.) [02-Q04]

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q02 Q04 Marks 4 4

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q3 P2-Q3 P2-Q2 P2-Q6 P2-Q2Marks 5 5 4 6 4


Type 3: Find equation from given slope Note: After each integration, a constant is added and have to be found.

Past Questions

1. The slope at any point (x, y) of a curve C is given by dydx = 4 – 2x

and C passes through the point (1, 0). (a) Find an equation of C. (b) Find the area of the finite region bounded by C and the x-axis. [91-P2-Q5]

2.The slope of the tangent to a curve C at any point (x, y) on C is x2 – 2. C passes through the point (3, 4).

(a) Find the equation of C. (b) Find the coordinates of the point on C at which the slope of the tangent is –2. [92-P2-Q4][Solution]

(a) Given dydx = x2 – 2

y = 13 x3 – 2x + C

Put x=3, y=4, C = 1

Equation of C : y = 13 x3 – 2x + 1

(b) For slope of tangent = -2, x2 – 2 = -2 x = 0 , y = 1 The coordinates of the point is (0, 1)

3. The slope of a curve C at any point (x, y) on C is 3x2 – 6x – 1. C passes through the point (1,0). (a) Find the equation of C. (b) Find the equation of the tangent to C at the point where C cuts the y-axis. [93-P2-Q6][Hint]

(a) The equation of C is y = x3 – 3x2 – x + 3 (b) It cuts the y-axis at (0,3), and the equation of the tangent is x + y – 3 = 0

4. The slope at any point (x, y) of a curve C is given by dydx = 8 – 10x and C passes through the

point A(1,13). (a) Find the equation of C. (b) Find the equation of the normal to C at the point where C cuts the y-axis. [94-P2-Q8]

[Hint] The equation of C is y = -5x2 + 8x + 10 (b) It cuts y-axis at (0,10), the slope of the tangent is 8.

The slope of the normal is -18 . The equation of the normal is x + 8y – 80 = 0

5. The slope at any point (x, y) of a curve C is given by dydx = 2x x2 + 1 and C cuts the y-axis at the

point (0,1). Find the equation of C. (Hint: Let u = x2 + 1.) [95-P2-Q1]

6. The slope at any point (x, y) of a curve is given by dydx = tan3x.sec x. If the curve passes through

the origin, find its equation. (Hint: Let u = sec x.) [96-P2-Q6] [Solution] y = ∫ tan3x.sec x.dx

Let u = sec x, du = tan x.sec x.dx y = ∫ (u2 – 1)du

= u3

3 – u + C

= 13 sec3x – sec x + C where C is a constant

Since the curve passes through O,

0 = 13 – 1 + C

C = 23

The equation of the curve C is y = 13 sec3x – sec x +

23

7. The slope at any point (x, y) of a curve is given by dydx = 6x +

1x2 , where x>0. If the curve cuts

the x-axis at the point (1,0), find its equation. [97-P2-Q5]

Year 1990 1989 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q5(a) P2-Q4 P2-Q6 P2-Q8 P2-Q1 P2-Q6 P2-Q5 P2-Q4 P2-Q6 P2-Q6 Marks 3 6 7 7 5 6 5 5 6 7

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q03 Q10 Q10 Marks 4 6 5


8. The slope at any point (x, y) of a curve is given by dydx = cos2x. If the curve passes through the

point (π2 , π), find its equation. [98-P2-Q4]

9. The slope at any point (x, y) of a curve is given by dydx = 3x2 – 2x + k. If the curve touches the

x-axis at the point (2,0), find (a) the value of k, (b) the equation of the curve. [99-P2-Q6][Hint]

The curve touches x-axis at (2,0), ie. slope of tangent at (2,0) is 0.

10. The slope at any point (x , y) of a curve C is given by dydx = 2x + 3. The line y = -x + 1 is a

tangent to the curve at point A. Find (a) the coordinates of A, (b) the equation of C. [00-P2-Q6]

11. The slope of any point (x , y) of a curve C is given by dydx = 3x2 + 1. If the x-intercept of C is 1,

find the equation of C. [04-Q03]

12. (a) Show that ddx [x(x + 1)n] = (x + 1)n - 1[(n + 1)x + 1], where n is a rational number.

(b) The slope at any point (x , y) of a curve C is given by dydx = (x + 1)2004(2006x + 1).

If C passes through the point (-1 , 1), find the equation of C. [05-Q10]

13. The slope at any point (x , y) of a curve is given by dydx = 3 + 2 cos 2x. If the curve passes

through the point (π4 ,

3π4 ), find its equation. [06-Q10]


Type 4: Finding area

Note: 1. Find the intersecting points. 2. Keep the “positive” value of the area.

Past Questions

1. The figure shows the graph of y = sin x for 0 < x < π. (a) Copy the figure into your answer book and sketch the graph of y = sin 2x for 0 < x < π on the same figure. Calculate the x-coordinates of the intersecting points of the two curves for 0 < x < π. (b) Find the area bounded by the two curves for 0 < x < π. [89-P2-Q8]

2. In the figure, the shaded area enclosed by the curves y = cos x,

y = k(x - π2 )2 and the y-axis is 2 squared units. Find the value of k.

[90-P2-Q4]

3. The slope at any point (x , y) of a curve C is given by dydx = 4 – 2x and C passes through the point

(1 , 0). (a) Find an equation of C. (b) Find the area of the finite region bounded by C and the x-axis. [91-P2-Q5] 4. The curve y = x3 – x2 – 2x cuts the x-axis at the origin and the points (a, 0) and (b, 0), as shown.

(a) Find the values of a and b. (b) Find the total area of the shaded parts. [92-P2-Q6]

5. The figure shows the curve of y = sin x and y = cos x, where 0 < x < 2π, intersecting at points A and B. (a) Find the coordinates of A and B. (b) Find the area of the shaded region as shown in the figure. [93-P2-Q5]

6. The figure shows two curves y = x3 and y = x3 – 6x2 + 12x intersecting at the origin and a point

A. (a) Find the coordinates of A. (b) Find the area of the shaded region in the figure. [94-P2-Q7]

7. The figure shows the curve C : y = sin x for 0 < x < π (a) Find the area of the finite region bounded by the curve C and the x-axis.

(b) A horizontal line L cuts C at two points A and B. A is the point (π6 ,

12 ).

(1) Write down the coordinates of B. (2) Find the area of the finite region bounded by C and L. [95-P2-Q5]

[Solution]

(a) Area = sin x dx ∫π

0

= 2

(b) (1) The coordinates of B are (5π6 ,

12 )

(2) Area = ∫ 65

6

π

π (sin x - 12 ) dx

= 3 - π3

8. The curve C : y = 4x – x2 cuts the x-axis at the origin O and the point A(4, 0) as shown.

(a) Find the area of the region bounded by C and the line segment OA. (b) In the other figure, the shaded region is enclosed by the curve C and the line segments OP and

PA, where P is the point (1,3). Using (a), find the total area of the shaded region. [96-P2-Q5] [Hint]

(a) Area of (a) = (4x – x∫4

02) dx

(b) Area of (b) = Area of (a) – Area of ∆OPA

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q8 P2-Q4 P2-Q5(b) P2-Q6 P2-Q5 P2-Q7 P2-Q5 P2-Q5 P2-Q8 P2-Q4Marks 6 5 4 6 6 6 6 6 6 5

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q06 Q09 Q13 Q06Marks 5 5 6 5


9. The line L : y = 3x – 2 and the curve C : y = x2 intersect at two points A(1,1) and B(2,4). Let S1denote the area of the region bounded by C and line segment AB, and S2 denote the area bounded by C, L and the y-axis. (a) Find S1. (b) Which of the following expressions represent(s) the total area S1 + S2?

(1) (3x – 2 – x∫2

02) dx

(2) (x∫1

02 – 3x + 2) dx + (3x – 2 – x∫

2

12) dx

(3) | 3x – 2 – x∫2

02 | dx

(4) | (3x – 2 – x∫2

02) dx | [98-P2-Q8]

[Hint]

(a) S1 = (3x – 2 – x∫2

12) dx

(b) Both (2) and (3) are correct. 10. In the figure, the line L : y = 6x and the curves C1 : y = 6x2 and C2 : y = 3x2 all pass through the

origin. L also intersects C1 and C2 at the points (1, 6) and (2, 12) respectively. Find the area of the shaded region. [99-P2-Q4]

[Hint]

Area of the shaded region = (6x∫1

02 – 3x2) dx + (6x – 3x∫

2

12) dx

11. The figure shows the curves y = sin x and y = cos x.

Find the area of the shaded region. [02-Q06] 12. In the figure, the curve C : y = x2 and line L : y = 4x – 4

intersect at the point (2 , 4). Find the area of the shaded region bounded by C, L and the x-axis. [03-Q09]

13. (a) Find ∫ sin πx dx.

(b) The Figure shows two curves y = f(x) and y = g(x) intersecting at three points (0 , 0), (1 , 2) and (2 , 0) for 0< x < 2. It is given that f(x) - g(x) = 2 sin πx. Find the area of the shaded region as shown in the Figure.

[05-Q13] 14. The following figure shows the graph of y = sin 2x and y = cos x. Find the area of the shaded

region.

[07-Q06]


Type 5a: Integrate hard trigonometric functions

Past Questions

Type 5b: Integrate differential equation Past Questions

Type 6a: Volume by revolving about axis

Past Questions

Type 6b: Volume by revolving about off-axis

Past Questions

1. In the Figure, the shaded region is bounded by the circle x2 + y2 = 9, the x-axis. the y-axis and the line y = 2. Find the volume of the solid generated by revolving the region about the y-axis.

[04-Q04]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q9 P2-Q9 P2-Q12 P2-Q10 P2-Q12a P2-Q9 P2-Q11 P2-Q9 P2-Q12 P2-Q8 Marks 16 16 16 16 16 16 16 16 16 16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Marks

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q8 P2-Q9 P2-Q8Marks 16 16 16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Marks

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q12 P2-Q11 P2-Q11 P2-Q12 P2-Q13 P2-Q12b P2-Q10 P2-Q12 P2-Q13 P2-Q11Marks 16 16 16 16 16 5 16 16 16 16

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q16 Marks 12

Year 1989 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q13 P2-Q12c P2-Q11 Marks 16 3 16

Year 2001 2002 2003 2004 2005 2006 2007P&Q 04 Q Marks 4


Coordinate Geometry Type 1: Straight Lines and Rectilinear Figures Note: Remember the formulae of finding distance between lines or point-to-line, and finding area. Past Questions

1. A straight line L1 : y = mx + c, where m and c are constants, make an angle of 45o with the line L2 : 17x – 7y + 14 = 0.

(a) Find the two values of m. (b) If the distance from the point (1 , 2) to L1 is 5, and m takes the greater of the two values

obtained in (a), find the two values of c. [89-P2-Q7] 2. In the figure, A(3, 0), B(0 , 5) and C(0 , 1) are three points and O is the origin. y B D C x O A D is a point on AB such that the area of ∆BCD equals half of the area of ∆OAB. Find the equation

of the line CD. [90-P2-Q7]

3. A straight line with slope m passes through the point (4 , 7). (a) Write down the equation of the line. (b) If the distance from the origin to the line is 1, find the two possible values of m. [92-P2-Q3]

4. Two lines pass through (4, 3) and each line makes an angle π4 with the line y =

13 x. Find the

equations of the two lines. [93-P2-Q4]

5. Let θ be the acute angle between the two lines L1 : y = 2x and L2 : y = 3x. (a) Find tan θ. (b) Find the equation of the line other than L2 which makes an angle θ with L1 and passes through

the origin. [94-P2-Q6] [Solution] (a) Slope of L1 = 2 and slope of L2 = 3,

tan θ = 3 - 2

1 + (3)(2)

= 17

(b) Let the slope of the required line be m.

2 - m

1 + 2m = 17

m = 139

ie. the equation of the required line is 13x – 9y = 0 6. Given three points A(0 , 2), B(4, 6) and C(3 , 0). P is a point on AB such that AP : PB = λ : 1,

where λ > 0. (a) Find the coordinates of P in terms of λ. (b) If the area of ∆PAC is 6, find the value(s) of λ. [97-P2-Q3] [Solution]

(a) The coordinates of P are (4λ

1 + λ , 6λ + 21 + λ )

(b) 6

201

261

40320

21

=++

+ λλ

λλ

λ = 32

(c) Let the equation of L3 be 2x + 2y + c = 0

23

22

)21(2)0(2

22=

+

++ c

| 1 + c | = 12 c = 11 or –13 (rej.) ie. the equation of L3 is 2x + 2y + 11 = 0

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q7 P2-Q7 P2-Q3 P2-Q4 P2-Q6 P2-Q3 P2-Q3 P2-Q3

P2-Q5 Marks 6 6 6 6 6 5 5 6

6

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q10 Q12 Q06

Q12 Q11 Q02

Q07 Marks 6 7 4

6 6 3

5


7. L1 : 2x + 2y – 1 = 0 and L2 : 2x + 2y – 13 = 0 are two parallel lines. (a) Find the y-intercept of L1. (b) Find the distance between L1 and L2. (c) L3 is another line parallel to L1. If the distance between L1 and L3 is equal to that between L1

and L2, find the equation of L3. [99-P2-Q3] [Solution]

(a) The y-intercept of L1 = 12

(b) Distance between L1 and L2 = 22 22

13)21(2)0(2

+

−+

= 3 2

8. Given an ellipse E : 1916

22

=+yx

. P(-4 , 0) and Q(4 cos θ , 3 sin θ) are points on E,

where 0 < θ < 2π

. R is a point such that the midpoint of QR is the origin O.

(a) Write down the coordinates of R in terms of θ. (b) If the area of ∆PQR is 6 squared units, find the coordinates of Q. [00-P2-Q3] 9. The coordinates of points A and B are (1 , 2) and (2 , 0) respectively. Point P divides AB internally

in the ratio 1 : r. (a) Find the coordinates of P in terms of r.

(b) Show that the slope of OP is 2r

2 + r .

(c) If ∠AOP = 45o, find the value of r. [00-P2-Q5] 10. Two lines L1 : x + y – 5 = 0 and L2 : 2x – 3y = 0 intersect at a point A. Find the equations of the

two lines passing through A whose distances from the origin are equal to 2. [01-Q10] 11. The Figure shows two lines L1 : y = -x + c and L2 : y = 2x, where c > 0. The two lines intersect at point P. (a) Let θ be the acute angle between L1 and L2. Find tan θ. (b) L1 intersects the x- and y-axes at the points A and B respectively. Find AP : PB. [04-Q12]

12. The figure shows the line L1 : 2x + y - 6 = 0 intersecting the x-axis at point P.

(a) Let θ be the acute angle between L1 and the x-axis. Find tan θ. (b) L2 is a line with positive slope passing through the origin O. If L1 intersects L2 at a point Q such

that OP = OQ, find the equation of L2. [05-Q06] 13. A (-1 , 0), B (4 , 2) and C (0 , 6) are three points on a rectangular coordinates system. (a) Find the area of ∆ABC.

(b) P is a point on the line segment BC such that area of ∆APCarea of ∆ABC =

14 . Find the coordinates of P.

[05-Q12] 14. Let L1 be the straight line y = 2x - 5. L2 and L3 are two straight lines

passing through the origin and each makes an angle of 45o with L1. (a) Find the equations of L2 and L3. (b) Find the area of the triangle bounded by L1, L2 and L3. [06-Q11] 15. It is given that the four points A (0 , -2), B (1 , -3), C (2 , 0) and D (k , k) form a quadrilateral ABCD.

Find the area of this quadrilateral. [07-Q02]

16. It is given the points A (2 , 1) and B (-2 , 4). C is a point on AB such that AC : CB = 1 : 2. (a) Find the coordinates of C. (b) Show that OC bisects ∠AOC, where O is the origin. [07-Q07]


Type 2: Family of Straight Lines Past Questions

1. A family of straight lines is given by the equation (2 – k)x + (1 + 2k)y – (4 + 3k) = 0, where k is any constant. Find equations of the two lines in the family whose distances from the origin are equal to 1. [91-P2-Q4]2. A family of straight lines is given by the equation 2x – 3y + 4 + k(4x + 2y – 1) = 0, where k is any constant.

(a) Find the equation of the line in the family which passes through the point (1, 0). (b) Find the equation of the line in the family with slope 2. [95-P2-Q2]

3. L is the line y = 2x + 3. (a) A line with slope m makes an angle of 45o with L. Find the value(s) of m. (b) A family of straight lines is given by the equation 2x – 3y + 2 + k(x – y – 1) = 0 where k is real. Using (a), find the equation of the line in the family with positive slope which

makes an angle of 45o with L. [97-P2-Q6] 4. Two lines L1 : 2x + y – 3 = 0 and L2 : x – 3y + 1 = 0 intersect at a point P. (a) Write down an equation of the family of straight lines passing through P. (b) Suppose L is a line passing through P and the origin, find (1) the equation of L, (2) the acute angle between L and L1 correct to the nearest degree. [98-P2-Q5] 5. A family of straight lines is given by the equation y – 3 + k(x – y + 1) = 0 where k is real. (a) Find the equation of a line L1 in the family whose x-intercept is 5. (b) Find the equation of a line L2 in the family which is parallel to the x-axis. (c) Find the acute angle between L1 and L2. [99-P2-Q5] 6. Given two lines L1 : 2x – 3y + 4 = 0 and L2 : x + y – 3 = 0. (a) Write down the equation of the family of straight lines passing through the point of intersection of

L1 and L2. (b) Find the equations of two lines in the family in (a) such that the distance from the origin to each

line is 1. [03-Q09]

Year 2001 2002 2003 2004 2005 2006 2007P&Q 08 QMarks 5

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q4 P2-Q2

P2-Q8P2-Q6 P2-Q5 P2-Q5

Marks 5 616

6 6 6


Type 3: Locus Past Questions 1. S and T are variable points on the lines y = 0 and x – y = 0 respectively, such that the length of ST is

always equal to 2 units. Find the equation of the locus of the mid-point of ST. [90-P2-Q8] 2. P(0, 4) and Q(2, 6) are two points and R(x, y) is a variable point.

(a) Find the area of ∆PQR in terms of x and y. (b) If the area of ∆PQR is 4 squared units, find the equation of the locus of R. [94-P2-Q4]

3. A(t , 12 t2) is a point on the parabola x2 = 2y. B is the point (2 , 2).

(a) Find the equation of the locus of the mid-point of AB as A moves along the parabola. (b) Sketch the locus in (a). [95-P2-Q3] 4. P(x , y) is a variable point such that the distance between P and the point (4 , 0) is always equal to

twice the distance from P to the line x – 1 = 0. (a) Find the equation of the locus of P. State whether the locus is a circle, ellipse, a hyperbola or a

parabola. (b) Sketch the locus of P. [96-P2-Q7] [Solution] (a) (x - 4)2 + y2 = 2|x – 1| x2 + y2 – 8x + 16 = 4x2 – 8x + 4 3x2 – y2 = 12 which is a hyperbola (b) y x -2 2

5. Given a point A(4 , 0), P(h , k) is a variable point on the circle C : x2 + y2 = 4. Let M be the midpoint of AP.

(a) Express the coordinates of M in terms of h and k. (b) Find the equation of the locus of M. [01-Q03] 6. P (x , y) is a variable point such that the distance from P to the line x – 4 = 0 is always equal to

twice the distance between P and the point (1 , 0). (a) Show that the equation of the locus of P is 3x2 + 4y2 – 12 = 0. (b) Sketch the locus of P. [02-Q05] 7. Let O be the origin and A be the point (3 , 4). P is a variable point such that the area of ∆OPA is

always equal to 2. Show that the locus of P is a pair of parallel lines. Find the distance between these two lines. [04-Q10] 8. The straight line y = x + k intersects the curve y = x2 at two points P and Q. It is known that

the locus of the mid-point of PQ, as k varies, lies on a straight line L. Find the equation of L. [06-Q05] 9. The curve C : y = x2 and the straight line L : y = mx – 2m intersect at two distinct points A and B. (a) Find the range of values of m. (b) (1) Show that the coordinates of the mid-point of AB are

⎟⎟⎠

⎞⎜⎜⎝

⎛ −2

4,2

2 mmm

(2) It is given that the straight line x + y = 5 bisects the line segment AB. Using (b)(1), or otherwise, find the value(s) of m. [07-Q13]


Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q8 P2-Q4 P2-Q3 P2-Q7 P2-Q10 P2-Q10Marks 6 5 6 6 16 16


Type 4: Circles and Family of Circles Past Questions 1. Given a line L : x – 7y + 3 = 0 and a circle C : (x – 2)2 + (y + 5)2 = a , where a is a positive number. (a) Find the distance from the centre of C to L. (b) If L is a tangent to C, find a. [98-P2-Q2]

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P2-Q10 P2-Q11 P2-Q9 P2-Q10 P2-Q11 P2-Q9 P2-Q10 P2-Q10 P2-Q2 P2-Q9Marks 16 16 16 16 16 16 16 16 4 16

Year 2001 2002 2003 2004 2005 2006 2007P&Q 14 Q Marks 12


Vector Note: 1. Obtain vectors through (a) addition or subtraction. (b) point of division. 2. DO NOT forget the vector sign for vectors.

3. ba ⋅ = θcos.ba where θ is the angle between the two vectors.

4. When a is perpendicular to b , ba ⋅ = 0

5. To prove A, B and C are collinear, check whether ACkAB = .

6. aaa ⋅=2

Past Questions

1. Let OA = i + 3j, OB = 4i – j and C be a point dividing AB internally in the ratio k : 1.

(a) Express OC in terms of k, i and j. (b) If OC is perpendicular to AB, find the value of k. [89-P1-Q2]

2. Given OA = 5j, OB = -i + 7j. P is a point such that AP = t AB .

(a) Express OP in terms of t. (b) If OP is perpendicular to AB, find (1) the value of t.

(2) OP . [90-P1-Q2]

3. In the figure, OAD is a triangle and B is the mid-point of OD. The line OE cuts the line AB at C such that AC : CB = 3 : 1.

D E B C O A

Let OA = a and OB = b.

(a) Express OC in terms of a and b.

(b) (1) Let OC : CE = k : 1. Express OE in terms of k, a and b.

(2) Let AE : ED = m : 1. Express OE in terms of m, a and b. Hence find k and m. [91-P1-Q5] [Solution]

(a) OC = 43ba +

(b) (1) OE = k + 1

k

Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000P&Q P1-Q2

P1-Q9 P1-Q2 P1-Q8

P1-Q5 P1-Q8

P1-Q1 P1-Q8

P1-Q6P1-Q8

P1-Q3P1-Q10

P1-Q7P1-Q8

P1-Q7P1-Q10

P1-Q7P1-Q9

P1-Q5P1-Q9

P1-Q7P1-Q10

P1-Q8 P1-Q9

Marks 5 16

6 16

7 16

5 16

7 16

6 16

8 16

6 16

7 16

6 16

6 16

7 16

OC

= k + 1

4k a + 3(k + 1)

4k b

(2) As OD = 2 b , OE = m

bma+

+1

2

= 1

1 + m a + 2m

1 + m b

Equating the vectors of (1) and (2),

⎪⎩

⎪⎨

⎧

+=

++

=+

mm

kk

mkk

12

4)1(3

11

41

Solving, m = 32 and k =

53

Year 2001 2002 2003 2004 2005 2006 2007P&Q Q08

Q14 Q06Q14

Q06 Q13

Q11 Q07 Q08

Marks 512

512

5 12

6 5 5


4. Given OA = 5i – j, OB = -3i + 5j and APB is a straight line.

(a) Find AB and | AB |. (b) If | AP | = 4, find AP .

[92-P1-Q1] [Solution]

(a) AB = OB - OA

= -8 i + 6 j

| AB | = (-8)2 + 62 = 10

(b) AP = 410 AB

= -3.2 i + 2.4 j

5. Given OA = 3i – 2j, OB = i + j. C is a point such that ∠ABC is a right angle.

(a) Find AB .

(b) Find AB ⋅ AB and AB ⋅ BC .

Hence find AB ⋅ AC . [93-P1-Q6]

6. P, Q and R are points on a plane such that OP = i + 2j, OQ = 3i + j and PR = -3i – 2j, where Ois the origin.

(a) Find PQ and | PQ |. (b) Find the value of cos ∠QPR. [94-P1-Q3]

7. Let OP = 2i + 3j and OQ = -6i + 4j. Let R be a point on PQ such that PR : RQ = k : 1, where k > 0.

(a) Express OR in terms of k, i and j.

(b) Express OP ⋅OR and OQ ⋅OR in terms of k. (c) Find the value of k such that OR bisects ∠POQ. [95-P1-Q7] [Solution]

(a) OR = 1+

+k

OQkOP

= 2 - 6kk + 1 i +

3 + 4kk + 1 j

(b) OP ⋅OR = 2(2 - 6k)

k + 1 + 3(3 + 4k)

k + 1

= 13

k + 1

OQ ⋅OR = -6(2 - 6k)

k + 1 + 4(3 + 4k)

k + 1

= 52k

k + 1

(c) If OR bisects ∠POQ, cos ∠POR = cos ∠QOR

OROP

OROP

⋅

⋅ = OROQ

OROQ

⋅

⋅

13| OQ | = 52k | OP | 13 52 = 52k 13

k = 12

8. Given OA = 4i + 3j and C is a point on OA such that | OC | = 165 .

(a) Find the unit vector in the direction of OA .

Hence find OC .

(b) If OB = i + 4j, show that BC is perpendicular to OA. [96-P1-Q7]


9. Let a and b be two vectors such that a = 2i + 4j, |b| = 5 and cos θ = 45 , where θ is the angle

between a and b. (a) Find |a|. (b) Find a⋅b. (c) If b = mi + nj, find the values of m and n. [97-P1-Q7] 10.

The figure shows the points A, B and C whose position vectors are i – j, 4i + 4j and –2i + 7j

respectively.

(a) Find the vectors AB and AC .

(b) By considering AB ⋅ AC , find ∠BAC to the nearest degree. [98-P1-Q5] 11. Let a, b be two vectors such that a = 3i + 4j and |b| = 4. The angle between a and b is 60o. (a) Find |a|. (b) Find a⋅b. (c) If the vector (ma + b) is perpendicular to b, find the value of m. [99-P1-Q7] [Solution]

(a) a = 32 + 42 = 5

(b) ba ⋅ = o60cosba ⋅ = (5)(4)cos 60o = 10

(c) If (m a + b ) is perpendicular to b ,

(m a + b ) ⋅b = 0

m ba ⋅ + bb ⋅ = 0 10m + 42 = 0

m = -1.6

12. B D O C

In the figure, OA = i, OB = j. C is a point on OA produced such that AC = k, where k > 0. D is a point on BC such that BD : DC = 1 : 2.

(a) Show that OD = 1 + k

3 i + 23 j.

(b) If OD is a unit vector, find (1) k, (2) ∠BOD, giving your answer correct to the nearest degree. [00-P2-Q8] 13. Let a and b be two vectors such that |a| = 4, |b| = 3 and the angle between a and b is 60o. (a) Find a.b (b) Find the value of k if the vectors (a + kb) and (a – 2b) are perpendicular to each other. [01-Q08] 14. The figure shows a parallelogram OABC. The position vectors of

the points A and C are i + 4j and 5i + 2j respectively.

(a) Find OB and AC . (b) Let θ be the acute angle between OB and AC. Find θ correct

to the nearest degree. [02-Q10] 15. In the figure, point P divides both line segments

AB and OC in the same ratio 3 : 1. Let OA= a,

OB = b.

(a) Express OP in terms of a and b.

(b) Express OC in terms of a and b. Hence show that OA is parallel to BC. [03-Q06] 16. In the Figure, OAB is a triangle. C is a point on AB such that

AC : CB = 1 : 2. Let → = a and →

= b. OA OB

(a) Express in terms of a and b. →

OC

(b) If | a| = 1, |b| = 2 and ∠AOB = 2π3 , find

→

OC . [04-Q06]

A


17. The figure shows two vectors a and b, where |a| = 3, |b| = 5, and the angle between the two vectors is 120o.

(a) Find a.b (b) Let c be a vector such that a + b + c = 0. Find |c| [05-Q11] 18. Let a and b be two vectors such that |a| = 3 , |b| = 2 and the angle between them is 150o. (a) Find a.b (b) Find |a + 2b| [06-Q07]

19. In the figure, OCA is a straight line and BC ⊥ OA. It is given that OA= 6i + 3j and

OB = 2i + 6j. Let →

= kOC OA .

(a) Express BC in terms of k, i and j. (b) Find the value of k. [07-Q08]

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