Mathematics Term 1 STPM Chapter 6 Vectors 6 245 Exercise 6.3 1. Find the value of a · b given that (a) a = 2i + 5j, b = –4i – 3j (b) a = i – 3j + k, b = 3i + j + 5k (c) a = –5i – j + k, b = 3j – 6k (d) a = –k, b = i – 6k (e) a = –2i + 5j, b = –4i + 2j + k 2. Determine the angle between a and b given that (a) a = i + 2j, b = 2i + 3j (b) a = 3i + j + 4k, b = i + j – k (c) a = i – 2j, b = 2i + 3k (d) a = 2i + j – 4k, b = 4i + 2j + k (e) a = 2i + j + k, b = 3i – j – 5k 3. Find the acute angle between the line joining (1, 3, 2) to (2, 5, –1) and the line joining (–1, 4, 3) to (3, 2, 1). 4. The vectors →PQ and →PR are 1 –2 6 –32 and 1 –2 –3 6 2 respectively. (a) Determine the scalar product →PQ · →PR. (b) Find the angle between the two vectors. 5. Three points A, B and C have position vectors given respectively by a = 7i + 4j – 2k b = 5i + 3j – 3k c = 6i + 5j – 4k find the angle BAC. 6. Determine whether the following pairs of vectors are orthogonal (perpendicular to each other). (a) u = 2i – 4j + 2k, v = i + 2j + 3k (b) u = 3i – j + 4k, v = 4j + k (c) u = 4i – 2j + k, v = j – 2k 7. If the vectors i + 3j and 6i + lj are perpendicular, find the value of l. 8. Show that the three vectors, 2i + j – 4k, i – 3j + 5k, and 3i – 2j + k, form a right-angled triangle. 9. O k j i A B M F E G D C N In the diagram, OABCDEFG is a cube in which each side has length 12 units. Unit vectors i, j and k are parallel to →OA, →OC and →OD respectively.
246 Mathematics Term 1 STPM Chapter 6 Vectors 6 The point N is such that →AN = 1 3 →AB. M is the midpoint of DF. (a) Express each of the vectors →OM and →NM in terms of i, j and k. (b) Determine the angle OMN, to the nearest 0.1°. 10. Find (i + 2j + 3k) × (2i + j – k). 11. The points P, Q and R have position vectors 3i – 2j + 4k, –4i – 2j + 3k and 5i + 2j – 6k respectively. Find (a) →PR × →PQ (b) →QR × →QP 12. Find the area of a parallelogram whose adjacent sides are 5i – 2j + k and –2i + j + 3k. 13. Find the area of the triangle formed by the points whose position vectors are a = 3i + j, b = 5i + 2j + k and c = i – 2j + 3k. 14. Find the area of the triangle having vertices A(1, 1, 1), B(0, 1, 2) and C(3, 2, 1). 15. Find a vector which is perpendicular to both a and b where a = 2i – 3j – 5k and b = i – 2j – 3k. 16. Find the scalar l so that the vector 2i + j – lk is perpendicular to the sum of the vectors 2i – j + 2k and i + 2j + k. 17. Find a unit vector perpendicular to both of the vectors m = 2i – 3j + k and n = 4i – 2j + k. 6.2 Vector Geometry Equation of a line As we know, to define the equation of a straight line in two dimensional xy-plane, we must have either two known points (coordinates) on the line or a single point (coordinates) with the direction (gradient) of the line. However, to determine the equation of a straight line, in three dimensional xyz-plane, we have to know (a) the position vector of a point which lies on the line and a vector which is parallel to the straight line or (b) the position vector of two points which lie on the straight line. Consider case (a): Equation of a straight line passing through a known point and parallel to another line. a r z x L y O R(x, y, z) A(x1, y1, z1) p = ai + bj + ck Figure 6.19
Mathematics Term 1 STPM Chapter 6 Vectors 6 247 In the diagram above, L is a straight line which is parallel to a vector p and passes through a point A(x1 , y1 , z1 ) with position vector a. R(x, y, z) is a general point on the straight line with position vector r. Since →AR // p AR is scalar multiple of p AR = lp where l is a scalar In ∆OAR; →AR = →OR – →OA = lp r – a = lp r = a + lp The vector equation of a straight line which passes through a point with position vector a and is parallel to a vector p is r = a + lp. For each value of scalar l, this vector equation gives the position vector of one point on the straight line. r is the position vector of a general point on the line. In terms of cartesian components; xi + yj + zk = (x1 i + y1 j + z1 k) + l(ai + bj + ck) xi + yj + zk = (x1 + a l)i + (y1 + bl)j + (z1 + c l)k Equating the corresponding components, x = x1 + al y = y1 + bl Parametric equations of a straight line z = z1 + cl Equating l in each of the parametric equations, x – x1 a = y – y1 b = z – z1 c . This is the cartesian equation of a straight line which passes through the point (x1 , y1 , z1 ) in the direction 1 a b c 2. Example 17 Find the vector and cartesian equation of the line that is parallel to the vector –i + 3j + 2k and passes through the point P, position vector 2i – j – k. Solution: Vector equation of the line is r = a + lb r = (2i – j – k) + l(–i + 3j + 2k) r = (2 – l)i + (–1 + 3l)j + (–1 + 2l)k thus x = 2 – l ⇒ 2 – x = l y = –1 + 3l ⇒ y + 1 3 = l z = –1 + 2l ⇒ z + 1 2 = l so, the cartesian equation is x – 2 –1 = y + 1 3 = z + 1 2 (= l)
248 Mathematics Term 1 STPM Chapter 6 Vectors 6 Example 18 Find the vector equation of the straight line that passes through the point with position vector 3i – 2j and is perpendicular to the line r = i – j + l(i – 2j). Solution: The given line is parallel to the vector i – 2j. If the required line is parallel to a vector ai + bj, it follows that (ai + bj) · (i – 2j) = 0 a – 2b = 0 a = 2b The required line is therefore parallel to any vector of form 2bi + bj or b(2i + j). Taking 2i + j as one such vector, the required equation is r = 3i – 2j + l(2i + j). Consider case (b): Equation of a straight line passing through two known points. b a r z x L y O R(x, y, z) A(x1, y1, z1) B(x2, y2, z2) Figure 6.20 In the diagram above, L is a straight line passing through two known points A(x1, y1, z1) and B(x2 , y2 , z2 ) whose position vectors are a and b respectively. R(x, y, z) is a general point on the straight line with position vector r. Since the points A, B and R are collinear, →AR = l →AB →OR – →OA = l( →OB – →OA) r – a = l(b – a) r = a + l(b – a)
Mathematics Term 1 STPM Chapter 6 Vectors 6 249 Example 19 Find the vector and cartesian equations of the line joining the points A(3, 1, 4) and B(–1, 2, 3). Solution: Vector equation of the line, r = 3i + j + 4k + l(–4i + j – k) r = (3 – 4l)i + (1 + l)j + (4 – l)k x = 3 – 4l ⇒ 3 – x 4 = l y = 1 + l ⇒ y – 1 = l z = 4 – l ⇒ 4 – z = l So, the cartesian equation is 3 – x 4 = y – 1 = 4 – z x – 3 –4 = y – 1 = z – 4 –1 Exercise 6.4 1. Find the vector equation of each of the following straight lines. (a) L1 passes through A(2, –1, 6) and is parallel to x-axis. (b) L2 passes through B(–4, 0, 3) and is parallel to y-axis. (c) L3 passes through C(0, 3, –1) and is parallel to z-axis. (d) L4 passes through D(2, –5, 0) and is parallel to the vector –i + 6j. (e) L5 passes through position vector –2i + j – 3k and parallel to the vector i + 2j – k. 2. Find the vector equation of the line joining each of the following pairs of points. (a) (2, 0, 3), (3, –1, 4) (b) (1, 2, 2), (2, –2, 0) (c) (3, 1, 4), (–1, 2, 5) 3. Find the cartesian equation of the line that is parallel to the vector –2i + 5j + 6k and passes through the point with position vector i – 3j + 2k. 4. Find the vector equation of a line with cartesian equation x – 1 –3 = y + 5 2 = z – 2 6 . 5. Find the vector equation of the line with cartesian equation x + 1 7 = y + 2 2 = z + 5 6 . 6. Find the vector equations of the lines with the following cartesian equations; (a) x – 2 3 = y – 1 = z + 1 6 (b) x – 6 = y + 2 5 = z – 1
250 Mathematics Term 1 STPM Chapter 6 Vectors 6 7. Find the vector and cartesian equations of the line (a) passing through (1, –3, 2) and parallel to r = 3i + 2j + l(i – j + 5k), (b) passing through (1, 2, 1) and parallel to r = i + j + k + l(i + 2j – 3k). 8. Find the vector and cartesian equations of the straight line joining the points A(2, –1, 5) and B(–1, 4, 3). 9. Given that A(p, 2, q) is a point on the straight line whose vector equation is L : r = 2i – j + k + l(i – j + 2k). Determine the values of p and q. 10. Find the vector equation of the straight line that passes through the point with position vector i – 3j and is perpendicular to the line r = 3i + 2j + l(i – 2j). Equation of a plane To determine the equation of a plane in space, we must first specify a unit vector, ^ n, that is normal to the plane (i.e. perpendicular to every vector lying on the plane) and a known point A(x1 , y1 , z1 ) lying on the plane. Let R(x, y, z) be a general point on the plane. a A(x1, y1, z1) R(x, y, z) r n ^ z x y O Figure 6.21 A and R are two points lying on the plane. So, →AR also lies on the plane. ^ n is a unit vector normal to the plane. \ ^ n →AR From the scalar product of two perpendicular vectors, →AR · ^ n = 0 ( →OR – →OA) · ^ n = 0 (r – a) · ^ n = 0 r · ^ n – a · ^ n = 0 r · ^ n = a · ^ n a · ^ n is a scalar. Let a · ^ n = d (a scalar) ⇒ r · ^ n = d
Mathematics Term 1 STPM Chapter 6 Vectors 6 251 This is the vector equation of the plane, where r is the position vector of a general point on the plane, ^ n is a unit vector perpendicular to the plane, and d is the distance of the plane from the origin. In terms of cartesian components; r = xi + yj + zk a = x1 i + y1 j + z1 k Let n = ai + bj + gk where n is a vector normal to the plane. Vector equation of the plane, r · n = a · n (xi + yj + zk) · (ai + bj + gk) = (x1 i + y1 j + z1 k) · (ai + bj + gk) ax + by + gz = ax1 + by1 + gz1 ax + by + gz = c (where c = ax1 + by1 + gz1 ) This is the cartesian equation of a plane with n = ai + bj + gk as a normal to the plane. Example 20 Find the vector equation and cartesian equation of the plane that passes through P(1, 3, –2) and is perpendicular to the vector i – j + 2k. Solution: Given that vector normal n = i – j + 2k r · n = a · n r ~ · (i – j + 2k) = (i + 3j – 2k) · (i – j + 2k) r ~ · (i – j + 2k) = 1 – 3 – 4 r ~ · (i – j + 2k) = –6 (vector equation of the plane) Let r = xi + yj + zk, (xi + yj + zk) · (i – j + 2k) = –6 x – y + 2z = –6 (Cartesian equation of the plane) Example 21 Find the vector equation of the plane that passes through three points P(1, –3, 2), Q(0, 4, –1) and R(–2, 0, 3). Deduce the cartesian equation of the plane. Solution: →PQ = →OQ – →OP = (4j – k) – (i – 3j + 2k) →PQ = –i + 7j – 3k →PR = →OR – →OP = (–2i + 3k) – (i – 3j + 2k) →PR = –3i + 3j + k →PQ and →PR are two vectors on the plane. →PQ × →PR is a vector normal to the plane. →PQ × →PR = u i –1 –3 j 7 3 k –3 1 u = (7 + 9)i – (–1 – 9)j + (–3 + 21)k →PQ × →PR = 16i + 10j + 18k
252 Mathematics Term 1 STPM Chapter 6 Vectors 6 The vector equation for the plane is r · (16i + 10j + 18k) = (i – 3j + 2k) · (16i + 10j + 18k) r · (16i + 10j + 18k) = 16 – 30 + 36 r · (16i + 10j + 18k) = 22 The cartesian equation of the plane is 16x + 10y + 18z = 22 Exercise 6.5 1. Find the vector equation and cartesian equation of the plane that passes through P(–1, 3, 1) and is perpendicular to the vector n = i – j – 3k. 2. Find the vector equation and cartesian equation of the plane that passes through A(–3, 4, 1) and is parallel to the plane x – 2y – z = 2. 3. Find the cartesian equation of the plane with vector equation r · (i –3j – 5k) = 7. 4. Find the vector equation and the cartesian equation of the plane that passes through three points, P(1, –2, 0), Q(–1, 0, 3) and R(4, 2, 1). 5. Find a vector equation of the plane containing the three points A, B and C whose position vectors are 2i + j – k, 3i + j + k and 5i – 2j + 3k respectively. 6. A plane contains a point A whose position vector is 2i – 3j – k and is perpendicular to the vector i + 2j + 2k. Find the vector and cartesian equation of the plane. 7. Find the vector and cartesian equation of the plane that is perpendicular to 2i – 3j + k and contains the point with position vector i – 3j – 2k. 8. Find the vector equation of the plane that passes through the origin and perpendicular to the line r = (2 + l)i + (1 – 3l)j + (3 + 2l)k. Hence, find the cartesian equation of the plane. Angle between two straight lines Let L1 and L2 be two straight lines in space having equations r = a1 + l1b1 and r = a2 + l2b2 respectively. The line L1 is in the direction of b1 and L2 is in the direction of b2 . Since the angle between any two lines is defined as the angle between their direction vectors, the angle between L1 and L2 is the angle between vectors b1 and b2 . b1 b2 z x O y Figure 6.22
Mathematics Term 1 STPM Chapter 6 Vectors 6 253 From the scalar product of two vectors, b1 · b2 = |b1 ||b2 | cos q cos q = b1 · b2 |b1||b2| In terms of cartesian components, L1 : r = x1 i + y1 j + z1 k + l1 (a1 i + b1 j + c1 k) L2 : r = x2 i + y2 j + z2 k + l2 (a2 i + b2 j + c2 k) The angle q between the two lines is the angle between the two direction vectors, (a1i + b1j + c1k) and (a2 i + b2 j + c2 k). From the scalar product of two vectors, (a1 i + b1 j + c1k) · (a2 i + b2 j + c2 k) = |a1 i + b1 j + c1k||a2 i + b2 j + c2 k| cos q cos q = a1 a2 + b1 b2 + c1 c2 (a1 2 + b1 2 + c1 2 · (a2 2 + b2 2 + c2 2 Example 22 Find the angle between the lines r = 2i + j + l1 (i + j – 3k) and r = –i – j + l2 (5i – j + k). Solution: A vector parallel to L1 is i + j – 3k and a vector parallel to L2 is 5i – j + k. Let q be the acute angle between lines L1 and L2 , cos q = (i + j – 3k) · (5i – j + k) [12 + 12 + (–3)2 ] · [52 + (–1)2 + (1)2 ] cos q = 5 – 1 – 3 11 · 27 \ q = 86°40´ Example 23 Find the angle between the lines x + 1 2 = y + 5 2 = z + 3 –1 and x – 4 1 = y + 1 2 = z – 3 3 . Solution: L1 : r = –i – 5j – 3k + l1 (2i + 2j – k) L2 : r = 4i – j + 3k + l2 (i + 2j + 3k) A vector parallel to L1 is 2i + 2j – k and a vector parallel to L2 is i + 2j + 3k.
254 Mathematics Term 1 STPM Chapter 6 Vectors 6 Let q be the angle between lines L1 and L2 , cos q = (2i + 2j – k) · (i + 2j + 3k) 22 + 22 + (–1)2 · 12 + 22 + 32 cos q = 2 + 4 – 3 9 · 14 q = 74°30´ Angle between a line and a plane n b q Figure 6.23 Given that the line r = a + lb and the plane r · n = d. The angle between the line and the plane is given by q where sin q = b · n |b||n| . Example 24 Find the acute angle between the line x + 1 4 = y – 2 = z – 3 –1 and the plane 3x – 5y + 4z = 6. Solution: Vector equation of the line is r = –i + 2j + 3k + l(4i + j – k). Vector equation of the plane is r · (3i – 5j + 4k) = 6. The line is parallel to 4i + j – k and the plane is perpendicular to 3i – 5j + 4k. If q is the angle between the line and the plane, (4i + j – k) · (3i – 5j + 4k) = |4i + j – k||3i – 5j + 4k | cos (90° – q) 3 = 18 · 50 sin q sin q = 0.1 \ q = 5°44´ Angle between two planes The angle between two planes, with normal vectors n1 and n2 respectively, is given by the angle between the two normal vectors, n1 and n2 .
Mathematics Term 1 STPM Chapter 6 Vectors 6 255 n2 n1 Plane 2 Plane 1 Figure 6.24 Angle between the two planes = Angle between their respective normal vectors From the scalar product of two vectors, n1 · n2 = |n1 ||n2 | cos q cos q = n1 · n2 |n1||n2| Example 25 Find the acute angle between the planes 2x + 4y – 2z + 7 = 0 and 3x – y + z – 8 = 0. Solution: The angle between two planes is the angle between the two corresponding normal vectors n1 and n2 . Vector normal to the plane 2x + 4y – 2z + 7 = 0 is n1 = 2i + 4j – 2k. Vector normal to the plane 3x – y + z – 8 = 0 is n2 = 3i – j + k. Let the angle between the two planes be q, then cos q = n1 · n2 |n1||n2| = (2i + 4j – 2k) · (3i – j + k) 22 + 42 + (–2)2 · 32 + (–1)2 + 12 cos q = 6 – 4 – 2 24 · 11 = 0 q = 90°
256 Mathematics Term 1 STPM Chapter 6 Vectors 6 Exercise 6.6 1. Find the angles between the lines, L1 and L2 . (a) L1 : r = 2i + j + l1 (i + j) (b) L1 : r = 2i + j – 3k + l1 (i – j + k) L2 : r = –3i – j + l2 (2i – j) L1 : r = 2i + j – 3k + l2 (–i + j – k) 2. Find the acute angle between the lines r = 3i – 2j + l(i + j + k) and x – 3 2 = y – 2 –3 = z – 1 4 . 3. Find the acute angle between the line x + 5 2 , 3 – y 2 = z + 4 and the z-axis. 4. Find the acute angle between the straight lines x + 2 = y – 3 2 = z + 5 –1 and x – 4 = y = z + 1 2 . 5. Find the acute angle between the line x – 3 2 = y – 1 3 = z + 4 and the plane 2x + 4y – z = 1. 6. Find the acute angle between the line x + 1 = y = z – 3 2 and the plane 2x + y + 3z – 5 = 0. 7. Find the acute angle between the planes r · (i – j + k) = 1 and r · (2i – j + 3k) = 5. 8. Find the angle between the planes 6x + 2y + 5z = 1 and x – 4y + 3z = 6. 9. Find the acute angle between the planes 2x + 4y – 2z + 5 = 0 and r · (3i – j + k) = 4. 10. Find the acute angle between the planes x – 2y + 3z + 1 = 0 and 2x + y + 3z + 1 = 0. Point of intersection of two lines Example 26 The line L1 and L2 have cartesian equations x – 2 3 = y – 1 2 = z – 3 and x + 3 4 = y – 1 = z 2 respectively. Show that L1 and L2 intersect, and find the coordinates of the point of intersection. Solution: If L1 and L2 intersect, there is a common point which lies on both lines. Let (x1 , y1 , z1 ) be the common point, ⇒ x1 – 2 3 = y1 – 1 2 = z1 – 3 ……… 1 x1 + 3 4 = y1 – 1 = z1 2 ……… 2 For equation 1, x1 – 2 3 = y1 – 1 2 i.e. 2x1 – 4 = 3y1 – 3 ……… 3 For equation 2, x1 + 3 4 = y1 – 1 i.e. x1 + 3 = 4y1 – 4 ……… 4
Mathematics Term 1 STPM Chapter 6 Vectors 6 257 From 4, x1 = 4y1 – 7 Substitute into equation 3, 2(4y1 – 7) – 4 = 3y1 – 3 8y1 – 14 – 4 = 3y1 – 3 5y1 = 15 y1 = 3 x1 = 4(3) – 7 = 5 From 2, z1 = 2y1 – 2 = 2(3) – 2 = 4 Thus, the lines L1 and L2 do intersect. The point of intersection has coordinates (5, 3, 4). Example 27 Find the point of intersection of the lines r = (i + 7j – 4k) + l1 (i + 3k) and r = (j – k) + l2 (i + 2j + k). Solution: Rearrange in terms of i, j and k, L1 : r = (1 + l1 )i + 7j + (–4 + 3l1 )k L2 : r = l2 i + (1 + 2l2 )j + (–1 + l2 )k At the intersection point, L1 = L2 \ 1 + l1 = l2 ; 7 = 1 + 2l2 and –4 + 3l1 = –1 + l2 From 1 + 2l2 = 7 2l2 = 6 l2 = 3 ⇒ 1 + l1 = 3 l1 = 2 Substitute the values of l1 and l2 into L1 or L2 r = 3i + 7j + 2k So the point of intersection has coordinates (3, 7, 2). Point of intersection between a line and a plane Example 28 Find the point of intersection between the line x 2 = y – 1 = z + 5 2 and the plane 3x – 2y + z = 5. Solution: Let x 2 = y – 1 = z + 5 2 = l ⇒ x = 2l y = l + 1 z = 2l – 5
258 Mathematics Term 1 STPM Chapter 6 Vectors 6 Substitute these parametric equations into equation of the plane 3x – 2y + z = 5 3(2l) – 2(l + 1) + (2l – 5) = 5 6l – 2l + 2l – 2 – 5 = 5 6l = 12 \ l = 2 when l = 2; x = 2(2) = 4 y = 2 + 1 = 3 z = 2(2) – 5 = –1 So the point of intersection has coordinates (4, 3, –1). Example 29 Find the coordinates of point where the line x + 1 4 = y + 1 2 = z + 3 5 cuts the plane r · (i – j + 3k) = 8. Solution: Cartesian equation of the plane is x – y + 3z = 8. Let x + 1 4 = y + 1 2 = z + 3 5 = l so x = 4l – 1 y = 2l – 1 z = 5l – 3 Substitute these parametric equation into the cartesian equation of the plane x – y + 3z = 8 (4l – 1) – (2l – 1) + 3(5l – 3) = 8 4l – 2l + 15l – 1 + 1 – 9 = 8 17l = 17 l = 1 with l = 1; x = 4(1) – 1 = 3 y = 2(1) – 1 = 1 z = 5(1) – 3 = 2 So the point of intersection has coordinates (3, 1, 2). Line of intersection of two planes Two non-parallel planes intersect. The intersection of two non-parallel planes is a line. The line of intersection is perpendicular to both vectors n1 and n2 normal to the planes. The line of intersection is also parallel to n1 × n2 .
Mathematics Term 1 STPM Chapter 6 Vectors 6 259 n2 n1 n1 fi n2 Plane 1 Plane 2 Figure 6.25 Example 30 Find, in both cartesian and vector form, the equation of the line of intersection of the planes 2x – y – z = 5 and x – 2y + z = 1. Solution: Eliminate z from the two equations, 3x – 3y = 6 x – y = 2 x = y + 2 Eliminate y from the two equations, 3x – 3z = 9 x = 9 + 3z 3 = z + 3 Cartesian equation of the line of intersection is x = y + 2 = z + 3 Vector equation of the line of intersection is r = –2j – 3k + l(i + j + k)
260 Mathematics Term 1 STPM Chapter 6 Vectors 6 Exercise 6.7 1. The lines L1 and L2 have cartesian equations x + 1 = y – 2 –2 = z + 2 5 and x – 3 2 = y + 1 = z – 1 –4 respectively. Show that L1 and L2 intersect, and find the coordinates of the point of intersection. 2. Lines L1 and L2 have cartesian equations x – 2 = y + 3 4 = z – 5 2 and x – 1 –1 = y – 8 = z – 18 3 respectively. Show that L1 and L2 intersect, and find the position vector of the point of intersection. 3. Find the point of intersection of lines L1 and L2 . (a) L1 : r = –2j + 8k + l1 (i – j + k) L2 : r = i + 3j + k + l2 (–2i – j + 2k) (b) L1 : r = i + 3j – k + l1 (2i – 8j + 5k) L2 : r = i – j + 2k + l2 (–i + 2j – k) 4. Lines L1 and L2 have vector equations r = (j – k) + l1 (i – 2j – k) and r (i + 2j – 3k) + l2 (i – cj) respectively. Given that lines L1 and L2 intersect, find (a) l1 and l2 , (b) the value of c, (c) the coordinates of the point of intersection. 5. Find the point of intersection between the line x + 3 2 = y + 7 = z + 5 3 and the plane 3x + 2y + z = 16. 6. Determine the coordinates of the point where the line x + 2 –1 = y – 1 2 = z – 4 cuts the plane 3x – y + 2z = 10. 7. Determine the coordinates of the point where the line x + 1 = y – 2 4 = z – 3 cuts the plane r · (2i – j + 3k) = 8. 8. Find, in both cartesian and vector form, the equation of the line of intersection of the planes 2x – 3y – z = 1 and 3x + 4y + 2z = 3. 9. Find the cartesian equation of the line of intersection of the planes 2x – y – z = 4 and 3x – 2y + z = 0. 10. Find the cartesian equation of the line of intersection of the planes x + y – z = 3 and 3x + 4y = 1.
Mathematics Term 1 STPM Chapter 6 Vectors 6 261 Summary 1. The magnitude of a vector xi + yj + zk is x2 + y2 + z2 . 2. b1 x y z 2 = 1 bx by bz 2 where l and b are two scalars. 3. 1 x1 y1 z1 2 + 1 x2 y2 z2 2 = 1 x1 + x2 y1 + y2 z1 + z2 2 4. The scalar product of two non-zero vectors a and b is a · b = | a||b| cos q, where q is the angle between a and b. 5. Properties of the scalar product: (a) a · b = b · a (b) a · (b + c) = a · b + a · c 6. If p = x1 i + y1 j + z1k and q ~ = x2 i + y2 j + z2k, then p · q = x1 x2 + y1 y2 + z1 z2 7. The vector product of two vectors a and b is a × b = (|a||b| sin q)n ^ , where q is the acute angle between a and b, and n ^ is the unit vector perpendicular to both a and b. 8. Properties of the vector product. (a) a × b = –b × a (b) a × (b + c) = (a × b) + (a × c) 9. If a = x1 i + y1 j + z1 k and b = x2 i + y2 j + z2 k, then a × b = u i x1 x2 j y1 y2 k z1 z2 u 10. Area of a parallelogram with a and b as adjacent sides = |a × b|. O C A B a a b b
262 Mathematics Term 1 STPM Chapter 6 Vectors 6 11. Equation of a straight line which passes through a known point A(x1 , y1 , z1 ) with vector position a and is parallel to a vector b. (a) Vector equation: r = a + lb (b) Parametric equations: x = x1 + al y = y1 + bl z = z1 + c l (c) Cartesian equation: x – x1 a = y – y1 b = z – z1 c 12. The equation of a straight line which passes through two known points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) is r = a + l (b – a). 13. Equation of a plane with A(x1 , y1 , z1 ), a known point with position vector a lying on the plane, n a vector normal to the plane and R(x, y, z) a general point with position vector r on the plane. (a) Vector equation: r · n = a · n, where a = x1 i + y1 j + z1 k. (b) Cartesian equation: ax + by + gz = 0, where n = ai + bj + gk. 14. The angle between two straight lines r = a1 + l1 b1 and r = a2 + l2 b2 is the angle between b1 and b2 . If the angle is q, then cos q = b1 · b2 |b1||b2| . 15. The angle q between the line r = a + lb and the plane r · n = a · n is given by sin q = b · n |b||n| . 16. The angle between the planes r · n1 = a · n1 and r · n2 = a · n2 is given by the angle between the two normal vectors n1 and n2 . If the angle is q, then cos q = n1 · n2 |n1||n2| . STPM PRACTICE 6 1. Given that u = –i – j – 2k and v = 2i – 3j – 5k, find (a) |2u – 3v| (b) |–4u + v| 2. Find the displacement vector →PQ in terms of i, j and k, if (a) P = (3, –1, –2), Q = (2, –5, –4) (b) P = (0, 6, –9), Q = (–5, –1, 2) 3. Find a unit vector in the same direction as the given vector. (a) 10i – 2j – 2k (b) –i – j + 3k 4. Find a vector with the given magnitude and in the same direction as the given vector. (a) magnitude 2, v = 2i + 2j – 2k (b) magnitude 5, w = –3i – j + k 5. Find a · b if (a) a = 3i – j – 4k, b = –2i – 2j + k (b) a = i – 3j – 2k, b = 2i – k
Mathematics Term 1 STPM Chapter 6 Vectors 6 263 6. Find the angle between the vectors (a) (3, 2, 1) and (–1, 1, 2) (b) (4, –3, 2) and (–1, 5, 1) 7. A tetrahedron has vertices at A(0, 2, 1), B(1, 3, 1), C(3, 0, 1) and D(2, 1, 0). Calculate, to the nearest 0.1°, the angle (a) ∠BAC (b) ∠BCD 8. The line r = 1 1 8 –12 + λ1 2 p 12 is perpendicular to the plane r = 1 3 0 –32 + s 1 1 –1 q 2 + t 1 0 1 –12, where p and q are constants and λ, s, t R. (a) Determine the values of p and q. (b) Find the position vector of the point of intersection of the line and the plane. 9. A pyramid has a square base OPQR and vertex V. With respect to an origin O, the position vector of P, Q, R and V are →OP = 2i, →OQ = 2i + 2j, →OR = 2j and →OV = i + j + 3k. Find the acute angle between →VQ and the plane OVP. 10. P R O k j i A B F E G D C Q The diagram shows a cube OABCDEFG in which the length of each side is 12 units. The unit vectors, i, j and k are in the direction of →OA, →OC and →OD respectively. P and Q are the midpoints of →OA and →DG respectively. R is the center of square ABFE. (a) Express each of the vectors →PQ and →PR in terms of i, j and k. (b) Find, to the nearest 0.1°, the acute angle QPR. 11. A tetrahedron OPQR has base OPQ and vertex R. Given that →OP = i – 3j + 2k, →OQ = 3i – j – 6k and →OR = 5i + 3j + 2k, where O is the origin. (a) Show that →OR is perpendicular to both →OP and →OQ. (b) Calculate the angle between the edge PR and base OPQ of the tetrahedron. (c) Calculate the area of the base OPQ. Hence, deduce the volume of the tetrahedron.
264 Mathematics Term 1 STPM Chapter 6 Vectors 6 12. Determine the vector product a × b if (a) a = i – 2j + k and b = 2i + k (b) a = 2j + k and b = 4i + 2j – k 13. Find two unit vectors orthogonal to a = 2i + k and b = –i + 2j – k. 14. Find the area of the parallelogram with adjacent edges formed by 2i + k and j – 3k. 15. The position vectors of points X, Y and Z relative to the origin are i – 3j + k, 5i – 2j – 2k and 4i – j + 6k respectively. The points X, Y and Z are the vertices of a triangle XYZ. (a) Find the vector which is perpendicular to the vectors →XY and →XZ. (b) Calculate the area of the triangle XYZ. 16. Referred to an origin O, points A and B have position vectors given respectively by →OA = i + 2j – 2k and →OB = 2i – 3j + 6k. The point P on →AB is such that AP : PB = l : 1 – l. Show the →OP = (1 + l)i + (2 – 5l)j + (–2 + 8l)k. Find the value of l for which (a) →OP is perpendicular to →AB, (b) angles AOP and POB are equal. 17. The position vectors of the points A, B and C are given by a = 2i + 3j – 4k, b = 5i – j + 2k and c = 11i + lj + 14k. Find (a) the unit vector parallel to →AB, (b) the position vector of the point D such that ABCD is a parallelogram, (c) the value of l if A, B and C are collinear. 18. Three vectors u = ai + bj, v = i – 5j and w = 3i – 4j are such that u and v are perpendicular and the scalar product of u and w is 22. (a) Determine the value of a and b. (b) Find the angle between u and w. 19. Two lines l 1 and l 2 have equations r = 1 1 0 –22 + a1 4 –1 1 2 and r = 1 1 3 –12 – b 1 5 –2 1 2 respectively, where a and b are parameters. (a) Show that the two lines l 1 and l 2 intersect. (b) Find the Cartesian equation of the plane which contains the lines l 1 and l 2 . 20. Determine whether the line x – 2 3 = y – 1 = 4 – z lies in the plane x – 2y + z = 4 or not. 21. The position vectors of four points, P, Q, R and S relative to the origin are 2i – j – 4k, 5i + j – 3k, 2j – 2k and –3i – 3k respectively. Point T divides the line PR in the ratio 3 : 5. (a) Show that PQRS is a parallelogram. (b) Calculate the area of the parallelogram PQRS. (c) Find the position vector of T. (d) Determine the value of cos /PTQ. 22. Show that the point P(1, –1, 0) lies on both planes x – y + 5z = 2 and 3x – 2y + z = 5. Hence, find the vector equation of the line of intersection of both planes.
Mathematics Term 1 STPM Chapter 6 Vectors 6 265 23. Find the point of intersection of the lines x – 1 2 = y – 2 3 = z – 3 4 and x – 2 = y – 4 2 = z + 1 –4 . 24. Find the acute angle between the planes 2x + 2y – z = 3 and x + 2y + z = 1. 25. The line l 1 which is parallel to the vector 2i – j + 2k, passes through the point A whose position vector is 4i + 6j – 8k. The line l 2 which is parallel to the vector i + 7j – 2k, passes through the point B whose position vector is –6i + 26j + 12k. The point P on l 1 and the point Q on l 2 are such that PQ is perpendicular to both l 1 and l 2 . Find the position vectors of P and Q. 26. The lines L1 and L2 have equations r = 4i + j – 2k + l1 (i + 2j – 3k) and r = 5i – 3j – k + l2 (3i – 5k) respectively. Show that L1 and L2 intersect. Find the point of intersection. 27. Given that point A has position vector 30i – 3j – 5k. The line l passes through origin O is parallel to the vector 4i – 5j – 3k. The point B on l is such that AB is perpendicular to l. Find (a) the length of →AB, (b) the position vector of B. 28. Diagram below shows a parallelepiped with four rectangular sides, OPQR, LMNS, PQNM and ORSL. →OP and →OR are two vectors parallel to the unit vectors i and j respectively and the unit vector k is perpendicular to the plane OPQR. S N R Q L M O P Given that →OP = 3i, →OQ = 3i + 2j and →OL = i + 5k. (a) Determine the value of cos /QMS. (b) Calculate the area of the triangle PMS. (c) Find the vector equation of the plane PMS. 29. A plane passes through three points, P(1, 1, –3), Q(0, 2, 0) and R(–1, –2, 1). Line r = 1 5 –4 –12 + λ 1 1 3 1 2, when λ R, meets the plane at point S. (a) Find the Cartesian equation of the plane. (b) Determine the coordinates of point S. (c) Calculate the acute angle between the plane and the line.
266 STPM Model Paper (954/1) MATHEMATICS (T) PAPER 1 (One and a half hours) Instruction to candidates: Answer all questions in Section A and any one question in Section B. Answers may be written in either English or Bahasa Malaysia. All necessary working should be shown clearly. Scientific calculators may be used. Programmable and graphic display calculators are prohibited. Section A [45 marks] Answer all questions in this section. 1. The functions f and g are defined as follows: f : x ↦ x2 – 6x, x < 3 g : x ↦ 4 – e–x , x ∈ R (a) Show that f –1 exists. Find f –1 and state its domain and range. [4 marks] (b) Show that composite function fg does not exist. Find the maximum domain of g such that fg exists. [3 marks] 2. Expand 1 + x 1 – 3x as a series in ascending powers of x, up to and including the term in x2 . State the range of x for which the expansion is valid. By taking x = 1 9 , use the first three non-zero terms of the series to estimate 15, giving your answer as a fraction in its lowest form. [6 marks] 3. A hyperbola has an equation of 27x2 – 9y2 + 54x + 72y – 360 = 0. (a) Express the equation of the hyperbola in the standard form. [3 marks] (b) Determine the centre, vertices, foci and asymptotes of the hyperbola. [4 marks] (c) Sketch the graph of the hyperbola. Label its centre, vertices, foci and asymptotes on the graph. [2 marks] 4. Find the inverse matrix of 1 3 –2 5 1 0 2 2 –1 42 by using the elementary row operations. [4 marks] Hence, solve the system of linear equations 3x – 2y + 5z = 45, x + 2z = 15, 2x – y + 4z = 32. [2 marks]
267 Mathematics Term 1 STPM Model Paper (954/1) 5. Express 3 + i in the polar form. [3 marks] Use de Moivre’s theorem, show that ( 3 + i)12 + ( 3 – i)12 = 213. [3 marks] Hence, find the roots of 2z3 – [( 3 + i)12 + ( 3 – i)12]i = 0 in the Cartesian form. [3 marks] 6. Relative to the origin O, the position vectors of A, B, C and D are i + 4j – 2k, 5i + 5j – k, 4i +3j + k and 3i – 6j + 2k respectively. (a) Find the parametric equations of the straight line L1 which passes through the point D and perpendicular to the plane containing the points A, B and C. [3 marks] (b) Another straight line L2 passes through the points E(5, 1, –3) and F(2, 2, 3), (i) show that the lines L1 and L2 intersect and find the coordinates of intersection point between the lines L1 and L2 . [3 marks] (ii) calculate the acute angle between the lines L1 and L2 . [2 marks] Section B [15 marks] Answer any one question in this section. 7. The polynomial p(x) = ax3 – (1 + 6a)x2 + 2(8a – b)x – 27 where a and b are real constants, has a factor (x – 3). (a) Show that 7a – 2b = 12. [2 marks] (b) Write p(x) in terms of a and x. Find the quotient in terms of a and x when p(x) is divided by (x – 3). [3 marks] (c) When p(x) is divided by (x – 4), it leaves a remainder of 13, find the value of a. [2 marks] (d) With this value of a, (i) show that p(x) = 0 has only one real root. Find the set values of x for which p(x) . 0. [4 marks] (ii) express 4x – 9 p(x) in partial fractions. [4 marks] 8. The diagram below shows a model of pillar. The rectangular top QRST is parallel to the horizontal rectangular base MPON. The rectangular vertical plane ONTS is perpendicular to the base MPON and the top QRST. S T R Q P M N O The unit vectors i, j and k are parallel to →ON, →OP and →OS respectively, where O is the origin. Given that SR : NM = 3 : 5 and the vectors →ON, →OS and →OR are 2i, 4k and 3j + 4k respectively. (a) Find the position vector of M. [2 marks] (b) Calculate the area of the triangle RTM. [4 marks] (c) Calculate the angle between TR and TM. [2 marks] (d) Find the Cartesian equation of the plane RTM. [3 marks] (e) Calculate the acute angle between the plane RTM and the plane OTM. [4 marks]
268 Chapter 1 Functions Exercise 1.1 1. (a) –9 (b) 7 (c) 0 (d) 0 2. (a) (i) 3 (ii) 1 (iii) 0 (b) (i) 24 (ii) 99 3. (a) [–2, 6] or –2 < f(x) < 6 f(x ) x 0 6 –1 3 –2 (b) [–1, 3] or –1 < f(x) < 3 f(x ) x 0 3 –1 3 –1 (c) 3– 9 4 , 44 or – 9 4 < f(x) < 4 f(x ) x 0 4 3 _9 4 –2 – 4. (a) No (b) Yes (c) No 5. (a) 1, 4, 1 2 , 64 (b) 4 (c) R+ 6. x + 2 7. 4 – x2 8. (a) cos (1 + x) (b) cos (cos x) (c) 1 + cos x 9. (a) 1 – x2 , –1 , x , 1 (b) 1 – x (c) 2x2 – x4 10. (a) 4x2 – 6x + 1 (b) 2x2 + 6x – 1 (c) 4x – 9 (d) x4 + 6x3 + 14x2 + 15x + 5 11. (a) 6 (b) 25 (c) 2a2 – 4a + 9 (d) 4x2 – 12x + 14 (e) 2x2 + 4x + 9 (f) x4 + 10x2 + 30 Exercise 1.2 1. (b) c, d (c) {1, 2, 3, 4, 5} (d) {a, b, c, d} (e) No 2. (a) –2 –1 0 1 2 3 1 7 X Y (b) {–2, –1, 0, 1, 2} (c) {1, 3, 7} (d) No 3. (a) 1 –5 –3 –1 3 5 7 11 15 3 5 7 9 X Y (b) –1, 15 (c) Yes 4. No. f(0) = f(1) = 0, 1 2 5. a 1 + a2 ; {g : 0 , y < 1 2 } 6. (a) f –1 : x ⟼ x + 2, x ∈ R (b) f –1 : x ⟼ x – 1 , x ∈ R, x > 1 (c) f –1 : x ⟼ x + 1, x > 0 (d) g–1 : x ⟼ 3 + x + 1 , x > –1 (e) g–1 : x ⟼ x + 9 , x . –9 (f) h–1 : x ⟼ 2 + 3x x , x ≠ 0 (g) h–1 : x ⟼ 2(x + 1) x – 1 , x ≠ 1 7. f –1(x) = 1 – x 2x – 1 , 1 2 , x < 1 9. (a) f(x ) x f 0 f –1 (b) f(x ) x f 0 f –1 (c) f(x ) x f 0 f –1 (d) f(x ) x f 0 f –1 ANSWERS
Mathematics Term 1 STPM Answers 269 Exercise 1.3 1. (a) 3 – –4 0 3 x y (b) 0 8 8 – 3 x y (c) 0 8 6 – –5 x y 2. (a) 0 3 – 2 x y (b) 0 fiff 1 – – 3 fiff 1 – 3 –1 x y (c) 0 2 x y 5 – 8 57 – 16 3. (a) 0 –1 x y 1 (b) 0 3 3 1 x y 1 –– fiff (c) 0 10 3 5 x y fiff 4. (a) y x –1 0 2 (b) y x 0 –1 –2 2 (c) x 0 -8 2 y (d) (e) y x 0 1 2 5. (a) -1 0 1 y x (b) 6. (a) -1 0 y x (b) x y 0 _ 2 _ 1 _ 2 _ 1 (c) –1 0 x y y x 0 –fi 2 fi 2 -1 0 1 y x
270 Mathematics Term 1 STPM Answers 7. (a) x y –1 0 1 (b) x y –1 –1 0 1 8. (a) x y 0 1 1_ 2 (b) x y 0 1 2 2 (c) x y –1 0 4 3 (d) x y 0 1 –4 5_ 2 – 9. 1 , x , 3 –1 0 1 3 x y 5 y = |3x – 5| y = |x + 1| 3 – 10. x , 4 – 10, 3 – 3 , x , 3 + 3, x . 4 + 10 0 6 1 6 x y y = |x| y = |x 2 – 7x + 6| 7_ 2 11. –4 –2 0 2 4 1 3 5 6 x y 12. –2 0 2 5 1 –2 –5 3 5 4 x y 13. 0 5 1 7 – – 3 3 – – 2 1 – 2 3 – 2 3 3 2 4 x y 14. 0 –4 4 –3 –2 –1 2 3 6 5 2 8 x y Exercise 1.4 1. (a) 1 (b) 2 1 2 (c) 13 2. (a) –7 (b) –2 5 8 (c) 11 3. (a) x2 + 5x + 1 (b) x2 – x – 2 (c) 5x2 + 4x – 3 (d) x3 + 2x2 – 1 (e) 2x3 + 3x2 – x – 1 4. (a) A = 1, B = 2 (b) A = 3, B = 4 (c) A = 3, B = –1 (d) A = –2, B = 5 3 , C = 7 3 (e) A = 3 2 , B = 3 2 , C = –2 5. (a) 2x3 + 3x2 + 5x + 14 (b) 6x4 – 13x3 + 16x2 – 17x + 3 (c) 4x4 + 11x3 – 10x2 + 5x – 2 (d) 5x5 – 7x3 + 3x2 + 2x – 3 (e) 9x5 – 4x3 + 18x2 – 8 6. (a) 3x – 5 ; 10 (b) 2x2 + x + 4 ; 13 (c) x2 – 3x + 9 ; –30 (d) x4 – 2x2 + 3x + 1 ; 0 (e) x2 – x ; 5x – 5 7. a = 3, b = –5 8. A = 3, B = 1 9. x2 + x – 1 10. a = 2, b = 0, c = 3
Mathematics Term 1 STPM Answers 271 Exercise 1.5 1. (a) 20 (b) 35 (c) 6 (d) 2 1 8 (e) 1 3 8 (f) –6 1 27 2. (a) Yes (b) No (c) Yes (d) Yes (e) Yes (f) No 3. (a) (x – 1)2 (2x + 1) (b) (x + 1)(x – 2)(3x + 1) (c) (x + 3)(x – 3)(x2 + 8) (d) x(1 – x + x2 )(1 + x + x2 ) (e) (x + 2)(2x – 1)(2x – 3) (f) (x + 1)(x – 2)(2x + 1)(2x – 1) 4. – 8 9 6. a = 2 3 , b = 7 3 7. p = 3, q = –6 8. a = –5, b = 7 9. a = 3, b = –5 10. 3, –2 11. k = 6, 0, –6 12. a = 3, 3(x – 3)(x + 1)2 ; a = –3, –3(x + 3)(x – 1)2 13. 1, –1 ± 3 2 14. 2; 1 2 , –1, –3 15. a = 3, b = –9, x = –1, 2, 2 16. a = 10, b = 1; 4 Exercise 1.6 1. (a) –1 < x < 2 (b) x , –5, x . 3 (c) –4 , x , 3 2 (d) x < – 1 2 , x > 1 4 (e) –10 < x < 3 (f) x , – 2 5 , x . 2 2. (a) {x : x < –3 or x > 3} (b) {x : x ∈ R, x ≠ –1} (c) {x : –3 < x < –2} (d) {x : – 2 5 < x < 1} (e) {x : –1 , x , 1 2 } (f) {x : x < – 1 2 , x > 3} 3. (a) x , –3, –2 , x , 1 (b) x < –1, x = 2 (c) x > 1, x = –2 (d) x , –2, – 1 2 , x , 1 (e) x < – 4 5 , 3 2 < x < 4 (f) –2 < x < 1 2 , x > 5 4. (a) –3 , x , –2, x . 1 (b) –1 , x , 1 2 (c) x < – 3 2 , 2 , x < 4 (d) x , –2, – 1 2 , x , 1, x . 3 (e) x , –3, x . 4, x = 1 (f) x , 1 2 (7 – 3 5 ), 1 2 , x , 2, x . 1 2 (7 + 3 5 ) 5. (a) 1 , x , 3 (b) x < –2, x > 8 (c) x , –3, x . 1 3 (d) –3 < x < 8 (e) x > 1 2 (f) 1 , x , 7 3 (g) –6 < x < 0 (h) x , – 5 6 , x . 5 2 (i) x . – 3 5 (j) x , –8, x . – 8 3 (k) –4 < x < –1, 1 < x < 4 (l) x , 0 Exercise 1.7 1. 2 x + 1 2. 2 x + 2 – 1 x – 1 3. 1 2(x – 3) – 1 2(x + 3) 4. 1 2(x + 1) + 1 2(x – 1) 5. 1 4(x + 2) + 3 4(x – 2) 6. 1 3(x + 4) + 2 3(x + 1) 7. 1 2 + x + 2 3 + x – 3 4 + x 8. – 2 x + 2 + 1 x – 1 + 1 x + 1 9. 7 24(x – 3) + 7 8(x + 1) – 2 3x 10. 3 2(x + 1) + 5 6(x + 3) – 4 3x 11. 1 1 + x – 1 (1 + x) 2 12. 1 x – 1 – 1 x + 2 – 3 (x + 2)2 13. – 1 3(2x – 1) + 5 3(x – 2) + 4 (x – 2)2 14. 2 x – 1 x2 – 3 2x + 1 15. – 1 x2 + 1 2(x – 1) – 1 2(x + 1) 16. 1 + 1 x – 1 – 1 x + 1 17. 1 – 1 x + 1 – 1 x + 2 18. 2 – 4 x + 2 + 1 x + 1 19. x – 2 + 4 x + 1 – 1 (x + 1)2 20. x + 2 x – 1 – 1 x + 1 21. x x2 + 1 – 1 x + 2 22. 1 x – x x2 + x + 1 23. 1 x – 1 + 1 – x x2 + x + 1 24. 1 x – 1 – x x2 – x + 1 25. 1 8(x + 1) – 1 8(x – 1) + 1 4(x – 1)2 + 1 2(x – 1)2 Exercise 1.8 1. (a) 0 y x 1 (b) 0 y x 1
272 Mathematics Term 1 STPM Answers (c) 0 –1 y x (d) 0 y x 1 2. (a) 0 y x y = –1 (b) 0 y x y = 1 (c) 0 1 y x y = –1 (d) 0 2 y x y = 1 3. (a) 0 1 y x (b) –1 0 y x (c) 0 2 y x x = 1 4. (a) 0 1 y x 1_ 2 (b) –1 0 1 y x (c) 0 y x x = –1 x = 1 – 2 2 5. (a) 0 2 y x (b) 0 1 1 y x (c) 0 1 1 y x –e (d) 0 1 e y x Exercise 1.9 1. (a) 1 (b) 6 (c) 1 7 (d) 1 27 (e) 8 (f) 1 000 (g) 8 (h) 6 (i) 4 (j) 6 (k) 3 (l) 2 2. (a) 5 2 (b) 3 2 (c) 5 – 2 6 (d) 66 – 24 6 (e) 18 (f) 2 5 (g) 7 (h) 13 3. (a) 2 5 (b) 1 3 6 (c) 2 – 2 (d) 5( 3 – 2 ) (e) 1 17 (2 15 + 3) (f) 1 2 (1 + 2 )(1 + 3 ) (g) 1 2 ( 3 + 2 2 )( 5 + 3 ) (h) 1 6 (4 – 3 3 )( 6 + 2 3 ) (i) 2 2 (j) 4 4. (a) 4 (b) 1 3 (c) – 1 2 (d) 1 2 (e) – 2 3 (f) –3 (g) 3 2 (h) –2 5. (a) log3 1 4 3 2 (b) 2 log2 3 (c) 0 (d) log3 1 3 4 2 (e) – 1 3 (f) –log (x – 1) (g) log |(x – 1)—1 2 (x – 2)2 | (h) log 1 p(p – q) q2 2 6. (a) 5 (b) 2 (c) 3 Exercise 1.10 1. (a) 3 (b) 2 (c) 3 4 (d) –2 (e) 4 3 (f) 5 2 2. (a) –5 (b) 1 (c) –1 (d) 2 3. (a) 0, 3 (b) 0, 2 (c) 1, 2 (d) –1, 2 (e) 1, 2 (f) ± 1 2 4. (a) 1 + 2a + b (b) 5a – b (c) a a + b 5. (a) 1 + 2p (b) 1 p (c) 1 2 p (d) – 1 p
Mathematics Term 1 STPM Answers 273 6. (a) 22.627 (b) 0.463 (c) 11.212 (d) 1.035 (e) 0.499 (f) 3.824 7. (a) 1.683 (b) –3.822 (c) –0.523 (d) –0.339 (e) –1.795 8. (a) 4 (b) 3 (c) 9, 1 9 (d) 2, 64 (e) 1 2 , 64 (f) 5, 25 9. (a) x , 3.170 (b) x . –0.183 (c) x . 8.638 (d) x . 1.853 (e) x . 6.261 10. (a) x , 1, x . 1.585 (b) x , 1.099 Exercise 1.11 1. (a) (b) y x 0 3 180 360 –3 y x 0 1 180 360 –1 (c) y x 0 1 180 360 –1 30 2. (a) 2π y x 0 1 –1 (b) y x 0 1 –1 2π (c) y x 0 2 –2 2π (d) y x 0 1 –1 2π (e) y x 0 3 –3 2π (f) y x 0 1 2π (g) y x 0 2 2π (h) 2π y x 0 1 –1 3. (a) (b) _ _ y x 0 3π π 4 π 2 4 π _ _ y x 0 π 3π π 4 π 2 4 (c) _ π y x 0 π _ 3π π 4 2 4 Exercise 1.12 1. (a) 3 5 , 3 4 , 5 4 (b) 8 17 , – 17 15 , – 15 8 (c) – 24 25, – 7 25 , – 25 24 (d) 5 13 , – 12 13 , 12 5 2. (a) cos q (b) tan q (c) sin q (d) tan4 q (e) sec2 q (f) tan q 3. (a) sin4 q (b) cos q + sin q (c) 1 + cos q (d) sec q – sin q Exercise 1.13 1. (a) 1 4 ( 6 + 2 ) (b) 1 4 ( 2 – 6 ) (c) 3 – 2 (d) 2 – 3 2. (a) 1 2 (b) 1 2 (c) 0 (d) 1 3. (a) 3 (b) 1 3 (c) 1 2 6 (d) 2 4. 56 65 , – 33 65 5. – 29 35 , – 8 6 35 6. 1, 225° 7. 4 1 2
274 Mathematics Term 1 STPM Answers 9. (a) sin 3x + sin x (b) cos 8q + cos 2q (c) 3(cos x – cos 5x) (d) 1 2 (sin 6q – sin 2q) (e) cos 2A + cos 2B (f) 1 4 (1 + 2 sin 2q) 10. (a) 2 sin 3x cos 2x (b) 2 cos 5A cos 2A (c) –2 cos 4x sin x (d) 2 sin 3A sin 2A (e) 2 sin 45° cos 15° (f) 2 cos 60° cos 10° (g) 2 cos x sin y (h) –2 sin 2A sin A 11. (a) cos x (b) 0 12. (a) cot 5q (b) –tan 3q (c) cot 2x (d) –tan2 x Exercise 1.14 1. (a) 23°35´, 156°25´ (b) 60°57´, 240°57´ (c) 138°35´, 221°25´ (d) 226°3´, 313°57´ 2. (a) –49°28´, 49°28´ (b) 56°6´, 123°54´ (c) –37°36´, 142°24´ (d) –111°36´, 111°36´ 3. (a) 36°52´, 216°52´ (b) 135°, 315° (c) 48°35´, 131°25´ (d) 48°11´, 311°49´ 4. (a) 60°, 120°, 240°, 300° (b) 60°, 90°, 270°, 300° (c) 0°, 78°28´, 180°, 281°32´, 360° (d) 270° (e) 0°, 30°, 150°, 180°, 360° (f) 56°19´, 90°, 236°19´, 270° 5. (a) – 3 4 π, –0.148π, π 4 , 0.852π (b) –0.920π, –0.080π (c) – π 3 , 0, π 3 (d) –0.621π, – π 4 , 0.379π, 3 4 π 6. (a) 41°25´, 120°, 240°, 318°35´ (b) 60°, 300° (c) 70°32´, 120°, 240°, 289°28´ (d) 19°28´, 30°, 150°, 160°32´ (e) 18°26´, 135°, 198°26´, 315° (f) 45°, 75°58´, 225°, 255°58´ 7. (a) 2 3 π, 4 3 π (b) 0.232π, 0.768π, 3 2 π (c) π 4 , 3 4 π, 5 4 π, 7 4 π (d) 0, 0.580π, π, 1.420π, 2π Exercise 1.15 1. (a) 135°, 161°34´, 315°, 341°34´ (b) 38°58´, 162°50, 218°58´, 342°50´ 2. (a) 2 cos (q + 45°); max. 2, q = 315°, min. – 2, q = 135° (b) 2 cos (q + 30°); max. 2, q = 330°; min. –2, q = 150° (c) 13 cos (q – 67°23´); max. 13, q = 67°23´; min. –13, q = 247°23´ (d) 5 cos (q – 26°34´); max. 5, q = 26°34´; min. – 5 , q = 206°34´ 3. (a) 74 sin (q – 0.951) (b) 3 5 cos (q – 0.464) (c) 5 5 cos (q + 0.464) (d) 65 sin (q + 0.519) 4. (a) 26°10´, 110°14´ (b) 41°36´; 244°40´ (c) 180°, 313°36´ (d) 97°58´, 205°54´ 6. (a) 10°54´, 231°1´ (b) 257°36´, 349°48´ 7. r = 5, s = 6, a = 36°52´ Exercise 1.16 1. –π , x , – 3 4 π, 1 4 π , x , π 2. – π 2 , x , 0.212π, 0.788π , x , π 3. –π , x , –0.732π, 0.268π , x , π 4. – 11 12 π , x , 7 12 π, – 5 12 π , x , – 1 12 π, 1 12 π , x , 5 12 π, 7 12 π , x , 11 12 π 5. –0.885π , x , –0.615π, –0.385π , x , –0.115π, 0.115π , x , 0.385π, 0.615π , x , 0.885π 6. – 1 4 π , x , 0, 1 4 π , x , π 7. –π , x , –0.580π, 0.580π , x , π 8. –π , x , – π 2 , 0 , x , π 2 STPM PRACTICE 1 1. f –1 : x ⟼ 1 x – 3 4 2 —1 3 ; graph of f–1 is the reflection of the graph of f about y = x. 2. fg : x ⟼ –ln x, f–1 : x ⟼ e x (a) graph of fg is the reflection of the graph of f about the x-axis. (b) graph of f–1 is the reflection of the graph of f about y = x 3. (a) x y 0 2 –2 6 –3 –2 –1 1 2 3 Range of f is [–2, 6] (b) g–1 : x ⟼ x – 4 – 1 4. (a) f –1(x) = 1 ± x – 1 (b) 4 3 < x < 3 2 (c) No 5. f is a one-to-one function [–1, 1], [– 1 2 π, 1 2 π], 1 6 π, 1 2 , 1 6 π 6. (a) g : x ⟼ x2 + 1 (b) g : x ⟼ 1 x 2 + 1 7. (a) f : x ⟼ (x + 1)2 (b) f : x ⟼ x2 – 1 8. f : No, [–1, 1] g : No, [1 2 ] h : Yes, [–∞, ∞] 9. (a) f –1(x) = x – 1 (b) g–1(x) = (x + 3)—1 3 y x f 0 f –1 y x g 0 g–1
Mathematics Term 1 STPM Answers 275 (c) h–1(x) = log2 x (d) k–1(x) = 10x y x h 0 h–1 y x k 0 k–1 10. (a) 0 1 2 f(x) = ex + 1 y = f(x) x 0 2 –4 y = h(x) h(x) = x2 – 4 x (b) (i) Any horizontal lines y = k cuts the graphs f(x) and h(x) only at one point, f(x) and h(x) are one-toone functions, therefore f–1 and h–1 exist. (ii) f –1 : x ↦ ln (x – 1) Domain f–1(x): {x : x . 1} h–1 : x ↦ x + 4 Domain h–1(x): {x : x > –4} (c) fh(x) = ex2 – 4 + 1 Domain f ° h: {x : x > 0} Range f ° h: {y : y > 1 + e–4} (d) g(x) = h ° f(x) = hf(x) 11. (a) (b) x y 0 1 –1 2 4 x y 0 1 –1 2 4 (c) (d) y x 0 1 2 2 4 y x 0 1 –1 2 4 (e) y x 0 1 –1 –2 2 12. (a) (i) 1 (ii) 1 (c) 2 3 < y < 2 (d) –1, 1 (e) y x 0 2 –1 1 13. (a) (b) y x 0 3 3 (2,–1) 1 y x 0 x = 1 x = 3 (2,–1) 14. (a) (b) y x x = –a x = a ( ) 0, – 1 a2 – y x x = –a x = a ( ) 0, – 1 a2 0 (c) 0 y x x = –a x = a 15. (a) a = 3, b = 1 (b) (x2 – 4)(3x2 + x + 6) + 5x – 2 (c) f(x) = 31x + 1 6 2 2 + 71 12 (i) Since 1x + 1 6 2 2 . 0 for ∀x ∈ R, therefore f(x) is always positive for all real values of x. (ii) Minimum value of f(x) = 71 12 when x = – 1 6 (d) {x : x , –2, x . 2, x ∈ R} 16. a = 7, b = 1; f(x) = (x – 1)(x + 1)2 (3x + 4); 1– 4 3 , –12 < (–1, 1) y 0 –1 1 y = f(x) –4 x 4 – – 3 y 0 –1 1 4 x 4 – – 3 y = |f(x)| 17. (a) 0 1+ 2 y = 2 – (1 – x) 2 (1, 2) 1 –1 In y = 2ex – 1 y = g(x) x 1 2 __
276 Mathematics Term 1 STPM Answers (b) {y : y < 2} (c) Any horizontal lines y = k for –1 , k , 2 will cut the graph at two points, therefore g is not a one-to-one function. 18. x , – 1 2 , x . 1 y = |2x – 1| x y —1 2 0 y = 1 |x| 19. x = – 1 3 , 1 5 ; x , – 1 3 , x . 1 5 y = 4|x | x y 0 1 y = |x–1| 1 – – 3 1– 5 20. A = 2, B = –3, C = 1 21. 3x 3 + 4x2 – 2x + 5 22. 1 1 – x + 2 2x + 1 – 3 (2x + 1)2 23. 1 x – 1 – 1 2(x – 1)2 – 2x + 1 2(x2 + 1) 24. 1 x – 3 – 1 x + 2 + 2 (x + 2)2 25. 14, x + 1 27. a = 4, b = –20 28. 2 – x 29. k = –7, (2x – 1)(x – 1)(x + 2)(2x + 3) 30. a = 2, b = 2 31. a = 5, b = 10, c = 0; x , –5 or x . 3 33. a = 1, x = 2, 4; a = –3, x = 2, 4 9 36. 3, –1 37. k = 2, a = 10 38. (a) k = 2; 1 2 , –1, –3 (b) –1, 1, 4, –1 39. a = 4, b = 3; x = –1, 2; {x : x , –1 or x . 2} 40. k . 1 41. (a) – 3 4 , x , 3 (b) – 3 , x , 1 or x . 3 43. p + q2 r + 2q pr 45. (a) –8 , x , 0 (b) –3 < x < – 5 7 (c) –1 , x , 0, 3 , x , 4 (d) x , –2, x . 2 46. 0 2 2 3 – 2 __ y = (2 – 3x) y = –(2 – 3x) y = 2 – x2 (0, 2) (1, 1) x y {x : x < 0 or x > 1} 47. (a) a = 1 4 (b) 0 1 2 1 2 h(x) = In(x – 1) y = x 1 4 _ f(x) = 1 + e4x x y 48. –1.14 49. –1, 1 50. 1.09 51. –2, 1 52. 1.31 55. (a) –2, –9 (b) 6 (c) –2 (d) 16 9 56. 3 57. x = 1 or x = 4 58. 3, 81 60. 101 61. a 2 , 3a 62. 88 64. (a) (b) y x 0 y x (c) y x –π 0 π π – 4 3– –π 4 65. 0 y x y x 0
Mathematics Term 1 STPM Answers 277 66. y x 0 1 –1 5 90 180 y = 5 sin x y = 1 – 2 sin 2x 6.5°, 143.5°, 366.5°, 503.5° 67. r = 13, a = 22.6° (a) 13, 337.4°, –13, 157.4° (b) y x 0 –13 13 –22.6 90 180 270 360 (c) 85.3°, 229.5° 68. y x y = cos 2x 1 0 __ 6 π __ 3 π __ 2 π __ 3 2π __ 6 5π __ 6 7π __ 3 4π __ 2 3π __ 3 5π __ 2π 6 π 11π –1 1 2 __ 1 2 __ – {x : 0 < x , π 6 , π 3 , x , 2π 3 , 5π 6 , x , 7π 6 , 4π 3 , x , 5π 3 , 11π 6 , x < 2π} 69. (a) 2 (b) m , – 1 2 70. 8 71. 15 cos (x – 0.6435) (a) min. –15, x = 3.7851 rad; max. 15, x = 0.6435 rad (b) x = 2.2143, 5.3559 (c) x y 0 0.6435 10 12 15 –15 y = 12 cos x + 9 sin x 2.2143 3.7851 5.3559 2π 1.485 6.086 {x : 1.485 < x < 2.214, 5.356 < x < 6.086} 72. tan 1 π 4 – x 2 2, 2 + 1, 2 – 3 73. (a) 1, 1 2 (b) 8 15 74. (a) 0°, 54°44´, 125°16´, 180°, 234°44´, 305°16´, 360° (b) 257°35´, 349°47´ 75. (a) 0°, 63°26´, 116°34´, 180°, 243°26´, 296°34´, 360° (b) 45°, 60°, 135°, 225°, 300°, 315° 76. 323 325 , – 36 325 77. 18.4°, 45°, 198.4°, 225° 78. 4 sin 1q + π 3 2 (a) min. –4, q = 7π 6 ; max. 4, q = π 6 (b) 0 2 –2 –4 4 __ 6 π __ 3 π __ π 2π 2 π __ 3 2π __ 6 5π __ 6 7π __ 3 4π __ 2 3π __ 3 5π __ 6 11π 2 3 y = 2 sin θ + 12 cos θ y x (c) q = π 2 , 5π 6 , 3π 2 , 11π 6 ; 5q : π 2 , q , 5π 6 , 3π 2 , q , 11π 6 6 79. 90°, 306.87° Chapter 2 Sequences and Series Exercise 2.1 1. (a) 25, 30; 5n (b) 16, 19; 3n + 1 (c) 14, 17; 3n – 1 (d) 32, 64; 2n (e) 3 625 , – 3 3 125 ; (–1)n – 1 3 5n – 1 (f) 9 11, 11 13 ; 2n – 1 2n + 1 (g) 625 16 , 3125 32 ; ( 5 2 ) n – 1 (h) 1 30 , 1 42 ; 1 n(n + 1) (i) 5, –6; (–1)n – 1n 2. (a) 1 2 , 2 3 , 3 4 , 4 5 (b) 1 1! , – 1 2! , 1 3! , – 1 4! (c) 1 12 , – 2x 32 , 4x2 52 , – 8x3 72 (d) – x 1.3 , x3 3.5 , – x5 5.7 , x7 7.9 (e) cos x x , cos 2x 2x2 , cos 3x 3x3 , cos 4x 4x4 3. (a) (–1)n (2n – 1) 3n + 2 (b) 1 – (–1)n 2 (c) n + 3 n + 5 . 1 – (–1)n 2 4. (a) 3, 2, 2, 3, 8, 35, … (b) 1, 3, 5, 11, 21, 43, … (c) 1, 3, 7, 15, 31, 63, 127, … 5. (a) Convergent, 1 (b) Divergent (c) Convergent, 2 (d) Convergent, 3 6. (a) 0 (b) 0 (c) ∞ (d) 1 (e) 2 5 (f) 2 3 7. (a) 10 ∑ r=1 r 3 (b) 50 ∑ r=1 (2r – 1) (c) 9 ∑ r=0 1 2r (d) 6 ∑ r=1 (–1)r – 11 1 3 2 r – 1 (e) 7 ∑ r=1 (14 – 3r) (f) 6 ∑ r=1 (3 + 2r ) (g) 10 ∑ r=1 (–1)r rxr (h) 8 ∑ r=1 r (r + 1)(r + 2)
278 Mathematics Term 1 STPM Answers 8. (a) 0 + 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90 (b) 1 + 1 4 + 1 9 + 1 16 + 1 25 + 1 36 + 1 49 + 1 64 (c) 5 + 8 + 11 + 14 +17 (d) 3 – 6 + 9 – 12 + 15 – 18 + 21 – 24 + 27 (e) 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 – 1 – 2 (f) 15 + 24 + 35 + 48 + 63 + 80 + 99 (g) 25 + 16 + 9 + 4 + 1 (h) –6 – 3 + 4 + 15 + 30 Exercise 2.2 1. (a) 31, 4n + 3 (b) –17, 25 – 7n (c) 31, 2n – 9 (d) 12 1 3 , 1 3 (2n + 7) 2. (a) 860 (b) 1225 (c) 437 (d) 291 3. (a) 819 (b) 745 (c) 54 (d) 120 4. 470 5. 9072 6. 16 734 7. 6 1 4 , 2600 8. 13; 22 9. 6, 11, 16 10. 4, 6 11. 7 12. 64 13. (a) 15 (b) –12 (c) 5 27 (d) log 9 Exercise 2.3 1. (a) 64, 2n – 2 (b) 2 3 , 1621 1 3 2 n – 1 (c) 25 32 , 2001– 1 4 2 n – 1 (d) – 81 16 , –1 3 2 2 n – 3 2. (a) 9, 127 3 4 (b) 9, 5 394 729 (c) 7, 198 7 16 (d) 8, –7 1541 2187 3. (a) 124 3124 3125 (b) 2 85 128 (c) 1 2 [1 – (–3)n ] (d) ak [a2n – 1] a2 – 1 4. 3 4 5. m = –5, n = 2 1 2 6. (a) 9 (b) 1 9 (c) 1 × 1015 7. p = 13 8. –53 1 3 ; –385 5 6 9. –40; –8 10. 1 24 (3n – 1); 8 11. 10 Exercise 2.4 1. (a) Yes (b) No (c) Yes (d) Yes (e) No (f) Yes 2. (a) 9 (b) 2 3 (c) 11 1 9 (d) 27 3. (a) 16 33 (b) 8 111 (c) 968 1665 (d) 1117 3330 (e) 9209 9990 4. (a) 1 1 – 2x ; |x| , 1 2 (b) 2 2 + x ; |x| , 2 (c) 3 1 + 2x ; |x| , 1 2 (d) 3x 3 – x ; |x| , 3 5. 4 3 , 4 9 , 4 27 , 2 6. 1 3 7. 13 m 8. 1 3 5 , 9 9. 243; 24331 – 1 5 9 2 n 4; 8 10. 1 + 1 4 + 1 16 + 1 64 ; all values of r, 1 + r 2 Exercise 2.5 1. (a) 4 + 14 + 36 + 76; 130 (b) 0 – 21 – 44; –65 (c) 17 + 21 + 25 + 29 + 33 + 37; 162 (d) 90 + 45 + 30 + 45 2 + 18; 205 1 2 (e) 3 2 + 3 2 + 0 – 3 2 – 3 2 + 0 + 3 2 + 3 2 ; 3 (f) 3 – 5 + 9 – 17 + 33 – 65; –42 2. (a) 1 2 n(n3 + 2n2 + 1) (b) 1 12 n(n + 1)(3n2 + 11n + 4) (c) 1 2 n(n3 + 8n2 + 17n – 2) (d) 1 4 n(n3 + 14n2 + 65n + 112) 3. (a) r(3r + 1); n(n + 1)2 (b) r 2 (r + 4); 1 12 n(n + 1)(3n2 + 19n + 8) (c) r(r + 2); 1 6 n(n + 1)(2n + 7) (d) (3r – 2)(3r + 1)(3r + 4); 1 12[(3n – 2)(3n + 1)(3n + 4)(3n + 7) + 56] (e) 1 r(r + 1)(r + 2) ; 1 2 3 1 2 – 1 (n + 1)(n + 2)4 (f) r(r + 1)(r + 2); 1 4 n(n + 1)(n + 2)(n + 3) 4. 22 100, 20 825 5. 88 200 6. (a) 1 2 n(6n2 + 3n – 1) (b) 1 3 n(2n + 1)(6n2 + 31n + 37) (c) 1 6 n(28n2 + 57n + 29) (d) 3 4 n(3n + 1)(9n2 + 3n + 2) 7. 1 4 n2 (n + 1)2 8. (a) n 2(n + 2) (b) 2n 3(n + 3) (c) 11n3 + 48n2 + 49n 18(n + 1)(n + 2)(n + 3) (d) n(3n + 1) 4(n + 1)(n + 2) 9. (a) 1 3 n(48n3 + 80n2 – 6n – 47) (b) 5n3 + 30n2 + 37n 36(n + 1)(n + 2)(n + 3) 10. r (r + 1)! , 1 – 1 (n + 1)!
Mathematics Term 1 STPM Answers 279 11. Ur = 6r – 1; 9n2 + 2n 12. 1 + 2r r 2 (r + 1)2 , n(n + 2) (n + 1)2 13. A = 1, B = –3, C = 2 Exercise 2.6 1. (a) 120 (b) 40 320 (c) 3 628 800 (d) 336 (e) 720 (f) 4896 (g) 285 (h) 8008 (i) 1260 (j) 252 2. (a) 10! 7! (b) 12! 10! (c) 22! 19! (d) 18! 2! 15! 5! (e) 21! 5! 18! 8! (f) n! (n – 3)! (g) (n + 2)! (n – 2)! (h) (2n + 5)! (2n + 2)! (i) n! (n – 1)! 3! (j) (n – 5)! 2! (n – 9)! 5! 3. (a) 11(9!) (b) 15(7!) (c) 291(7!) (d) (n + 2)n! (e) (n2 + n – 1)(n – 1)! (f) n(n2 + n + 2)n! (g) (n + 1)2 (n – 1)! (h) 31(7!) 30(3! 2!) (i) 263(7!) 210(3! 2!) (j) (n – 1)! (r + 1)! (nr + n + 1) 4. (a) 70 (b) 120 (c) 1140 (d) 1 (e) 200 (f) 560 (g) 15 4 (h) 26 45 (i) 6188 (j) 84 Exercise 2.7 1. (a) p4 + 4p3 q + 6p2 q2 + 4pq3 + q4 (b) m7 – 7m6 n + 21m5 n2 – 35m4 n3 + 35m3 n4 – 21m2 n5 + 7mn6 – n7 (c) 64 + 192k 2 + 240k4 + 160k6 + 60k8 + 12k10 + k12 (d) 32a5 – 80a4 b + 80a3 b2 – 40a2 b3 + 10ab4 – b5 (e) x 3 + 3x + 3 x + 1 x3 (f) y 8 – 4y6 + 7y4 – 7y2 + 35 8 – 7 4y2 + 7 16y4 – 1 16y6 + 1 256y8 2. (a) 210x 4 (b) 16 128x2 (c) 1760 a3 b9 (d) –945p4 q6 (e) –20 (f) 36 x3 3. 64x 5 + 160 x + 20 x7 4. 14 5. 15 6. 2 7. n = 8, a = – 1 2 8. (a) 1 + 7x + 21x 2 + 35x3 (b) 1 + 7x + 14x 2 – 7x3 9. 1 + 16x + 136x 2 + 784x 3 10. 1 + 9x + 24x 2 11. (a) –16 464 (b) –151 164 (c) –1760 (d) 26 Exercise 2.8 1. (a) 1 – 2x + 3x 2 – 4x3 (b) 1 + x + x 2 + x3 (c) 1 – 6x + 24x 2 – 80x3 (d) 1 4 – 1 4 x + 3 16 x 2 – 1 8 x3 (e) 1 3 – 2 9 x + 4 27 x 2 – 8 81 x3 2. (a) 1 – 1 4 x + 3 32 x 2 – 5 128 x3 ; |x| , 2 (b) 1 – 2 3 x – 4 9 x 2 – 40 81 x3 ; |x| , 1 2 (c) 1 2 – 1 4 x + 1 8 x 2 – 1 16 x3 ; |x| , 2 (d) 1 – 1 2 x + 3 8 x 2 – 5 16 x3 ; |x| , 1 (e) 1 – 1 2 x – 5 8 x 2 + 3 16 x3 ; |x| , 1 (f) –1 + 2x – 2x 2 + 2x3 ; |x| , 1 (g) 2 + 4x + 33 4 x 2 + 81 4 x3 ; |x| , 1 3 (h) – 1 2 + 3 4 x – 13 8 x 2 + 51 16 x3 ; |x| , 1 2 (i) 1 – x + x 3 ; – 1 2 (1 + 5 ) , x , 1 2 ( 5 – 1) 3. 1.1255 4. 30.43168 5. 1 + 30x + 420x 2 + 3 640x3 ; 1.03042 6. 1 – x 2 + 3 2 x4 – 5 2 x 6 7. 1 – 1 2 x – 1 8 x 2 – 1 16 x3 ; 0.9487 8. 1.732 9. 3.1623 (5 sig.fig.) 10. 1 4 + 1 4 x + 3 16x 2 + 1 8 x3 + 5 64x4 ; 0.309 11. 1 + x – 1 2 x 2 + 1 2 x3 ; 2.236 12. 1 – 7x + 28x 2 – 112x 3 ; |x| , 1 4 13. 1 – 5x + 30x 3 – 15x 4 ; 0.950 030 STPM PRACTICE 2 1. n(n – 1); 15th 2. (a) 3, 4 1 2 , 4 1 6 , 4 + 1 n(n – 1), n . 1 (b) 2, 4, 18, n(n!), n . 1 3. (a) 250 (b) 510 (c) n(n + 1) (d) (n + 1)! – 1 6. n(n + 2) 3(2n + 1)(2n + 3) 7. b2 8. 11, 20 9. Explicit formula: un = 4n – 1 Recursive formula: un = un – 1 + 4 for n = 2, 3, 4, … and u1 = 3 or un + 1 = un + 4 for n = 1, 2, 3, 4, … and u1 = 3 10. (a) wn = 8n + 4 (b) First term = 12; Common difference = 8 (c) Sn = un + wn – 3 11. 11 12. (a) U2 = 231 + 1 3 5 24, U3 = 231 + 1 3 5 2 + 1 3 5 2 2 4, U4 = 231 + 1 3 5 2 + 1 3 5 2 2 + 1 3 5 2 3 4 (b) lim Ur = 5 r → ∞ 14. (a) (n + 1)(2n + 1) (b) –975 15. (a) mn (b) mn – m + 1; Sn = 1 2 m(2mn – m + 1) 16. (a) ur = 5 24r + 1
280 Mathematics Term 1 STPM Answers (b) Sn = 1 6 31 – 1 1 16 2 n 4; lim Sn = S∞ = 1 n → ∞ 6 17. (a) 138 (b) 1 9 (10n – 1); 1 81 (10n + 1 – 9n – 10) 18. (a) 676 2475 (b) 493 825 (c) 1168 1665 19. (b) un + 1 = 1 2 un (c) S∞ = lim n → ∞ Sn = 3 2 20. 5 2 x + 15 4 x 2 + 65 8 x3 21. (a) 752 22. 1 + apx + 1 2 a2 p(p – 1)x2 + 1 6 a3 p(p – 1)(p – 2)x3 ; – 3 4 , – 8 3 ; |x| , 4 3 . 23. n = 4; 1 – 1 2 x – 1 8 x 2 ; 1249 200 24. p = 3, q = 2, n = 8 25. 1 + x + 1 2 x 2 26. 1 6 1 1 n 2 4 , 0.98952 27. (a) 1 3 n(28n2 – 1) (b) 1 999 , 4 37 28. (a) 167 167; 111 445 (b) A = 6 13, B = 1; 6 13 29. (a) 1 + 1 4 x – 3 32x 2 (b) 20x 4 + 160x2 + 64; 1364 30. – 1 1 + x – 2 1 – x + 3 (1 – x) 2 ; 0, 5, 6; 3r + 1 + (–1)r + 1 31. (a) Sn – 1 = 2x2 3 (1 – xn – 1) 1 – x 4 (b) 15 ∑ r = 1 (2r – 1)3r = 602 654 097 14 ∑ r = 1 (2r + 1)3r + 2 = 1 807 962 291 Chapter 3 Matrices Exercise 3.1 1. (a) 1 3 –6 12 0 0 6 2 (b) 1 0 –3 1 –5 3 –12 (c) 1 1 1 3 5 –3 3 2 (d) 1 1 –5 5 –5 3 1 2 (e) 1 2 –13 11 –15 9 1 2 2. (a) 1 24 33 30 24 –6 18 24 –39 9 2 (b) 1 16 18 9 16 5 –2 16 –31 28 2 3. A2 = 1 3 –5 5 8 2 , AB = 1 2 –1 2 13 8 3 2 BC = 1 3 4 –3 16 2 –22 , BD = 1 –10 2 –1 2 2 C2 = 1 5 –2 –4 1 16 2 3 –5 –32 , CD = 1 5 11 –1 10 –1 –7 2 DA = 1 6 4 –7 –11 2 0 2 , DB = 1 1 2 –3 –17 –2 7 –5 6 –72 4. (a) 1 –4 10 2 9 2 (b) 1 –7 7 –4 12 2 (c) 1 9 1 –1 10 4 2 2 (d) 1 –8 8 –2 15 –4 6 2 5. x = 2, y = 4, z = 1, w = 3 7. a = 1, b = –3 8. (a) 1 50 0 10 –50 0 –10 100 0 20 2 (b) 1 0 0 0 –15 –15 30 –30 –30 60 2 (c) (57) (d) (5) (e) 1 12 2 –2 –6 –1 1 36 6 –62 (f) (17) 9. (a) 1 38 –16 27 –4 2 (b) 1 6 11 1 8 16 –1 16 24 12 2 10. (a) 1 –1 –10 0 –4 31 104 2 (b) 1 –1 2 –3 –4 8 –12 –12 24 –36 2 (c) 1 –11 –18 –11 –18 2 (d) 1 2 –5 –32 2 –5 –32 2 (e) 1 11 –22 33 4 –8 12 –135 270 –405 2 (f) 1 2 –5 –32 8 –20 –128 24 –60 –384 2 Exercise 3.2 1. (a) 9 (b) 2 (c) x2 + y2 – 2 2. (a) 0 (b) 0 (c) –14 3. (a) x(x – 1) (b) 8a(1 – 2a) (c) axy(2 – ay) 4. (a) –3 (b) 3 5 (c) 3 5. (a) –96 (b) 192 (c) 75 6. (a) 18 5 (b) 2 (c) – 1 2 7. (a) 0, –9 (b) 0, –2 (c) –3, 0, 2 Exercise 3.3 1. (a) 1 0 1 –1 1 2 (b) 1 –3 2 5 –3 2 (c) 1 7 –4 –5 3 2 (d) 1 3 7 –2 –5 2 (e) None (f) 1 –2 3 –1 2 2 (g) None (h) 1 —1 3 0 0 –—1 2 2 2. (a) 1 0 1 –1 4 2 (b) 1 3 2 1 1 2 (c) 1 3 0 –2 2 2 3. (a) A–1B (b) BA–1 (c) B (d) A–1BA
Mathematics Term 1 STPM Answers 281 5. 1 –3 4 4 –5 2 6. 1 –2 6 –2 5 2 7. 1 9 –25 –5 14 2; m = –5, n = 1; 1 3 5 1 2 2, 1 43 –120 –24 67 2 8. 1 8 0 0 0 8 0 0 0 8 2 , 1 8 1 5 –6 15 –7 10 –21 1 2 –52 10. (a) 1 8 1 4 –10 –2 8 2 –2 –2 1 1 2 (b) – 1 23 1 3 –2 –5 16 –3 4 –20 –2 –5 2 (c) inverse does not exist (d) 1 3 1 1 1 –7 –2 1 –1 0 0 3 2 (e) – 1 60 1 7 –29 13 –13 11 –7 9 –3 –92 11. (a) 1 –3 0 0 3 3 3 17 11 13 2 (b) 1 8 –3 1 13 –11 –5 5 12 16 2 (c) 1 6 1 –8 6 2 7 –3 –1 5 –3 1 2 (d) 1 3 1 –5 13 –3 –11 25 –6 10 –23 6 2 (e) 1 18 1 116 –60 –26 233 –123 –53 –211 111 49 2 (f) 1 6 1 –2 0 0 –4 13 –3 6 –11 3 2 Exercise 3.4 1. (a) x = 1, y = –2 (b) x = –2, y = 1 (c) x = 0, y = –1 (d) p = –1, q = –2 (e) r = 4, s = 7 (f) u = 2, v = –3 2. (a) (–1, 2) (b) None (c) (–1, 1) (d) (3, –1) 3. (a) x = 1 k , y = k + 1 k , k ≠ 0 (b) x = 2 k + 2 , y = – 1 k + 2 ; k ≠ ±2 4. (a) x = 1, y = –1, z = –1 (b) x = –1, y = 2, z = 2 (c) x = –15, y = 8, z = 2 5. (a) unique solution, (1, –2, –2) (b) no solution (c) infinitely many solutions 6. x = 3 4 , y = 5 4 , z = 1 2 7. AB = 1 6 0 0 0 6 0 0 0 6 2 (a) x = 1, y = 1, z = 2 (b) x = –1, y = –2, z = 2 8. A–1 = – 1 10 1 1 3 5 –2 4 0 –5 –5 –5 2 ; 1 x y z 2 = 1 1 –—5 2 8 –—5 –2 9. B–1 = 1 8 1 2 3 –5 2 7 –1 0 –4 4 2 ; 1 x y z 2 = 1 —3 2 2 —3 4 —7 4 10. (a) x = 3, y = 2, z = –1 (b) x = –2, y = 1, z = –1 (c) x = 3, y = 5, z = –2 (d) x = 3, y = 12, z = 9 STPM PRACTICE 3 2. (a) 1 4 –1 6 5 8 13 2 (b) 1 –3 1 2 3. (a) –2 1 7 5 – 3 5 (b) None (c) – 2 3 –1 5 3 2 (d) – 1 11 1 5 –3 –7 3 2 4. x = –4, y = 9 5. (a) 1 –2 –1 1 1 –4 –5 4 –7 –2 2 (b) Q–1 = 1 9 (Q2 + 2Q – 8I) = – 2 9 – 1 9 1 9 1 9 – 4 9 – 5 9 4 9 – 7 9 – 2 9 6. (b) det B = –4 det A = –4(p – q)(q – r)(p – r) 7. a + d, ad – bc 9. 1 4 4 –7 –2 –3 4 –1 –1 2 2 10. W–1 = 1 –1 –3 2 2 5 –4 –2 –5 3 2 ; x = 2, y = 7, z = –10 11. (a) a = 3, b ≠ 5 (b) a = 3, b = 5 (c) a ≠ 3, b = any real values (real numbers) 12. (a) 1 3 9 6 2 6 4 1 4 –2 21 x y z 2 = 1 1 a2 a 2 (b) augmented matrix 1 3 9 6 2 6 4 1 4 –2 1 a2 a 2 ; row-echelon form 1 3 0 0 2 0 0 1 1 0 1 a2 – 3 4a2 + a – 14 2 Since row 3 has all zero entries, the system of linear equations does not have a unique solution. (c) a = 7 4 , –2 ; x = t, y = – 3 2 t, z = 1 where t ∈ R (d) 5a : a ≠ 7 4 , a ≠ –2, a ∈ R6 13. (a) augmented matrix 1 1 2 3 3 5 m 1 m 13 n 2n 4n 2 ; row-echelon from 1 1 0 0 3 –1 0 1 m – 2 m2 – 11m + 28 n 0 n 2 (b) (i) m = 4, 7 and n ≠ 0 (ii) m ≠ 4, 7 and n ∈ R (c) m = 4, 7 and n = 0 ; x = –16t, y = 5t, z = t where t ∈ R
282 Mathematics Term 1 STPM Answers Chapter 4 Complex Numbers Exercise 4.1 1. (a) 2 + 2i (b) 9 – 5i (c) 2 + 2 2i (d) –3 – 7 2i (e) 4 + 4 5i 2. (a) 8 + 2i (b) 1 – 5i (c) 3 + 3i (d) –5 + 3i 3. (a) –4 + 3i (b) 4 + 2i (c) –5 – 12i (d) 11 + 3i (e) –1 + 2 2i (f) –3 + i 4. (a) 1 – i (b) 2 5 + 1 5 i (c) i (d) i (e) 18 13 – 1 13 i (f) 7 25 – 1 25 i 5. (a) 2 (b) 5 (c) – 3 5 – 4 5 i (d) –3 + 4i (e) –3 + 4i (f) 1 5 6. (a) 7 – 7i (b) 7 + 7i (c) 25i (d) 24 25 – 7 25 i (e) 24 25 + 7 25 i 7. (a) x = 9, y = –7 (b) x = – 3 2 , y = 7 2 (c) x = 3, y = –5 (d) x = 2, y = 4 (e) x = –5, y = 3 8. (a) ±(2 – i) (b) ±(4 + 3i) (c) ±(3 + 2i) (d) ±(3 – i) (e) ±(7 + 5i) 9. – 11 10 + 3 10 i 10. a = b = 5 2 11. z = 3 – 2i 12. z = 3 – 4i Exercise 4.2 1. (a) –1 ± 2i (b) 1 2 (3 ± 7i) (c) – 1 4 (7 ± 31i) (d) 1 2 (5 ± 3i) (e) 1 3 (2 ± 2i) 2. (a) z2 – 2 2z + 3 = 0 (b) z2 – 4z + 7 = 0 (c) z2 – 3z + 1 = 0 (d) 25z2 – 30z + 11 = 0 (e) 7z2 + 8z + 3 = 0 3. (a) z = 2, 1 ± i 2 (b) z = – 1 2 , –2 ± i 3 (c) z = 3, 3 2 ± i 2 2 (d) z = – 2 5 , 5 2 ± i 3 2 (e) z = 1 3 , 2 ± i 4. z2 + 4z + 2; z = – 1 2 , –2 ± i 2 5. (z2 – 2z + 5)(z2 + z + 1); 1 ± 2i, – 1 2 ± i 3 2 6. (a) ± 2 , ±2 2i (b) – 1 2 , 1, ±i (c) –3, 2i, 2i, –2i, –2i (d) i 3, –i 3, 2i, –2i, –2, 2 (e) ± 2, ±i 2, ±2i, ±2 (f) –2, ±1, 1 ± i 3 , 1 2 ± i 3 2 , – 1 2 ± i 3 2 Exercise 4.3 1. (a) y x 0 θ P(0, 3) (b) y x 0 θ P(–2, 1) |z| = 3, q = π 2 |z| = 5, q = 2.678 rad (c) y x 0 θ P(2, –4) (d) y x 0 θ P(5, 7) |z| = 2 5, |z| = 74, q = –1.107 rad q = 0.951 rad (e) y x 0 θ P(–4, –4) (f) y x 0 θ P(–4, 3) |z| = 4 2, q = – 3 4 π |z| = 5, q = 2.498 rad 2. (a) z = 2 3cos π 4 – i sin π 4 4 (b) z = 23cos π 6 + i sin π 6 4 (c) z = 5[cos(0.927) + i sin(0.927)] (d) z = 13[cos(–1.176) + i sin(–1.176)] (e) z = 7 [cos(–0.714) + i sin(–0.714)] (f) z = 43 [cos(0.867) + i sin(0.867)] 3. (a) z3 = 12 + 5i (b) |z3 | = 13, arg z3 = 0.395 rad, z3 = 13 [cos(0.395) + i sin(0.395) 4. (a) z = –1 + 3i (b) |z| = 10, arg z = –1.249 rad, z = 10 [cos(–1.249) + i sin(–1.249) 5. 8 6. 6 5 + 8 5 i, 2, 0.927, 2 [cos(0.927) + i sin(0.927)] 7. (a) 2, π 4 (b) 2 2 , 2 2 8. –0.977, 2.16 Exercise 4.4 1. (a) cos 2π 3 + i sin 2π 3 (b) cos 1– 4π 3 2 + i sin 1– 4π 3 2 (c) cos 5π 6 + i sin 5π 6 (d) cos 3π 2 – i sin 3π 2 2. (a) (cos q + i sin q) 7 (b) (cos q + i sin q) –3 (c) 1cos q 3 + i sin q 3 2 –5 (d) 1cos q 6 + i sin q 6 2 13
Mathematics Term 1 STPM Answers 283 3. (a) 2 3cos π 4 + i sin π 4 4, 4 + 4i (b) 2 3 3cos π 3 + i sin π 3 4; –144( 3 – 3i) (c) cos 1– π 6 2 + i sin 1– π 6 2, – 1 2 ( 3 + i) (d) 13 cos(2.4) + i sin(2.4)], 135 [cos(5.88) + i sin(5.88)] (e) 4 3cos 1– π 3 2 + i sin 1– π 3 24, 512(1 + 3i) (f) 23 [cos(0.485) + i sin(0.485)], 23—5 2 [cos(2.425) + i sin(2.425)] 4. (a) cos 23 12 q + i sin 23 12 q (b) cos 8q + i sin 8q (c) cos 5q + i sin 5q (d) cos 29 12 q + i sin 29 12 q (e) cos 11 12 q + i sin 11 12 q 5. 1 2 3cos π 4 + i sin π 4 4, i 32 6. (a) ±2—1 4 3cos π 8 + i sin π 8 4 (b) ± 5 [cos (–0.464) + i sin (–0.464)] (c) ± 13 [cos (–0.588) + i sin (–0.588)] (d) ± 2 3cos 1– π 122 + i sin 1– π 1224 (e) ±2—3 4 3cos 1– π 8 2 + i sin 1– π 8 24 (f) ±3—1 4 3cos π 8 + i sin π 8 4 7. (a) 2 —1 6 3cos 1 2k 3 + 1 122π + i sin 1 2k 3 + 1 122π4, k = 0, 1, 2 (b) 5 —1 3 3cos 1 2k 3 π – 0.3092 + i sin 1 2k 3 – 0.30924, k = 0, 1, 2 (c) 13—1 3 3cos 1 2k 3 π – 0.3922 + i sin 1 2k 3 π – 0.39224, k = 0, 1, 2 (d) 2 —1 3 3cos 1 2k 3 – 1 182π + i sin 1 2k 3 – 1 182π4, k = 0, 1, 2 (e) 2 —1 3 3cos 1 2k 3 – 1 122π + i sin 1 2k 3 – 1 122π4, k = 0, 1, 2 (f) 3 —1 6 3cos 1 2k 3 + 1 122π + i sin 1 2k 3 + 1 122π4, k = 0, 1, 2 8. |z| = 2, arg z = π 3 , zn = 2n 3cos nπ 3 + i sin nπ 3 4 9. z2 = 1 4 3cos 2 3 π + i sin 2 3 π4, z3 = 1 8 [cos π + i sin π] 10. 23cos π 6 + i sin π 6 4, 2 2 3cos 3 4 π + i sin 3 4 π4 z = 1 2 —1 6 3cos 1 2k 3 – 7 362π + i sin1 2k 3 – 7 362π4, k = 0, 1, 2 11. q = π 3 , π 4 STPM PRACTICE 4 1. (a) p = 7 5 , q = – 4 5 (b) p = 2 – i, q = 2 + i 3. (1, 2), (–1, –2) 4. |z1 | = 2 , arg z1 = – 1 4 π |z2 | = 2 2 , arg z2 = – 3 4 π |z3 | = 1, arg z3 = – π 3 5. ±(1 + 2i), ±(1 – 2i) 6. (a) 2 1cos π 4 + i sin π 4 2, 2 3cos 1– π 4 2 + i sin 1– π 4 24; 1 2 , π (b) 6 5 + 8 5 i; 2, 0.927 rad 7. ±1.23 rad. 8. sin 3 8 π, sin 5 8 π, sin 7 8 π, – 1 2 2 + 2 9. (a) a = 1 2 (3 + 21), b = 1 2 (–3 + 21) (b) (i) –1, – 3 (ii) 5 4 , π 12 10. 4 cos2 q – 2; ±i, 1 2 + 3 2 i, 1 2 – 3 2 i 11. 2 3cos π 6 + i sin π 6 4, 2 2 3cos 1– π 4 2 + i sin1– π 4 24 ; 1 2 —1 6 3cos 1 2k 3 + 5 362π + i sin 1 2k 3 + 5 362π4, k = 0, 1, 2 13. a = 5, b = –10, c = 1; 2 cos π 5 , 2 cos 2 5 π, 2 cos 3 5 π, 2 cos 4 5 π 14. (a) |z| = 4 ; arg z = – π 6 (c) 1 32(–1 + 3i) 15. 83cos 2π 3 + i sin 2π 3 4 ; 1.5321 + 1.2856i, –1.8794 + 0.6840i, 0.3473 – 1.9696i 16. (a) w = 1 + i ; real part = 1, imaginary part = 1 (b) w = 2 3cos π 4 + i sin π 4 4 (c) 1.0842 + 0.2905i, –0.7937 + 0.7937i, –0.2905 – 1.0842i 17. z = 4 2 3cos1– π 4 2 + i sin 1– π 4 24, w = 2 3 1cos 5π 6 + i sin 5π 6 4 ; w8 z6 = 81 128 3cos π 6 + i sin π 6 4 18. (a) 3 – i, 1 + i 3, – 3 + i, –1 – i 3 19. (a) real part = 32, imaginary part = –32 3 (b) z1 = 83cos 1– π 6 2 + i sin 1– π 6 24 = 4 3 – 4i z2 = 83cos 5π 6 + i sin 5π 6 4 = –4 3 + 4i Chapter 5 Analytic Geometry Exercise 5.1 1. 3x2 + 3y2 – 16y + 16 = 0 2. y2 = 3a(2x – 3a) 3. x2 + y2 + 4y = 0 4. 16x + 10y – 15 = 0 5. x2 + y2 – 5x = 0 6. (a) (y – 3)2 = x + 1 (b) 27y2 = 4x3 (c) (x – 4)y = 9 (d) y = (x – 1)(x – 2) (e) x2 = 12(y – 1) (f) (y – 1)2 = (x + 1)3 (g) y(x – 1)2 = 8 (h) x2 (y + 1) = 64
284 Mathematics Term 1 STPM Answers Exercise 5.2 1. (a) (b) 0 y x (2, 0) x = –2 0 y x (–3, 0) x = 3 (c) (d) 0 9 8 9 8 y x fi–, 0ff x = – – 0 2 3 2 3 y x fi– –, 0ff x = – 2. (a) (3, 0) (b) (1, –2) 0 y x (3, 0) 0 y x (1, –2) (c) (–1, 1) (d) (–4, 3) 0 y x (–1, 1) 0 y x (–4, 3) 3. (a) y2 = 8x (b) y2 = 6x (c) y2 = 12x (d) 3y2 = 20x 4. (a) x = 3 2 t 2 , y = 3t (b) x = 9 4 t 2 , y = 9 2 t (c) x = 15 8 t 2 , y = 15 4 t + 1 (d) x = 2 3 t 2 + 1, y = 4 3 t (e) x = 2(2t 2 + 1), y = 8t + 1 5. (a) y = x + 2; y + x = 6 (b) y = x + 3; y + x = 9 (c) 2y + 2x + 5 = 0; 2y = 2x – 15 (d) 2y = 2x + 3a; 2y + 2x = 9a 6. y = 3x + 3, 3y + x + 81 = 0, (–9, –24) 7. ty = x + t 2 , tx + y = t 3 + 21 ; (8, –4 2 ) 8. 8 2 a, 32a2 9. p + q = 1, pq = –2 ; 1 5 2 a, a2 10. x = –a 12. 2ax = y2 – y Exercise 5.3 1. (a) Major axis: 4 Minor axis: 8 3 0 (2, 0) x y 4 (0, —) 3 (b) 0 x y ( ) ( ) 3 ,0 4 2 3 0, 2 – Major axis: 3 Minor axis: 3 2 2 (c) 0 x y ( ) 3 0, 2 – ( ) 5 2 – ,0 Major axis: 5 Minor axis: 3 (d) 0 2 (2, 0) 2fi2 y x Major axis: 4 2 Minor axis: 4 (e) 0 3 (1, 3) x y 4 Major axis: 8 Minor axis: 6 (f) y x 0 2 3 3 2 (–, –2) Major axis: 6 Minor axis: 4 2. (a) x2 16 + y2 9 = 1 (b) x2 4 + (y – 1)2 9 = 1 (c) 4(x – 2)2 81 + (y – 1)2 18 = 1 (d) x2 4 + y2 3 = 1 , (x – 2)2 4 + y2 3 = 1
Mathematics Term 1 STPM Answers 285 (e) (x + 4)2 36 + (y – 1)2 20 = 1 , (x – 6)2 36 + (y – 1)2 20 = 1 3. (a) 4x2 + y2 = 4 (b) x2 16 + y2 9 = 1 (c) (x – 5)2 4 + (y – 2)2 9 = 1 (d) 4(x – 1)2 + (y + 1)2 = 4 4. (a) x = 2 + 3 cos q, y = 3 + 2 sin q (b) x = –2 + 2 cos q, y = 1 + 3 sin q (c) x = 2 + 2 cos q, y = –1 + sin q (d) x = 1 + 10 cos q, y = –1 + 6 sin q 5. (a) 2x + 3 3y = 12 (b) 5y = 3x + 15 2 (c) 3x + 4 3y + 24 = 0 6. (a) x + y = 5; y = x – 1 (b) x – 2y = 4; 2x + y = 3 (c) 9y = 2x – 5; 9x + 2y + 20 = 0 7. (a) (2, 1) (b) (3, 2) (c) (2, 6) (d) (2, 1) 9. 1 2 , –2; 2y = x + 8, (–6, 1); y + 2x = 14; (48 7 , 2 7 ) 10. 7m2 + 12m + 5 = 0; outside 11. y = x ± 13 , y + x = ± 13 12. y + x = ±5 13. 35°10´ 14. (x + 8)2 15 + y2 9 = 1 15. 3y + 2x = ± 43 Exercise 5.4 1. (a) 0 y x (–4, 0) (4, 0) 3 4 y = –x 3 4 y = – –x (b) 0 y x (–fi10, 0) (fi10, 0) 1 y = –x fi2 1 y = – –x fi2 (c) 0 x y (d) 0 x y (e) –2 0 x y y = 1 (f) 0 –1 x y x = 2 (g) 0 x y (h) 0 y x x = 1 y = 2 2. (a) (i) 5 2 , (± 5 , 0) (ii) x = ± 4 5 5 , y = ± 1 2 x (b) (i) 2 3 3 , (±4, 0) (ii) x = ±3, y = ± 3 3 x
286 Mathematics Term 1 STPM Answers (c) (i) 5 2 , (±4 5 , 0) (ii) x = ± 16 5 5 , y = ± 1 2 x (d) (i) 6 2 , (±3 6 , 0) (ii) x = ±2 6 , y = ± 2 2 x (e) (i) 13 3 , (1 ± 13, 0) (ii) x = 1 ± 9 13 , y = ± 2 3 (x – 1) (f) (i) 14 3 , (1 ± 70 , 2) (ii) x = 1 ± 3 14 70 , (y – 2) = ± 5 7 (x – 1) 3. (a) xy = 9 (b) 4xy = 25 (c) xy + 4 = 0 (d) 9xy + 4 = 0 (e) xy = y + 1 (f) xy – x + 16 = 0 4. (a) x = 4t, y = 4 t (b) x = 5 3 t, y = 5 3t (c) x = 1 4 t, y = – 1 4t (d) x = 8t, y = 8 t (e) x = 1 + 2t, y = 2 t (f) x = 2 + t, y = 1 t 5. (a) t 2 y + x = 6t ; ty – t 3 x = 3(1 – t 4 ) (b) 9y + 4x = 24 ; 12y – 27x + 65 = 0 (c) t 2 y + x = 3t ; 2ty – 2t 3 x = 3(1 – t 4 ) (d) 6y + x = 9 ; 4y – 24x + 105 = 0 6. (a) 4y + 3x + 8 3 = 0, 4y + 3x – 8 3 = 0 (b) y + 4x = 8 7. 1– 3 8 , – 242, 51 8 17 8. QR = 10 ; Q´R´ = 35 12 9. x + 3y = 9 10. 1 2 , –4 ; 1 16 7 , – 8 7 2 11. 1 4 k , 4k2 12. tan–1 1 35 132 or 69°37´ 14. p = 1 3 15. ± b2 + c 2 a , y = ± b a x. STPM PRACTICE 5 1. (a) (x + 2)2 16 + (y – 3)2 12 = 1; ellipse (b) Centre, C = (–2, 3); Foci, F1 = (–4, 3) and F2 = (0, 3) (c) 0 6 –6 –5 –4 –3 –2 –1 1 2 F1 (–4, 3) F2 C(–2, 3) (0, 3) y x 2. (a) (x – 9)2 a2 + (y + 12)2 b2 = 1 ; ellipse (b) The conic is a circle with centre (9, –12) and radius 15. 0 –24 18 (9, –12) y x 4. m2 x´ – my´ + a = 0 5. (a) (p + q)y = 2x + 8pq ; py = x + 4p2 (b) M(–4p2 , 0) (c) A(4, 0) 6. y x 0 2 4 x 2 + 4y 2= 16 x 2 + 4y 2= 2x 1, 1 2 – 4 –2 2 ( ) – 1, 1 2 ( ) – – 7. (2, 1), (–1, 3) 8. (a) (x – 1)2 9 + (y – 4)2 16 = 1 (b) C = (1, 4) ; F1 = (1, 4 – 7 ) ; F2 = (1, 4 + 7 ) (c) Vertices = (1, 0) and (1, 8) 0 –3 –2 –1 1 2 3 4 5 1 2 3 4 5 6 7 8 (1, 8) (–2, 4) C(1, 4) (4, 4) F2 (1, 4 + 7) F1 (1, 4 – 7) y x 12. (b) q2 y + x = 2cq 13. (a) (–5, 0), (5, 0) (b) y = – 3 4 x, y = 3 4 x 17. (a) (y – 3)2 9 – x2 9 = 1 ; centre = (0, 3) ; vertices = (0, 0), (0, 6) (b) y = x + 3, y = –x + 3 (c) –3 0 3 (0, 3) (0, 6) y = –x + 3 y = x + 3 y x
Mathematics Term 1 STPM Answers 287 Chapter 6 Vectors Exercise 6.1 1. (a) 4 5 i – 3 5 j (b) 5 13 i + 12 13 j (c) – 2 5 i – 1 5 j 2. 12i – 5j 3. (a) 2 7 i + 3 7 j – 6 7 k (b) –4i – 6j + 12k 4. 5i – 2j + k ; 30 units 5. –4 26 i + 1 26 j + 3 26 k 6. 6i + 3j + 7k 7. (a) 21 units (b) 5 units (c) 3 units 8. (a) 6 untis (b) 7 units (c) 186 units 9. (a) No (b) Yes (c) Yes 11. (a) 2i – 2j + 2k, i – 9k (b) 3(i – j + k) Exercise 6.2 1. (a) –2i – 8j (b) 16i + 36j (c) 7 4 (3i + 2j) (d) –35i – 14j (e) 6i + 3j 2. u – 2v 3. (a) i + j + 5k (b) 1 2 (–i + j + 23k) (c) –12i – 7j + 10k 4. (a) (–1, 3, 0) (b) (–2, 0, –8) (c) (–3, 3, –8) (d) (2, –4, 3) 5. (a) (5, –6, –1) (b) (0, 2, –3) (c) (–1, 0, 5) (d) (0, 0, –4) (e) (0, 9, 0) 7. l1 = 4, l2 = 1, l3 = –2 9. 17i – 3j – 10k 398 10. 3i – k 10 Exercise 6.3 1. (a) –23 (b) 5 (c) –9 (d) 6 (e) 18 2. (a) 7°8´ (b) 90° (c) 75°38´ (d) 73°24´ (e) 90° 3. 80°24´ 4. (a) –32 (b) 130°46´ 5. 60° 6. (a) orthogonal (b) orthogonal (c) orthogonal 7. –2 9. (a) →OM = 6i + 6j + 12k; →NM = –6i + 2j + 12k (b) 53.0° 10. –5i + 7j – 3k 11. (a) –4i + 72j + 28k (b) 4i – 72j – 28k 12. 339 unit2 13. 29 unit2 14. 6 2 unit2 15. –i + j – k 16. l = 7 3 17. –i + 2j + 8k 69 Exercise 6.4 1. (a) L1 : r = 2i – j + 6k + li (b) L2 : r = –4i + 3k + lj (c) L3 : r = 3j – k + lk (d) L4 : r = 2i – 5j + l(–i + 6j) (e) L5 : r = –2i + j – 3k + l(i + 2j – k) 2. (a) r = 2i + 3k + l(i – j + k) (b) r = i + 2j + 2k + l(i – 4j – 2k) (c) r = 3i + j + 4k + l(–4i + j + k) 3. x – 1 –2 = y + 3 5 = z – 2 6 (=l) 4. r = i – 5j + 2k + l(–3i + 2j + 6k) 5. r = –i – 2j – 5k + l(7i + 2j – 3k) 6. (a) r = 2i + j – k + l(3i + j + 6k) (b) r = 6i – 2j + k + l(i + 5j + k) 7. (a) r = i – 3j + 2k + l(i – j + 5k); x – 1 1 = y + 3 –1 = z – 2 5 (b) r = i + 2j + k + l(i + 2j – 3k); x – 1 1 = y – 2 2 = z – 1 –3 8. r = (2i – j + 5k) + l(–3i + 5j – 2k); x – 2 –3 = y + 1 5 = z – 5 –2 (=l) 9. p = –1, q = –5 10. r = i – 3j + l(2i + j) Exercise 6.5 1. r · (i – j – 3k) = –7, x – y – 3z = –7 2. r · (i – 2j – k) = –12, x – 2y – z = –12 3. r · (i – 3j – 5k) = 7 4. r · (–10i + 10j – 14k) = –32; –10x + 10y – 14z = –32 5. r · (6i + 2j – 3k) = 17 6. r · (i + 2j + 2k) = –6, x + 2y + 2z = –6 7. r · (2i – 3j + k) = 9, 2x – 3y + z = 9 8. r · (i – 3j + 2k) = 0, x – 3y + 2z = 0 Exercise 6.6 1. (a) 71°34´ (b) 0° 2. 71°14´ 3. 70°32´ 4. 80°24´ 5. 61°1´ 6. 79°6´ 7. 22°12´ 8. 71°34´ 9. 90° 10. 49°59´ ≈ 50°
288 Mathematics Term 1 STPM Answers Exercise 6.7 1. (1, –2, 5) 2. (4, 5, 9) 3. (a) (–3, 1, 5) (b) (3, –5, 4) 4. (a) l1 = 2, l2 = 1 (b) c = –5 (c) (2, –3, –3) 5. (5, –3, 7) 6. (1, –5, 1) 7. (2, 14, 6) y + 5 2 z – 13 2 8. x = 2y + 5 7 = 2z – 13 –17 or x = = , 7 2 – 17 2 r = – 5 2 j + 13 2 k + l(i + 7 2 j – 17 2 k) y + 4 3 z + 8 3 9. x 1 = = 5 3 1 3 10. x – 11 4 = y + 8 –3 = z STPM PRACTICE 6 1. (a) 474 units (b) 46 units 2. (a) –i – 4j – 2k (b) –5i – 7j + 11k 3. (a) 1 27 (5i – j – k) (b) 1 11 (–i – j + 3k) 4. (a) 2 3 (i + j – k) (b) 5 11 (–3i – j – k) 5. (a) –8 (b) 4 6. (a) 83°44´ (b) 127°25´ 7. (a) 78.7° (b) 36.8° 8. (a) p = 1 ; q = –1 (b) 1 –1 7 –2 2 9. 35° 10. (a) →PQ = –6i + 6j + 12k (b) 61.9° →PR = 6i + 6j + 12k 11. (b) 58.7° (c) 152 unit2 , 25 1 3 unit3 12. (a) –2i + j + 4k (b) –4i + 4j – 8k 13. ± 1 21 (–2i + j + 4k) 14. 41 unit2 15. (a) 11i – 29j + 5k (b) 1 2 987 unit2 16. (a) 5 18 (b) 3 10 17. (a) 1 61 (3i – 4j + 6k) (b) 8i + (4 + l)j + 8k (c) l = –9 18. (a) a = 10, b = 2 (b) 64.4° 19. (b) x + y – 3z = 7 20. Yes 21. (b) 3 26 unit2 (c) 5 4 i + 1 8 j – 13 4 k (d) 35 16 201 16 201 22. 1 1 –1 0 2 + t 1 9 14 1 2 where t ∈ R 23. (1, 2, 3) 24. 47°7´ 25. 8i + 4j – 4k, –8i + 12j + 16k 26. 2i – 3j + 4k 27. (a) 22 units (b) 12i – 15j – 9k 28. (a) 7 390 390 (b) 1 2 329 unit2 (c) r . (–10i – 15j + 2k) = –30 29. (a) 13x – 2y + 5z = –4 (b) (–1, –22, –7) (c) 14.9° STPM Model Paper (954/1) 1. (a) f : x ↦ x2 – 6x = (x – 3)2 – 9 0 3 y = f(x) y x y = k (3, –9) Any horizontal line y = k is drawn horizontally with x-axis, will cut the graph at only one point. Therefore f is a one-to-one function and f–1 exists. (shown) Let f –1(x) = y x = f(y) x = (y – 3)2 – 9 ± x + 9 = y – 3 y = 3 – x + 9 as y < 3 f –1(x) = 3 – x + 9 f –1 : x ↦ 3 – x + 9 Df –1 = [–9, ∞) Rf –1 = (–∞, 3] (b) 0 3 4 y x g(x) = 4 – e–x Rg = (–∞, 4) Rg = (–∞, 4) ⊄ Df = (–∞, 3] Therefore fg does not exist. (shown) We need Rg = (–∞, 3] 4 – e–x < 3 x < 0 Therefore the maximum domain of g for which fg exists = (–∞, 0]
Mathematics Term 1 STPM Answers 289 2. 1 + x 1 – 3x = (1 + x) 1 —2 (1 – 3x) – 1 —2 = 31 + 1 2 (x) + 1 1 2 21 1 2 – 12 2! (x) 2 + …4 31 – 1 2 (–3x) + 1– 1 2 21– 1 2 – 12 2! (–3x) 2 + …4 = 31 + 1 2 x – 1 8 x2 + …431 + 3 2 x + 27 8 x2 + … 4 = 1 + 3 2 x + 27 8 x2 + … + 1 2 x + 3 4 x2 + … – 1 8 x2 + … = 1 + 2x + 4x2 + … Expansion is valid for : 5x : – 1 3 , x , 1 3 6 By taking x = 1 9 , 1 + x 1 – 3x = 1 + 1 9 1 – 31 1 9 2 = 5 3 5 3 ≈ 1 + 21 1 9 2 + 41 1 9 2 2 5 × 3 3 × 3 ≈ 1 + 21 1 9 2 + 41 1 9 2 2 15 3 ≈ 1 + 21 1 9 2 + 41 1 9 2 2 15 ≈ 331 + 21 1 9 2 + 41 1 9 2 2 4 ≈ 103 27 Alternative solution: 5 3 ≈ 1 + 21 1 9 2 + 41 1 9 2 2 5 × 5 3 × 5 ≈ 1 + 21 1 9 2 + 41 1 9 2 2 5 15 ≈ 1 + 21 1 9 2 + 41 1 9 2 2 15 ≈ 5 1 + 21 1 9 2 + 41 1 9 2 2 ≈ 405 103 3. (a) 27x2 – 9y2 + 54x + 72y – 360 = 0 27(x2 + 2x) – 9(y2 – 8y) – 360 = 0 27[(x + 1)2 – 1] – 9[(y – 4)2 – 16] – 360 = 0 27(x + 1)2 – 9(y – 4)2 – 27 + 144 – 360 = 0 27(x + 1)2 – 9(y – 4)2 = 243 (x + 1)2 9 – (y – 4)2 27 = 1 (b) a = 3, c = a2 + b2 = 9 + 27 = 6 Centre = (–1, 4) Vertices = (–4, 4) and (2, 4) Foci = (–7, 4) and (5, 4) Equation of asymptotes: (y – 4) = ± 3 3 3 (x + 1) y = – 3x – 3 + 4 and y = 3x + 3 + 4 (c) –10 1 1 2 3 4 5 6 –7 –6 –5–4–3 –2 2 3 4 5 (–7, 4) (–4, 4) (–1, 4) (2, 4) (5, 4) y x y = – 3x – 3 + 4 y = 3x + 3 + 4 4. 1 3 1 2 –2 0 –1 5 2 4 u 1 0 0 0 1 0 0 0 1 2 R1 ↔ R2 → 1 1 3 2 0 –2 –1 2 5 4 u 0 1 0 1 0 0 0 0 1 2 R2 = –3R1 + R2 → R3 = –2R1 + R3 1 1 0 0 0 –2 –1 2 –1 0 u 0 1 0 1 –3 –2 0 0 1 2 R2 ↔ R3 → 1 1 0 0 0 –1 –2 2 0 –1 u 0 0 1 1 –2 –3 0 1 0 2 R2 = –R2 → 1 1 0 0 0 1 –2 2 0 –1 u 0 0 1 1 2 –3 0 –1 0 2 R3 = 2R2 + R3 → 1 1 0 0 0 1 0 2 0 –1 u 0 0 1 1 2 1 0 –1 –22 R3 = –R3 → 1 1 0 0 0 1 0 2 0 1 u 0 0 –1 1 2 –1 0 –1 2 2 R1 = –2R3 + R1 → 1 1 0 0 0 1 0 0 0 1 u 2 0 –1 3 2 –1 –4 –1 2 2 The inverse matrix of 1 3 1 2 –2 0 –1 5 2 4 2 = 1 2 0 –1 3 2 –1 –4 –1 2 2 1 3 1 2 –2 0 –1 5 2 4 21 x y z 2 = 1 45 15 32 2 1 2 0 –1 3 2 –1 –4 –1 2 21 3 1 2 –2 0 –1 5 2 4 21 x y z 2 = 1 2 0 –1 3 2 –1 –4 –1 2 21 45 15 32 2 1 x y z 2 = 1 90 + 45 – 128 0 + 30 – 32 –45 – 15 + 64 2 1 x y z 2 = 1 7 –2 4 2 Hence, x = 7, y = –2, z = 4.
290 Mathematics Term 1 STPM Answers 5. Let z = 3 + i |z| = ( 3) 2 + (1)2 = 2 Arg z = tan–1 1 3 = π 6 z = 21cos π 6 + i sin π 6 2 ( 3 + i)12 = 321cos π 6 + i sin π 6 2412 ; ( 3 – i)12 = 321cos 1– π 6 2 + i sin 1– π 6 22412 ( 3 + i)12 + ( 3 – i)12 = 321cos π 6 + i sin π 6 2412 + 321cos 1– π 6 2 + i sin 1– π 6 22412 = 2123cos 12π 6 + i sin 12π 6 4 + 2123cos 1– 12π 6 2 + i sin 1– 12π 6 24 = 212[cos 2π + i sin 2π] + 212[cos(–2π) + i sin (–2π)] = 212[1 + i(0)] + 212[1 + i(0)] = 213 (shown) 2z3 – [( 3 + i)12 + ( 3 – i)12]i = 0 2z3 – 213i = 0 z3 = 212i z3 = 212(0 + i) z3 = 2121cos π 2 + i sin π 2 2 z = (212) 1 —3 3cos 1 π 2 + 2kπ2 + i sin 1 π 2 + 2kπ24 1 —3 z = 16 3cos 1 π 6 + 2kπ 3 2 + i sin 1 π 6 + 2kπ 3 24 , k = 0, 1, 2 k = 0, z = 161cos π 6 + i sin π 6 2 = 161 3 2 + i 1 2 2 = 8( 3 + i) k = 1, z = 161cos 5π 6 + i sin 5π 6 2 = 161– 3 2 + i 1 2 2 = 8(– 3 + i) k = 2, z = 161cos 3π 2 + i sin 3π 2 2 = 16(0 – i) = –16i The roots are 8( 3 + i), 8(– 3 + i) and –16i. 6. (a) →AB = 1 5 5 –1 2 – 1 1 4 –2 2 = 1 4 1 1 2 →AC = 1 4 3 1 2 – 1 1 4 –2 2 = 1 3 –1 3 2 n1 = →AB × →AC = i 4 3 j 1 –1 k 1 3 = (3 + 1)i – (12 – 3)j + (–4 – 3)k = 4i – 9j – 7k (Normal vector n1 = Parallel vector of the line L1 ) Vector equation of L1 : r = (3i – 6j + 2k) + λ(4i – 9j – 7k) Parametric equations of L1 : x = 3 + 4λ, y = –6 – 9λ, z = 2 – 7λ (b) (i) →EF = 1 2 2 3 2 – 1 5 1 –3 2 = 1 –3 1 6 2 Vector equation of L2 : r = (5i + j – 3k) + µ(–3i + j + 6k) Parametric equations of L1 : x = 3 + 4λ, y = –6 – 9λ, z = 2 – 7λ Parametric equations of L2 : x = 5 – 3µ, y = 1 + µ, z = –3 + 6µ Equating the two lines: 3 + 4λ = 5 – 3µ 3µ + 4λ = 2 ......................a –6 – 9λ = 1 + µ µ = –7 – 9λ .....................b 2 – 7λ = –3 + 6µ .....................c Substitute b into a: 3(–7 – 9λ) + 4λ = 2 –21 – 27λ + 4λ = 2 23λ = –23 λ = –1 When λ = –1, from b: µ = –7 – 9(–1) = 2, checking from equation c: λ = –1 , LHS = 2 – 7λ = 2 – 7(–1) = 9 µ = 2 , RHS = –3 + 6µ = –3 + 6(2) = 9 = LHS The equations are consistent with λ = – 1 and µ = 2, RHS = LHS, therefore the two lines L1 and L2 intersect. (shown) Position vector of point of intersection = –i + 3j + 9k Coordinates of point of intersection = (–1, 3, 9) (ii) cos q = n1 . n2 |n1 ||n2| = (4i – 9j – 7k) . (–3i + j + 6k) 146 . 46 = –63 6 716 q = 140.2° Acute angle = 180° – 140.2° = 39.8° 7. (a) p(x) has a factor (x – 3), p(3) = a(3)3 – (1 + 6a)(3)2 + 2(8a – b)(3) – 27 = 0 27a – 9 – 54a + 48a – 6b – 27 = 0 21a – 6b = 36 7a – 2b = 12 (shown) (b) From 7a – 2b = 12, substitute –2b = 12 – 7a into p(x) = ax3 – (1 + 6a)x2 + (16a – 2b)x – 27, p(x) = ax3 – (1 + 6a)x2 + (16a + 12 – 7a)x – 27 p(x) = ax3 – (1 + 6a)x2 + 3(3a + 4)x – 27
Mathematics Term 1 STPM Answers 291 Long division: ax2 – (3a + 1)x + 9 x – 3 ax3 – (1 + 6a)x2 + 3(3a + 4)x – 27 ax3 – 3ax2 –(3a + 1)x2 + 3(3a + 4)x –(3a + 1)x2 + 3(3a + 1)x 9x – 27 9x – 27 Alternative solution: ax3 – (1 + 6a)x2 + 3(3a + 4)x – 27 = (x – 3)(ax2 + px + 9) Comparing coefficients of x2 , –(1 + 6a) = p – 3a p = 3a – 1 – 6a = –3a – 1 = –(3a + 1) The quotient = ax2 – (3a + 1)x + 9 (c) Remainder Theorem: p(4) = 13 p(4) = a(4)3 – (1 + 6a)(4)2 + 3(3a + 4)(4) – 27 = 13 64a – 16 – 96a + 36a + 48 – 27 = 13 4a = 8 a = 2 (d) (i) p(x) = (x – 3)(2x2 – 7x + 9) = 0 (x – 3) = 0 x = 3 (real root) (2x2 – 7x + 9) = 0 x = –(–7) ± (–7)2 – 4(2)(9) 2(2) = 7 ± –23 4 (complex roots) Therefore p(x) = 0 has only one real root. (shown) Alternative solution: p(x) = (x – 3)(2x2 – 7x + 9) = 0 (x – 3) = 0 x = 3 (real root) (2x2 – 7x + 9) = 0 b2 – 4ac = (–7)2 – 4(2)(9) = –23 < 0 (no real root) Therefore p(x) = 0 has only one real root. (shown) 2x2 – 7x + 9 = 21x2 – 7 2 x + 9 2 2 = 231x – 7 4 2 2 – 49 16 + 9 2 4 = 21x – 7 4 2 2 + 23 8 For (x – 3)(2x2 – 7x + 9) . 0, (x – 3)321x – 7 4 2 2 + 23 8 4 . 0. Since 1x – 7 4 2 2 > 0, 321x – 7 4 2 2 + 23 8 4 . 0 for x ∈ R. Therefore (x – 3) . 0. The set values of x: x ∈ (3, ∞). (ii) 4x – 9 p(x) = 4x – 9 (x – 3)(2x2 – 7x + 9) = A x – 3 + Bx + C 2x2 – 7x + 9 4x – 9 = A(2x2 – 7x + 9) + (Bx + C)(x – 3) When x = 3, 4(3) – 9 = A[2(3)2 – 7(3) + 9] 3 = 6A A = 1 2 Coefficients of x2 : 0 = 2A + B 0 = 21 1 2 2 + B B = –1 Constants: –9 = 9A – 3C –9 = 91 1 2 2 – 3C C = 9 2 Hence, 4x – 9 p(x) = 1 2(x – 3) + –x + 9 2 2x2 – 7x + 9 or 4x – 9 p(x) = 1 2(x – 3) + 9 – 2x 2(2x2 – 7x + 9) . 8. Given that →ON = 2i, →OS = 4k and →OR = 3j + 4k (a) Given that SR : NM = 3 : 5 →OM = →ON + →NM →OM = →ON + 5 3 →SR →OM = 1 2 0 0 2 + 5 3 1 0 3 4 2 – 1 0 0 4 2 →OM =1 2 5 0 2 = 2i + 5j Alternative solution: →OR = 3j + 4k →OS + →SR = 4k + 3j \ →SR = 3j Since →SR parallel to →NM. SR : NM = 3 : 5, therefore →NM = 5j →OM = →ON + →NM = 2i + 5j (b) →OT = →ON + →NT = →ON + →OS = 1 2 0 0 2 + 1 0 0 4 2 = 1 2 0 4 2 →TR = →OR – →OT = 1 0 3 4 2 – 1 2 0 4 2 = 1 –2 3 0 2
292 Mathematics Term 1 STPM Answers →TM = →OM – →OT = 1 2 5 0 2 – 1 2 0 4 2 = 1 0 5 –4 2 →TR × →TM = i –2 0 j 3 5 k 0 –4 = (–12 – 0)i – (8 – 0)j + (–10 – 0)k = –12i – 8j – 10k = –2(6i + 4j + 5k) Area of triangle RTM = 1 2 →TR × →TM = 1 2 × 2 × 62 + 42 + 52 = 77 (c) Let the acute angle between TR and TM = q Method 1 : Cross Product →TR × →TM = →TR →TM sin q →TR × →TM = 2 × 77 From the answer in (b) sin q = 2 77 22 + 32 . 52 + 42 = 2 77 13 . 41 q = 49.5° Method 2 : Scalar Product →TR . →TM = →TR →TM cos q cos q = 1 –2 3 0 2 . 1 0 5 –4 2 22 + 32 . 52 + 42 = 15 13 . 41 q = 49.5° (d) n1 = 6i + 4j + 5k From the answer in (b) = a parallel vector of the normal vector Choose a point of plane RTM, point R(0, 3, 4) Equation of the plane: r . n1 = D r . 1 6 4 5 2 = 1 0 3 4 2 . 1 6 4 5 2 r . 1 6 4 5 2 = 0 + 12 + 20 r . (6i + 4j + 5k) = 32 Cartesian equation: 6x + 4y + 5z = 32 (e) Normal vector of the plane RTM: n1 = 6i + 4j + 5k From the answer in (b) Normal vector of the plane OTM: →OT × →OM = i 2 2 j 0 5 k 4 0 = (0 – 20)i – (0 – 8)j + (10 – 0)k = –20i + 8j + 10k = –2(10i – 4j – 5k) n2 = 10i – 4j – 5k Let the acute angle between the plane RTM and the plane OTM = a cos a = n1 . n2 n1 n2 = 1 6 4 5 2 . 1 10 –4 –5 2 62 + 42 + 52 . 102 + 42 + 52 = 60 – 16 – 25 77 . 141 = 19 77 . 141 a = 79.5°
STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 954/1 Mathematics (T) Paper 1 Written test Section A Answer all 6 questions of variable marks. Section B Answer 1 out of 2 questions All questions are based on topics 1 to 6. 60 (26.67%) 45 15 1—1 2 hours Central assessment Second Term 954/2 Mathematics (T) Paper 2 Written test Section A Answer all 6 questions of variable marks. Section B Answer 1 out of 2 questions. All questions are based on topics 7 to 12. 60 (26.67%) 45 15 1—1 2 hours Central assessment Third Term 954/3 Mathematics (T) Paper 3 Written test Section A Answer all 6 questions of variable marks. Section B Answer 1 out of 2 questions. All questions are based on topics 13 to 18. 60 (26.67%) 45 15 1—1 2 hours Central assessment First, Second and Third Terms 954/4 Mathematics (T) Paper 4 Coursework 3 assignments, each based on topics 1 to 6, topics 7 to 12 and topics 13 to 18. 180 (20%) Through-out the three terms School-based assessment for school candidates Assessment by appointed assessor for private candidates
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