Instructions to candidates
- The paper contains two sections: Section I and Section II.
- Answer all the questions in Section I and only five questions from Section II
- All answers and working must be written on the question paper in the spaces provided below each question.
- Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
- Non – programmable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.
SECTION 1 (50 MARKS)
Answer all questions in this section.
- Solve for x in the equation 2Sin2x - 1 = Cos2x + Sinx for 0°≤x≤360°(3marks)
-
- Expand (1+3/x)5 up to the fifth term (2marks)
- Hence use your expansion to evaluate the value of (2.5)5 to 3 d.p. (2 marks)
- Complete the table below for y = 8 − 2x − x2 for −4 ≤ x ≤ 2.
x −4 −3 −2 −1 0 1 2 y - Make p the subject of the formula (3 marks)
- The figure below shows a rectangular based right pyramid. Find the angle between the planes ABCD and ABV. (3marks)
- An object A of area 10cm2 is mapped onto its image B of area 60cm2 by a transformation whose matrix is given by P= . Find the possible values of x. (3 marks)
- Find the value of x in the equation (3marks)
log105 − 2 + log10(2x + 10) = log10(x − 4) - The data below shows marks obtained by 10 students in a test.
71, 55, 69, 45, 65, 57, 71, 82, 55, 50. Calculate the standard deviation using an assumed mean of 60. (3marks) - Evaluate by rationalizing the denominator and leaving your answer in surd form. (3marks)
√8
1 + Cos45° - The position vectors fof points A and B are 5i + 4j − 6k and 2i − 2j respectively. A point X divides AB in the ratio −3: 5. Find the coordinates of X. (3marks)
- A closed box has a square base of side x metres and its height h metres. The total surface area of the box is 24m2.
- Find the expression of h in terms of x.
- Hence find the value of x that would make the volume of the box maximum. (4 marks)
- M varies directly as D and as the cube of V. Calculate the percentage change in M when V is increased by 10% and D is reduced by 10%. (3marks)
- Find the value of t if the gradient of the graphs of the functions y = x2 − x3 and y = x − tx2 are equal at x = 1/3 . (3marks)
- The image of a point A, under the transformation represented by the matrix T = is A’ (-2, 4)Find the coordinates of A (3marks)
- In the figure below, ABC is a tangent at B and CDE is a straight line. ∠BED=50°, ∠DEF=35° and ∠ECB=25°.
Calculate the values of x and z. (2marks) - The equation of a circle is given by 4x2 + 4y2 − 8x + 2y − 7 = 0
Determine the coordinates of the centre of the circle. (3marks)
SECTION II (50 MARKS)
Answer only five questions from this section
-
- Using a ruler and a pair of compasses only construct triangle ABC in whichAB = 6cm, BC = 5.5cm and angle ABC = 60°. Measure AC. (3marks)
- On the same side of AB as C, determine the locus of a point P such that angle APB = 60° (2marks)
- Construct the locus of R such that AR = 3cm (1mark)
- Identify the region T such that AR≥ 3cm and angleAPB≥ 60° by shading the unwanted part. (2marks)
- Determine point Q such that area of AQB is half the area of ABC and that Angle AQB = 60°. (2marks)
- A sequence is formed by adding corresponding terms of an AP and GP. The first, second and third terms of the sequence formed are 14, 34 and 78 respectively.
- Given that the common ratio of the GP is 3;
- Find the first term of the AP and GP and the common difference of the AP. (2marks)
- Find the sixth term and the sum of the first six terms of the sequence. (3marks)
- The second and third terms of a geometric progression are 24 and 12(x + 1) respectively.
Find the whole number value of x and hence the first term given the sum of the first three terms of the progression is 76. (5marks)
- Given that the common ratio of the GP is 3;
- Income tax rate are as shown below.
Income (k£ p.a) Rate (Ksh per £) 1- 4200 2 4201 - 8000 3 8001 - 12600 5 12601 – 16800 6 16801 and above 7 - his gross tax p.a in Ksh (2marks)
- his taxable income in k£ p.a (4marks)
- his basic salary in Ksh. p.m (2marks)
- his net salary per month (2marks)
- The diagram below shows a sketch of the line y = 3x and the curve y = 4 − x2 intersecting at point P and Q.
- Find the co-ordinates of P and Q (4marks)
- Given that QN is perpendicular to the x-axis at N, calculate
- the area bounded by the curve y = 4 − x2 , the x-axis and line QN. (2marks)
- the area of the shaded region that lies below the x - axis (2marks)
- the area of the region enclosed by the curve y = 4 − x2, the line y = 3x and the y-axis (2marks)
- The positions of two towns A and B are (500N, 450W) and (500N, K0W) respectively. It takes a plane 5 hours to travel from A to B at an average speed of 800knots. The same plane takes 1½ hours to travel from B to another town C at the same average speed. Given that C is to the north of B, calculate to the nearest degree,
- The value of K (4marks)
- The latitude of C (3marks)
- If the plane started from A at 9.00am and flew to C through B, find the local time at C when the plane arrived there. (3marks)
-
- Complete the table below for the equation y = x3 + 2x2 to 2 d.p(2marks)
x −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2x2 18 12.5 8 4.5 2 0 1 4.5 x3 −27 −8 −1 0 1 y −9 0 1 0 2 - On the grid provided, draw the graph of y = x3 + 2x2 for −3 ≤ x ≤ 1.5. Take a scale of 2cm to represent 1 unit on the x- axis and 1cm to represent 1unit on the y – axis. (3marks)
- Use your graph to solve
- x3 + 2x2 = 0 (2marks)
- x3 + 2x2 − x − 2 = 0 (3marks)
- Complete the table below for the equation y = x3 + 2x2 to 2 d.p(2marks)
- The table below shows the marks obtained by 47 students in a mathematics test.
Marks 31 – 35 36 – 40 41 – 45 46 – 50 51 – 55 56 - 60 No. of candidates 4 6 12 15 8 2 - On the grid provided, draw a cumulative frequency curve. (3marks)
Use your graph to estimate- The median (2marks)
- The semi interquartile range (2marks)
- In order to pass the test a candidate had to score more than 40 marks. Calculate the percentage of candidates who passed. (3marks)
- On the grid provided, draw a cumulative frequency curve. (3marks)
- In the triangle OPQ below, OP = p and OQ = q. R is a point on PQ such that PR: RQ = 1 : 3 and 5OS = 2 OQ. PS intersects OR at T.
- Express in term of p and q
- OS (1mark)
- PQ (1mark)
- OR (2mark)
- Given that OT= hOR and PT = kPS. Determine the values of h and k. (6marks)
- Express in term of p and q
MARKING SCHEME
SECTION 1 (50 MARKS)
Answer all questions in this section.
- Solve for x in the equation 2Sin2x - 1 = Cos2x + Sinx for 0°≤x≤360°(3marks)
2Sin2x − 1 = 1 − Sin2x + sin x
2Sin2x + Sin2x − sinx −1 − 1 =0
3sin2x − sin x − 2 = 0
3sin2x − 3 sinx + 2sinx − 2 = 0
3Sinx(Sinx − 1) + 2 (Sinx − 1) = 0
(Sinx − 1) (3Sinx + 2) = 0
Sinx − 1 = 0
Sin x = 1
x = 90°
OR
3Sinx = −2
3 3
Sin x = −2/3
x = 180 + 41.81 = 138.19
x = 360 − 41.81 = 318.19 -
- Expand (1+3/x)5 up to the fifth term (2marks)
1 5 10 10 5
1 + 5(3/x) + 10(3/x)2 + 10(3/x)3 + 5(3/x)4
1 + 15 + 90 + 270 + 405
x x2 x3 x4 - Hence use your expansion to evaluate the value of (2.5)5 to 3 d.p. (2 marks)
2.5 = 1 + 3/x
2.5 − 1 = 3/x
1.5 = 3/x
1.5 = 3
1.5 1.5
x = 2
1 + 15 + 90 + 270 + 405
2 4 8 16
= 90.0625
= 90.063
- Expand (1+3/x)5 up to the fifth term (2marks)
- Complete the table below for y = 8 − 2x − x2 for −4 ≤ x ≤ 2.
x −4 −3 −2 −1 0 1 2 y 0 5 8 9 8 5 0
A = ½ × 1 {0+0+2(5+8+9+8+5)}
½ × 2(35)
= 35 - Make p the subject of the formula (3 marks)
e² = p − 3u
1 y − 3xp
e²(y −3xp) = p − 3u
e²y −3e²xp = p − 3u
e²y + 3u = p + 3e²xp
p + 3e²xp = e²y + 3u
p(1 + 3e2x) = e2y + 3u
p = e2y + 3u
1 + 3e2x - The figure below shows a rectangular based right pyramid. Find the angle between the planes ABCD and ABV. (3marks)
VM = √(82 − 22)
= √60 = 7.746
CosM = 1.5
7.746
Cos M = 0.19365
Cos−1 0.19365
= 78.83° - An object A of area 10cm2 is mapped onto its image B of area 60cm2 by a transformation whose matrix is given by P= . Find the possible values of x. (3 marks)
Area scale factor = 60/10 = 6
Determinant
x(x + 3) − 12 = x2 + 3x − 12
x2 + 3x − 12 = 6
x2 + 3x − 12 − 6 = 0
x2 + 3x − 18 = 0
x2 + 6x − 3x − 18 = 0
x(x + 6) − 3(x + 6) = 0
(x + 6) (x−3) = 0
x + 6 = 0 x − 3 = 0
x = −6 x = 3 - Find the value of x in the equation (3marks)
log105 − 2 + log10(2x + 10) = log10(x − 4)
log105 − 2log1010 + log10(2x + 10) = log10(x − 4)
log105(2x + 10) − log10100 = log10(x − 4)
log10(10x + 50) = log10(x − 4)
100
10x + 50 = x − 4
100 1
10x + 50 = 100(x − 4)
10x + 50 = 100x − 400
50 + 400 = 100x − 10x
450 = 90x
90 90
x = 5 - The data below shows marks obtained by 10 students in a test.
71, 55, 69, 45, 65, 57, 71, 82, 55, 50. Calculate the standard deviation using an assumed mean of 60. (3marks)
= √(121.6 − 4)
= √117.6
= 10.84 - Evaluate by rationalizing the denominator and leaving your answer in surd form. (3marks)
√8
1 + Cos45°
(√8) √2
(1 + 1/√2)√2
√16 = 4
√2+1 √2+1
4 × √2 − 1
√2+1 √2 − 1
4√2− 4
2 − 1
= 4√2 − 4 - The position vectors fof points A and B are 5i + 4j − 6k and 2i − 2j respectively. A point X divides AB in the ratio −3: 5. Find the coordinates of X. (3marks)
X ( 9.5, 13, −15) - A closed box has a square base of side x metres and its height h metres. The total surface area of the box is 24m2.
- Find the expression of h in terms of x.
S.A = 2x2 + 4xh
2x2 + 4xh = 24
4xh = 24 − 2x2
h = 24 − 2x2
4x
h = 24 − 2x²
4x 4x
h = 6/x − x/2 - Hence find the value of x that would make the volume of the box maximum. (4 marks)
V = b.a × h
= x2 × h
=hx2
V = hx2
x2 = 6/x − x/2
V = 6x − x³
2
dv/dx = 6 − 3x2
2
6 − 3x2 = 0
2
6 = 3x2
2
6 − 3x2 = 0
2
12 = 3x2
3 2
√4 = √x2
2 = x
- Find the expression of h in terms of x.
- M varies directly as D and as the cube of V. Calculate the percentage change in M when V is increased by 10% and D is reduced by 10%. (3marks)
M = KDV3
M1 = K(0.9D) (1.1V)3
M1 = 0.9 × 1.13KDV3
M1 = 1.1979KDV3
= 1.1979M
%Change = 1.1979M − M × 100
M
= 0.1979 × 100
= 19.79% - Find the value of t if the gradient of the graphs of the functions y = x2 − x3 and y = x − tx2 are equal at x = 1/3 . (3marks)
y = x2 − x3
dy/dx = 2x − 3x2
y = x − tx2
dy/dx = 1 − 2tx
x = 1/3
dy/dx = 2(1/3) − 3(1/3)2
= 2/3 − 1/3 = 1/3
dy/dx = 1 − 2t(1/3)
= 1 − 2/3t
1/3 = 1 − 2/3t
2/3t = 1 − 1/3
2/3t = 2/3
t = 1 - The image of a point A, under the transformation represented by the matrix T = is A’ (-2, 4)Find the coordinates of A (3marks)
- In the figure below, ABC is a tangent at B and CDE is a straight line. ∠BED=50°, ∠DEF=35° and ∠ECB=25°.
Calculate the values of x and z. (2marks)
z = 50°
x = 50 + 25 = 75° - The equation of a circle is given by 4x2 + 4y2 − 8x + 2y − 7 = 0
Determine the coordinates of the centre of the circle. (3marks)
4x2 + 4y2 − 8x + 2y − 7
4 4 4 4 4
x2 − 2x + (−2/2)2 + y2 + (1/4)2 = 7/4 + (−2/2)2 + (¼)2
(x − 1)2 + (y + ¼)2 = 7/4 + 1 + 1/16
(x−1)2 + (y + ¼)2 = 28 + 16 + 1
16
(x−1)2 + (y + ¼)2 = 45/16
Centre (1,− ¼)
SECTION II (50 MARKS)
Answer only five questions from this section
-
- Using a ruler and a pair of compasses only construct triangle ABC in which AB = 6cm, BC = 5.5cm and angle ABC = 60°. Measure AC. (3marks)
- On the same side of AB as C, determine the locus of a point P such that angle APB = 60° (2marks)
- Construct the locus of R such that AR = 3cm (1mark)
- Identify the region T such that AR≥ 3cm and angleAPB≥ 60° by shading the unwanted part. (2marks)
- Determine point Q such that area of AQB is half the area of ABC and that Angle AQB = 60°. (2marks)
Area of Δ ABC = ½ × 6 × 5.5 × Sin 60
= 14.289
14.2894 = ½ × 6 × h
3 3
h = 4.76 ≈ 5cm
- Using a ruler and a pair of compasses only construct triangle ABC in which AB = 6cm, BC = 5.5cm and angle ABC = 60°. Measure AC. (3marks)
- A sequence is formed by adding corresponding terms of an AP and GP. The first, second and third terms of the sequence formed are 14, 34 and 78 respectively.
- Given that the common ratio of the GP is 3;
- Find the first term of the AP and GP and the common difference of the AP. (2marks)
AP: a a+d a+2d
GP x xr xr2 +
x 3x 9x
a + x =14
2(a+d+3x=34)
a+2d+9x = 78
2a+2d+6x = 68
a+2d+9x = 78 −
a − 3x = −10
a + x = 14
−4x = −24
−4 −4
x = 6
a + x = 14
a + 6 = 14
a = 8 - Find the sixth term and the sum of the first six terms of the sequence. (3marks)
14, 34, 78
T6 = a + 5d + 35x
a+d+3x=34
8+d+18=34
d=34−26
d=8
T6 = a + 5d + 243x
= 8 + 5(8) + 243 × 6
=1506
T5 = a + 4d + 81x
= 8 + 4(8) + 81 × 6
= 526
T4 = a + 3d + 27x
8 + 3(8) + 27 × 6 = 194
S6
14+34+78+194+526+1506
= 2352
- Find the first term of the AP and GP and the common difference of the AP. (2marks)
- The second and third terms of a geometric progression are 24 and 12(x + 1) respectively.
Find the whole number value of x and hence the first term given the sum of the first three terms of the progression is 76. (5marks)
ar = 24
ar2 = 12(x+1)
r = 12(x+1) = x+1
24 2
ar = 24
a(x+1) = 24
2
a = 24 × 2 = 48
x+1 x+1
(48 + 24 + 12(x+1) = 76) (x+1)
x+1
48+24(x+1) +12(x+1)2 = 76(x+1)
48+24x+24+12(x2+2x+1) = 76x+76
72 + 24x + 12x2 + 24x + 12 − 76x − 76 = 0
12x² − 28x + 8 = 0
4 4 4
3x2 − 7x + 2 = 0
3x2 − 6x − x + 2 = 0
3x(x−2) −1(x−2) = 0
(x−2)(3x−1) = 0
x = 2 or x = 1/3
x = 2
- Given that the common ratio of the GP is 3;
- Income tax rate are as shown below.
Income (k£ p.a) Rate (Ksh per £) 1- 4200 2 4201 - 8000 3 8001 - 12600 5 12601 – 16800 6 16801 and above 7 - his gross tax p.a in Ksh (2marks)
4000
+1100
5100 × 12 = 61,200 - his taxable income in k£ p.a (4marks)
4200 × 2 = 8400
3800 × 3 = 11400
4600 × 5 = 23000
42800
x × 6 = 18400
x = 18400 = 3066.67
6
12600
+3066.67
£15666.67 - his basic salary in Ksh. p.m (2marks)
15666.67 × 20
12
26111.11
−10,800
15311.11 - his net salary per month (2marks)
26,111.11
− 4000
22,111.11
- his gross tax p.a in Ksh (2marks)
- The diagram below shows a sketch of the line y = 3x and the curve y = 4 − x2 intersecting at point P and Q.
- Find the co-ordinates of P and Q (4marks)
4 − x2 = 3x
0 = x2 + 3x −4
x2 + 4x − x − 4 = 0
x(x+4) −1(x+4) = 0
(x+4)(x−1) = 0
x = −4 x =1
y =3(−4) y = 3(1)
=−12 =3
Q(−4, −12) P(1,3) - Given that QN is perpendicular to the x-axis at N, calculate
- the area bounded by the curve y = 4 − x2 , the x-axis and line QN. (2marks)
- the area of the shaded region that lies below the x - axis (2marks)
- the area of the region enclosed by the curve y = 4 − x2, the line y = 3x and the y-axis (2marks)
- the area bounded by the curve y = 4 − x2 , the x-axis and line QN. (2marks)
- Find the co-ordinates of P and Q (4marks)
- The positions of two towns A and B are (500N, 450W) and (500N, K0W) respectively. It takes a plane 5 hours to travel from A to B at an average speed of 800knots. The same plane takes 1½ hours to travel from B to another town C at the same average speed. Given that C is to the north of B, calculate to the nearest degree,
- The value of K (4marks)
D=S×T
=800×5
=4000nm
1° = 60Cos50°
? 4000
4000 × 1
60Cos50°
= 103.7°
K − 45 = 103.7
K = 103.7 + 45
= 148.7°W - The latitude of C (3marks)
D = S × T
= 800 × 3/2 = 1200nm
1° = 60nm
? 1200nm
1200 × 1 =20°
60
x − 50 = 20
x = 70°W - If the plane started from A at 9.00am and flew to C through B, find the local time at C when the plane arrived there. (3marks)
Distance A − B − C
= 4000nm + 1200
= 5200nm
T = D/S
= 5200
800
= 6½hrs
θ = 103.7°
1° = 4min
103.7 = 103.7 × 4 = 414.8 = 6hrs 55min
60
9:00
−6:55
02:05a.m
+6:30
8:35a.m at C
- The value of K (4marks)
-
- Complete the table below for the equation y = x3 + 2x2 to 2 d.p(2marks)
x −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2x2 18 12.5 8 4.5 2 0.5 0 0.5 1 4.5 x3 −27 −15.63 −8 −3.38 −1 −0.13 0 0.13 1 3.38 y −9 −3.13 0 1.12 1 0.37 0 0.63 2 7.88 - On the grid provided, draw the graph of y = x3 + 2x2 for −3 ≤ x ≤ 1.5. Take a scale of 2cm to represent 1 unit on the x- axis and 1cm to represent 1unit on the y – axis. (3marks)
- Use your graph to solve
- x3 + 2x2 = 0 (2marks)
- x3 + 2x2 − x − 2 = 0 (3marks)
- Complete the table below for the equation y = x3 + 2x2 to 2 d.p(2marks)
- The table below shows the marks obtained by 47 students in a mathematics test.
Marks 31 – 35 36 – 40 41 – 45 46 – 50 51 – 55 56 - 60 No. of candidates 4
46
1012
2215
378
452
47- On the grid provided, draw a cumulative frequency curve. (3marks)
Use your graph to estimate- The median (2marks)
47/2 = 23.5 46 - The semi interquartile range (2marks)
- The median (2marks)
- In order to pass the test a candidate had to score more than 40 marks. Calculate the percentage of candidates who passed. (3marks)
- On the grid provided, draw a cumulative frequency curve. (3marks)
- In the triangle OPQ below, OP = p and OQ = q. R is a point on PQ such that PR: RQ = 1 : 3 and 5OS = 2 OQ. PS intersects OR at T.
- Express in term of p and q
- OS (1mark)
→
OS = 2/5 OQ = 2/5q - PQ (1mark)
PQ = q − p - OR (2mark)
OR = ¼q + ¾p
- OS (1mark)
- Given that OT= hOR and PT = kPS. Determine the values of h and k. (6marks)
- → →
OT = h OR
= h(¼q + ¾p)
= h/4q + 3h/4p - → → →
OT = OP + PT
= p + k (−p + 2/5q)
= p − kp + 2k/5q
= p(1−k) + 2k/5q
h/4q + 3h/4p = p(1−k) + 2k/5q
3h/4 = 1 − k
h/4 = 2k/5
h = 8/5k
¾(8/5k) = 1− k
6/5k + 5/5k = 1
11/5k = 1
k = 5/11
h = 8/5(5/11)
h = 8/11
- → →
- Express in term of p and q
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