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vedanta Excel In Opt. Mathematics - Book 10 (d) In the given circle, O is the centre and P is the point of contact. What is A the relation between OP and AB ? OP (e) What are the meanings of similarity and line of symmetry ? B 11.7 Inversion Transformation Transformation is a type of function. So, it may also have its inverse. The inverse of a transformation is called its inverse. If T is a transformation on a set A to itself which carries each element 'a' of A into a corresponding element 'b' of A. Then, the transformation that reverses transformation T, by carrying each element 'b' of A back into its original element 'a' of A is called inverse transformation. Then, it is denoted by T–1. We can write: T:aob T–1 : b o a The inverse of reflection is the reflection on the same line of reflection. Similarly, the inverse of translation is a transformation in the same magnitude but opposite in direction. The size 1 of transformation of magnitude p and p , having the same centre are inverse to each other. Inversion transformation does not preserve collinearity and distance. It can be generalized as reflection in a line. Concept of Inversion Circle and Inversion Point O P=P' OP P' O P' P Fig (i) Fig (ii) Fig (iii) In all figure O is the centre of circle, r is radius. In figure (i), the points P and P' are on the circumference of the circle. In figure (ii) P lies inside the circle and P' lies outside the circle. In figure (iii) P' lies inside the circle and P lies outside the circle. In each of above cases OP. OP' = r2 is true. We can observe that P o P', P' o P the points P and P' are called relative inverse of each other. Inversion in a circle is a method of transformation of A geometric figure into other figure. It is like process of D reflection on a line. Since the reflection on a line depends O P' P' on the particular line chosen, inversion in a circle depends on the particular circle. Let D be a circle with centre O and radius r and P are any point other than origin. Then, P' is B called inversion of P. Inversion is the process of transformation of point P to corresponding set of points P' called in the inverse points. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 325

vedanta Excel In Opt. Mathematics - Book 10 Let D be a fixed circle. Then, in the above figure, O is the inversion centre, D is called inversion circle, AB is called polar and P is the pole. For every point P other than O, we define a point P' is the inverse point of P. It is to be noted that if P is inside the circle P' is outside of the circle and vice-versa. Features of Inversion Circle The following are the some important features of inversion circle. (a) To each point P of the plane other than O, there corresponds an inverse point P'. (b) If P lies on the circumference of the inversion D, then P = P' (c) If OP < r, i.e. if P lies inside there circle P' lies outside the circle EFG. Then OP' > r. Conversely if OP > r, i.e. if P lies outside the circle, P' and inverse point P' lies inside the circle EFG. (d) The point P and the inverse point P' can always be interchanged. (e) If P' is the image of P, P is the image of the P'. Hence, the point and its inverse point can be always be interchanged. This property can be referred as symmetric feature of inversion. i.e., (P')' = P 11.8 Construction of Inverse Point Geometrically There may arise three cases. (a) When the point P lies on the circumference of the circle then the point O P' is the inverse point itself. P' = P (b) When the point P is inside the circle D. Draw D U P' o OP a ray OP passing through O and P. A chord UV V is drawn perpendicular to the ray OoP passing through the point P. Then tagents UP' and VP' are drawn to the circle at the points U and V. OU and OV are joined. The point of intersection of the tangents UP' and VP' is P' which lies on the ray OoP. Join OU and OV. Then ‘OPU and ‘OUP' will be right angles. In right angled triangles ∆OPU and ∆OP'U, we have, ‘OPU = ‘OUP' (Both are right angles) ‘UOP = ‘UOP' (Common angle) ‘OUP = ‘UP'O (the remaining angles) Hence 'OPU and 'OP'U are similar. Therefore, from similar triangles ∆OPU and ∆OP'U. we have, OP = OU or. OP.OP' = OU2 OU OP' ? OP.OP' = r2(i.e. OU = r) Hence, the point P' is the inverse of point P. 326 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 U M E P (c) Let the P be outside the inversion circle O P' D. Join OP. Let M be the mid point of V OP. Draw another circle E about centre M and radius OM = MP. Then the two circles D and E intersect at the points U and V. Join UP and VP. Then ‘OUP and ‘OVP are inscribed in semi-circle E. D Hence, they are right angles. Again UP and VP are tangents to the circle D. The chord UV meets OP at P'. The point P' is the inverse of point P. Hence, the point P' is the inverse of the point P relative to the circle D. 11.9 Use of Coordinates in Inversion Circle (a) Centre at the origin (Standard form): P (x, y) y The inverse of a point P(x, y) with respect to a circle with centre at the origin and the radius r is a point P'(x', y') such that x' = r2x , y = r2y P'(x',y') x2 + y2 x2 + y2 Proof: Sx x' OR Let O (0, 0) be the centre and r be the radius y' of inversion circle. Then equation of the circle is x2 + y2 = r2. Let P'(x', y') be the inverse point of P(x, y). Then, by definition, OP.OP' = r2, where, OP = x2 + y2 OP' = x'2 + y'2 From P and P' draw PSAOX and P'RAOX. Since the points O, P', P are collinear, the triangles OPS and OP'R are similar. In similar triangles the corresponding sides are proportional. OR = P'R = OP' OS PS OP x' y' OP' OP or, xx' = yy' = OP . OP or, x = y = OP.OP' = x2 r2 y2 OP2 + Then, and x' = r2x y' = x2 r+2y y2 x2 + y2 Hence the equation of inverse of any point P(x, y) with respect to a circle with centre at the origin and the radius r is a point P'(x', y) such that Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 327

vedanta Excel In Opt. Mathematics - Book 10 x' = r2x and y' = r2y x2 + y2 x2 + y2 Note : 1. When r = 0, x2 + y2 = 0, is the equation of inversion circle. Then x' and y' do not have any real values, then the center of inverse is a point at infinity. 2. For any point on the circumference of the inversion circle x2 + y2 = r2, its inverse point P'(x', y') is given by x = x' and y = y'. (b) Centre at any Point (Central form) y P(x, y) The equations of inverse of P'(x', y')P'(x', y') S any point P(x, y) with respect C(h, k) r x' to a circle with centre at C(h, k) and radius r is a point R W P'(x', y') such that O x' = h + (x – r2 (x – h) k)2 UV h)2 + (y – y' and y' = k + r2 (y – k) (x – h)2 + (y – k)2 Proof: x Let C(h, k) be the centre and r be the radius of the circle. Then equation of the inversion circle is given by (x – h)2 + (y – k)2 = r2 Let P'(x', y') be the inverse point of P(x, y) Then by definition, CP.CP' = r2. Now, CP = (x – h)2 + (y – k)2 CP' = (x' – h)2 + (y' – k)2 Draw CU A OX, P'V A OX, PW A OX. Then also draw CS A PW, CS cuts PV at R. Since the points C, P' and P are collinear, 'CPS and 'CP'R are similar. In similar triangles, the corresponding sides are proportional CR = P'R = CP' CS PS CP x' – h y' – k CP'.CP or, x–h = y–k = CP2 or, x' – h = y' – k = r2 x–h y–k (x – h)2 + (y – k)2 r2 (x – h) r2 (y – k) Then, x' = h + (x – h)2 + (y – k)2 and y' = k+ (x – h)2 + (y – k)2 Hence, the equation of inverse of any point P(x, y) with respect to a circle with centre at (h, k) and radius r is a point P'(x', y') such that x' = h + r2 (x – h) and y' = k + r2 (y – k) (x – h)2 + (y – k)2 (x – h)2 + (y – k)2 328 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Worked out Examples Example 1: From the adjoining figure give the following concepts: Solutions: (a) Inversion circle (b) Inversion radius (c) Centre of inversion (d) Relation between OP, OP' and OA. From the above figure, N D (a) Circle D is the inversion circle A (b) OA = r, radius of inversion circle OP P' R (c) O is the centre of inversion circle M (d) By definition of inversion circle OP.OP' = OA2 or, OP.OP' = r2 Example 2: Find the inverse of the point P, Q, R which are at distances 2, 4, 8 units from Solutions: the centre of O of the circle with radius 4 units.  Let OP = 2, OQ = 4, OR = 8 units.  Let P', Q', and R' the inverse points of P, Q, and R' respectively. Now, for inverse point of P. OP.OP' = r2, r = 4 R or, 2.OP' = 42 or, OP' = 16 Q 2 Q' ? OP' = 8 units R' i.e., P' is at 8 units from O. OP Similarly, for the inverse point of Q, P' OQ.OQ' = r2 or, 4 × OQ' = 16 or, OQ'= 4 units i.e., Q' is at 4 units from O. Here, OQ' = OQ = r. The point Q itself is its inverse point. For the inverse point of R. OR.OR'= r2 or, OR' = 42 = 16 =2 OR 8 ? R' is 2 units from O. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 329

vedanta Excel In Opt. Mathematics - Book 10 Example 3: Find the inverse of the point (i) A(3, 5) (ii) B(2, – 5) with respect to the circle Solutions: x2 + y2 = 1. Example 4: Equation of given circle is x2 + y2 = 1. Solutions: radius (r) = 1 unit, centre of circle = O (0, 0) (i) To find inverse of A(3, 5). r = 1, x = 3, y = 5. Let A'(x', y') be inverse of A. x' = r2x = 1×3 = 3 = 3 x2 + y2 32+ 52 9 + 25 34 r2y 12 × 5 5 5 y' = x2 + y2 = 32+ 52 = 9 + 25 = 34 Hence, the required inverse point is A'( 3 , 354 ) 34 (ii) To find the inverse of B(2, –5) r = 1, x = 2, y = 5 Let B'(x', y') be inverse image of B. Then, x' = r2x y' x2 + y2 = 12 × 2 = 2 22+ (–5)2 29 = r2y = 12 × (–5) = – 5 x2 + y2 22+ (–5)2 29 Hence, the required inverse of point B is B' 2 , – 5 . 29 29 Find the inverse of the point (i) P(3, 2), (ii) Q(3, 3) with respect to the circle with centre (2, 3) and radius of inverse circle 2 units. According question, radius (r) = 2, centre (h, k) = (2, 3) (i) To find inverse of point P(3, 2). r = 2, (h, k) = (2, 3), (x, y) = (3, 2) Let, P'(x', y') be required inverse of point P. Then, =h+ r2 (x – h) =2+ 22 (3 – 2) x' (x – h)2 + (y – k)2 (3 – 2)2 + (2 – 3)2 and y' = 2 + 22 × 1 = 2 + 4 = 2 + 2 = 4 (1)2 + (–1)2 2 =k+ r2 (y – k) =3+ 22 (2 – 3) (x – h)2 + (y – k)2 (3 – 2)2 + (2 – 3)2 = 3 + 22 × 1 = 3 + –4 = 3 – 2 = 1. (1)2 + (–1)2 2 Hence, the inverse point of P(3, 2) is P'(4, 1). 330 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (ii) To find inverse of point Q(3, 3) r = 2, (h, k) = (2, 3), (x, y) = (3, 3) x – h = 3 – 2 = 1, y – k = 3 – 3 = 0 Let Q' (x', y') be required inverse of point Q. x' = h+ r2 (x – h) =2+ 22 × (3 – 2) (x – h)2 + (y – k)2 (3 – 2)2 + (3 – 3)2 22 × 1 = 2 + (1)2 + (0)2 =2+4 =6 and y' = k+ (x – r2 (y – k) k)2 = 3 + 22 × (3 – 3) h)2 + (y – (3 – 2)2 + (3 – 3)2 = 3 + 22 × 0 = 3 + 0 = 3. (1)2 + (0)2 Hence, the inverse point of Q(3, 3) is Q'(6, 3). Exercise 11.6 Very short questions A P' 1. From the given figure, give the following concepts: OP (a) inversion circle D (b) inversion radius (c) centre of inversion (d) relation among OP, OP' and r. 2. Fill in the gaps in the following table: Distance of the Distance of the inverse point from S.N. given point from the Radius of inversion centre of inversion centre of inversion Circle circle circle .......... .......... (a) 1 2 1 (b) 1 3 2 9 2 (c) 18 .......... 3 625 (d) 16 .......... 1 (e) 3 .......... 4 (f) .......... 12 (g) .......... 25 9 (h) .......... 24 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 331

vedanta Excel In Opt. Mathematics - Book 10 Short question 3. Find the inverse points of the following points with respect to the inversion circle with centre at the origin. S.N. Given points Radius of inversion circle a P(2, 3) 5 b Q(1, 2) 2 c R(4, 5) 4 d S(3, 0) 3 4. (a) Find the distance of inverse of the points A, B and C which are at distances 3, 6, 9 units respectively from the centre O of the circle with radius 6 units. (b) Find the inverse of the points P, Q and R which are at distances 2, 4, 8 respectively units from the centre O of the circle with radius 8 units. (c) Find the inverse of the points M, N and P which are at distances 2.5, 5 and 10 units from the centre O of the circle with radius 5 units. 5. (a) Find the inverse of the point (2, –5) with respect to the circle x2 + y2 = 1. (b) Find the inverse of the point (3, 4) with respect to the circle x2 + y2 = 4. (c) Find the inverse of the point (4, 5) with respect to the circle x2 + y2 = 25. 6. (a) Find the inverse of the point (6, 2) with respect to a circle centre at the point (1, 2) of radius 3 units. (b) Find the inverse of the point (4, 2) with respect to a circle centre at the point (5, 6) of radius 5 units. (c) Find the inverse of the point (5, 10) with respect to a circle with centre at the point (3, 4) of radius 6 units. Long Questions 7. (a) Find inversion line segment of AB with points A(1, 2), B(3, 3) with respect to inversion circle x2 + y2 + 8x + 8y + 24 = 0. (b) Find the inverse line segment of PQ where P(4, 5) and Q(4, 3) with respect to the circle x2 + y2 – 4x – 6y = 3, show both of the line segments on the same graph with the inversion circle. 2. (a) 4 (b) 81 (c) 3 (d) 4 2 (e) 3 (f) 1, (g) 324 (h) 48 (d) (9, 0) 3. (a) 50 , 75 (b) 4 , 8 (c) 64 , 80 13 13 5 5 41 41 4. (a) 12, 6, 4 (b) 32, 16, 8 (c) 10, 5, 2.5 5. (a) 2 , – 5 (b) 12 , 16 (c) 100 , 125 29 29 25 25 41 41 6. (a) (2.8, 2) (b) 60 , 2 (c) 24 , 47 17 17 5 5 7. (a) A'B', A' - 204 , - 196 , B' –24 , 24 (b) P'(6, 7), Q'(10, 3), P'Q' 61 61 7 7 332 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 11.10 Use of Matrices in Transformation Matrices can be used in most of the elementary transformation. Let us take a point P(x. y) on the plane. ( ) (Let it be pre-multiplied by ) ( ) ( )0 1 -1 0 then -1 x -x 0 1 0 y =y Also, P(x, y) y - axis P'(-x, y) ( )-x Above two images y and (-x, y) are same image of the point P(x, y). ( )Hence, -1 0 represents a transformation matrix that represents a transformation 0 1 matrix that reflects a given point on Y-axis. Definition : The matrix which is used to transform a point to get its image is called transformation matrix. Translation Translation by using 2 × 1 matrix aa ( ) ( )Let us consider T = b , a column vector. Here, T is a 2 × 1 matrix. This matrix b represents a translation of 'a' unit in the direction of X-axis and 'b' units in the direction of Y-axis. Y Let P(x. y) be any point on the plane. P'(x + a, y + b) a P(x, y) a b X' O X ( )Let us translate P(x, y) by T = b . Y' P T P' ( )x ( a ) ( )x + a y+ b = y+b ( )We get that P'(x + a, y + b) is the image of P due to translation T = a . b Example: ( )Find the image of P(4, 5) under translation vector -2 . Solution: 2 ( )Given point P(4, 5) can be written as 4 5 ( )and -2 translation vector T = 2 which is 2 × 1 matrices. Now, P T P' ( ) ( ) ( )4 -2 2 5 + 2 =7 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 333

vedanta Excel In Opt. Mathematics - Book 10 Alternatively, Let P(x', y') be image of P (4, 5) under translation T we get, P T P' ( ) ( ) ( ) ( )x' 4 -2 2 y' = 5 + 2 = 7 Hence, P'(2, 7) is the required image of P(4, 5). Transformation using 2 × 2 matrices x [ ]A point (x, y) can be written in the matrix form as y . [ ]a b Let it be pre-multiplied by a matrix c d of order 2 × 2. Then, we get the resulting matrix of order 2 × 1. [ ] [ ] [ ] [ ]a b i.e. c d x ax + by ax + by y = cx + dy cx + dy is the image of P [a ]b under transformation matrix. c d is the transformation matrix of order 2 × 2. Example : Let P(3, 4), Q(0, 3), and R(3, 0) the vertices of ∆PQR. Find the image of ∆PQR Solution: [ ]0 -1 using 2 × 2 matrix -1 0 . The vertices of given ∆PQR can be written in the matrix form as, P QR [ ]3 0 3 4 30 [ ]Transformation matrix is 0 -1 Now, Image = transformation-1mat0rix × object matrix P QR [= P' Q' R' [ ] [ ]0 -1 -1 0 0-4 ]0-3 0+0 3 03 -3+0 4 30 0 + 0 -3 + 0 P' Q' R' [ ]= -4 -3 0 -3 0 -3 Hence, the coordinates of image vertices are P'(-4, -3), Q'(-3, 0) and R'(0, -3). ∆ P'Q'R' is the image of ∆PQR. Reflection using 2 × 2 matrices. Let P(x, y) be a point and P'(x' y') its image under certain geometrical transformation. 334 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (a) Reflection on X-axis. Since we have P(x, y) P'(x, -y) Which shows that x' = x and y' = - y. We can write, x' = 1 . x + 0 . y y' = 0 . x + (-1) . y Writing above linear equations in matrix form. [ ] [ ] [ ]1 0 0 -1 xx y = -y [-1 ]0 Hence, 0 1 is the transformation matrix of reflection on y-axis. (b) Reflection in y - axis Since P(x, y) y-axis P'(-x, y) which shows that x' = -x and y' = y , We can write x' = x (-1).x + 0.y y' = y = 0.x + 1.y Writing above linear equations in matrix form. [ ] [ ] [ ] [ ]x' -1 0 y' = 0 1 x -x y= y [-1 ]0 Hence 0 1 is the transformation matrix of reflection on y-axis. (c) Reflection on the line y = x. Since we have, P(x, y) y= x P'(y, x) i.e. x' = y and y' = x x' = y = 0.x + 1.y y' = x = 1.x + 0.y Writing these linear equation in the matrix form, [ ] [ ] [ ] [ ]x' 0 1 y y' = 1 0 x. x y= [0 ]1 The matrix 1 0 is the transformation matrix of reflection on the line y = x Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 335

vedanta Excel In Opt. Mathematics - Book 10 (d) Reflection on the line y = -x Since we have, P(x, y) y=-x P' (-y, -x) i.e. x' = -y, y' = -x x' = -y = 0. x + (-1) y y' = -x = (-1)x + 0.y Writing these equations in the matrix form [ ] [ ] [ ] [ ]x' 0 -1 y' = -1 0 x -y y = -x . [ ]0 -1 Hence, the matrix -1 0 describes reflection on the line y = -x. Rotations Using 2 × 2 Matrices (a) Rotation of +90° about origin O. Since we have, P(x, y) P(-y, x) i.e. x' = -y and y' = x x' = 0.x + (-1)y y' = x = 1.x + 0.y Writing these equation in matrix form, we write [ ] [ ] [ ] [ ]x' 0 -1 y' = 1 0 x -y y= x. [0 ]-1 Hence, the matrix 1 0 represents a transformation matrix of rotation of +90° about origin. (b) Rotation of 180° about origin. Since we have, P(x, y) R[0,180°] P'(-x, -y) i.e. x' = -x and y' = -y x' = - x = (-1) . x + o.y y' = -y = o.x + (-1). y Writing there equations in matrix form, we get, [ ] [ ] [ ] [ ]x' -1 0 y' = 0 -1 x -x y = -y . 336 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 [ ]Hence, the matrix -1 0 represents the rotation of 180° about the origin. 0 -1 (c) Rotation of +270° about origin Since we have, P(x, y) R[0,270°] P'(y, -x) i.e. x' = y and y' = -x x' = y = 0.x + 1. y y' = - x = (-1).x + o.y Writing these equations in the matrix form, [ ] [ ] [ ] [ ]x' 0 1 y' = -1 0 xy y = -x . [ ]Hence, the matrix 01 -1 0 represents rotation of + 270° or -90° about the origin. Enlargement Using 2 × 2 Matrices Let O be the centre of enlargement and k the scal factor. Let it be denoted by E[(0, 0), k]. Now, P(x, y) E[(0,0),k] P'(kx, ky) i.e x' = kx = k.x + 0.y y' = ky = 0.x + k.y There equations can be written in the matrix form [ ] [ ] [ ] [ ]x' k y' = 0 0 x kx k= y= ky . [k ]0 Hence, 0 k represents enlargement matrix with centre O and scale factor k. Unit Square Y B(1, 1) The square having vertices 0(0, 0), (1, 0), (1, 1) and (0, 1) is C called unit square. (0, 1) In the figure, OABC is a unit square. X' O Y' i.e. OA = OC = AB = CB = 1 unit X In matrix form, unit square is represented by A(1, 0) [ ]0 1 1 0 00 1 1 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 337

vedanta Excel In Opt. Mathematics - Book 10 Transformations in terms of matrices can be tabulated as follows: S.N Transformations Objects Images Equations Matrices 1. Reflection on x-axis (x, y) (x, -y) x'=1.x + 0.y [ ]1 0 2. Reflection on y-axis (x, y) (-x, y) y'=0.x+(-1)y 0 -1 x'=(-1)x+o.y 3. Reflection on y=x (x, y) (y, x) [ ]-1 0 y'=o.x+1.y 01 4. Reflection on line y=-x (x, y) (-y, -x) x'=o.x+1.y [ ]0 1 5. Rotation of +90° or (x, y) (-y,x) y'=1.x+o.y 10 +270° about 0 (y, -x) x'=o.x+(-1)y (-x, -y) [ ]0 -1 6. Rotation of -90° or (x, y) (kx, ky) y'=(-1)x +o.y -1 0 +270° about 0 (x, y) x'=o.x+(-1)y [ ]0 -1 7. Rotation of 180° about (x, y) y'=1.x+o.y 10 origin x'=o.x+1.y [ ]0 1 8. Enlargement with centre (x, y) y'=(-1)x+o.y -1 0 origin and scale factor k x'=(-1)x+o.y [ ]-1 0 9. Identity transformation (x, y) y'=o.x+(-1y) 0 -1 x'=k.x+o.y [ ]k 0 y'=o.x+k.y 0k x'=1.x+o.y [ ]1 0 o.x+1.y 01 Worked out Examples Example 1. [ ]Find the image of P(2, 3) using matrix which is translated by the vector Solution: -3 T= 1 Example 2. Solution: [ ]-3 Here, the translation vector T 1 Translating P by translation vector T, we get, [ ] [ ] [ ] [ ] [ ]x' 2 -1 -3 2 - 3 y' = 3 + 1 = 3 + 1 = 4 . [0 ]-1 Which transformation is associated with matrix -1 0 ? Use it to transform the point (6, 4). Let P(x, y) be any point on the plane. [ ] [ ] [ ] [ ]Then -y 0 -1 x 0 + (-y) .= -x . -1 0 y= -x + 0 338 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 P(x, y) P'(-y, -x) This shows that P is reflected on line y = -x. [ ]0 -1 Hence, -1 0 is associated with transformation of reflection on line y = -x. [ ] [ ] [ ]Again, 0 -1 6 -4 -1 0 4 = -6 . ? (-4, -6) is the image of (6, 4). [find the image of R(3, 2). ]0 ? Use it to Example 3. Which transformation is associated with the matrix 3 Solution: 3 0 Let P(x, y) be any point on the plane. 3 [ ] [ ] [ ]Transformation matrix M = 0 0 x 3x 3 y = 3y . ? P(x. y) P'(3x, 3y) Hence, the given matrix M is associated with enlargement about O with scale factor 3. R' [ ][ ] [ ]Also, 30 39 03 2= 6 Example 4. Write down the 2 × 2 transformation matrix which represents enlargement Solution: with centre (0, 0) by scale factor 2. Use it find the image of point (4, 3). Transformation matrix representing enlargement about O and scale factor 2 is [ ]2 0 M= 0 2 . Let given point be P(4, 3) M P P' [ ] [ ] [ ]2 0 4 8 Now, 0 2 = 3 = 6 ? The image of P(4, 3) is P'(8, 6) Example 5. [4 ]0 To which transformation is the matrix P = 0 4 associated ? What types of transformation do P2 and P3 represent? Solution: [4 ]0 Here, P = 0 4 is associated with the enlargement about centre origin O(0, 0) and scale factor 3. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 339

vedanta Excel In Opt. Mathematics - Book 10 40 4 [ ] [ ] [ ]Now, P2 = P × P = 0 40 16 0 0 4 = 0 16 Hence P2 represents an enlargement about centre origin O(o,o) with scale factor 16. 16 0 [ ][ ][ ]Again. P3 = P2. P = 0 16 40 64 0 04 0 64 . Hence P3 represents an enlargement matrix about centre origin O(0, 0) with scale factor 64. Example 6. The vertices of a unit square are O(0, 0), A(1, 0), B(1, 1) and C(0, 1), The Solution: square OABC is reflected on the line y = x. Find the coordinates of image using 2 × 2 matrix. Given unit square OABC can be written in the matrix form. OA B C [ ]0 1 1 0 00 11 01 0 [ ]2 × 2 matrix of reflection on line y = x is M = 1 OA B C O' A' B' C' [ ][ ] [ ]Now, 1 0 0 0 1 1 = 0 1 1 0 . 01 0 1 1 0 00 11 Hence, the coordinates of image of unit square are O(0, 0), A(0, 1), B(1, 1), and C'(1, 0). Example 7. Find the image of ∆ABC with vertices A(2, 3), B(-1, 2), and C(2, -3) using [ ] [ ]the matrix 0 -1 0 -1 -1 0 . Which transformation is the matrix -1 0 associated with ? Solution: Writing the coordinates of vertices A(2, 3), B(-1, 2) and C(2, -3) of ∆ABC in the AB C [ ]2 -1 2 matrix form , 3 2 -3 0 -1 [ ]Given transformation matrix is (M) = -1 0 AB C A' B' C' [ ] [ ] [0 -1 -3 -2 ]3 Now, -1 0 2 -1 2 -2 3 2 -3 = -2 1 Hence, the coordinates of image of the ∆ABC are A'(-3, -2), B'(-2, 1) and C'(3, -2). [ ]The matrix 0 -1 is associated with reflection on the line y = -x. -1 0 340 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 8. [Transform a unit square under the matrix 3 ]2 Solution: coordinates of the image. 1 1 and write the Let the vertices be O(0, 0), A(1, 0), B(1, 1) and C(0, 1) of a unit square. OA B C 01 10 [ ]Writing the given unit square in the matrix form 0 0 1 1 [ ]3 2 Transformation matrix, M = 1 1 . Now, M × object OABC = Image O'A'B'C'. OA B C O' A' B' C' [ ][ ] [ ]i.e., 1 1 0 0 1 1 = 0 1 2 1 32 0 1 1 0 03 52 Hence the coordinates of image vertices are O'(0, 0). A(3, 1), B'(5, 2), C(2, 1) Example 9. ∆PQR whose vertices are P(3, 6), Q(4, 2), and R(1, 1) maps onto ∆P'Q'R' Solution: with the vertices P'(-6, 3), Q'(-2, 4), and R'(-2, 2). Which is the single transformation for this mapping? Also find a 2 × 2 matrix that represents this transformation. Here, ∆PQR with vertices P(3, 6), Q(4, 2), and R(1, 1) is mapped onto ∆P'Q'R' with the vertices P'(-6, 3), Q(-2, 4), and R(-1, 1) Here, P(3, 6) P'(-6, 3) i.e. (x, y) (-y, x) Hence, the point (x, y) is transformed to (-y, x) which represents the rotation of +90° about the origin. [a ]b LetM = c d be the required 2×2 transformation matrix which transform ∆PQR to ∆P'Q'R' . Then, AB C P' Q' R' [ ] [ ] [ ]a b 2 -1 2 -6 -2 -1 c d . 3 2 -3 = 3 4 1 . [ ] [ ]or, 3c + 6d 4c + 2d c + d = 3 4 1 . 3a + 6b 4a + 2b a + b -6 -2 -1 Equating the corresponding elements, we get, 3a + 6b = -6 or, a + 2b = -2........ (i) 4a + 2b = -2 or, 2a + b = -1........ (ii) a + b = -1 ........ (iii) 3c + 6d = 3 or, c + 2d = 1 ....... (iv) 4c + 2d = 4 or, 2c + d = 2....... (v) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 341

vedanta Excel In Opt. Mathematics - Book 10 c + d = 1 ......... (vi) Solving equation (i) and (iii), we get, a + 2b = -2 a + b = -1 -- + b=-1 Put the value of 'a' in equation (iii), we get, a=0 Again solving equations (iv) and (vi), we get, c + 2d = 1 c+ d = -1 -- + d=0 Put the value of d in equation (vi), we get, c = 1 [ ]0 -1 ? M = 1 0 which is the required matrix. [Example 10. Find the 2 × 2 matrix which transforms the unit square 0 1 1 ]0 [ ]to a parallelogram03 4 1 0 0 1 0 1 3 2 1 . Solution: [Let M = a ]b be the required matrix. Then by question c d [ ][ ] [a b 0 1 1 0 = 0 3 4 ]1 cd 0 0 1 1 0 1 3 [ ] [ ]or, 2 0 a a+b b = 0 3 4 1 0 c c+d d 0 1 3 2 Equating the corresponding elements of equal matrices, we get, a = 3, b = 1, d = 2, c = 1 [ ].Required transformation matrix is M =3 1 1 2 Example 11. Prove by matrix method that the reflection in x-axis followed by rotation about (0, 0) through 180° is equivalent to the reflection on y-axis. Solution: The 2 × 2 transformation matrix representing the reflection on x-axis 0 1 -1 0 -1 0 -1 [ ] [ ]0 is and rotation about O(0, 0) through 180° is Now, the reflection followed by rotation about O(0, 0) through 180° is given 342 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 by [ ] [ ] [ ]-1 01 0 = -1 0 ............................. (i) 0 -1 0 -1 0 1 The 2 × 2 transformation matrix representing reflection in y-axis is [ ]-10 .............. (ii) 0 1 From (i) and (ii), we conclude that the reflection on x-axis followed by rotation about (0, 0) through 180° is equivalent to the reflection on y-axis. Example 12. A unit square is rotated about origin through 180°. Find the image of unit square using 2 × 2 matrix. Plot the graphs of unit square and its image on the same graph paper. Solution: Let O(0, 0). A(1, 0). B(1, 1) and C(0, 1) be the vertices of unit square OABC. OA B C [ ]Theunitsquarecanbeexpressedinthematrixformas. 0 1 1 0 0 0 1 1 -1 ]0 [The 2 × 2 matrix of rotation about origin through 180° is 0 -1 . C' OA B C O' A' B' [ ][ ] [-1 0 0 1 1 0 ]0 0 -1 -1 Now, 0 -1 0 0 1 1 = 0 0 -1 -1  ? The required image vertices of unit square are O'(0, 0), A'(-1, 0), B'(-1. -1). C'(0, -1). Graphs of unit square OABC and its image O'A'B'C' are drawn in the graph given below. Scale 20 Y lines - 1 unit B' C(0, 1) B(1, 1) X' A' X O A(1, 0) C' Y' Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 343

vedanta Excel In Opt. Mathematics - Book 10 Exercise 11.6 Very short Questions 1. Let P(x, y) be any point on the plane and P(x', y') be the image of P. Write the transformation matrices under the following cases: (a) (x', y') = (-y, - x) (b) (x', y') = (-x, -y) (c) (x', y') =(-x, -y) (d) (x', y') = (-x, y) (e) (x', y') = (y, x) (f) (x' , y') =(kx, ky) (g) (x', y') = (x + a, y + b) (h) (x', y') = (3x, 3y) 2. Write the type of transformation represented by each of the following transformation matrices: ( )1 0 ( )1 0 ( )0 1 (a) 0 1 (b) 0 -1 (c) 1 0 ( )0 -1 ( )0 -1 ( )0 1 (d) -1 0 (e) 1 0 (f) -1 0 ( )-1 0 ( )2 0 (4 )0 (g) 0 -1 (h) d 2 (i) 0 4 Find the image of P(4, 5) using above matrices in each care. Short Questions 3. Write down 2 × 2 transformation matrices in each of the following transformation and using the matrix find image of given point. (a) Reflection on y-axis, A(3, 4) (b) Reflection on x-axis. B(6, 3) (c) Reflection on line y = x , C(-4, -2) (d) Reflection on line x + y = 0, D(2, 3) (e) Rotation about origin through, +90°, E(-4, -4) (f) Rotation about origin through 180°, F(6, 7) (g) Enlargement with centre at the origin by scale factor 3, G(4, 7) [ ]4. (a) If a matrix value of a. a0 3 2 maps the point (3, 4) on to the point (15, 17), find the (b) If a point (x, y) is transformed into (y, x) by a 2 × 2 matrix, find the 2 × 2 transformation matrix. (c) Find a 2 × 2 matrix which transforms point (8, -4) to P'(4, 8). 2 [ ]5. (a) Find the image of A(6, 7), under the translation matrix 4 followed by 344 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 [ ]3 translation 2 . 34 [ ] [ ](b) Find the image of M(2. 3) under translation 2 followed by translation 2 . 2 ]0 [6. (a) To which transformation is the matrix M= 0 2 associated with ? What types of transformation do M2 and M3 represent ? [1 ]0 (b) What does the matrix 0 -1 represent? Find the image of P (4, 5) using the matrix. Long Questions 7. (a) P(2, 1), Q(5, 1), R(5, 4) and S(2, 4) are the vertices of a square PQRS. The square [1 ]2 is transformed by matrix 1 -2 into a parallelogram. Find the vertices the parallelogram. (b A triangle with vertices P(2, 1), Q(5, 3), and R(6, 7) is transformed by the matrix [ ]0 -1 -1 0 . Find the image of the triangle. (c) A square with vertices A(1, 1), B(6, 1), C(6, 6), and D(1, 6) is transformed by the [transformation matrix 3 ]0 0 3 . Find the image of the square ABCD. (d) A quadrilateral PQRS with vertices P(2, 1) Q(5, 2), R(6, 5) and S(3, 5) is transformed [ ]-1 0 by matrix 0 -1 find the image of the quadrilateral PQRS. 8. (a) A line PQ having P(5, 1), Q(8, 6) map's onto P'Q' having P'(-5, 1), Q(-8, 6) so that the image P'Q' is formed. What is the single transformation for the mapping ? Also find the 2 × 2 matrix. (b) A square ABCD with vertices A(2, 0), B(5, 1) C(4, 4) and D(1, 3) is mapped onto parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of the parallelogram are A'(2, 2), B'(7,3) C'(12, - 4) and D'(7, -5). Find the 2 × 2 matrix. (c) ∆ABC with the vertices A(2, 7), B(2, 9), and C(6, 7) is mapped onto ∆A'B'C' whose vertices are A'(7, 2), B(9, 2) and C'(7, 6). Which is the single transformation for the mapping? Also find the 2 × 2 matrix of transformation ? (d) ∆PQR with vertices P(5, 1), Q(12, 4) and R(4, 5) maps onto the ∆P'Q'R' with the vertices P'(-5, -1) Q'(-12, -4) and R'(-4, -5). Which is the single transformation for this mapping ? Also find the 2 × 2 matrix that represents the transformation. 9. (a) A unit square having vertices A(o. o), B(1, 0), C(1, 1) and D(0, 1) is mapped to the parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of parallelogram are A'(o,o), B'(3, 0), C(4, 1) and D'(1, 1). Find the 2 × 2 transformation matrix. (b) Find the 2 × 2 matrix which the unit square is transformed to parallelogram Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 345

vedanta Excel In Opt. Mathematics - Book 10 ( )0 3 4 1 . 0 1 3 2 [ ](c) 0 1 1 0 Find the 2 × 2 matrix which transforms the unit square 0 0 1 1 [ ]onto a parallelogram 0 4 6 2 . 0 1 3 2 [ ](d) A rectangle 02 20 is mapped onto a rectangle 00 -1 -1 [ ] [ ]0 0 2 2 0 a b , find the matrix of 0 1 1 by the 2 × 2 matrix c d transformation. (e) A unit square OMNP having vertices O(o, o), M(1, 0), N(1, 1), and P(0, 1) is transformed to unit, square O'M'N'P' through reflection y = -x. Write the vertices of the image unit square so formed. Draw the graphs of both of the units square on the same graph paper. 10. (a) Verify by the matrix method that the reflection on the line y = x followed by the reflection on the y-axis is equivalent to rotation about origin through +90°. (b) Verify by the matrix method that a reflection in the line X - axis followed by the Y - axis is the rotation about the origin through 180°. (c) Prove by the matrix method that the reflection on the line y = x followed by the rotation about the origin +90° is the reflection on y-axis. 3. (a) A'(-3, 4) (b) B'(6, -3) (c) C'(-2, -4) (d) D'(-3, -2) (e) E'(4, -4) (f) F'(-6, -7) (g) G'(12, 21) 4. (a) 5 5. (a) A'(11, 13) (b) 01 (c) 0 -1 10 10 (b) M'(9, 7) 6. (a) E[(0, 0), 2], M2 represents E[(0, 0), 4], M3 represents E[(0, 0), 8] (b) Reflection in X-axis, P'(4, -5) 7. (a) P'(4, 0), Q'(7 3), R'(13, -3), S'(10, -5) (b) A'(-1, -2), Q'(-3, -5), R'(-7, -6) (c) A'(3, 3), B'(18, 3), C'(18, 18), D'(3, 18) (d) P'(-2, -1), Q'(-5, -2) R'(-6, -5), S'(-3, -5) 8. (a) -1 0 (b) 12 (c) reflection on y = x, 0 1 0 1 1 -2 1 0 (d) Rotation of 180° about origin, -1 0 0 -1 9. (a) 3 1 (b) 31 (c) 42 (d) 10 0 1 12 12 0 -1 (e) O'(0, 0), M'(0, -1), N(-1, -1) 346 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 12Statistics 12.0 Measure of Dispersion Introduction Measures of central tendency gives us an idea of the concentration of observations about the central part of the distribution. It tells nothing about how each variate value of a set of data is scattered from an average value of those items. Two or more two data may have equal mean, median or mode but with more scattered than the other. It may be clear from the following two set of data. Averages X Md Mo Set I 63 64 65 66 66 67 68 69 66 66 66 Set II 64 65 66 66 66 66 67 68 66 66 66 The above two data sets have the same means, the same medians and same mode, all equal to 66. They have the same number of observations, n = 8. But these two data set are different. What is the main difference between them ? The two data sets have the same measure of central tendency (as measured by any of the three measures of central tendency - mean, median and model) but they have a different variability. In particular, we can see that data set I is more scattered than data set II. The values of data set I are more spread out: they lie farther away from their mean than do those data set II. Hence, the certain measures are evolved which reflect on the scattering of values of numerical terms are known as measures of dispersion. In statistics, the term dispersion is used commonly to mean scatteredness, variation, diversity, deviation, fluctuations, spread, heterogeneity, etc. Dispersion may be defined as the degree of scatteredness or variation of a variable about a central value. Less the value of dispersion more data will be representative. Dispersion of the data are measured in terms of the following measures of dispersion. 1. Range 2. Quartile Devication 3. Mean Deviation 4. Standard Deviation In this chapter we study about quartile deviation, mean deviation, and standard deviation for continuous series of data. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 347

vedanta Excel In Opt. Mathematics - Book 10 12.1 Quartile Deviation (Q.D) Quartile deviation is a measure of dispersion based on the upper quartile (Q3) and the lower quartile (Q1) of a set of data. The difference between the upper quartile (Q3) and the lower quartile (Q1) is called interquartile range. Interquartile range = Q3 - Q1 Half of the interquartile range is called semi-interquartile range or quartile deviation. It is denoted Q.D. and is given by the formula. Quartile Deviation (Q.D.) = 1 (Q3 - Q1) 2 It is an absolute measure of dispersion. It has the same unit as that of original data. For comparative study of two or more data sets, we calculate coefficient of quartile deviation. It is the relative measure of dispersion based the quartiles Q3 and Q1 Coefficient of Quartile Deviation = Q3 - Q1 Q3 + Q1 The coefficient of Q.D. is unitless. It is used to compare variability of two or more data sets. The data having less value of coefficient of quartile deviation is less variable or less heterogeneous or more uniform or more stable than other and vice-versa. Calculation of Quartile Deviation for continuous Series. For continuous data in the form of frequency distribution, the following formulas are used. For lower quartile or the first quartile (Q1) N Find the value of 4 for the first quartile class. N - c.f 4 Q1 = L + ×i f Where, (i) L is the lower limit of the class in which the first quartile lies. (ii) f is the frequency of the first quartile class. (iii) c.f. is the cumulative frequency of the class preceding the first quartile class. (iv) i is the length of class interval (or class - size). For the upper quartile or the third quartile (Q3) Find the value of 3N to find the third quartile class. 4 3N 4 - c.f Q3 = L + f ×i Where, (i) L is the lower limit of the class in which the third class lies. (ii) f is the frequency of the third quartile class. (iii) c.f. is the cumulative frequency preceding the third quartile class. (iv) i is the length of class interval (or class size). 348 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Worked Out Examples Example 1. The quartile deviation of a continuous data is 20. If the first quartile is 45, Solution: find the third quartile. Example 2. Given, Quartile Deviation (Q.D.) = 20 Solution: The first quartile (Q1) = 45 Example 3. Now, Q.D. = 1 (Q3 - Q1) 2 1 or, 20 = 2 (Q3 - Q1) or, 20 = 1 (Q3 - 45) or, 40 = Q3 - 45 2 ? Q3 = 85 The coefficient of quartile deviation of a grouped data is 0.25. If the upper quartile is 75, find the lower quartile. Coefficient of Q.D. = 0.25 The upper quartile (Q3) = 75 Now, coefficient of Q.D. = Q3 - Q1 Q3 + Q1 5041Q.215===27727755555++-- QQQQ1111 or, or, 75 + Q1 = 300 - 4Q1 ? Q1 = 45 or, or, Calculate the quartile deviation and its coefficient from the following data : Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 1 18 25 28 30 33 22 15 22 Solution: To calculate quartile deviation and its coefficient. Marks obtained No. of students c. f. 0-10 11 11 10-20 18 29 20-30 25 54 30-40 28 82 40-50 30 112 50-60 33 145 60-70 22 167 70-80 15 182 80-90 22 204 N = 204 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 349

vedanta Excel In Opt. Mathematics - Book 10 To find Q1 = N = 204 = 51 4 4 c.f., just greater 51 is 54. So, Q1 lies in the class (20 - 30). i.e. (20-30) is the first quartile class. L = 20, c.f = 29, f = 25, i = 10 N - c.f 4 Now, Q1 =L+ ×i f 51 - 29 = 20 + 25 × 10 = 20 + 8.8 = 28. 8 marks. To find Q3 , 3N = 3 × 204 = 153 4 4 c.f. just greater 153 is 167. Q3 lies in the class (60 - 70) i.e. (60 - 70) is the third quartile class. L = 60, c.f. = 145, f = 22, i = 10 3N - c.f 4 Now, Q3 =L+ ×i f Dev=iat6io0n+(Q.1D5)3=2-21214(5Q3×- Q110) = 60 + 3.64 = 63.64 marks Quartile = 1 (63.64 - 28.8) =17.42 2 Q3 - Q1 63.64 - 28.8 coefficient Quartile Deviation (Q.D) = 2 = 63.64 + 28.8 = 0.38 Example 4. Calculate the quartile deviation and its coefficient. x less than 10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80 and above f3 4684783 5 Solution: The given data is open ended. We can calculate deviation. x f cf Less than 10 3 3 4 7 10-20 6 13 20-30 8 21 30-40 4 25 40-50 7 32 50-60 8 40 60-70 3 43 70-80 5 48 80 and above N = 48 To find the first quartile (Q1) N = 48 = 12 4 4 350 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 c.f., just greater than 12 is 13. So, Q1 lies in (20 - 30) L = 20, c.f. = 7, f = 6, i = 10 N - c.f 12 - 7 50 4 6 6 Q1 =L+ × i = 20 + × 10 = 20 + f = 20 + 8.33 = 28.33 To find the upper quartile (Q3), 3N = 3 × 48 = 36 4 4 c.f. just greater 36 is 40. So, Q3 lies in (60 - 70) L = 60, c.f. = 32, f = 8, i = 10 3N - c.f 36 - 32 × 10 4 8 - 28.33) = Q3 =L+ ×i = 60 + Quartile f 65 Q1) = = 60 + 4 × 10 = (Q3 - 8 1 Deviation (Q.D) = 2 1 (65 18.34 2 Q3 - Q1 65 - 28.33 coefficient of quartile diviation = Q3 + Q1 = 65 + 28.33 = 0.393 Example 5. Calculate the quartile deviation and its coefficient from the given data : x 20-29 30-39 40-49 50-59 60-69 70-79 80 - 89 f 4 6 8 12 7 6 5 Solution: To calculate quartile deviation and its coefficient, given inclusive data is changed into exclusive. The correction factor is given by using the formula. Correction factor = Lower limit of the 2nd class – upper limit of first class 30 - 29 2 2 = = 0.5 So, the exclusive class of above data are given by xf cf 4 19.5 - 29.5 4 10 18 29.5 - 39.5 6 30 37 39.5 - 49.5 8 43 48 49.50 - 59.5 12 59.5 - 69.5 7 69.5 - 79.5 6 79.5 - 89.5 5 N = 48 To find the lower quartile (Q1), N = 48 = 12 4 4 c.f., just greater 12 is 18. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 351

vedanta Excel In Opt. Mathematics - Book 10  ? Q1 lies in class (39.5 - 49.5) L = 39.5, c.f. = 10, f = 8, i = 10 N - c.f 12 -10 4 8 Now, Q1 =L+ × i. = 39.5 + × 10 f 2 = 39.5 + 4 × 10 = 39.5 + 5 = 44.5 To find upper quartile (Q3) , 3N = 3 × 48 = 36 4 4 c.f., just greater 36 is 37. Hence, Q3 lies in the class (59.5 - 69.5) L = 59.5, c.f. = 30, f = 7, i = 10 3N - c.f 36 - 30 4 7 Now, Q3 =L+ × i. = 59.5 + × 10 f = 59.5 + 8.57 =68.07 Q.D. = 1 (Q3 - Q1) = 1 (68.,07 - 44.5) = 11.785 2 2 Q3 - Q1 coefficient of Q.D. = Q3 + Q1 = 68.07 - 44.5 = 0.2094 68.07 + 44.5 Exercise 12.1 Short Questions 1. (a) Define measure of dispersion. (b) List methods of measure of dispersion. (c) Define quartile deviation. (d) Write down formula to calculate Q.D. (e) Write down formula to calculate coefficient of Q.D. (f) Write a difference between Q.D. and its coefficient. 2. Calculate quartile deviation from following: (a) Q1 = 25, Q3 = 75 (b) Q1 = 25, Q3 = 65 (c) Q1 = 40, Q3 = 80 (d) Q1 = 5, Q3 = 75 3. Calculate the coefficient of quartile deviation from the following: (a) Q1 = 20, Q3 = 40 (b) Q1 = 40, Q3 = 80 (c) Q1 = 2, Q3 = 22 (d) Q1 = 2.5, Q3 = 22 4. (a) In a set of continuous data, if Q1 = 35, and quartile deviation is 20, find Q3. (b) The coefficient of quartile diviation of a grouped data is 0.25 and the upper quartile is 60. Find the value of lower quartile. 352 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (c) The coefficient of quartile deviation for continuous data is 0.39 and upper quartile 49.75. Find the first quartile. Long Questions : 5. Calculate the quartile deviation and its coefficient from the following data : (a) Marks 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 4 12 16 10 8 6 2 (b) Marks 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 3 5 6 8 4 4 3 (c) Volume of water (l) 5-10 10-15 15-20 20-25 25-30 30-40 No. of families 4 12 16 6 2 1 (d) Age (years) 0-5 5-10 10-15 15-20 20-25 25-30 No. of persons 5 20 15 20 10 2 (e) Age (years) 60-65 65-70 70-75 75-80 80-85 85-90 No. of students 7 5 8 4 3 3 (f) x 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 30-40 f 6 10 18 30 15 12 10 6 5 6. Calculate quartile deviation and its coefficient from the following data: (a) Marks Less than 30 30-40 40-50 50-60 60-70 70 and above No. of students 3 6954 2 (b) x Below 25 25-30 30-35 35-40 40-45 45 and above f 5 12 22 25 20 9 (c) x Below 30 30-40 40-50 50-60 60-70 70 and above f 2 4686 4 (d) Income (Rs.) Below 50 50-70 70-90 90-110 110- 130-150 150 above 130 Workers 5 10 20 25 18 12 10 7. Calculate the quartile deviation and its coefficient from the given data: (a) Class interval 20-29 30-39 40-49 50-59 60-69 70-79 Frequency 100 80 75 95 70 40 (b) Marks 0-9 10-19 20-29 30-39 40-49 50-59 No. of students 1 4 8 10 5 3 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 353

vedanta Excel In Opt. Mathematics - Book 10 8. (a) Calculate the semi-interquartile range and its coefficient of the following data: (prepare a frequency distribution table taking class interval 10) 70 55 51 42 57 45 60 47 63 53 33 65 39 82 55 64 58 61 65 42 50 52 54 45 45 25 36 59 63 39 65 45 49 54 64 75 42 41 52 35 30 35 15 48 26 20 40 55 46 18 (b) Prepare a frequency distribution table taking (0 - 4) as one of the class. The calculate the quartile deviation. 7, 3, 10, 4, 1, 9, 11, 18, 8, 5, 6,4, 13, 17,6,8, 12,17,19, 5, 3, 17, 16, 3, 2, 14, 13, 4, 10 2. (a) 25 (b) 20 (c) 20 (d) 35 3. (a) 0.33 (b) 0.33 (c) 0.83 (d) 0.7959 4. (a) 75 (b) 36 (c) 21.83 5. (a) 11.56, 0.23 (b) 13.18, 0.25 (c) 3.51, 0.22 (d) 5.13, 0.38 (f) 5.5, 0.27 (e) 6.31, 0.088 (b) 5.008, 0.137 (c) 10.83, 0.20 (d) 23.33, 0.22 6. (a) 10.21, 0.215 (b) 8.035, 0.251 7. (a) 13.79, 0.30 (b) 3.97, 0.3683 8. (a) 9.62, 0.1924 12.2 Mean Deviation Mean deviation is a measure of dispersion that is based on all the values of a set of data. It is defined as the arithmetic mean of the absolute deviations taken from central value (mean, median or mode). It shows the variation of the items from an average. Therefore, it is also called average deviation. Mean deviation is denoted by MD. While computing mean deviation, the deviation of items can be taken from mean, median or mode. However, it gives the best results when deviations are taken from the median. In calculation of mean deviation, negative sings are ignored and all the values are treated as positive. Calculation of Mean Deviation for Continuous Series : Let m be the mid values of corresponding classes in continuous series. Then, Mean Deviation from mean = ∑ f|m- x| Mean Deviation from median = ∑ Nf|m-Md| N ∑ f|m-Mo| Mean Deviation from mode = N Here, we study only mean deviation from mean or median 354 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Mean deviation is an absolute measure of dispersion. For computing two or more series having different units, we calculate coefficient of mean deviation. The relative measure of dispersion based on mean deviation is called coefficient of mean deviation. Coefficient of M.D. from mean = M.D. from mean mean M.D. from median Coefficient of M.D. from median = median Worked out Examples Example 1. In a grouped data if ∑f|m-x| =400, x = 20, N = 50, find M.D. and its Solution: coefficient. Example 2. Here, N = 50, x = 20, ∑f|m-x| =400 ∑ f|m - x | 400 Now, Mean Deviation from mean = 50 = 50 = 8 M.D. from mean 8 coefficient of M.D. from mean = = 20 = 0.4 Mean Calculate the mean deviation and it coefficient from the following data : (i) deviations from mean (ii) deviations from median. Class 0 - 10 10 -20 20 -30 30 -40 40 -50 Frequency 5 8 15 10 6 Solution: (i) To calculate Mean Deviation from mean. Class Mid-values Frequency (f) fm |m-x| f|m - x| (m) 104.5 0-10 5 5 25 20.9 87.2 13.5 10-20 15 8 120 10.9 91.0 114.6 20-30 25 15 375 0.9 6f|m - x| = 410.9 30-40 35 10 350 9.1 40-50 45 6 270 19.1 N = 44 6fm = 1140 Mean ( x ) = ∑fm = 1140 = 25.9 N 44 ∑f|m-x | Mean deviation from mean = 410N.9 = = 44 9.34 Coefficient of M.D. from mean = M.D. from mean mean 9.34 = 25.9 = 0.36 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 355

vedanta Excel In Opt. Mathematics - Book 10 (ii) Calculation of mean deviation from median: Class Mid-values Frequency (f) cf |m-md| f|m -md| (m) 0-10 5 5 5 21 105 10-20 15 8 13 11 88 20-30 25 15 28 1 15 30-40 35 10 38 9 90 40-50 45 6 44 19 114 N = 44 6f|m - md| = 412 To calculate median, N = 44 = 22 2 2 c.f just greater than 22 is 28. Hence median class is (20 -30). L = 20, c.f. = 13, f = 15, i = 10 N - c.f 2 Median (Md) =L+ ×i f 22 - 13 = 20 + 15 × 10 = 20 + 6 = 26 Now, M.D. from median = ∑f |m-Md| N 412 = 44 = 9.36 Exercise 12.2 Short Question 1. (a) Define Mean Deviation (b) Write down the formula to compute mean deviation for continuos series of data: (i) from mean (ii) from median (c) Write down the formula to compute the coefficient of mean deviation: (i) from mean (ii) from median 2. (a) In a grouped data, if ∑ f|m- x | = 400, N = 80, x = 40, find the M.D. and its coefficient . (b) In a continuous series of data, if ∑f |m-Md| = 250, Md = 22, N = 25, find the M.D. and its coefficient. 356 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Long questions 3. Calculate the mean, deviation from (i) mean (ii) median for the following data : (a) Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 3 4 9 2 4 (b) Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 10 12 25 35 40 50 (c) Marks 0-10 10-20 20-30 30-40 40-50 Number of students 9 6 4 12 9 (d) Class interval 0≤x≤10 10≤x≤20 20≤x≤30 30≤x≤40 Number of students 15 12 7 8 4. Calculate mean deviation from median and its coefficient for the following grouped data : (a) Marks 20-30 30-40 40-50 50-60 60-70 No. of students 5 7 8 6 4 (b) Marks 0-20 20-40 40-60 60-80 80-100 No. of students 5 6 7 10 12 (c) Marks 0-10 10-20 20-30 30-40 40-50 No. of students 5 8 15 16 6 (d) Profit (Rs) 0-10 10-20 20-30 30-40 40-50 50-60 No. of persons 10 12 25 35 40 50 5. Compute mean deviation and its coefficient from mean of the following date : (a) Mid point 2 6 10 14 18 22 26 Frequency 7 13 10 5 8 4 3 (b) Mid-value 5 15 25 35 45 55 No. of students 2 46864 6. Construct a frequency distribution table taking a class interval of 10 and calculate the M.D. from median. 28, 49 37, 5, 18, 14, 24, 7, 38, 46, 30, 21, 16, 31, 45, 27, 10, 4, 17, 29, 35, 36, 41, 47, 44, 33, 34, 17, 18, 20. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 357

vedanta Excel In Opt. Mathematics - Book 10 7. Compute the mean deviation its coefficients, from (i) mean and (ii) median from the given data : Age (years) Number of people Less then 10 10 Less than 20 25 Less than 30 40 Less than 40 45 Less than 50 70 Less than 60 85 Less than 70 100 Less than 80 110 Less than 90 120 2. (a) 5, 0.125 (b) 10, 0.45 (b) (i) 12.57, 0.33 (ii) 12.45, 0.28 3. (a) (i) 11.46, 0.35 (ii) 11.25, 0.34 (d) (i) 9.59, 0.56 (ii) 9.05, 0.60 (ii) 13.21, 0.43 (c) 9.56, 0.34 (d) 12.45, 0.30 (c) (i) 13.43. 0.51 (b) 23.60, 0.37 4. (a) 10.58, 0.24 (b) 11.73, 0.36 (ii) 19.75, 0.43 5. (a) 6.05, 0.36 7.(i) 16.32, 0.37 6. 12.67, 0.63 12.3 Standard Deviation Standard deviation is the positive square root of the mean of the square of the deviations about mean. It is also known as \"Root mean square deviation\" Standard deviation is denoted by Greek letter V (read as sigma). Among all the methods of finding out dispersion, standard deviation is regarded as the best because of the following reasons. 1. Its value of based on all the variate values. 2. The deviation of each variate is taken from the mean. 3. Algebraic sign of each deviation is considered. Calculation of Standard Deviation for Continuous Series : To calculate standard deviation for a continuous series, the following methods can be applied. (a) Direct Method In this method, the formulae used is ( )V = ∑fm2 - ∑fm 2 N N ..............................(I) where, m is mid-value of each class. 358 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 To use the above formula, follow the following steps. 1. Find the mid value of each class which is denoted by m. 2. Find ∑f = N 3. Find the products fm and ∑fm 4. Multiply fm by m to get fm2 and ∑fm2, 5. Substitute all above values of ∑fm, ∑fm2, N in formula (I), we get standard deviation of given data. (b) Actual Mean Method In this method, deviations of each mid point of class intervals are taken from the arithmetic mean. The formula used in this method is V = ∑f(m - x )2 ...(II) N Note : Above formulae (I) and (II) represent same direct method To use this formula, follow the steps below: 1. Find the mid point of each class which is denoted by m 2. Find arithmetic mean, x = ∑fm where N= ∑f N 3. Take deviations of each mid point from the arithmetic mean. i.e. (m - x ) 4. Square each of (m - x ), to get (m - x )2 5. Multiply each of (m - x )2 with their corresponding frequencies to get f(m - x )2 and obtain ∑f(m - x )2 6. Substitute the values of each of ∑f (m- x )2 and N in formulae (II), we get standard deviation. (c) Short-cut Method In this method, deviation of each of mid-point the class is taken from assumed mean. The fromula used in this method is ( )V = ∑fd2 - ∑fd 2 N N ......... (III) Where, d = m - a a = assumed mean m = mid point of each class . The following step is used in this method: 1. Find the mid-point of each class which is denoted by m. 2. Take middle figure of mid points as an assumed mean and denote it by a. Take deviations of each mid point from the assumed mean. Denote it by d i.e. d = m - a 3. Multiply each of d with the corresponding frequency f, to get fd and obtain ∑fd. 4. Multiply each of fd with the corresponding d to get fd2 and then find ∑fd2. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 359

vedanta Excel In Opt. Mathematics - Book 10 5. Substitute all the values of ∑fd2 , ∑fd, N in formula (III), we get standard deviation. (d) Step Deviation Method : This method is used mostly in the cases, when each of variety value has a common factor. In case of continuous series, the common factor, is h, the class-size. The formula used in this method is ( )V = ∑fd'2 - ∑fd' 2 N N × h ........... (IV) Where, d' = m -a h a = assumed mean m = mid point of each class h = class size The following steps are used in this method: 1. Find the mid point of each class which is denoted by m. 2. Take an assumed mean 'a' from the mid figure of mid-points. 3. Find deviation of each mid point from the assumed mean i.e. (m - a) 4. Divide each of (m - a) by h, denote it by d' = m -a h 5. Multiply each of d' by the corresponding frequency f to get fd' and find ∑fd' 6. Multiply each of fd' by d' and denote thus by fd'2 and find ∑fd'2 7. Substitute the values ∑fd' , ∑fd', ∑fd'2, N in formula (IV) to get standard deviation. Coefficient of Standard Deviation The relative measure of standard deviation is known as the coefficient of standard deviation. ? Coefficient of S.D. = standard deviation = V mean x If the coefficient of standard deviation is multiplied by 100%, then it is called coefficient of variation (C.V.). ? Coefficient of variation (c.v) = V × 100% x Varience The square of standard deviation is called varience. If standard deviation is V, the varience is V Example : If standard deviation of a set of data is 4, then the varience is V2 = 16. i,e, S.D. (V)= 4, varience (V2) = 42 = 16 360 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Worked out Examples Example 1. Calculate the standard deviation from given data by direct method. Class 40-50 50-60 60-70 70-80 80-90 90-100 20 14 Frequency 6 10 20 30 Solution: To calculate standard deviation by direct method : Example 2. Class Mid-value f fm fm2 Solution: 12150 40-50 45 6 270 30250 84500 50-60 55 10 550 168750 144500 60-70 65 20 1300 126350 6fm2 = 566500 70-80 75 30 2250 80-90 85 20 1700 90-100 95 14 1330 N = 100 6fm = 7400 ( )∑fm2- ∑fm 2 N N Standard deviation (V) = ( )= 566500 - 7400 2 100 100 = 5665 - 5476 = 13.74 Find the standard deviation from the following data by (a) Actual mean (or direct Method) (b) Short -cut method (c) Step deviation method Also find the coefficient of S.D. and coefficient of variation. x 0 -10 10-20 20-30 30-40 40-50 50-60 F 4 6 10 20 6 4 (a) Actual mean method (or direct method) To calculate standard deviation by actual mean method (or direct method). x Mid-value f fm |m-x| |m-x|2 f|m - x|2 4 0-10 5 6 20 -26 676 2704 10-20 15 10 90 -16 256 1536 20-30 25 20 250 -6 36 360 30-40 35 6 700 4 16 320 40-50 45 4 270 14 196 1176 50-60 55 220 24 576 2304 N = 50 6fm = 6fm = 6f|m - x|2 7400 1550 = 8400 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 361

vedanta Excel In Opt. Mathematics - Book 10 Mean (x ) = ∑fm = 1550 = 31 N 50 ∑f(m - x )2 Standard deviation = N = 8400 168 = 12.96 50 = coefficient of S.D. (V) = V × 100% x = 12.96 × 100% = 41.81% 31 (b) To calculate standard deviation by shortcut method. Let = 35 x Mid-value f d=m–d fd fd2 0-10 5 4 -3 -120 3600 -120 2400 10-20 15 6 -20 -100 1000 20-30 25 10 -10 30-40 35 20 0 00 40-50 45 6 10 60 600 50-60 55 4 20 80 1600 N = 50 6fd = -200 6fm = 1550 ( )∑fd2– ∑fd 2 N N Standard deviation (V) = = ( )9200– – 200 2 50 50 = 184 – 16 = 168 = 12.96 (c) To calculate standard deviation by step-deviation method x Mid-value (m) f m – 35 fd' fd'2 10 0-10 5 4 -3 -12 36 10-20 15 6 -2 12 24 20-30 25 10 -1 -10 10 30-40 35 20 0 0 0 40-50 45 61 6 6 50-60 55 4 2 8 16 N = 50 6fd' = -20 6fd'2 = 92 We have from above table, N = 50, ∑fd' = –20, ∑fd'2 Standard deviation (V) = ( )∑fd'2–∑fd' 2 N N ×h 362 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 ( )= 92 – –20 2 × 10 50 50 = 46 – –4 × 10 25 25 = 42 × 10 = 1.296 × 10 = 12.96 25 Example 3. Calculate the standard deviation from the given data Class 0 -10 10-20 20-30 30-40 40-50 Frequency 4 10 15 18 20 Solution: To calculate standard deviation, the given data can be written in the following table: x Mid-value f d = m– a fd fd2 h 16 0-10 5 4 -2 -8 6 0 10-20 15 6 -1 -6 3 8 20-30 25 = a 5 0 0 6fd2 = 38 30-40 35 31 3 40-50 45 22 4 N = 50 6fd = -7 standard deviation (V) = ( )∑fd2 – ∑fd 2 NN ×h where, h = 10, ∑fd'2 = 33, ∑fd' = –7 ( )= 33 – –7 2 20 20 × 10 ? V = 1.65 – 0.1225 × 10 = 12.36 Exercise 12.3 Very Short Question 1. (a) Define standard deviation (b) Write the formulas to calculate standard devition for continuous series of data by (i) direct method (ii) actual mean method (iii) Short-cut method (iv) step-deviation method (c) Define coefficient of variation with formulae (d) Define varience Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 363

vedanta Excel In Opt. Mathematics - Book 10 Short questions 2. Calculate standard deviation and its coefficient for continuous series, if (a) ∑fm2 = 566500, N = 100, ∑fm = 7400, x = 31 (b) ∑fd2 = 993, N = 60, x = 37, ∑fd = –130 (c) ∑fd' = –20, ∑fd'2 = 92, h = 10, x = 37, N = 100 (d) ∑fd' = –4, ∑fd'2 = 28, N = 29, x = 24.5, i = 10 Long Questions 3. Calculate the standard deviation and its coefficient from the following data by (i) direct method (ii) short-cut method (iii) step-deviation (a) Class Interval 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 10 8 32 40 22 18 (b) Marks 0-10 10-20 20-30 30-40 40-50 No. of students 7 12 24 10 7 (c) Wages (Rs.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of Workers 12 18 35 42 50 45 20 8 (d) Age (yrs.) 20-25 25-30 30-35 35-40 40-45 45-50 No. of persons 170 110 80 45 40 45 4. Calculate the standard deviation and coefficient of variation: (a) Mark Secured 0-20 20-40 40-60 60-80 80-100 No. of Students 2 8 16 10 4 (b) Height (cm) 0-8 8-16 16-24 24-32 32-40 No. of plants 6 7 10 8 9 5. Calculate the standard deviation and coefficient of variation from the following: (a) x 0≤x<10 10≤x<20 20≤x<30 30≤x<40 40≤x<50 f 7 10 14 12 6 (b) Mid-value 4 8 12 16 20 24 Frequency 10 15 11 16 14 5 (c) x less than 10 less than 20 less than 30 less than 40 less than 50 f 12 19 24 33 40 (d) x above 20 above 40 above 60 above 80 above 100 and less than 120 f 50 42 30 18 7 364 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (e) Marks 0-10 0-20 0-30 0-40 0-50 54 No. of students 7 18 30 42 6. From the data given below which series is more variable (inconsistent). Variable 10-20 20-30 30-40 40-50 50-60 60-70 Section A 10 18 32 40 22 18 Frequency 18 22 40 32 20 10 Section B Project work List the marks obtained by the students in your class in optional mathematics and compulsory mathematics. Find the standard deviation, coefficient of standard deviation and coefficient of variation (C.V.). Compare the marks obtained in these two subjects in terms of C.V. 2. (a) 13.74, 0.4432 (b) 3.44, 0.093 (c) 9.34, 0.2535 (d) 9.7, 0.3959 3. (a) 13.72, 0.3156 (b) 11.39, 0.4617 (c) 17.25, 0.426 (d) 8.22, 0.269 4. (a) 20.27, 38.24 (b) 10.86, 50.75 5. (a) 12.29, 49.16% (b) 6.07, 45.47% (c) 15.03, 75.15% (d) 25.79, 37.41% (e) 12.87, 50.27% CV(A') = 33.34%, CV(B) = 37.02% 6. V(A) = 14.05, V(B) = 14.10 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 365

vedanta Excel In Opt. Mathematics - Book 10 Curriculum Development Centre, Sanothimi, Bhaktapur Syllabus of Optional Mathematics - Grade 10 S.N. Content Course Title Tentative 1. Algebra periods 2. Continuity Functions: Algebraic and trigonometrical functions with 35 3. Matrix 4. Coordinate their graphs of the types y = mx + c, y = ax2, a ≠ 0; y = ax3, Geometry a ≠ 0; y = sinA, y = cosA, y = tanA(–2S ≤ A ≤ 2S). Composite function of two functions and inverse functions with their presentation in arrow diagrams. Polynomials: Simple operation of polynomials, Synthetic division, Remainder and Factor theorem and their applications; Solutions of polynomial equations upto degree three using Remainder and Factor theorems. Sequence and Series: Arithmetic sequence - Introduction, General term, Mean, Sum of the terms of an arithmetic sequence, Sum of the first n natural numbers including odd and even natural numbers. Geometric Sequence: Introduction, General term, Mean, Sum of the finite terms of a geometric sequence. Linear Programming: Introduction, Linear inequalities, Detriment of inequality by graph, Maximization and minimization of linear programming problems. Quadratic Equations and Graphs: Graphs of quadratic 10 and cubic functions, Solution of quadratic equations using graph, Solution of simultaneous linear and 15 quadratic equations by graphical and substitution 30 method. Simple Concept of Continuity: Investigation of continuity in set of real numbers, Investigation of continuity and discontinuity in different types of set of numbers, Investigation of the continuity of the function using graph, Symbolic representation of the continuity of the function. Determinant of matrix of order 2 × 2, Inverse of a 2 × 2 matrix, Solution of simultaneous linear equations using matrix method, Cramer's rule, and its application upto a determinant of order 2 × 2. Straight Line: Angle between two straight lines, Condition for two lines to be parallel and perpendicular. Pair of Straight Lines: Equations of the lines given by the homogeneous equation of degree two and the angle between them. Conditions for the two lines to be coincident and perpendicular. 366 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Conic Section: Introduction, Different types of conic section formed by the intersection of a cone and a plane. Circle: Definition, Equation of the circle of the types: x2 + y2 = r2, (x – h)2 + (y – k)2 = r2, (x – x1) (x – x2) + (y – y1) (y – y2) = 0 and x2 + y2 + 2gx + 2fy + c = 0 and the related problems. 5. Trigonometry Trigonometric Identities: Trigonometric ratios of 35 18 multiple and submultiple angles (sine, cosine and 15 tangent only), Transformation of identities into the sum, 12 difference and the product form (in sine and cosine only) Conditional Identities: Solution of the trigonometric identities under the condition A + B + C = Sc Trigonometric Equations: Solution of trigonometric equation (upto degree two) (0 ≤ T < 2Sc) Height and Distance: Word problems of height and distance consisting of two angle - angle of elevation and angle of depression. 6. Vectors Scalar product of two vectors (dot product), Condition for two vectors to be perpendicular. Vector Geometry: Mid-point theorem, Section formula. Theorems: (i) The straight line joining the middle points of two sides of a triangle is parallel to and half of the third side. (ii) The straight line joining the vertex and the middle point of the base is perpendicular to the base. (iii) The straight lines joining the middle points of the sides of a quadrilateral taken in order is a parallelogram. (iv) The diagonals of a parallelogram bisect each other. (v) The diagonals of a rectangle are equal. (vi) The diagonals of a rhombus bisect each other at right angles. (vii) The angle in the semi-circle is a right angle. (viii)The middle point of the hypotenuse of a right angled triangle is equidistant from the vertices. 7. Transformation Coposite transformation of any of two transformations. Reflection, Rotation, Translation and Enlargement, Inversion transformation, Inverse circle, Use of matrix in transformation. 8. Statistics Dispersion: Quartile deviation and its coefficient (continuous series only), Mean deviation from mean and median and its coefficient (continuous series only), Standard deviation and its coefficient and analysis (continuous series only), coefficient of variation (C.V.). Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 367

vedanta Excel In Opt. Mathematics - Book 10 SEE Specification Grid 2076 Issued by CDC K U A HA SN Contents Topics Each Each Each Each TQ TM of 1 of 2 of 4 of 5 Mark Marks Marks Marks 1. Algebra Function Polynominals 2 3 2 1 8 21 Sequence and Series Quadratic Equation and Graph 2. Limit and Numbers and Continuity Continuity Discontinuity in Graph 1 - 1 - 25 Notational Representation of Continuity 3. Matrix Determinant and Inverse of Matrix Solving Equations by Matrix 12 1 - 49 Method Cramer's Rule 4. Coordinate Angle between two lines Geometry Pair of straight lines 2 2 1 1 6 15 Conic Sections Circle 5. Trigonometry Multiple and sub-multiple angles Transformation of Trigonometric Identies Conditional Trigonometric 23 3 - 8 20 Identities Trigonometric Equations Height and Distance 6. Vectors Scalar Product 1 2 - 1 4 10 Vector Geometry 7. Transformation Combined Transformation Inversion Transformation and 1 - 1 1 3 10 Inversion Circle Matrix Transformation 8. Statistics Quartile Deviation Mean Deviation -12 - 3 10 Standard Deviation and Coefficient of Variation Total 10 13 11 4 38 100 INDEX U = Understanding A = Application K = Knowledge HA = Higher Ability TQ = Total no. of Questions TM = Total Marks 368 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 SEE New Model Questions 2076 Issued by CDC-2076 (2019) Group A (10 × 1 = 10) 1. (a) Define trigonometric function. (b) What is the arithmetic mean between two numbers 'a' and 'b'? 2. (a) Write a set of numbers which is continuous in number line. (b) If matrix A = p q , what is the value of |A|? r s 3. (a) If the slopes of two straight lines are m1 and m2 respectively and T be the angle between them, write the formula for tanT. (b) Which geometric figure is formed if a plane intersects a cone parallel to its base? 4. (a) Express sin2A in terms of tanA. (b) Define angle of elevation. 5. (a) What is the scalar product of two vectors oa and ob if the angle between them is T? (b) In an inversion transformation if P' is image of P and r is radius of inversion circle with centre O, write the relation of OP.OP' and r. Group B (13 × 2 = 26) 6. (a) Find f-1(x) if f(x) = 4x + 5. (b) If g(x) = 2x – 1 and f(x) = 4x, find the value of gof(x). (c) What are the points of intersection of the curve f(x) = x2 – 1 and f(x) = 3? 7. (a) If A = 2 –1 , find |A| and write if A–1 is defined. 3 1 (b) According to Cramer's rule, find the values of D1 and D2 for ax + by = c and px + qy = r. 8. (a) Find the slopes of two straight lines 3x + 4y + 5 = 0 and 6x + 8y + 7 = 0 and write the relationship between them. (b) Find the single equation for the pair of lines represented by 3x + 2y = 0 and 2x – 3y = 0. 9. (a) Convert sin6A . cos4A into sum or difference of sine or cosine. (b) Express 1 sinA in terms of sub-multiple angle of tangent. + cosA (c) If 2sin2T = 3, find the value of T. (0° ≤ T ≤ 180°) 10. (a) Find the angle between two vectors oa and ob if |oa | = 2, |ob | = 12 and oa .ob = 12. A (b) From the adjoining figure, find AoP and express op in terms of oa and ob . oa op P (c) If the standard deviation of a set of data is 0.25, find its O ob B variance. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 369

vedanta Excel In Opt. Mathematics - Book 10 Group C (11 × 4 = 44) 11. Solve : x3 – 3x2 – 4x + 12 = 0 12. Optimize P = 5x + 4y under the given constraints : x – 2y ≤ 1, x + y ≤ 4, x ≥ 0, y ≥ 0 13. For a real valued function f(x) = 2x + 3 (a) Find the values of f(2.95), f(2.99), f(3.01), f(3.05), and f(3). (b) Is this function continuous at x = 3? 14. By using matrix method, solve the following system of equations: 3x + 5y = 11, 2x – 3y = 1 15. Find the single equation of pair of straight lines passing through the origin and perpendicular to the lines represented by 2x2 – 5xy + 2y2 = 0. 16. Find the value of : sin20° . sin30° . sin40° . sin80° 17. If A + B + C = Sc, prove that sin2A – sin2B + sin2C = 2sinA.cosB.sinC 18. From a point at the ground level infront of a tower, the angle of elevations of the top and bottom of flagstaff 6 m high situated at the top of a tower are observed 60° and 45° respectively. Find the height of the tower and the distance between the base of the tower and point of observation. 19. Find the 2 × 2 matrix which transforms a unit square to a parallelogram 0 34 1 . 0 01 1 20. Find the mean deviation from mean and its coefficient from given data: Marks obtained 0-10 10-20 20-30 30-40 40-50 No. of students 2 3 6 5 4 21. Find the standard deviation and coefficient of variation from given data: Age 0-4 4-8 8-12 12-16 16-20 20-24 No. of students 7 7 10 15 7 6 Group D (4 × 5 = 20) 22. A contractor on construction job specifies a penalty for delay of completion beyond a certain date as: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day and so on. The penalty for each succeeding day is Rs. 50 more than that of the preceding day. How much money does the contractor have to pay as penalty, if he delays the work by 30 days? 23. On a wheel, there are three points (5, 7), (–1, 7) and (5, –1) located such that the distance from a fixed point to these points is always equal. Find the coordinate of the fixed point and then derive the equation, representing the locus that contains all three points. 24. By using vector method, prove that the quadrilateral formed by joining the mid-points of adjacent sides of a quadrilateral is a parallelogram. 25. The coordinates of vertices of a quadrilateral ABCD are A(1, 1), B(2, 3), C(4, 2), and D(3, –2). Rotate this quadrilateral about origin through 180°. Reflect this image of quadrilateral about y = –x. Write the name of transformation which denotes the combined transformation of above two transformations. 370 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur