北京梓桢国际教育内部教学资料报班电话 Pure 1 真题分类 主编吴梓桢 北京梓桢国际教育中心 公司网站 : ( 登录网站可下载更多学习资料 ) 公司地点 : 第一教室 : 北京市海淀区中关村

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1 北京梓桢国际教育内部教学资料报班电话

2 北京梓桢国际教育内部教学资料报班电话 Pure 1 真题分类 主编吴梓桢 北京梓桢国际教育中心 公司网站 : ( 登录网站可下载更多学习资料 ) 公司地点 : 第一教室 : 北京市海淀区中关村东路 13 号都市网景 号楼 103( 地铁 10 号线知春里,13 号线大钟寺, 知春路站,4 号线人民大学站步行 10 分钟左右均可到达 ) 第二教室 : 北京市通州区潞河区名苑 1 号楼联系方式 : 手机 或 固话 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 1 -

3 北京梓桢国际教育内部教学资料报班电话 公司简介 北京市梓桢国际是北京最早 (005 年 ) 从事 A-Level 教学研究 A Level 教学书籍的编写出版 A Level 教育培训的机构, 同时我们也为 A-Level 学生提供科学而专业的教学咨询与评估. 005 年以来, 我们一直为 打造国内精品的 A-Level 私塾 而努力 : 筹集 A Level 教学资源 研究 A Level 课程 亲自实践 A Level 教学 聘用并培训 A Level 教师 编写真正适合大陆学生的 A Level 教学讲义和书籍 我们萃取各校优秀外教的讲义 / 习题 / 练习册与从国外购的学习资料中优秀的部分, 编了近 0 本大小的习题及讲义, 这是国内第一批 A-Level 教学资料 : 包括 A-Level 教材注解 A-Level 考试考点精讲 历年 CIE 高考真题分类汇编 A-Level 教科书经典题集粹 等等 基于多年对 A-Level 课程的研究并结合公司的内部讲义 资料, 我们形成了一套精辟的教学模式, 对于有一定语言基础的学生, 每门理论课只需上 8 至 10 次, 每次 3 小时, 就能学透国内一整年的 A-Level 课程 多年来, 我们接触了大量背景不同却持着同一个留学梦的孩子们, 有很多国内 A-Level 在读生来中心课外辅导 ; 也有去英国读 A-Level 的孩子在出国前来提前预习 ; 还有英国 A-Level 在读生利用圣诞节 复活节来中心进行 A-Level 高考的考前辅导 ; 还有不少成绩较好的国内在读高中生利用寒暑假来中心学 A-Level 课程,3 至 4 个月的时间学完 AS/A 并充分备考, 也考入了英国理想的大学 为了能为更多的孩子圆留学梦, 在培训辅导部稳步拓展的同时, 北京梓桢国际教育又与北京市私立树人 瑞贝学校 ( 成立于 1993 年 ) 合作, 开设了全日制住宿教学部与全脱产 / 半脱产住宿教学部, 现已面向全国招生, 旨在建立一个以培训准备赴英 美 加 澳等国家就读大学本科课程的高中学生为对象的国际教育考试优质培训基地, 基本课程包括 A-Level 课程 SAT 课程 AP 课程及雅思 托福课程, 国际高中采用国外高中教材, 理论课同时采用中教汉英双语与外教全英语授课 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - -

4 北京梓桢国际教育内部教学资料报班电话 目录 Section A Coordinates,Points and Lines(Chapter 1)... 4 Section B Functions and Graphs (Chapter 3) Section C Quadratics (Chapter 4) Section D Differentiations and Their Applications Section E Sequences ( Chapter 8 & 14 )... 9 Section F The Binomial Theorem ( Chapter 9 ) Section G Trigonometry ( Chapter 10 ) Section H Combining and Inverting Functions (Chapter 11 ) Section I Vectors (Chapter 13 ) Section J Integration and Its Application ( Chapter 16 & 17 ) Section K Radians ( Chapter 18 ) Answer Section A Section B Section C Section D Section E Section F Section G Section H Section I Section J Section K 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 3 -

5 北京梓桢国际教育内部教学资料报班电话 Section A Coordinates,Points and Lines(Chapter 1) 1 The diagram shows a rectangle ABCD, where A is (3, ) and B is (1, 6). (i) Find the equation of BC. [4] (0w) Given that the equation of AC is y = x - 1, find (ii) the coordinates of C, [] (iii) the perimeter of the rectangle ABCD. [3] The line L 1 has equation x + y = 8. The line L passes through the point A(7, 4) and is perpendicular to L 1. (03s) (i) Find the equation of L. [4] (ii) Given that the lines L 1 and L intersect at the point B, find the length of AB. [4 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 4 -

6 北京梓桢国际教育内部教学资料报班电话 The diagram shows a trapezium ABCD in which BC is parallel to AD and angle BCD = 90. The coordinates of A, B and D are (, 0), (4, 6) and (1, 5) respectively. (03w) (i) Find the equations of BC and CD. [5] (ii) Calculate the coordinates of C. [] 6 4 The curve y = 9 - and the line y + x = 8 intersect at two points. Find x (i) the coordinates of the two points, [4] (04s) (ii) the equation of the perpendicular bisector of the line joining the two points. [4] 5 The equation of a curve is y = x 4x + 7 and the equation of a line is y + 3x = 9. The curve and the line intersect at the points A and B. (04w) (i) The mid-point of AB is M. Show that the coordinates of M are ( 1 1, 7 ) [4] (ii) Find the coordinates of the point Q on the curve at which the tangent is parallel to the line y + 3x = 9. [3] (iii) Find the distance MQ. [1] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 5 -

7 北京梓桢国际教育内部教学资料报班电话 The diagram shows a rhombus ABCD. The points B and D have coordinates (, 10) and (6, ) respectively, and A lies on the x-axis. The mid-point of BD ism. Find, by calculation, the coordinates of each of M, A and C. [6](05s) 7 Three points have coordinates A (, 6), B (8, 10) and C (6, 0). The perpendicular bisector of AB meets the line BC at D. Find (05w) (i) the equation of the perpendicular bisector of AB in the form ax + by = c, [4] (ii) the coordinates of D. [4] 8 The curve y = 1x intersects the line 3y = 4x + 6 at two points. Find the distance between the two points. [6] (06s) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 6 -

8 北京梓桢国际教育内部教学资料报班电话 The diagram shows a rectangle ABCD. The point A is (, 14), B is (, 8) and C lies on the x-axis. Find (07s) (i) the equation of BC, [4] (ii) the coordinates of C and D. [3] 10 Three points have coordinates A(,5), B(10,9) and C(6,). Line L 1 passes through A and B. Line L passes through C and is perpendicular to L 1. Find the coordinates of the point of intersection of L 1 and L. (01w) 11 The three points A(1, 3), B(13, 11) and C(6, 15) are shown in the diagram. The perpendicular from C to AB meets AB at the point D. Find (06w) (i) the equation of CD, [3] (ii) the coordinates of D. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 7 -

9 北京梓桢国际教育内部教学资料报班电话 The three points A (3, 8), B (6, ) and C (10, ) are shown in the diagram. The point D is such that the line DA is perpendicular to AB and DC is parallel to AB. Calculate the coordinates of D. [7] (07w) 13 The equation of a curve C is y = x 8x + 9 and the equation of a line L is x + y = 3. (08s) (i) Find the x-coordinates of the points of intersection of L and C. [4] (ii) Show that one of these points is also the stationary point of C. [3] 14 In the diagram, the points A and C lie on the x- and y-axes respectively and the equation of AC is y + x = 16. The point B has coordinates (, ). The perpendicular from B to AC meets AC at the point X. (08s) (i) Find the coordinates of X. [4] The point D is such that the quadrilateral ABCD has AC as a line of symmetry. (ii) Find the coordinates of D. [] (iii) Find, correct to 1 decimal place, the perimeter of ABCD. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 8 -

10 北京梓桢国际教育内部教学资料报班电话 The equation of a curve is y = 5 x 8. (08w) (i) Show that the equation of the normal to the curve at the point P(, 1) is y + x = 4. [4] This normal meets the curve again at the point Q. (ii) Find the coordinates of Q. [3] (iii) Find the length of PQ. [] 16 The diagram shows points A, B and C lying on the line y = x + 4. The point A lies on the y-axis and AB = BC. The line from D(10, 3) to B is perpendicular to AC. Calculate the coordinates of B and C. [7] (09s/8) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 9 -

11 北京梓桢国际教育内部教学资料报班电话 The diagram shows a triangle ABC in which A is (3, ) and B is (15, ). The gradients of AB, AC and BC are m, m and m respectively, where m is a positive constant. (10s/11/8) (i) Find the gradient of AB and deduce the value of m. [] (ii) Find the coordinates of C. [4] The perpendicular bisector of AB meets BC at D. (iii) Find the coordinates of D. [4] 18 In the diagram, A is the point ( 1, 3) and B is the point (3, 1). The line L 1 passes through A and is parallel to OB. The line L passes through B and is perpendicular to AB. The lines L 1 and L meet at C. Find the coordinates of C. [6] (10s/1/4) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

12 北京梓桢国际教育内部教学资料报班电话 The diagram shows a rhombus ABCD in which the point A is ( 1, ), the point C is (5, 4) and the point B lies on the y-axis. Find (10s/13/8) (i) the equation of the perpendicular bisector of AC, [3] (ii) the coordinates of B and D, [3] (iii) the area of the rhombus. [3] 0 The line L 1 passes through the points A(, 5) and B(10, 9). The line L is parallel to L 1 and passes through the origin. The point C lies on L such that AC is perpendicular to L. Find (11s/1/7) (i) the coordinates of C, [5] (ii) the distance AC. [] x y 1 The line 1, where a and b are positive constants, meets the x-axis at P a b and the y-axis at Q. 1 Given that PQ = (45) and that the gradient of the line PQ is, find the values of a and b. [5] (11s/13/3) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

13 北京梓桢国际教育内部教学资料报班电话 The diagram shows a rectangle ABCD. The point A is (0, ) and C is (1, 14). The diagonal BD is parallel to the x-axis. (09w/1/9) (i) Explain why the y-coordinate of D is 6. [1] The x-coordinate of D is h. (ii) Express the gradients of AD and CD in terms of h. [3] (iii) Calculate the x-coordinates of D and B. [4] (iv) Calculate the area of the rectangle ABCD. [3] 3 The diagram shows part of the curve y and the line y = 3x + 4. The curve 1 x and the line meet at points A and B. (10w/1/8) (i) Find the coordinates of A and B. [4] (ii) Find the length of the line AB and the coordinates of the mid-point of AB. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 1 -

14 北京梓桢国际教育内部教学资料报班电话 Points A, B and C have coordinates (, 5), (5, 1) and (8, 6) respectively. (10w/13/) (i) Find the coordinates of the mid-point of AB. [1] (ii) Find the equation of the line through C perpendicular to AB. Give your answer in the form ax + by + c = 0. [3] 5 (11w/11/9) A line has equation y kx 6and a curve has equation y x 3x k, where k is a constant. (i) for the case where k=. the line and the curve intersect at points A and B. find the distance AB and the coordinates of the mid-point of AB. [5] (ii) find the two values of k for which the line is a tangent to the curve. [4] 6 (11w/1/9) The diagram show a quadrilateral ABCD in which the point A is (-1,-1), the point B is (3,6) and the point C is (9,4). The diagonals AC and BD intersect at M. Angle BMA 90and BM MD. Calculate (i) the coordinates of M and D.[7] (ii) the ratio AM : MC.[] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

15 北京梓桢国际教育内部教学资料报班电话 Section B Functions and Graphs (Chapter 3) 1 The line x + y = 9 intersects the curve xy + 18 = 0 at the points A and B. Find the coordinates of A and B. [4] (0s) Find the coordinates of the points of intersection of the line y + x = 11 and the curve xy = 1. [4] (03w) 3 Find the set of values of k for which the line y = kx 4 intersects the curve y = x x at two distinct points. [4] (09s/1) 4 (i) Express x 4x 1 in the form a( x b) c and hence state the coordinates of the minimum point, A, on the curve y = x 4x 1. [4] The line x y + 4 = 0 intersects the curve y = x 4x + 1 at points P and Q. It is given that the coordinates of P are (3, 7). (ii) Find the coordinates of Q. [3] (iii) Find the equation of the line joining Q to the mid-point of AP. [3] (11s/11/10) 5 Find the set of values of m for which the line y = mx + 4 intersects the curve y = 3x 4x + 7 at two distinct points. [5] (11s/13/) 6 (11w/1/4) The equation a curve is y x13and the equation of a line is y x k, where k is a constant. (i) In the case where k=8, find the coordinates of the points of intersection of the line and the curve [4] (ii) Find the value of k for which the line is a tangent to the curve. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

16 北京梓桢国际教育内部教学资料报班电话 Section C Quadratics (Chapter 4) 1 The equation of a curve is xy = 1 and the equation of a line l is x + y = k, where k is a constant. (05w) (i) In the case where k = 11, find the coordinates of the points of intersection of l and the curve. [3] (ii) Find the set of values of k for which l does not intersect the curve. [4] (iii) In the case where k = 10, one of the points of intersection is P (, 6). Find the angle, in degrees correct to 1 decimal place, between l and the tangent to the curve at P. [4] 18 1 Find the real roots of the equation 4 x 4 x 3 (i) Express x - 1x + 11 in the form a(x + b) + c. (01w) (ii) Given that f:x x - 1x + 11, for the domain x 0, find the range of f. 4 Determine the set of values of the constant k for which the line y = 4x + k does not intersect the curve y = x. [3] (07w) 5 The equation x px q 0, where p and q are constants, has roots 3 and 5. (11s/1/3) (i) Find the values of p and q. [] (ii) Using these values of p and q, find the value of the constant r for which the equation x px q r 0 has equal roots. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

17 北京梓桢国际教育内部教学资料报班电话 (11w/13/3) The diagram shows the curve and the line y y x x xintersecting at points A,O,B. (i) Show that the x-coordinates of A and B satisfy the equation 4 x 3x 0.[] (ii) Solve the equation 4 x 3x 0 giving your answers in an exact form. [3] and hence find the coordinates of A and B, 7 (11w/13/7) (i) (ii) A straight line passes through the point (,0) and has gradient m. Write down the equation of the line.[1] Find the two values of m for which the line is a tangent to the curve y x x 4 5. For each value of m, find the coordinates of the point where the line touches the curve. [6] (iii) Express x 4x5in the form ( x a) b and hence, or otherwise, write down the coordinates of the minimum point on the curve. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

18 北京梓桢国际教育内部教学资料报班电话 Section D Differentiations and Their Applications ( Chapter 6 & 7 & 1 & 15 ) 1 A curve has equation y =x 3 + 3x - 9x + k where k is a constant.(0w) dy (i) Write down an expression for.[] dx (ii) Find the x-coordinates of the two stationary points on the curve. [] (iii) Hence find the two values of k for which the curve has a stationary point on the x-axis. [3] A hollow circular cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is 19π cm. The cylinder has a radius of rcm and a height of h cm. (0s) (i) Express h in terms of r and show that the volume, V cm3, of the cylinder is 1 given by V = (19r - r 3 ). [4] Given that r can vary, (ii) find the value of r for which V has a stationary value, [3] (iii) find this stationary value and determine whether it is a maximum or a minimum. [3] 6 3 (a) Differentiate 4x + x 6 (b) Find ( 4x ) dx x with respect to x. [] (03s) 4 A solid rectangular block has a base which measures x cm by x cm. The height of the block is y cm and the volume of the block is 7 cm 3. (03w) (i) Express y in terms of x and show that the total surface area, A cm, of the 16 block is given by A 4x. [3] x Given that x can vary, (ii) find the value of x for which A has a stationary value, [3] (iii) find this stationary value and determine whether it is a maximum or a minimum. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

19 北京梓桢国际教育内部教学资料报班电话 The diagram shows a glass window consisting of a rectangle of height hm and width rm and a semicircle of radius r m. The perimeter of the window is 8 m. (04s) (i) Express h in terms of r. [] (ii) Show that the area of the window, A m, is given by A = 8r r 1 r. [] Given that r can vary, (iii) find the value of r for which A has a stationary value, [4] (iv) determine whether this stationary value is a maximum or a minimum. [] 1 6 Find the gradient of the curve y = x 4x at the point where x = 3. [4] (05s) 7 The equation of a curve is y = x 3x + 4. (05s) (i) Show that the whole of the curve lies above the x-axis. [3] (ii) Find the set of values of x for which x 3x + 4 is a decreasing function of x. [1] The equation of a line is y + x = k, where k is a constant. (iii) In the case where k = 6, find the coordinates of the points of intersection of the line and the curve. [3] (iv) Find the value of k for which the line is a tangent to the curve. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

20 北京梓桢国际教育内部教学资料报班电话 The diagram shows the cross-section of a hollow cone and a circular cylinder. The cone has radius 6 cm and height 1 cm, and the cylinder has radius r cm and height h cm. The cylinder just fits inside the cone with all of its upper edge touching the surface of the cone. (05w) (i) Express h in terms of r and hence show that the volume, V cm 3, of the cylinder is given by V = 1πr πr 3. [3] (ii) Given that r varies, find the stationary value of V. [4] 9 A curve has equation y = x k. Given that the gradient of the curve is 3 when x =, find the value of the constant k. [3] (06s) 10 Find the value of the constant c for which the line y = x + c is a tangent to the curve y = 4x. [4] (07s) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

21 北京梓桢国际教育内部教学资料报班电话 The diagram shows a rectangular block of ice, x cm by x cm by 3x cm. (i) Obtain an expression, in terms of x, for the total surface area, A cm, of the da block and write down an expression for. dx (01w) (ii) Given that the ice is melting in such a way that A is decreasing at a constant rate of 0.14cm s -1, calculate the rate of decrease of x at the instant when x =. 1 Find the value of the constant k for which the line y + x = k is a tangent to the curve y = x -6x (01w) 13 The diagram shows an open container constructed out of 00 cm of cardboard. The two vertical end pieces are isosceles triangles with sides 5x cm, 5x cm and 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 0 -

22 北京梓桢国际教育内部教学资料报班电话 x cm, and the two side pieces are rectangles of length y cm and width 5x cm, as shown. The open top is a horizontal rectangle. (06w) (i) Show that y = 00 4x 10x. [3] (ii) Show that the volume, V cm3, of the container is given by V = 40x 8.8x 3. [] Given that x can vary, (iii) find the value of x for which V has a stationary value, [3] (iv) determine whether it is a maximum or a minimum stationary value. [] 14 Find the area of the region enclosed by the curve y = x, the x-axis and the lines x = 1 and x = 4. [4] (07w) 15 The equation of a curve is y = (x 3) 3 6x. (07w) dy d y (i) Express and dx dx in terms of x. [3] (ii) Find the x-coordinates of the two stationary points and determine the nature of each stationary point. [5] 16 A wire, 80 cm long, is cut into two pieces. One piece is bent to form a square of side x cm and the other piece is bent to form a circle of radius r cm (see diagram). The total area of the square and the circle is Acm. (08w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 1 -

23 北京梓桢国际教育内部教学资料报班电话 (i) Show that A = ( 4) x 160x [4] (ii) Given that x and r can vary, find the value of x for which A has a stationary value. [4] 18 The diagram shows part of the curve y, which crosses the x-axis at x 3 A and the y-axis at B. (10s/11/7) The normal to the curve at A crosses the y-axis at C. (i) Show that the equation of the line AC is 9x + 4y = 7. [6] (ii) Find the length of BC. [] 18 A solid rectangular block has a square base of side x cm. The height of the block is h cm and the total surface area of the block is 96cm. (10s/1/8) (i) Express h in terms of x and show that the volume, V by V 1 3 4x x. [3] 3 cm, of the block is given Given that x can vary, (ii) find the stationary value of V, [3] (iii) determine whether this stationary value is a maximum or a minimum.[] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - -

24 北京梓桢国际教育内部教学资料报班电话 The equation of a curve is y (x 3) 4x. (10s/1/10) 6 (i) dy Find. [3] dx (ii) Find the equation of the tangent to the curve at the point where the curve intersects the y-axis.[3] (iii) Find the set of values of x for which of x. [3] 3 (x 3) 4x is an increasing function The volume of a spherical balloon is increasing at a constant rate of 50 cm per second. Find the rate of increase of the radius when the radius is 10 cm. [Volume of a sphere = r.] [4] (11s/11/) 1 The variables x, y and z can take only positive values and are such that z = 3x + y and xy = 600. (11s/11/6) 100 (i) Show that z 3x. [1] x (ii) Find the stationary value of z and determine its nature. [6] The equation of a curve is y = x 4 + 4x + 9. (09w/11/4) (i) Find the coordinates of the stationary point on the curve and determine its nature. [4] (ii) Find the area of the region enclosed by the curve, the x-axis and the lines x = 0 and x = 1. [3] 1 3 The equation of a curve is y. (09w/11/7) x 3 dy (i) Obtain an expression for. [] dx (ii) Find the equation of the normal to the curve at the point P(1, 3). [3] (iii) A point is moving along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.01 units per second. Find the rate of change of the y-coordinate as the point passes through P. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 3 -

25 北京梓桢国际教育内部教学资料报班电话 (i) The diagram shows the line y = x + 5 and the curve y = x 4x + 7, which intersect at the points A and B. Find (09w/1/10) (a) the x-coordinates of A and B, [3] (b) the equation of the tangent to the curve at B, [3] (c) the acute angle, in degrees correct to 1 decimal place, between this tangent and the line y = x + 5. [3] (ii) Determine the set of values of k for which the line y = x + k does not intersect the curve y x 4x 7.[4] 5 The diagram shows a metal plate consisting of a rectangle with sides x cm and y cm and a quarter-circle of radius x cm. The perimeter of the plate is 60 cm. (10w/11/8) (i) Express y in terms of x. [] (ii) Show that the area of the plate, Acm, is given by A = 30x x. [] Given that x can vary, 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 4 -

26 北京梓桢国际教育内部教学资料报班电话 (iii) find the value of x at which A is stationary, [] (iv) find this stationary value of A, and determine whether it is a maximum or a minimum value. [] 6 The equation of a curve is y 3 4x x (10w/11/10) (i) Show that the equation of the normal to the curve at the point (3, 6) is y = x + 9. [4] (ii) Given that the normal meets the coordinate axes at points A and B, find the coordinates of the mid-point of AB. [] (iii) Find the coordinates of the point at which the normal meets the curve again. [4] 7 The length, x metres, of a Green Anaconda snake which is t years old is given approximately by the formula x = 0.7 (t 1), (10w/1/3) where 1 t 10. Using this formula, find dx (i),[] dt (ii) the rate of growth of a Green Anaconda snake which is 5 years old. [] 8 A curve has equation y = kx + 1 and a line has equation y = kx, where k is a non-zero constant. (10w/1/6) (i) Find the set of values of k for which the curve and the line have no common points. [3] (ii) State the value of k for which the line is a tangent to the curve and, for this case, find the coordinates of the point where the line touches the curve. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 5 -

27 北京梓桢国际教育内部教学资料报班电话 The diagram shows an open rectangular tank of height h metres covered with a lid. The base of the tank has sides of length x metres and 1 x metres and the lid is a rectangle with sides of length 5 x metres and 4 x metres. When full the 4 5 tank holds 3 4m of water. The material from which the tank is made is of negligible thickness. The external surface area of the tank together with the area of the top of the lid is A m. 3 4 (i) Express h in terms of x and hence show that A x.. [5] x (ii) Given that x can vary, find the value of x for which A is a minimum, showing clearly that A is a minimum and not a maximum. [5] 1 30 A curve has equation y x. (10w/13/5) x 3 dy d y (i) Find and. [] dx dx (ii) Find the coordinates of the maximum point A and the minimum point B on the curve. [5] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 6 -

28 北京梓桢国际教育内部教学资料报班电话 (11w/11/7) The diagram shows the dimensions in metres of an L-shaped garden. The perimeter of garden is 48 m. (i) find an expression for y in terms of x [1] (ii) given that the area of the garden is A m, show that (iii) A x x 48 8 [] given that x can vary, find the maximum area of the garden, showing that this is a maximum value rather than a minimum value. [4] 3 (11w/11/4) A function f is defined for xrand is such that f '( x) x 6. The range of the function is given by f( x) 4 (i) State the value of x for which f(x) has a stationary value. [1] (ii)find an expression for f(x) in terms of x. [4] 33(11w/11/) 3 A curve has equation y 3x 6x 4x. Show that the gradient of the curve is never negative. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 7 -

29 北京梓桢国际教育内部教学资料报班电话 w/13/8 A curve y=f(x) has a stationary point at P(3,-10). It is given that f '( x) x kx 1,where k is a constant. (i) Show that k and hence find the x-coordinate of the other stationary point, Q.[4] (ii) Find f ''( x) and determine the nature of each of the stationary points P and (iii) Q[] Find f(x).[4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 8 -

30 北京梓桢国际教育内部教学资料报班电话 Section E Sequences ( Chapter 8 & 14 ) 1 A progression has a first term of 1 and a fifth term of 18. (0s) (i) Find the sum of the first 5 terms if the progression is arithmetic. [3] (ii) Find the 13th term if the progression is geometric. [4] A geometric progression, for which the common ratio is positive, has a second term of 18 and a fourth term of 8. Find (0w) (i) the first term and the common ratio of the progression, [3] (ii) the sum to infinity of the progression. [] 3 In an arithmetic progression, the 1st term is 10, the 15th term is 11 and the last term is 41. Find the sum of all the terms in the progression. [5] (03s) 4 (a) A debt of $376 is repaid by weekly payments which are in arithmetic progression. The first payment is $60 and the debt is fully repaid after 48 weeks. Find the third payment. [3] (03w) (b) Find the sum to infinity of the geometric progression whose first term is 6 and whose second term is 4. [3] 5 A geometric progression has first term 64 and sum to infinity 56. Find (i) the common ratio, [] (ii) the sum of the first ten terms. [] (04s) 6 Find (04w) (i) the sum of the first ten terms of the geometric progression 81, 54, 36,..., [3] (ii) the sum of all the terms in the arithmetic progression 180, 175, 170,..., 5. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 9 -

31 北京梓桢国际教育内部教学资料报班电话 A geometric progression has 6 terms. The first term is 19 and the common ratio is 1.5. An arithmetic progression has 1 terms and common difference 1.5. Given that the sum of all the terms in the geometric progression is equal to the sum of all the terms in the arithmetic progression, find the first term and the last term of the arithmetic progression. [6] (05s) 8 A small trading company made a profit of $ in the year 000. The company considered two different plans, plan A and plan B, for increasing its profits. (05w) Under plan A, the annual profit would increase each year by 5% of its value in the preceding year. Find, for plan A, (i) the profit for the year 008, [3] (ii) the total profit for the 10 years 000 to 009 inclusive. [] Under plan B, the annual profit would increase each year by a constant amount $D. (iii) Find the value of D for which the total profit for the 10 years 000 to 009 inclusive would be the same for both plans. [3] 9 Each year a company gives a grant to a charity. The amount given each year increases by 5% of its value in the preceding year. The grant in 001 was $5000. Find (06s) (i) the grant given in 011, [3] (ii) the total amount of money given to the charity during the years 001 to 011 inclusive. [] 10 The second term of a geometric progression is 3 and the sum to infinity is 1. (07s) (i) Find the first term of the progression. [4] An arithmetic progression has the same first and second terms as the geometric progression. (ii) Find the sum of the first 0 terms of the arithmetic progression. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

32 北京梓桢国际教育内部教学资料报班电话 A precious metal is extracted from a mine. In the first year of operation, 000kg of the metal was extracted. In each succeeding year, the amount extracted was 90% of the previous year s amount. (01w) Find (i) the amount of metal extracted in the 10 th year of operation (ii) the total amount of metal extracted in the first 0 years of operation, (iii) the total amount of metal that would be extracted over a very long period of time. 1 (a) Find the sum of all the integers between 100 and 400 that are divisible by 7. [4] (06w) (b) The first three terms in a geometric progression are 144, x and 64 respectively, where x is positive. Find (i) the value of x, (ii) the sum to infinity of the progression. [5] 13 The 1st term of an arithmetic progression is a and the common difference is d, where d 0. (07w) (i) Write down expressions, in terms of a and d, for the 5th term and the 15th term. [1] The 1st term, the 5th term and the 15th term of the arithmetic progression are the first three terms of a geometric progression. (ii) Show that 3a = 8d. [3] (iii) Find the common ratio of the geometric progression. [] 14 The first term of a geometric progression is 81 and the fourth term is 4. Find (08s) (i) the common ratio of the progression, [] (ii) the sum to infinity of the progression. [] The second and third terms of this geometric progression are the first and fourth terms respectively of an arithmetic progression. (iii) Find the sum of the first ten terms of the arithmetic progression. [3] 15 The first term of an arithmetic progression is 6 and the fifth term is 1. The progression has n terms and the sum of all the terms is 90. Find the value of n. [4] (08w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

33 北京梓桢国际教育内部教学资料报班电话 (a) Find the sum to infinity of the geometric progression with first three terms 0.5, and [3] (09s/7) (b) The first two terms in an arithmetic progression are 5 and 9. The last term in the progression is the only term which is greater than 00. Find the sum of all the terms in the progression. [4] 17 The ninth term of an arithmetic progression is and the sum of the first 4 terms is 49. (10s/11/3) (i) Find the first term of the progression and the common difference. [4] The nth term of the progression is 46. (ii) Find the value of n. [] 18 (a) Find the sum of all the multiples of 5 between 100 and 300 inclusive.[3] (10s/1/7) (b) A geometric progression has a common ratio of 3 and the sum of the first 3 terms is 35. Find (i) the first term of the progression, [3] (ii) the sum to infinity. [] 19 The first term of a geometric progression is 1 and the second term is 6. Find (10s/13/1) (i) the tenth term of the progression, [3] (ii) the sum to infinity. [] 0 A television quiz show takes place every day. On day 1 the prize money is $1000. If this is not won the prize money is increased for day. The prize money is increased in a similar way every day until it is won. The television company considered the following two different models for increasing the prize money. (11s/11/8) Model 1: Increase the prize money by $1000 each day. Model : Increase the prize money by 10% each day. On each day that the prize money is not won the television company makes a 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 3 -

34 北京梓桢国际教育内部教学资料报班电话 donation to charity. The amount donated is 5% of the value of the prize on that day. After 40 days the prize money has still not been won. Calculate the total amount donated to charity (i) if Model 1 is used, [4] (ii) if Model is used. [3] 1 (a) A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector. [6] (11s/1/10) (b) The first, second and third terms of a geometric progression are k + 3, k + 6 and k, respectively. Given that all the terms of the geometric progression are positive, calculate (i) the value of the constant k, [3] (ii) the sum to infinity of the progression. [] (a) A geometric progression has a third term of 0 and a sum to infinity which is three times the first term. Find the first term. [4] (11s/13/6) (b) An arithmetic progression is such that the eighth term is three times the third term. Show that the sum of the first eight terms is four times the sum of the first four terms. [4] 3 The first term of an arithmetic progression is 8 and the common difference is d, where d 0. The first term, the fifth term and the eighth term of this arithmetic progression are the first term, the second term and the third term, respectively, of a geometric progression whose common ratio is r. (09w/11/8) 3 (i) Write down two equations connecting d and r. Hence show that r 4 and find the value of d.[6] (ii) Find the sum to infinity of the geometric progression. [] (iii) Find the sum of the first 8 terms of the arithmetic progression. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

35 北京梓桢国际教育内部教学资料报班电话 A progression has a second term of 96 and a fourth term of 54. Find the first term of the progression in each of the following cases: (09w/1/3) (i) the progression is arithmetic, [3] (ii) the progression is geometric with a positive common ratio. [3] 5 (a) The fifth term of an arithmetic progression is 18 and the sum of the first 5 terms is 75. Find the first term and the common difference. [4] (10w/11/6) 7 (b) The first term of a geometric progression is 16 and the fourth term is 4. Find the sum to infinity of the progression. [3] 6 (a) The first and second terms of an arithmetic progression are 161 and 154 respectively. The sum of the first m terms is zero. Find the value of m. [3] (10w/1/5) (b) A geometric progression, in which all the terms are positive, has common ratio r. The sum of the first n terms is less than 90% of the sum to infinity. Show that r n > 0.1. [3] 7 (a) A geometric progression has first term 100 and sum to infinity 000. Find the second term. [3] (10w/13/9) (b) An arithmetic progression has third term 90 and fifth term 80. (i) Find the first term and the common difference. [] (ii) Find the value of m given that the sum of the first m terms is equal to the sum of the first (m + 1) terms. [] (iii) Find the value of n given that the sum of the first n terms is zero. [] 8(11w/11/6) (a) the sixth term of an arithmetic progression is 3 and the sum of the first ten terms is 00. find the seventh term. [4] (b) a geometric progression has first term 1 and common ratio r. A second geometric progression has first term 4 and common ratio 1 4 r. The two progressions have the same sum to infinity, S. find the values of r and S 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

36 北京梓桢国际教育内部教学资料报班电话 (11W/1/10) (a) An arithmetic progression contains 5 terms and the first terms is -15. The sum of all the terms in the progression is 55. Calculate (i) the common difference of the progression [] (ii) the last term in the progression.[] (iii) the sum of all the positive terms in the progression. [] (b) A college agrees a sponsorship deal in which grants will be received each year for sports equipment. This grant will be in 01 and will increase by 5% each year. Calculate (i) the value of the grant in 0.[] (ii) the total amount the college will receive in the years 01 to 0 inclusive.[] 30(11w/13/) The first and second terms of a progression are 4 and 8 respectively. Find the sum of the first 10 terms given that the progression is (i) an arithmetic progression. [] (ii) a geometric progression. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

37 北京梓桢国际教育内部教学资料报班电话 Section F The Binomial Theorem ( Chapter 9 ) 1 Find the value of the term which is independent of x in the expansion of [3] (0w) 4 3 x. x Find the value of the coefficient of in the expansion of x.[3] x x (03s) 3 Find the coefficient of x in the expansion of 5 3x. [4] (04w) x 4 Find the coefficient of x 3 in the expansion of (04s) (i) (1 + x) 6, [3] (ii) (1 3x)(1 + x) 6. [3] 5 (i)find the first 3 terms in the expansion of ( x) 6 in ascending powers of x. [3] (05s) (ii) Find the value of k for which there is no term in x in the expansion of (1 + kx)( x) 6. [] 6 The first three terms in the expansion of ( + ax) n, in ascending powers of x, are 3 40x + bx. Find the values of the constants n, a and b. [5](06s) 7 Find the coefficient of x in the expansion of ( x 6 x ). [3] (06w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

38 北京梓桢国际教育内部教学资料报班电话 (i)find the first three terms in the expansion of ( + u) 5 in ascending powers of u. [3] (07w) (ii) Use the substitution u = x + x in your answer to part (i) to find the coefficient of x in the expansion of ( + x + x ) 5. [] 9 (i)find the first 3 terms in the expansion, in ascending powers of x, of ( + x ) 5. [3] (08s) (ii) Hence find the coefficient of x 4 in the expansion of (1 + x ) ( + x ) 5. [3] 10 Find the value of the coefficient of x in the expansion of x 6. [3] ( ) x (08w) 11 (i) Find the first 3 terms in the expansion of ( x 5 3 ) in ascending powers of x. [3] (09s/3) (ii) Hence find the value of the constant a for which there is no term in x in the 5 expansion of (1 + ax) ( 3x). [] 1 (i) Find the first 3 terms in the expansion of (10s/11/) 5 3 x in descending powers of x.[3] x 3 (ii) Hence find the coefficient of x in the expansion of 1 x x x 5 [] 13 (i) Find the first 3 terms in the expansion of ( 1 ax 5 ) in ascending powers of x. [] (10s/1/6) (ii) Given that there is no term in x in the expansion of ( 1 x)(1 ax) 5, find the value of the constant a. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

39 北京梓桢国际教育内部教学资料报班电话 (iii) For this value of a, find the coefficient of x in the expansion of ( 1 ax 5 x)(1 ). [3] 14 (i) Find the first three terms, in descending powers of x, in the expansion of x x 6 [3] (10s/13/) (ii) Find the coefficient of 4 x in the expansion of 1 x x. [] x 6 15 (i) Find the terms in x and 3 3 x in the expansion of 1 6 x. [3] (11s/1/) (ii) Given that there is no term in value of the constant k. [] 3 3 x in the expansion of ( k x) 1 6 x, find the 16 The coefficient of 3 x in the expansion of 5 ( a x) (1 x ) 6, where a is positive, is 90. Find the value of a. [5] (11s/13/1) 17 (i) Find the first 3 terms in the expansion of ( x 6 ) in ascending powers of x. [3] (09w/11/3) (ii) Given that the coefficient of x in the expansion of ( 1 x 6 x ax )( ) is 48, find the value of the constant a. [3] 18 (i) Find, in terms of the non-zero constant k, the first 4 terms in the expansion of 8 ( k x) in ascending powers of x. [3] (09w/1/) (ii) Given that the coefficients of value of k. [] x and 3 x in this expansion are equal, find the 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

40 北京梓桢国际教育内部教学资料报班电话 In the expansion of ( 1 ax 6 ), where a is a constant, the coefficient of x is 30. Find the coefficient of 3 x. [4] (10w/11/) 0 (i) Find the first 3 terms in the expansion, in ascending powers of x, of ( 1 x ) 8. [] (10w/1/1) (ii) Find the coefficient of 4 x in the expansion of ( x 8 x )(1 ). [] 1 Find the term independent of x in the expansion of 9 1 x. [3] (10w/13/1) x Find the coefficient of x in the expansion of 7 x. [3] (11s/11/1) x 3 (11w/11/1) Find the term independent of x in the expansion of 1 ( x ) x 6 [3] 3 (11w/1/1) (i) Find the first 3 terms in the expansion of ( ) 5 y in ascending powers of y. [] (ii) Use the result in part (i) to find the coefficient of x in the expansion of ( ( )) 5 x x [3] 4. (11w/13/1) The coefficient of constant k. [3] x in the expansion of ( k x) is 30. Find the value of the 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

41 北京梓桢国际教育内部教学资料报班电话 Section G Trigonometry ( Chapter 10 ) 1 cos x 1 (i)show that sin x tan x may be written as. [1] cos x (ii) Hence solve the equation sin x tan x = 3, for 0 x 360. [4](0s) The function f, where f(x) = a sin x + b, is defined for the domain 0 x π. Given 1 that f 3 = and that f = -8, (0s) (i) find the values of a and b, [3] (ii) find the values of x for which f(x) = 0, giving your answers in radians correct to decimal places, [] (iii) sketch the graph of y = f(x). 3 (i)show that the equation 3tanθ = cosθ can be expressed as (0w) sin θ + 3 sinθ- = 0. [3] (ii) Hence solve the equation 3tanθ = cosθ, for 0 θ 360. [3] 4 (i)sketch the graph of the curve y = 3 sin x, for π x π. [] The straight line y = kx, where k is a constant, passes through the maximum of this curve for π x π. (03s) (ii) Find the value of k in terms of π. [] (iii) State the coordinates of the other point, apart from the origin, where the line and the curve intersect. [1] 5 Find all the values of x in the interval 0 x 180 which satisfy the equation sin 3x + cos3x = 0. [4] (03s) 6 (i)show that the equation 4 sin 4 θ + 5 = 7 cos θ may be written in the form 4x + 7x = 0, where x = sin θ. [1] (03w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

42 北京梓桢国际教育内部教学资料报班电话 (ii) Hence solve the equation 4 sin 4 θ + 5 = 7cos θ, for 0 θ 360. [4] 7 (i) Sketch and label, on the same diagram, the graphs of y = sin x and y = cosx, for the interval 0 x π. [4] (04w) (ii) Hence state the number of solutions of the equation sin x = cos x in the interval 0 x π. [1] 8 The function f : x 5 sin x + 3cos x is defined for the domain 0 x π. (04w) (i) Express f(x) in the form a + b sin x, stating the values of a and b. [] (ii) Hence find the values of x for which f(x) = 7 sinx. [3] (iii) State the range of f. [] 9 (i)show that the equation sinθ + cosθ = (sinθ cosθ) can be expressed as tanθ = 3. [] (05s) (ii) Hence solve the equation sinθ + cosθ = (sinθ cosθ), for 0 θ 360. [] 10 In the diagram, ABED is a trapezium with right angles at E and D, and CED is a straight line. The lengths of AB and BC are d and ( 3 )d respectively, and 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

43 北京梓桢国际教育内部教学资料报班电话 angles BAD and CBE are 30 and 60 respectively. (i) Find the length of CD in terms of d. [] (ii) Show that angle CAD = tan 1 3. [3] (05w) 11 Solve the equation 3 sin θ - cosθ - 3 = 0, for 0 θ 180. [4] 1 Solve the equation (06s) sin x + 3cosx = 0, for 0 x 180 [4] 13 In the diagram, ABC is a triangle in which AB = 4 cm, BC = 6 cm and angle ABC = 150. The line CX is perpendicular to the line ABX. (06s) (i) Find the exact length of BX and show that angle CAB = tan 1 4 (ii) Show that the exact length of AC is cm. [] [4] 14 Prove the identity 1 tan 1 tan x x 1 sin x. [4] (07s) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 4 -

44 北京梓桢国际教育内部教学资料报班电话 The function f is defined by f(x) = a bcosx, for 0 x π. It is given that f(0) = 1 1 and f = 7. (07s) (i) Find the values of a and b. [3] (ii) Find the x-coordinates of the points where the curve y = f(x) intersects the x-axis. [3] (iii) Sketch the graph of y = f(x). [] 16 It is given that = sin + cos and b = cos - sin, where (01w) (i) Show that + b is constant for all values of. (ii) Given that = 3b, show that tan = 7 4 and find the corresponding values of. 17 (i) Show that the equation sin + 3sin cos = 4cos can be written as a quadratic equation in tan. (ii) Hence, or otherwise, solve the equation in part (i) for Given that x = sin 1 ( 5 ), find the exact value of (06w) (i) cos x, [] (ii) tan x. [] 19 (i)show that the equation 3 sin x tan x = 8 can be written as 3 cos x + 8cos x 3 = 0. [3] (07w) (ii) Hence solve the equation 3 sin x tan x = 8 for 0 x 360. [3] 0 In the triangle ABC, AB = 1 cm, angle BAC = 60 and angle ACB = 45. Find the exact length of BC. [3] (08s) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

45 北京梓桢国际教育内部教学资料报班电话 (i)show that the equation tan θ cosθ = 3 can be written in the form cos θ + 3cosθ - = 0. [] (08s) (ii) Hence solve the equation tan θ cosθ = 3, for 0 θ 360. [3] Prove the identity (08w) 1 sin x cosx. [4] cosx 1 sin x cosx 3 The function f is such that f(x) = a b cos x for 0 x 360, where a and b are positive constants. (08w) The maximum value of f(x) is 10 and the minimum value is. (i) Find the values of a and b. [3] (ii) Solve the equation f(x) = 0. [3] (iii) Sketch the graph of y = f(x). [] sin x sin x 4 Prove the identity tan x. (09s/01/1) 1 sin x 1 sin x 5 (09s/01/4) The diagram shows the graph of y = a sin(bx) + c for 0 x p. (i) Find the values of a, b and c. [3] (ii) Find the smallest value of x in the interval 0 x p for which y = 0. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

46 北京梓桢国际教育内部教学资料报班电话 The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k, (10s/11/1) (i) tan( x), [1] 1 (ii) tan ( x), [1] (iii) sin x. [] 7 The function f is such that f ( x) sin x 3cos x for 0 x. (10s/11/5) (i) Express f (x) in the form a bcos x (ii) State the greatest and least values of f(x). [] (iii) Solve the equation f(x) + 1 = 0. [3] 8 (i) Show that the equation (10s/1/1), stating the values of a and b.[] 3( sin x cos x) = (sin x 3 cos x) can be written in the form tan x 3 4. [] (ii) Solve the equation 3( sin x cos x) = (sin x 3 cos x), for 0 x 360. [] 9 The function f : x. a + b cos x is defined for 0 x p. Given that f(0) = 10 and that f, find (10s/13/3) 3 1 (i) the values of a and b, [] 5 (ii) the range of f f, [1] (iii) the exact value of. [] 6 30 (i) Show that the equation sin x tan x + 3 = 0 can be expressed as cos x 3cosx 0. [] (10s/13/4) (ii) Solve the equation sin x tan x + 3 = 0 for 0 x 360. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

47 北京梓桢国际教育内部教学资料报班电话 (i) Show that the equation tan sin 1 can be written in the form 4 tan sin 1 0. [] (11s/11/5) (ii) Hence solve the equation tan sin 1 for 0 q 360. [4] 3 (i) Prove the identity cos tan (1 sin) 1 1. [3] (11s/1/5) sin cos (ii) Hence solve the equation 4 tan (1 sin) for [3] 33 (i) Prove the identity 1 1 sin tan 1 cos. [3] (11s/13/8) 1 cos 1 1 (ii) Hence solve the equation 5, for [4] sin tan 34 Solve the equation 3 tan(x + 15 ) = 4 for 0 x 180. [4] (09w/11/1) 35 The equation of a curve is y = 3 cos x. The equation of a line is x + y =. On the same diagram, sketch the curve and the line for 0 x. [4] (09w/11/) (i) Prove the identity (sin x cosx)(1 sin x cosx) sin x cos x. [3] (09w/1/5) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

48 北京梓桢国际教育内部教学资料报班电话 (ii) Solve the equation (sin x cosx)(1 sin x cosx) 9sin x for 0 x 360. [3] sin x tan x 1 37 (i) Prove the identity 1. [3] (10s/11/4) 1 cosx cosx sin x tan x (ii) Hence solve the equation 0, for 0 x 360. [3] 1 cosx 38 Prove the identity (10w/1/) tan x sin x tan xsin x. [4] 39 Solve the equation 15 sin x = 13 + cos x for 0 x 180. [4] (10w/13/3) 40 (i) Sketch the curve y = sin x for 0 x. [1] (10w.13.4) (ii) By adding a suitable straight line to your sketch, determine the number of real roots of the equation sin x = x. State the equation of the straight line. [3] 41 11w/11/3 1 (i) Sketch, on a single diagram, the graphs of y cos and y for 0.[3] (ii) Write down the number of roots of the equation cos 1 0 interval 0 [1] (iii) Deduce the number of roots of the equation cos 10 in the interval 10 0 [1] in the 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

49 北京梓桢国际教育内部教学资料报班电话 (11w/1/5) (i) Sketch, on the same diagram, the graphs of y sin x and y cosx for [3] (ii) Verify that x 30 is a root of the equation sin x cosx, and state the other root of this equation for which 0 x 180[] (iii) Hence state the set of value of x, for 0 x 180, for which sin x cosx [] 0 x (11w/13/5) (i) Given that, show that, for real values of x, cos x.[3] 3sin x 8cos x (ii) Hence solve the equation 3sin ( 70 ) 8cos( 70 ) 7 0 for [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

50 北京梓桢国际教育内部教学资料报班电话 Section H Combining and Inverting Functions (Chapter 11 ) 1 The functions f and g are defined by (0s) f : x 3x +, x R, 6 g : x, x R, x x 3 (i) Find the value of x for which fg(x) = 3. [3] (ii) Sketch, in a single diagram, the graphs of y = f(x) and y = f -1 (x), making clear the relationship between the two graphs. [3] (iii) Express each of f -1 (x) and g -1 (x) in terms of x, and solve the equation f -1 (x) = g -1 (x). [5] (i) Express x + 8x - 10 in the form a(x + b) + c. [3] (0w) (ii) For the curve y = x + 8x - 10, state the least value of y and the corresponding value of x. [] (iii) Find the set of values of x for which y 14. [3] Given that f : x x + 8x - 10 for the domain x k, (iv) find the least value of k for which f is one-one, [1] (v) express f -1 (x) in terms of x in this case. [3] 3 The function f is defined by f : x ax + b, for x R, where a and b are constants. It is given that f() = 1 and f(5) = 7. (i) Find the values of a and b. [ (ii) Solve the equation ff(x) = 0. [3] (03S) 4 The equation of a curve is y = 8x x. (03S) (i) Express 8x x in the form a (x + b), stating the numerical values of a and b. [3] (ii) Hence, or otherwise, find the coordinates of the stationary point of the curve. [] (iii) Find the set of values of x for which y 0. [3] The function g is defined by g : x 8x x, for x 4. (iv) State the domain and range of g 1. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

51 北京梓桢国际教育内部教学资料报班电话 (v) Find an expression, in terms of x, for g 1 (x). [3] 5 Functions f and g are defined by (03w) f : x x 5, x R, 4 g : x, x R, x. x (i) Find the value of x for which fg(x) = 7. [3] (ii) Express each of f 1 (x) and g 1 (x) in terms of x. [3] (iii) Show that the equation f 1 (x) = g 1 (x) has no real roots. [3] (iv) Sketch, on a single diagram, the graphs of y = f(x) and y = f 1 (x), making clear the relationship between these two graphs. [3] 6 The functions f and g are defined as follows: (04s) f : x x x, x R, g : x x + 3, x R. (i) Find the set of values of x for which f(x) > 15. [3] (ii) Find the range of f and state, with a reason, whether f has an inverse. (iii) Show that the equation gf(x) = 0 has no real solutions. [3] (iv) Sketch, in a single diagram, the graphs of y = g(x) and y = g 1 (x), making clear the relationship between the graphs. [] 7 The function f : x x a, where a is a constant, is defined for all real x. (04w) (i) In the case where a = 3, solve the equation ff(x) = 11. [3] The function g : x x 6x is defined for all real x. (ii) Find the value of a for which the equation f(x) = g(x) has exactly one real solution. [3] The function h : x x 6x is defined for the domain x 3. (iii) Express x 6x in the form (x p) q, where p and q are constants. [] (iv) Find an expression for h 1 (x) and state the domain of h 1. [4] 8 A function f is defined by f : x (x 3) 3 8, for x 4. (05w) (i) Find an expression, in terms of x, for f (x) and show that f is an increasing 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

52 北京梓桢国际教育内部教学资料报班电话 function. [4] (ii) Find an expression, in terms of x, for f 1 (x) and find the domain of f 1. [4] 9 A function f is defined by f : x 3 sinx, for 0 x 360. (i) Find the range of f. [] (05s) (ii) Sketch the graph of y = f(x). [] A function g is defined by g : x 3 sinx, for 0 x A, where A is a constant. (iii) State the largest value of A for which g has an inverse. [1] (iv) When A has this value, obtain an expression, in terms of x, for g 1 (x). [] 10 Functions f and g are defined by (06s) f : x k x for x R, where k is a constant, g : x 9 x for x R, x. (i) Find the values of k for which the equation f(x) = g(x) has two equal roots and solve the equation f(x) = g(x) in these cases. [6] (ii) Solve the equation fg(x) = 5 when k = 6. [3] (iii) Express g 1 (x) in terms of x. [] 11 6 The diagram shows the graph of y = f(x), where f : x for x 0. x 3 (i) Find an expression, in terms of x, for f (x) and explain how your answer shows 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

53 北京梓桢国际教育内部教学资料报班电话 that f is a decreasing function. [3] (07s) (ii) Find an expression, in terms of x, for f 1 (x) and find the domain of f 1. [4] (iii) Copy the diagram and, on your copy, sketch the graph of y = f 1 (x), making clear the relationship between the graphs. [] The function g is defined by g : x (iv) Solve the equation fg(x) = 3. [3] 1 x for x 0. 1 (i) Sketch and label, on the same diagram, the graphs of y = cosx and y = cos3x for the interval 0 x. (01w) (ii) Given that f: x cosx, for the domain 0 x k, find the largest value of k for which f has an inverse. 13 The function f is defined by f : x x 3x for x R. (06w) (i) Find the set of values of x for which f(x) > 4. [3] (ii) Express f(x) in the form (x a) b, stating the values of a and b. [] (iii) Write down the range of f. [1] (iv) State, with a reason, whether f has an inverse. [1] The function g is defined by g : x x 3 x for x 0. (v) Solve the equation g(x) = 10. [3] 14 The function f is defined by f : x x 8x + 11 for x R. (07w) (i) Express f(x) in the form a(x + b) + c, where a, b and c are constants. [3] (ii) State the range of f. [1] (iii) Explain why f does not have an inverse. [1] The function g is defined by g : x x 8x + 11 for x A, where A is a constant. (iv) State the largest value of A for which g has an inverse. [1] (v) When A has this value, obtain an expression, in terms of x, for g 1 (x) and state the range of g 1. [4] 15 The function f is such that f(x) = (3x + ) 3 5 for x 0. (08s) (i) Obtain an expression for f (x) and hence explain why f is an increasing function. [3] (ii) Obtain an expression for f 1 (x) and state the domain of f 1. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 5 -

54 北京梓桢国际教育内部教学资料报班电话 Functions f and g are defined by (08s) f : x 4x k for x R, where k is a constant, 9 g : x for x R, x. x (i) Find the values of k for which the equation fg(x) = x has two equal roots. [4] (ii) Determine the roots of the equation fg(x) = x for the values of k found in part (i). [3] 17 The function f is defined by (08w) f : x 3x for x R. (i) Sketch, in a single diagram, the graphs of y = f(x) and y = f 1 (x), making clear the relationship between the two graphs. [] The function g is defined by g : x 6x x for x R. (ii) Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9. [5] The function h is defined by h : x 6x x for x 3. (iii) Express 6x x in the form a (x b), where a and b are positive constants. [] (iv) Express h 1 (x) in terms of x. [3] 18 The function f is defined by f : x x 1x 13 for 0 x A, where A is a constant. (09s/01/10) (i) Express f(x) in the form a( x b) c, where a, b and c are constants.[3] (ii) State the value of A for which the graph of y = f(x) has a line of symmetry.[1] (iii) When A has this value, find the range of f. [] The function g is defined by g : x x 1x 13 for x 4. (iv) Explain why g has an inverse. [1] (v) Obtain an expression, in terms of x, for g 1 (x). [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

55 北京梓桢国际教育内部教学资料报班电话 The function f is defined by f : x x 1x 7 for x R. (10s/11/9) (i) Express f(x) in the form a( x b) c.[3] (ii) State the range of f. [1] (iii) Find the set of values of x for which f(x) < 1. [3] The function g is defined by g : x x k for x R. (iv) Find the value of the constant k for which the equation gf(x) = 0 has two equal roots. [4] 0 The functions f and g are defined for x R by (10s/1/3) f : x x 4x, g : x 5x 3. (i) Find the range of f. [] (ii) Find the value of the constant k for which the equation gf(x) = k has equal roots. [3] 1 The function f : x 4 3 sin x is defined for the domain 0 x. (10s/1/11) (i) Solve the equation f(x) =. [3] (ii) Sketch the graph of y = f(x). [] (iii) Find the set of values of k for which the equation f(x) = k has no solution. [] 1 The function g : x 4 3 sin x is defined for the domain x A. (iv) State the largest value of A for which g has an inverse. [1] (v) For this value of A, find the value of g 1 (3). [] The function f : x x 8x 14 is defined for x R. (10s/13/10) (i) Find the values of the constant k for which the line y + kx = 1 is a tangent to the curve y = f(x).[4] (ii) Express f(x) in the form a( x b) c, where a, b and c are constants.[3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

56 北京梓桢国际教育内部教学资料报班电话 (iii) Find the range of f. [1] The function g : x x 8x 14 is defined for x A. (iv) Find the smallest value of A for which g has an inverse. [1] (v) For this value of A, find an expression for g 1 (x) in terms of x. [3] 3 Functions f and g are defined for x R by (11s/11/11) f : x x 1, g : x x. (i) Find and simplify expressions for fg(x) and gf(x). [] (ii) Hence find the value of a for which fg(a) = gf(a). [3] (iii) Find the value of b (b a) for which g(b) = b. [] (iv) Find and simplify an expression for f 1 g(x). [] The function h is defined by h : x x, for x 0. (v) Find an expression for h 1 (x). [] x 3 4 The function f is defined by f : x x R, x 1. (11s/1/6) x 1 (i) Show that ff(x) = x. [3] (ii) Hence, or otherwise, obtain an expression for f 1 (x). [] k 5 The function f is such that f ( x) 3 4cos x, for 0 x, where k is a constant. (11s/1/9) (i) In the case where k =, (a) find the range of f, [] (b) find the exact solutions of the equation f(x) = 1. [3] (ii) In the case where k = 1, (a) sketch the graph of y = f(x), [] (b) state, with a reason, whether f has an inverse. [1] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

57 北京梓桢国际教育内部教学资料报班电话 Functions f and g are defined by (11s/13/10) f : x 3x 4, x R, g : x ( x 1) 3 8, x > 1. (i) Evaluate fg(). [] (ii) Sketch in a single diagram the graphs of y = f(x) and y = f 1 (x), making clear the relationship between the graphs. [3] (iii) Obtain an expression for g (x) and use your answer to explain why g has an inverse. [3] (iv) Express each of f 1 (x) and g 1 (x) in terms of x. [4] 7 Functions f and g are defined by (09w/11/10) f : x x 1, x R, x 1 g : x, x R, x 3. x 3 (i) Solve the equation gf(x) = x. [3] (ii) Express f 1 (x) and g 1 (x) in terms of x. [4] (iii) Show that the equation g 1 (x) = x has no solutions. [3] (iv) Sketch in a single diagram the graphs of y = f(x) and y = f 1 (x), making clear the relationship between the graphs. [3] 8 The function f is defined by f : x 5 3 sin x for 0 x. (09w/1/4) (i) Find the range of f. [] (ii) Sketch the graph of y = f(x). [3] (iii) State, with a reason, whether f has an inverse. [1] 9 Functions f and g are defined for x R by (10w/11/3) f : x x 3, g : x x x. Express gf(x) in the form a( x b) c, where a, b and c are constants. [5] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

58 北京梓桢国际教育内部教学资料报班电话 A function f is defined by : x 3 tan x (i) State the range of f. [1] (ii) State the exact value of f for 0 x <. (10w/11/7) f. [1] (iii) Sketch the graph of y = f(x). [] (iv) Obtain an expression, in terms of x, for f 1 (x). [3] 3 31 The function f is defined by (10w/1/7) f ( x) x 4x 7 for x >. (i) Express f(x) in the form (x a) + b and hence state the range of f.[3] (ii) Obtain an expression for f 1 (x) and state the domain of f 1. [3] The function g is defined by g( x) x for x >. The function h is such that f = hg and the domain of h is x > 0. (iii) Obtain an expression for h(x). [1] 3 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

59 北京梓桢国际教育内部教学资料报班电话 The diagram shows the function f defined for 0 x 6 by (10w/13/7) x for 0 x, 1 x 1 x x 1 for < x 6. (i) State the range of f. [1] (ii) Copy the diagram and on your copy sketch the graph of y = f 1 (x). [] (iii) Obtain expressions to define f 1 (x), giving the set of values of x for which each expression is valid. [4] 33 (11w/11/11) Functions f and g are defined by f : x x 8x 10 for 0 x. g : x x for 0 x 10. (i) Express f(x) in the form a( x b) c, where a, b and care constants. [3] (ii) State the range of f. [1] 1 (iii) State the domain of f [1] (iv) Sketch on the same diagram the graphs of y f ( x), y g( x) and y f 1 ( x), making clear the relationship between the graphs. [4] (v) Find an expression for y f 1 ( x) [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

60 北京梓桢国际教育内部教学资料报班电话 (11w/1/) The function f and g are defined for xr by f : x 3x a g : b x Where a and b are constants. Given that ff()=10 and g 1 () 3, find (i) the value of a and b. [4] (ii) an expression for fg(x). [] 35 (11w/13/9) Functions f and are defined by f : x 3 for x 0 g x x x : 6 for 3 (i) express f 1 ( x) in terms of x and solve the equation 1 f ( x) f ( x). [3] (ii) on the same diagram sketch the graphs of y=f(x) and y f 1 ( x), showing the coordinates of their point of intersection and the relationship between the graphs.[3] (iii)find the set of values of x which satisfy gf ( x) 16. [5] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

61 北京梓桢国际教育内部教学资料报班电话 Section I Vectors (Chapter 13 ) The diagram shows a solid cylinder standing on a horizontal circular base, centre O and radius 4 units. The line BA is a diameter and the radius OC is at 90 to OA. Points O, A, B and C lie on the upper surface of the cylinder such that OO, AA, BB and CC are all vertical and of length 1 units. The mid-point of BB is M. (0s) Unit vectors i, j and k are parallel to OA, OC and OO respectively. (i) Express each of the vectors MO and (ii) Hence find the angle OMC. [4] MC ' in terms of i, j and k. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

62 北京梓桢国际教育内部教学资料报班电话 The diagram shows a triangular prism with a horizontal rectangular base ADFC, where CF = 1 units and DF = 6 units. The vertical ends ABC and DEF are isosceles triangles with AB = BC = 5 units. The mid-points of BE and DF are M and N respectively. The origin O is at the mid-point of AC. (03W) Unit vectors i, j and k are parallel to OC, ON and OB respectively. (i) Find the length of OB. [1] (ii) Express each of the vectors MC and MN in terms of i, j and k. [3] (iii) Evaluate MC. MN and hence find angle CMN, giving your answer correct to the nearest degree. [4] 3 The points A and B have position vectors i + 7j + k and 5i + 5j + 6k respectively, relative to an origin O. (04W) (i) Use a scalar product to calculate angle AOB, giving your answer in radians correct to 3 significant figures. [4] (ii) The point C is such that AB = BC. Find the unit vector in the direction of OC. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

63 北京梓桢国际教育内部教学资料报班电话 Relative to an origin O, the position vectors of the points A and B are given by (05S) OA = i + 3j k and OB = 4i 3j + k. (i) Use a scalar product to find angle AOB, correct to the nearest degree. [4] (ii) Find the unit vector in the direction of AB. [3] (iii) The point C is such that OC = 6j + pk, where p is a constant. Given that the lengths of AB and AC are equal, find the possible values of p. [4] 5 Relative to an origin O, the position vectors of points P and Q are given by (05W) OP = 3 1 and OQ = 1, q where q is a constant. (i) In the case where q = 3, use a scalar product to show that cospoq = 7 1. [3] (ii) Find the values of q for which the length of PQ is 6 units. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 6 -

64 北京梓桢国际教育内部教学资料报班电话 The diagram shows the roof of a house. The base of the roof, OABC, is rectangular and horizontal with OA = CB = 14m and OC = AB = 8m. The top of the roof DE is 5m above the base and DE = 6m. The sloping edges OD, CD, AE and BE are all equal in length. (06S) Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. (i) Express the vector OD in terms of i, j and k, and find its magnitude. [4] (ii) Use a scalar product to find angle DOB. [4] 7 Relative to an origin O, the position vectors of the points A and B are given by (07S) OA = and OB =. 4 (i) Given that C is the point such that AC = AB, find the unit vector in the direction ofoc. [4] 1 The position vector of the point D is given by OD = 4, where k is a constant, k 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

65 北京梓桢国际教育内部教学资料报班电话 and it is given that OD = m OA + n OB, where m and n are constants. (ii) Find the values of m, n and k. [4] 8 The diagram shows a prism with cross-section in the shape of a right-angled triangle OAC where OA = 6cm and OC = 8cm. The cross-section through E is the triangle BED. The length of the prism is 16cm. M is the mind-point of AE and N is the mind-point of AC. (01w) Unit vectors i, j and k are parallel to OA, OB and OC respectively as shown. (i) Express each of the vectors MN and MD in terms of i, j and k. (ii) Evaluate MN. MD and hence find the value of angle NMD, giving your answer to the nearest degree. 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

66 北京梓桢国际教育内部教学资料报班电话 Given that a =, b = and c = (i) the angle between the directions of a and b, (ii) the value of p for which b and c are perpendicular. p p, find (0w) p 1 10 The points A, B, C and D have position vectors 3i + k, i - j + 5k, j + 7k and -i + 10j + 7k respectively. (03s) (i) Use a scalar product to show that BA and BC are perpendicular. (ii) Show that BC and AD are parallel and find the ratio of the length of BC to the length of AD. 11 Relative to an origin O, the position vectors of the points A, B, C and D are given by (04s) OA = 1 3, OB = OC = 3 where p and q are constants. Find (i) the unit vector in the direction of AB, 4 p (ii) the value of p for which angle AOC = 90, and OD = (iii) the values of q for which the length of AD is 7 units q 1 The position vectors of points A and B are and 1 4 respectively, relative to an origin O. (i) Calculate angle AOB. [3] (06w) (ii) The point C is such that AC = 3 AB. Find the unit vector in the direction of OC. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

67 北京梓桢国际教育内部教学资料报班电话 The diagram shows a cube OABCDEFG in which the length of each side is 4 units. The unit vectors i, j and k are parallel to OA, OC and OD respectively. The mid-points of OA and DG are P and Q respectively and R is the centre of the square face ABFE. (07w) (i) Express each of the vectors PR and PQ in terms of i, j and k. [3] (ii) Use a scalar product to find angle QPR. [4] (iii) Find the perimeter of triangle PQR, giving your answer correct to 1 decimal place. [3] 14 Relative to an origin O, the position vectors of points A and B are i + j + k and 3i j + pk respectively. (08s) (i) Find the value of p for which OA and OB are perpendicular. [] (ii) In the case where p = 6, use a scalar product to find angle AOB, correct to the nearest degree.[3] (iii) Express the vector AB is terms of p and hence find the values of p for which the length of AB is 3.5 units. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

68 北京梓桢国际教育内部教学资料报班电话 The diagram shows a semicircular prism with a horizontal rectangular base ABCD. The vertical ends AED and BFC are semicircles of radius 6 cm. The length of the prism is 0 cm. The mid-point of AD is the origin O, the mid-point of BC is M and the mid-point of DC is N. The points E and F are the highest points of the semicircular ends of the prism. The point P lies on EF such that EP = 8 cm. (08w) Unit vectors i, j and k are parallel to OD, OM and OE respectively. (i) Express each of the vectors PA and PN in terms of i, j and k. [3] (ii) Use a scalar product to calculate angle APN. [4] 16 Relative to an origin O, the position vectors of the points A and B are given by (09S/01/6) OA i 8 j 4k and OB 7 i j k (i) Find the value of OA.OB and hence state whether angle AOB is acute, obtuse or a right angle.[3] (ii) The point X is such that AX 5 AB. Find the unit vector in the direction of OX. [4] 17 Relative to an origin O, the position vectors of the points A and B are given by 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

69 北京梓桢国际教育内部教学资料报班电话 (10s/1/5) 4 OA 3 and OB 1 1 p (i) Find the value of p for which OA is perpendicular to OB. [] (ii) Find the values of p for which the magnitude of AB is 7. [4] 18 Relative to an origin O, the position vectors of the points A, B and C are given by (10s/13/6) OA i j 4k, OB 3i j 8k, OC i j 10k. (i) Use a scalar product to find angle ABC. [6] (ii) Find the perimeter of triangle ABC, giving your answer correct to decimal places. [] 19 (11s/11/4) The diagram shows a prism ABCDPQRS with a horizontal square base APSD with sides of length 6 cm. The cross-section ABCD is a trapezium and is such 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

70 北京梓桢国际教育内部教学资料报班电话 that the vertical edges AB and DC are of lengths 5 cm and cm respectively. Unit vectors i, j and k are parallel to AD, AP and AB respectively. (i) Express each of the vectors CP and CQ in terms of i, j and k.[] (ii) Use a scalar product to calculate angle PCQ. [4] 0 Relative to the origin O, the position vectors of the points A, B and C are given by (11s/1/8) OA 3, 5 4 OB and 3 10 OC 0 6 (i) Find angle ABC. [6] The point D is such that ABCD is a parallelogram. (ii) Find the position vector of D. [] 1 (11s/13/5) In the diagram, OABCDEFG is a rectangular block in which OA = OD = 6 cm and AB = 1 cm. The unit vectors i, j and k are parallel to OA, OC and OD respectively. The point P is the mid-point of DG, Q is the centre of the square face CBFG and R lies on AB such that AR = 4 cm. (i) Express each of the vectors PQ and RQ in terms of i, j andk. [3] (ii) Use a scalar product to find angle RQP. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

71 北京梓桢国际教育内部教学资料报班电话 Relative to an origin O, the position vectors of the points A, B and C are given by (09w/11/9) OA 3, 6 (i) Find angle AOB. [4] 0 OB 6 and 8 OC 5 (ii) Find the vector which is in the same direction as AC and has magnitude 30. [3] (iii) Find the value of the constant p for which OA + pob is perpendicular to OC. [3] 3 In the diagram, OABCDEFG is a cube in which each side has length 6. Unit vectors i, j and k are parallel to OA, OC and OD respectively. The point P is such that AP = 1 AB and the point Q is the mid-point of DF. 3 (i) Express each of the vectors OQ and PQ in terms of i, j and k.[3] (ii) Find the angle OQP. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

72 北京梓桢国际教育内部教学资料报班电话 The diagram shows a pyramid OABC with a horizontal base OAB where OA = 6cm, OB = 8 cm and angle AOB = 90. The point C is vertically above O and OC = 10 cm. Unit vectors i, j and k are parallel to OA, OB and OC as shown. Use a scalar product to find angle ACB. [6] (10w/11/5) The diagram shows a pyramid OABCP in which the horizontal base OABC is a square of side 10 cm and the vertex P is 10 cm vertically above O. The points D, E, F, G lie on OP, AP, BP, CP respectively and DEFG is a horizontal square of side 6 cm. The height of DEFG above the base is a cm. Unit vectors i, j and k are parallel to OA, OC and OD respectively. (10w/1/9) (i) Show that a = 4. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

73 北京梓桢国际教育内部教学资料报班电话 (ii) Express the vector BG in terms of i, j and k. [] (iii) Use a scalar product to find angle GBA. [4] 6 The diagram shows triangle OAB, in which the position vectors of A and B with respect to O are given by (10w/13/10) OA i j 3k and OB 3i j 4k C is a point on OA such that OC poa, where p is a constant. (i) Find angle AOB. [4] (ii) Find BC in terms of p and vectors i, j and k. [1] (iii) Find the value of p given that BC is perpendicular to OA. [4] 7 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 7 -

74 北京梓桢国际教育内部教学资料报班电话 The diagram shows the parallelogram OABC. Given that OA i + 3j + 3k and OC 3i j + k, find (10s/11/10) (i) the unit vector in the direction of OB, [3] (ii) the acute angle between the diagonals of the parallelogram, [ (iii) the perimeter of the parallelogram, correct to 1 decimal place. [3] 8(11w/11/8) Relative to an origin O, the point A has position vector 4i 7jpk and the point B has position vector 8i jpk, where p is a constant. (i) Find OA OB. [] (ii) Hence show that there are no real values of p for which OA and OB are perpendicular to each other. [1] (iii) Find the values of p for which angle AOB 60.[4] 9 (11w/1/3) Relative to an origin O, the position vectors of points A and B are given by OA5i j k and OB i7j pk Where p is a constant. 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

75 北京梓桢国际教育内部教学资料报班电话 (i) Find the value of p for which angle AOB is 90. [3] (ii) In the case where p=4, find the vector which has magnitude 8 and is in the same direction as AB. [4] 30 (11w/13/6) Relative to an origin O, the position vectors of points 3i 4jk and 5i j 3k Aand Bare respectively. (i) Use a scalar product to find angle BOA.[4] The point C is he mid-point of AB. The point D is such that OD OB. (ii) Find DC. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

76 北京梓桢国际教育内部教学资料报班电话 Section J Integration and Its Application ( Chapter 16 & 17 ) 1 The diagram shows the curve y = 3 x and the line y = x intersecting at O and P. Find (i) the coordinates of P, [1] (ii) the area of the shaded region. [5] (0s) The diagram shows the points A(1, ) and B(4, 4) on the curve y = x. The line BC is the normal to the curve at B, and C lies on the x-axis. Lines AD and BE are perpendicular to the x-axis. (0w) (i) Find the equation of the normal BC. [4] (ii) Find the area of the shaded region. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

77 北京梓桢国际教育内部教学资料报班电话 The equation of a curve is y = 5x 4. (03s) (i) Calculate the gradient of the curve at the point where x = 1. [3] (ii) A point with coordinates (x, y) moves along the curve in such a way that the rate of increase of x has the constant value 0.03 units per second. Find the rate of increase of y at the instant when x = 1. [] (iii) Find the area enclosed by the curve, the x-axis, the y-axis and the line x = 1. [5] dy 4 A curve is such that = 3x 4x + 1. The curve passes through the point (1, 5). dx (03w) (i) Find the equation of the curve. [3] (ii) Find the set of values of x for which the gradient of the curve is positive.[3] 5 8 The diagram shows points A(0, 4) and B (, 1) on the curve y =. The tangent 3x to the curve at B crosses the x-axis at C. The point D has coordinates (, 0). (03w) (i) Find the equation of the tangent to the curve at B and hence show that the area 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

78 北京梓桢国际教育内部教学资料报班电话 of triangle BDC is 3 4. [6] (ii) Show that the volume of the solid formed when the shaded region ODBA is rotated completely about the x-axis is 8π. [5] 6 A curve is such that dx 6 dy = 4x 3 and P (3, 3) is a point on the curve. (i) Find the equation of the normal to the curve at P, giving your answer in the form ax + by = c. [3] (04w) (ii) Find the equation of the curve. [4] 7 A curve has equation y = x + x. (04w) dy d y (i) Write down expressions for and. [3] dx dx (ii) Find the coordinates of the stationary point on the curve and determine its nature. [4] (iii) Find the volume of the solid formed when the region enclosed by the curve, the x-axis and the lines x = 1 and x = is rotated completely about the x-axis. [6] 8 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

79 北京梓桢国际教育内部教学资料报班电话 The diagram shows part of the graph of y = and the normal to the curve at P x (6, 3). This normal meets the x-axis at R. The point Q on the x-axis and the point S on the curve are such that PQ and SR are parallel to the y-axis. (i) Find the equation of the normal at P and show that R is the point (4 1, 0). [5] (04s) (ii) Show that the volume of the solid obtained when the shaded region PQRS is rotated through 360 about the x-axis is 18π. [4] dy 9 A curve is such that = x 5. Given that the point (3, 8) lies on the curve, find dx the equation of the curve. [4] (05s) 10 A curve has equation y = x 4. (05s) (i) The normal to the curve at the point (4, ) meets the x-axis at P and the y-axis at Q. Find the length of PQ, correct to 3 significant figures. [6] (ii) Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x = 4. [4] dy A curve is such that = dx 3 x, and (1, 4) is a point on the curve. (05w) (i) Find the equation of the curve. [4] (ii) A line with gradient 1 is a normal to the curve. Find the equation of this normal, giving your answer in the form ax + by = c. [4] (iii) Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x =. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

80 北京梓桢国际教育内部教学资料报班电话 The diagram shows the curve y = x 3 3x 9x + k, where k is a constant. The curve has a minimum point on the x-axis. (06s) (i) Find the value of k. [4] (ii) Find the coordinates of the maximum point of the curve. [1] (iii) State the set of values of x for which x 3 3x 9x + k is a decreasing function of x. [1] (iv) Find the area of the shaded region. [4] 8 13 The equation of a curve is y = x + x dy d y (i) Obtain expressions for and. [3] dx dx. (07s) (ii) Find the coordinates of the stationary point on the curve and determine the nature of the stationary point. [3] (iii) Show that the normal to the curve at the point (, ) intersects the x-axis at the point ( 10, 0). [3] (iv) Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x =. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

81 北京梓桢国际教育内部教学资料报班电话 The diagram shows the curve y = 3 x 4. The shaded region is bounded by the curve, the x-axis and the lines x = 1 and x = 4. Find the volume of the solid obtained when this shaded region is rotated completely about the x-axis, giving your answer in terms of π. [4] (07s) 15 The gradient at any point (x, y) on a curve is 1 x. The curve passes through the point (4, 11). Find (i) the equation of the curve, [4] (ii) the point at which the curve intersects the y-axis. [] 16 A curve is such that dx 1 dy = x 1 and P(1, 5) is a point on the curve. (i) The normal to the curve at P crosses the x-axis at Q. Find the coordinates of Q. [4] (0w) (ii) Find the equation of the curve. [4] (iii) A point is moving along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the y-coordinate when x = 1. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

82 北京梓桢国际教育内部教学资料报班电话 A curve is such that dy dx 4 6 x, and P (1, 8) is a point on the curve. (i) The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR. [5] (06s) (ii) Find the equation of the curve. [4] dy 4 18 A curve is such that = - 3. (01w) dx 3 x (i) Given that the curve passes through the point(1,16), find the equation of the curve. (ii) Find the coordinates of the stationary point on the curve. 19 The diagram shows the curve y = 8x 1 and the tangent at the point P(3,5) on the curve. This tangent meets the y-axis at A. Find (i) the equation of the tangent at P, (ii) the coordinates of A, (iii) the area of the shaded region. (01w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

83 北京梓桢国际教育内部教学资料报班电话 The diagram shows the curve y = x(x 1)(x ), which crosses the x-axis at the points O(0, 0), A(1, 0) and B(, 0). (06w) (i) The tangents to the curve at the points A and B meet at the point C. Find the x-coordinate of C. [5] (ii) Show by integration that the area of the shaded region R 1 is the same as the area of the shaded region R. [4] 6 1 The equation of a curve is y =. (06w) 5 x (i) Calculate the gradient of the curve at the point where x = 1. [3] (ii) A point with coordinates (x, y)moves along the curve in such a way that the rate of increase of y has a constant value of 0.0 units per second. Find the rate of increase of x when x = 1. [] (iii) The region between the curve, the x-axis and the lines x = 0 and x = 1 is rotated through 360 about the x-axis. Show that the volume obtained 1 is. [5] 5 dy A curve is such that 4 x and the point P(, 9) lies on the curve. The dx normal to the curve at P meets the curve again at Q. Find (07w) (i) the equation of the curve, [3] (ii) the equation of the normal to the curve at P, [3] (iii) the coordinates of Q. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 8 -

84 北京梓桢国际教育内部教学资料报班电话 dy k The diagram shows a curve for which, where k is a constant. The curve 3 dx x passes through the points (1, 18) and (4, 3). (08s) 16 (i) Show, by integration, that the equation of the curve is y =. [4] x The point P lies on the curve and has x-coordinate 1.6. (ii) Find the area of the shaded region. [4] 4 The diagram shows the curve y = ( 3x 1) and the points P (0, 1) and Q(1, ) on the curve. The shaded region is bounded by the curve, the y-axis and the line y =. (08w) (i) Find the area of the shaded region. [4] (ii) Find the volume obtained when the shaded region is rotated through 360 about the x-axis. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

85 北京梓桢国际教育内部教学资料报班电话 Tangents are drawn to the curve at the points P and Q. (iii) Find the acute angle, in degrees correct to 1 decimal place, between the two tangents. [4] The diagram shows part of the curve y. (09s/01/9) 3 x (i) Find the gradient of the curve at the point where x =. [3] (ii) Find the volume obtained when the shaded region is rotated through 360 about the x-axis, giving your answer in terms of p. [5] The diagram shows the curve y x 3 6x 9x for x 0. The curve has a maximum point at A and a minimum point on the x-axis at B. The normal to the 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

86 北京梓桢国际教育内部教学资料报班电话 curve at C (, ) meets the normal to the curve at B at the point D. (09s/01/11) (i) Find the coordinates of A and B. [3] (ii) Find the equation of the normal to the curve at C. [3] (iii) Find the area of the shaded region. [5] 7 The diagram shows the curve y 6x x and the line y = 5. Find the area of the shaded region. [6] (10s/11/4) dy 1 8 A curve is such that 3x 6and the point (9, ) lies on the curve. (10s/11/6) dx (i) Find the equation of the curve. [4] (ii) Find the x-coordinate of the stationary point on the curve and determine the nature of the stationary point. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

87 北京梓桢国际教育内部教学资料报班电话 a The diagram shows part of the curve y, where a is a positive constant. Given that x the volume obtained when the shaded region is rotated through 360 about the x-axis is 4, find the value of a.[4] (10s/1/) 30 The diagram shows the curve y ( x ) and the line y + x = 7, which intersect at points A and B. Find the area of the shaded region. [8] (10s/1/9) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

88 北京梓桢国际教育内部教学资料报班电话 The equation of a curve is such that dy dx through the point P(, 11), find (10s/13/5) (i) the equation of the normal to the curve at P, [3] (ii) the equation of the curve. [4] 6. Given that the curve passes (3x ) 3 4 The diagram shows part of the curve y x which has a minimum point at M. x The line y = 5 intersects the curve at the points A and B. (10s/13/9) (i) Find the coordinates of A, B and M. [5] (ii) Find the volume obtained when the shaded region is rotated through 360 about the x-axis. [6] 33 (i) Sketch the curve y ( x ). [1] (11s/11/3) (ii) The region enclosed by the curve, the x-axis and the y-axis is rotated through 360 about the x-axis. Find the volume obtained, giving your answer in terms of. [4] dy 3 34 A curve is such that and the point (1, 1 ) lies on the curve. dx (1 x) (11s/11/7) (i) Find the equation of the curve. [4] (ii) Find the set of values of x for which the gradient of the curve is less than 31. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

89 北京梓桢国际教育内部教学资料报班电话 Find x dx x 3. [3] (11s/1/1) A curve has equation y and P(, ) is a point on the curve. (11s/1/4) 3 x 4 (i) Find the equation of the tangent to the curve at P. [4] (ii) Find the angle that this tangent makes with the x-axis. [] 37 The diagram shows part of the curve y 4 x x. The curve has a maximum point at M and meets the x-axis at O and A. (11s/1/11) (i) Find the coordinates of A and M. [5] (ii) Find the volume obtained when the shaded region is rotated through 360 about the x-axis, giving your answer in terms of. [6] dy 38 A curve is such that 1 dx x and P(9, 5) is a point on the curve. (11s/13/9) (i) Find the equation of the curve. [4] (ii) Find the coordinates of the stationary point on the curve. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

90 北京梓桢国际教育内部教学资料报班电话 d y (iii) Find an expression for and determine the nature of the stationary point. dx [] (iv) The normal to the curve at P makes an angle of tan 1 k with the positive x-axis. Find the value of k. [] dy 39 A curve is such that k x, where k is a constant. (09w/11/6) dx (i) Given that the tangents to the curve at the points where x = and x = 3 are perpendicular, find the value of k. [4] (ii) Given also that the curve passes through the point (4, 9), find the equation of the curve. [3] dy 40 The equation of a curve is such that 3 x. Given that the curve passes dx x through the point (4, 6), find the equation of the curve. [4] (09w/1/1) 3 41 The function f is such that f ( x) for x R, x.5. (09w/1/8) x 5 (i) Obtain an expression for f (x) and explain why f is a decreasing function. [3] (ii) Obtain an expression for f 1 (x). [] (iii) A curve has the equation y = f(x). Find the volume obtained when the region bounded by the curve, the coordinate axes and the line x = is rotated through 360 about the x-axis. [4] 1 4 Find x dx. [3] (10w/11/1) x 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

91 北京梓桢国际教育内部教学资料报班电话 The equation of a curve is y. (10w/11/11) x dy (i) Find an expression for and determine, with a reason, whether the curve dx has any stationary points. [3] (ii) Find the volume obtained when the region bounded by the curve, the coordinate axes and the line x = 1 is rotated through 360 about the x-axis. [4] (iii) Find the set of values of k for which the line y = x + k intersects the curve at two distinct points.[4] 44 The diagram shows part of the curve and the line x = 5 at B. (10w/1/11) y 1 (3x 1) 1 4. The curve cuts the y-axis at A (i) Show that the equation of the line AB is y 1 x 1. [4] (ii) Find the volume obtained when the shaded region is rotated through 360 about the x-axis. [9] 10 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

92 北京梓桢国际教育内部教学资料报班电话 A curve has equation y = f(x). It is given that f '( x) 3x x 5. (10w/13/6) (i) Find the set of values of x for which f is an increasing function. [3] (ii) Given that the curve passes through (1, 3), find f(x). [4] The diagram shows parts of the curves y 9 x and y and their points of 3 x intersection P and Q. The x-coordinates of P and Q are a and b respectively. (10w/13/11) (i) Show that x = a and x = b are roots of the equation x 6 9x Solve this equation and hence state the value of a and the value of b. [4] (ii) Find the area of the shaded region between the two curves. [5] (iii) The tangents to the two curves at x = c (where a < c < b) are parallel to each other. Find the value of c. [4] 47 (a) Differentiate x 3 5 x with respect to x. [3](11s/13/4) (b) Find a 3x 5 dx and hence find the value of x dx. [4] 0 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

93 北京梓桢国际教育内部教学资料报班电话 (11w/11/10) The diagram shows the curve y 1x meeting the x-axis at A and the y-axis at B. the y-coordinate of the points C on the curve is 3. (i) find the coordinates of B and C [] (ii) find the equation of the normal to the curve at C [4] (iii) find the volume obtained when the shaded region is rotated through 360about the y-axis. [5] 49. (11w/1/7) dy 8 A curve is such that 5. The line 3yx17 is the normal to the curve at dx x the point P on the curve. Given that the x-coordinate of P is positive, find (i) the coordinates of P [4] (ii) the equation of the curve [4] 50. (11w/1/8) The equation pf a curve is (i) (ii) y x x 8. Find an expression for dy, and the coordinates of the stationary point on the curve. dx [4] the volume obtained when the region bounded by the curve and the x-axis is rotated through 360about the x-axis. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 9 -

94 北京梓桢国际教育内部教学资料报班电话 w/13/10 The diagram shows the line yx1and the curve y x 1, meeting at (-1,0) and (0,1). (i) Find the area of the shaded region. [5] (ii) Find the volume obtained when the shaded region is rotated through 360about the y-axis. [7] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

95 北京梓桢国际教育内部教学资料报班电话 Section K Radians ( Chapter 18 ) The diagram shows the circular cross-section of a uniform cylindrical log with centre O and radius 0 cm. The points A, X and B lie on the circumference of the cross-section and AB = 3 cm. (0s) (i) Show that angle AOB = radians, correct to 3 decimal places. [] (ii) Find the area of the sector AXBO. [] The section AXBCD, where ABCD is a rectangle with AD = 18 cm, is removed. (iii) Find the area of the new cross-section (shown shaded in the diagram). [3] In the diagram, OPQ is a sector of a circle, centre O and radius r cm. Angle QOP = θ radians. The tangent to the circle at Q meets OP extended at R. (i) Show that the area, Acm, of the shaded region is given (0w) by A = 1r (tanθ-θ). [] (ii) In the case where θ = 0.8 and r = 15, evaluate the length of the perimeter of the shaded region. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

96 北京梓桢国际教育内部教学资料报班电话 In the diagram, triangle ABC is right-angled and D is the mid-point of BC. Angle DAC = 30 and angle BAD = x. Denoting the length of AD by l,(0w) (i) express each of AC and BC exactly in terms of l, and show that 1 AB = l 7, [4] (ii) show that x = tan [] 3 4 The diagram shows a semicircle ABC with centre O and radius 8 cm. Angle AOB = θ radians. (03s) (i) In the case where θ = 1, calculate the area of the sector BOC. [3] (ii) Find the value of θ for which the perimeter of sector AOB is one half of the perimeter of sector BOC. [3] (iii) In the case where θ = 3 1 π, show that the exact length of the perimeter of triangle ABC is ( ) cm. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

97 北京梓桢国际教育内部教学资料报班电话 The diagram shows the sector OPQ of a circle with centre O and radius r cm. The angle POQ is θ radians and the perimeter of the sector is 0 cm. 0 (i) Show that θ =. [] (03w) r (ii) Hence express the area of the sector in terms of r. [] (iii) In the case where r = 8, find the length of the chord PQ. [3] 6 In the diagram, OCD is an isosceles triangle with OC = OD = 10 cm and angle COD = 0.8 radians The points A and B, on OC and OD respectively, are joined by an arc of a circle with centre O and radius 6 cm. Find (04s) (i) the area of the shaded region, [3] (ii) the perimeter of the shaded region. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

98 北京梓桢国际教育内部教学资料报班电话 In the diagram, AC is an arc of a circle, centre O and radius 6 cm. The line BC is 1 perpendicular to OC and OAB is a straight line. Angle AOC = π radians. Find the 3 area of the shaded region, giving your answer in terms of π and 3. [5] (04w) 8 In the diagram, ABC is a semicircle, centre O and radius 9 cm. The line BD is perpendicular to the diameter AC and angle AOB =.4 radians. (05s) (i) Show that BD = 6.08 cm, correct to 3 significant figures. [] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

99 北京梓桢国际教育内部教学资料报班电话 In the diagram, OAB and OCD are radii of a circle, centreoand radius 16 cm. AngleAOC =α radians. AC and BD are arcs of circles, centre O and radii 10 cm and 16 cm respectively. (05w) (i) In the case where α = 0.8, find the area of the shaded region. [] (ii) Find the value of α for which the perimeter of the shaded region is 8.9 cm. [3] 10 The diagram shows a circle with centre O and radius 8 cm. Points A and B lie on the circle. The tangents at A and B meet at the point T, and AT = BT = 15 cm. (06s) (i) Show that angle AOB is.16 radians, correct to 3 significant figures. [3] (ii) Find the perimeter of the shaded region. [] (iii) Find the area of the shaded region. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

100 北京梓桢国际教育内部教学资料报班电话 In the diagram, OAB is a sector of a circle with centre O and radius 1 cm. The lines AX and BX are tangents to the circle at A and B respectively. Angle AOB = 3 1 π radians. (07s) (i) Find the exact length of AX, giving your answer in terms of 3. [] (ii) Find the area of the shaded region, giving your answer in terms of π and 3. [3] 1 The diagram shows an equilateral triangle OPQ, of side 1cm, and the points S such that OS = PS = QS. The are PXQ has centre O and radius 1cm. Find the perimeter of the shaded region, giving your answer in terms of and 3. (01w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

101 北京梓桢国际教育内部教学资料报班电话 In the diagram, AOB is a sector of a circle with centre O and radius 1 cm. The 1 point A lies on the side CD of the rectangle OCDB. Angle AOB = radians. 3 Express the area of the shaded region in the form ( 3) - bπ, stating the values of the integers a and b. [6] (06w) 14 In the diagram, AB is an arc of a circle, centre O and radius r cm, and angle AOB = θ radians. The point X lies on OB and AX is perpendicular to OB. (i) Show that the area, Acm, of the shaded region AXB is given by 1 A = r (θ- sinθcosθ). [3] 1 (ii) In the case where r = 1 and θ =, find the perimeter of the shaded region 6 AXB, leaving your answer in terms of 3 and π. [4] (07w) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

102 北京梓桢国际教育内部教学资料报班电话 The diagram shows a circle with centre O and radius 5 cm. The point P lies on the circle, PT is a tangent to the circle and PT = 1 cm. The line OT cuts the circle at the point Q. (08s) (i) Find the perimeter of the shaded region. [4] (ii) Find the area of the shaded region. [3] 16 In the diagram, the circle has centre O and radius 5 cm. The points P and Q lie on the circle, and the arc length PQ is 9 cm. The tangents to the circle at P and Q meet at the point T. Calculate (08w) (i) angle POQ in radians, [] (ii) the length of PT, [3] (iii) the area of the shaded region. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

103 北京梓桢国际教育内部教学资料报班电话 The diagram shows a circle with centre O. The circle is divided into two regions, R1 and R, by the radii OA and OB, where angle AOB = q radians. The perimeter of the region R 1 is equal to the length of the major arc AB. (09s/01/5) (i) Show that 1. [3] (ii) Given that the area of region R 1 is 30cm, find the area of region R, correct to 3 significant figures. [4] 18 The diagram shows ametal plate ABCDEF which has been made by removing the two shaded regions from a circle of radius 10 cm and centre O. The parallel edges AB and ED are both of length 1 cm. (10s/13/7) (i) Show that angle DOE is 1.87 radians, correct to 4 significant figures.[] (ii) Find the perimeter of the metal plate. [3] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

104 北京梓桢国际教育内部教学资料报班电话 (iii) Find the area of the metal plate. [3] 19 In the diagram, OAB is an isosceles triangle with OA = OB and angle AOB = radians. Arc PST has centre O and radius r, and the line ASB is a tangent to the arc PST at S. (11s/11/9) (i) Find the total area of the shaded regions in terms of r and. [4] (ii) In the case where 3 1 p and r = 6, find the total perimeter of the shaded regions, leaving your answer in terms of 3 and. [5] 0 In the diagram, AB is an arc of a circle, centre O and radius 6 cm, and angle AOB = 3 1 radians. The line AX is a tangent to the circle at A, and OBX is a straight line. (11s/13/7) (i) Show that the exact length of AX is 6 3 cm. [1] Find, in terms of and 3, (ii) the area of the shaded region, [3] (iii) the perimeter of the shaded region. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

105 北京梓桢国际教育内部教学资料报班电话 The diagram shows a semicircle ABC with centre O and radius 6 cm. The point B is such that angle BOA is 90 and BD is an arc of a circle with centre A. Find (09w/11/5) (i) the length of the arc BD, [4] (ii) the area of the shaded region. [3] A piece of wire of length 50 cm is bent to form the perimeter of a sector POQ of a circle. The radius of the circle is r cm and the angle POQ is q radians (see diagram). (09w/1/7) (i) Express q in terms of r and show that the area, Acm, of the sector is given by A 5r r. [4] (ii) Given that r can vary, find the stationary value of A and determine its nature. [4] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

106 北京梓桢国际教育内部教学资料报班电话 The diagram shows two circles, C 1 and C, touching at the point T. Circle C 1 has centre P and radius 8 cm; circle C has centre Q and radius cm. Points R and S lie on C 1 and C respectively, and RS is a tangent to both circles. (10w/11/9) (i) Show that RS = 8 cm. [] (ii) Find angle RPQ in radians correct to 4 significant figures. [] (iii) Find the area of the shaded region. [4] 4 The diagram shows a rhombus ABCD. Points P and Q lie on the diagonal AC such that BPD is an arc of a circle with centre C and BQD is an arc of a circle with 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

107 北京梓桢国际教育内部教学资料报班电话 centre A. Each side of the rhombus has length 5 cm and angle BAD = 1. radians. (10w/13/8) 5 (i) Find the area of the shaded region BPDQ. [4] (ii) Find the length of PQ. [4] The diagram shows points A, C, B, P on the circumference of a circle with centre O and radius 3 cm. Angle AOC = angle BOC =.3 radians. (10w/1/4) (i) Find angle AOB in radians, correct to 4 significant figures. [1]54 (ii) Find the area of the shaded region ACBP, correct to 3 significant figures. [4] 6. (11W/11/5) The diagram represents a metal plate OABC, consisting of a sector OAB of a circle with centre O and radius r, together with a triangle OCB which is right-angled at C. Angle AOB radians and OC is perpendicular to OA. 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

108 北京梓桢国际教育内部教学资料报班电话 (i) Find an expression in terms of r and for the perimeter of the plate. [3] 1 (ii) For the case where r 10 and = 5, find the area of the plate. [3] 7 (11w/1/6) The diagram shows a circle C1touching a circle C at a point X. Circle C1 has centre A and radius 6cm, and circle C has centre B and radius 10cm. points D and E lie on 1 C1 and C respectively and DE is parallel to AB. Angle DAX radians and angle 3 EBX (i) radians. By considering the perpendicular distances of D and E from AB, show that the exact value of is sin ( ).[3] 10 (ii) Find the perimeter of the shaded region, correct to 4 significant figures. [5] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

109 北京梓桢国际教育内部教学资料报班电话 (11w/13/4) In the diagram, ABCD is a parallelogram with AB=BD=DC=10cm and angle ABD=0.8 radians. APD and BQC are arcs of circles with centres B and D respectively. (i) Find the area of the parallelogram ABCD.[] (ii) Find the area of the complete figure ABQCDP.[] (iii) Find the perimeter of the complete figure ABQCDP. [] 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

110 北京梓桢国际教育内部教学资料报班电话 Answer Section A Coordinates,Points and Lines(Chapter 1) 1 (i) y - 6 = 1 ( x 1) or y = x +11 (ii) Solution C(13,1) (iii) 35.8 or 35.7 (i) Eqn of L y 4 = 1 (x 7) (ii) AB = 0 or (i) Eqn of BC y-6= 1 (x-4) eqn of CD y-5=-(x-1) (ii) C (10,9) 4 (i) (, 6) and ( 3, 11) (ii) y 8 1 = 1 (x + 1 ) (or y = x + 9) 5 (i) Mid point is M ( (ii) ( 1, 5 1 ) 4 (iii) Distance = 1 4 1, 7 1 ) 6 M (4,6) A ( 8, 0) C (16, 1) 7 (i) y + 3x = 31 (ii) D (7,5) 8 Distance = y-8 = ( x ) or 3y+x=0 3 C (10, 0) D (14, 6) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

111 北京梓桢国际教育内部教学资料报班电话 x = 4 (4, 6) 11 (i) m of AB = 8/1 m of perpendicular = -1/8 3 eqn of CD y - 15 = - ( x 6) (ii) eqn of AB y - 3 = ( x 1) 3 sin eqns y + 3x = 48 and 3y = x + 7 D(10, 9) 1 sim eqns (6., 9.6) 13 (i) Eliminates y to get x - 7x + 6 = 0 or y 5y +3 = 0 (x 3)(x ) = 0 x = or 1 dy (ii) = 4x 8 dx = 0 x = or completes the square and states stationary at x =. 14 (i) Gradient of AC = ½ Perpendicular gradient = Eqn of BX is y = (x ) Sim Eqns y + x = 16 with y = x (4, 6) (ii) X is mid-point of BD, D is (6, 10) (iii) AB = (14 ) 00 BC = ( 6 ) 40 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

112 北京梓桢国际教育内部教学资料报班电话 Perimeter = Perimeter = y = 5 - x 8, P(, l) 16. dy 8 (i) dx x m of tan = m of normal = ½ Eqn of normal y 1= - 1 (x ) y + x = 4 (ii) Sim eqns y + x = 4, y = 5 - x 8 x + 6x 16 = 0 or y 7y + 6 = 0 ( 8, 6) (iii) Length = (10 5 ) (accept 15 or 5 5 etc) Eqn BD y x 17 B(6,5) C(1,8) 17. ( i) m 1, ( ii) Sim eqns y+x=4, y=x+7 C(-1,6) (iii) M is (9,10) Perp gradient is, D(5,1) Equation of L is y-3= ( x 1), Equation of L is y-1=(x-3) (5,5) (i) y3 3( x ), (ii) D(4,-3) (iii) AC= 40 BD= 160 Area=40 0. (i) C (3.6, 1.8) (ii) d=3.58 x y a , P(a,0) and Q(0,b), Distance a b 45, a 6, b=3 a b b. (i) y-coordinate same as the y-coordinate of the mid-point of AC (ii) D(16,6) and B(-4, 6) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

113 北京梓桢国际教育内部教学资料报班电话 (iii) Product of gradients = -1 h 1h 64 0 h 16 or -4, so xd 16 and xb 4 Area (i) (-1,1), (/3,6) (ii) AB=5.7 mid-point=(-1/6, 7/) 7 4. (i) (,) (ii)x-y+4=0 5. (i)(1,8),(,), AB 6.71or 45 (ii) k=3 or 11 1 (,5) 6 (i) Gradient of AC= 1, Gradient of BD=- Eqn of BD is y-6=-(x-3) Eqn of AC is y+1= 1 ( x 1) Sim eqns M(5,) D(7,-) (ii) Ratio of AM : MC 45 : 0 3: 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

114 北京梓桢国际教育内部教学资料报班电话 Section B Functions and Graphs (Chapter 3) 1 x = 1,y = -1.5 and x = -3, y = 6 (1.5,8) and (4,3) 3. k or k 6 (i) A(1, 1) (ii) x=, y= (iii) mid-point of AP=(, 3), gradient of line=,equation is y 3 ( x ) 5. m or m 10 6 (i) (,3) and (6,1) dy (ii) k or 1 1 y, x 4, k 8 dx y 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

115 北京梓桢国际教育内部教学资料报班电话 (i) (1.5, 8) and (4, 3) (ii) - 96 < k < 96 or k < 96 (iii) 8.1 or 8. x = 1.5 or x = (i) (x - 3) - 7 (ii) Range f(x) -7 4 y = 4x + k and y = x x - 4x - k = 0 b - 4ac < k < 0 k < -4 x=4 M1 y=4 k=-4 M1A1 5. ( i) p, q 15 (ii) r=16 6. (i) Section C Quadratics (Chapter 4) (ii) 1 x only 1 1 (, ),( ), 7. (i) y=m(x-) oe (ii) m m x 3, m x 1 (3,), (1,) (iii) ( x ) 1,(,1) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

116 北京梓桢国际教育内部教学资料报班电话 Section D Differentiations and Their Applications 1 (i) dy/dx = 3x +6x+9 (ii) x = -3 or x = 1 (iii) k = -7 or k = 5 ( Chapter 6 & 7 & 1 & 15 ) (i) 略 (ii) r = 8 (iii) V = 1610 or (51π) Maximum 3 (a) dy/dx = 4 1x-3 (b) x 6x -1 + c 4 (i) y = 7 (x) or 36 x A = 4x+6xy A = 4x + 16 x (ii) x = 3 (iii) Stationary value = 108 cm d A/dx = x 3 Positive when x=3 Minimum. 1 5 (i) h = 4 - r - r (ii) A = 8r - r 1 - r (iii) r = 1.1 (iv) d A/dr = -4 - π,this is negative Maximum (i) dy/dx = x 3 = 0 when x = 1.5, y = 1.75 This is a minimum point, 1.75 > 0 Curve lies above the x - axis. (ii) Decreasing function for x < 1.5. (iii) ( 1, 8) and (, ) 3 (iv) (i) 略 (ii) r = 4 V = 64π ( or 01 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

117 北京梓桢国际教育内部教学资料报班电话 k = 1 10 c = 1 11 (i) da/dx = 8x (ii) (-) k = (i) Height = 3x. 10xy + 1.8x.3x. = 00 y = 00 4x 10x. (ii) v = 1.8x.3xy = 40x - 8.8x 3 (iii) dv = x dx -0 when x = -1 3 d V (iv) = -17.8x dx -ve Maximum 14 Area = integral of x attempted. x Uses limits 1 to 4 correctly or 9.33 or y = (x - 3) 3-6x dy (i) 3 (x 3) dx d y 1(x 3) dx -6 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

118 北京梓桢国际教育内部教学资料报班电话 dy (ii) s.p = 0 (x - 3) = 1 dx x = or x = 1 If x =, nd diff = + ve MIN If x = 1, nd diff = - ve MAX 16 (i) 4x + πr = 80 A = x + πr (ii) A = ( 4) x 160x 1600 da ( 4) x 160 dx = 0 when x = 160 ( 4) or (i) A is (3,0) 4y+9x=7 (ii) normal meets y-axis at (0, 6 ), curve meets y-axis at (0, -4) BC= x x x x (i) h= V=x h 4 (ii) V 64 dv (iii) 3x Max dx dy dx, (ii) y+9=10x (iii) x 1, x 1 (i) (x 3) 4 dv 4r dr dr dv dt dv dt dr dz 100 d z 400 (ii) 3 0 x 0, z=10, 0 min 3 dx x dx x ( i)( 1,6) nd differential 1 x, positive min 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

119 北京梓桢国际教育内部教学资料报班电话 x 5 5 ( ii) A x 9 x limits from 0 to dy dx 3 (i) 1( x 3) x ( ii) y 3 ( x 1) ( iii ) ( i) (a) x 3 or x (b) y 4 ( x 3) (c) m= angle of 63.4, m= angle of 6.6, angle btween=37 ( ii) k ( i) y 30 x x 4 1 (iii) x=15 (iv) max 6. (i) y x 9 7. ( i)0.7(t-1) 9 9 point is (-, ) (iii) ( 1, 4 4 ) 4, ii 3 mid 1 (ii) ( i)0 k 4 (ii) k=4 only (x-1) 0 (, ) 1 9. (ii) x=, min 30. dy 1 d y ( i) 1 dx ( x 3) dx ( x 3) 3 (ii) x 4, y 5,1 x 4,min, x,max 31. (i) 1 y 6(48 8 x) oe A 4xy xy or 3xy 3xy 6xy (ii) A x(48 8 x) 48x 8x 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

120 北京梓桢国际教育内部教学资料报班电话 y 48 16x x (iii) A 7 A 16( 0) Maximum x 3 (i) 3 f ( x) x 6 x( c) (ii) Subst(3, 4) 33. c f x x x 5 ( ) 6 5 y x 9x 1x 4 (3x ) (i) f '(3) k1 0 k ( x 3)( x ) 0 x,( Allow also =3) (ii) f''(x)=4x- f''(3)>0 hence min at P f''(-)<0 hence max at Q 3 Sub (3,-10) c=17 3 (iii) f(x)= x x 1 x( c) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

121 北京梓桢国际教育内部教学资料报班电话 (i) S 5 = 750 (ii) 13 th term = 40.5 or 40.6 (i) a = 7 r = 3 (ii) Answer = S n = (i) (ii) 18 5 (i) r = 4 3 (ii) S 10 = 4 6 (i) 39 (ii) a = 175 1st term = 05 8 (i) 9 th = (ii) S 10 = (iii) D = (i) 8144 (ii) (i) a = 6 (ii) S 0 = (i) 775kg (ii) 17600kg (iii) 0000kg 1 (a) a = 105 Either f = 399 or d = Section E Sequences ( Chapter 8 & 14 ) 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

122 北京梓桢国际教育内部教学资料报班电话 (b) r = 64/144 r = 3 (i) Either x = ar x = x or x 64 x = 96 (ii) Use of S = 1 r (i) + 4d and + 14d (ii) a + 4d = ar, a + 14d = ar 4d or or ac = b 4d 14d 3a = 8d 4d 14d (iii) r = or. 5 4d 14 (i) a = 81, ar 3 = 4 r 3 = 4/81 r = ⅔ or (ii) S = 81 ⅓ = 43 1 r (iii) nd term of GP = ar = 81 ⅔ = 54 3rd term of GP = ar = 36 3d = 18 (d = 6) S 10 = 5 (108 54) = st term = = 6 5 th term = + 4d = 1 d = 1.5 S n = n (1 + (n - 1)1.5) = n + 7n - 10 = 0 n = 8 ( a) a 0.5, r=0.5 S / 3( or0.667) ( b) 50 terms in the progression, ( id ) 1.5, a=10 (ii) n=5 18. ( a) S 800 (b) (i) a=45, (ii) S a (i) ar (ii) S r 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑

123 北京梓桢国际教育内部教学资料报班电话 (i) (ii) (a) arc length=5, perimeter=1.1 (b) (i) k=1, (ii) S 81. (a) a=45 (b) a=d, S 8=3d or 64a, S 4=8d or 16a 3. (i) r=3/4 d=-1/ (ii) S 3 (iii) S (i) d=-1 a=117 (ii) a= (a) a=1,d=3/ (b) a=16 and ar, r=, Sum to infinity = (a) d= (a) 19 r, ar 95 0 (b) (i) d 5, a 100 (ii) m 0 (iii) n (a) d=6 a= -7, 9 4 (b) r oe S (a) a=-15, n=5 (i) Use of ds 3 n (ii) Last term=a+4d 57 (iii) Positive terms are 3, 6, 57 Either a=0 or 3, n=19 or (b) r=1.05 (i) 11 th term= 30. (i) 0 (ii) 409 教室地址 :1 都市网景, 远洋沁山水,3 潞河名苑 - 1 -