Additional Mathematics CXC 2014 Solved

Recall that x x1  x2  b  b 2  4ac 2a Problem 1a(iii) 1 x 3 2 1 y  x 3 2 1 x  y 3 2 1 x3 y 2 2  x  3  y g ( x)  b  b 2  4ac b  b 2  4ac  2a 2a b  b 2  4ac  b  b 2  4ac 2a b  b x1  x2  2a 2b x1  x2  2a b x1  x2  a therefore x1  x2   g 1 ( x)  2  x  3 1b Problem 1a (i) Show that it is not a one to one. 1 x  0 2 x2  1 1  x  2 2  x 1 x  1 We can see that when x =  1, we get a result of Zero, this would represent 2 a -5 -2 1 +2 -1.5 4 -3 -3.5 a3 a Many to one and not a one-to-one. Problem 1(c) 1 2 1 2 x  x  x  x  48 2 2 x 2  2 x  48  0 x 2  8 x  6 x  48  0 x 2  8 x   6 x  48   0 x  x  8  6  x  8  0  x  8  x  6   0  x  8 and x  6 1 ( 8) 2  32 2 1 y  ( 6) 2  18 2 y Problem 2 (a) Problem 1a (ii) f ( x)  2 x 2  12 x  9 expressed in Standard Form Given : f ( x)  1  x 2 1 g ( x)  x  3 2 ; Standard Form f ( x)  a( x  h) 2  k . f ( x)  2 x 2  12 x  9  2( x 2  6 x)  9 factor a -2 from the first two terms. Evaluate  2  x 2  6 x  (32 )   9  2(32 ),Complete the square, 2 1  fg ( x)  1   x  3  2  1  1   1  x  3  x  3 2  2  note we added half of 6 squared then subtracted that same amount to balance the equation.  2  x  3  9  18 2 3 3 1   1   x2  x  x  9  2 2 4  1 3 3  1  x 2  x  x  9 4 2 2 1   x 2  3x  8 x  4 Problem 2b 3  5x  2 x2  0  2 x 2  6 x  1x  3  0 2  6 x   1x  3  0 2 x  x  3  1 x  3  0  2 x  1 x  3  0  2 x  1  0  x  3  0 x3 2 Therefore the vertex is at (h, k )  (3,9). Problem 3a (i) Given x  3y  6 kx  2 y  12 Express both lines in the slope -intercept form.  2 x2  5x  3  0  2 x  2  x  3   9 1 x 2 x 1 y    2 ; note that the gradient would be  . 3 3  kx y 6 2 Therefore if they are perpendicular, their gradients must be negative reciprocals..  k 3 2 k  6 P roblem 2c(i) Problem 3a (ii) If the Series is Geometric then it must have a common ratio Simultaneous Equations Elimination Method 6(x  3 y  6) Mutliply by 6 Find the ratio, using the recursive Formula: -6x  2 y  12 a1  an r 0.02  0.2 r 6x +18y =36 0.02 0.2 r  0.1 -6x + 2y =12 Add vertically r Proof: 0.2 0.1  0.02 20y = 48 y = 2.4 Therefore x  3y  6 0.02 0.1  0.002 x  3(2.4)  6 0.02 0.1  0.0002 x  6  7.2 P roblem 2c(ii) a S 1 1 r 0.2 S 1  0.1 0.2 S 0.9 2 S  10 9 10 S  2  10 10 9 2 S 9 Problem 5(a) y  3  4 x  x 2 , Point P(3, 6) lies on the curve. To find the equation of tangent. y  - x2  4 x  3 y '  2 x  4 y '(3)  2(3)  4 y '(3)  2 , Gradient of tangent Equation of tangent line. y  mx  c y  2 x  c 6  2(3)  c c  12  y  2 x  12 In the form of ax  by  c  0  2 x  y  12  0 x  1.2 The center of the Circle would be at the intersection of both lines , that is at (1.2, 2.4). Equation of the line : ( x - (-1.2)) 2  ( y - 2.4) 2  r 2 ( x + 1.2) 2  ( y - 2.4) 2  r 2 Problem 6(b)  Evaluate 3   4 cos x  2sin x dx 0   3 3   4 cos x  dx    2sin x dx 0 0  3  3 4   cos x  dx  2   sin x dx 0 0  4 sin x   2   cos x    4sin x  2 cos x 0 3       4sin  2 cos    4sin 0  2 cos 0 3 3    3  1    4    2      4(0)  2(1)   2     2   2 3 1 0  2  2 3 1 Problem 6(c) Problem 5 (b) f ( x)  2 x  9 x  24 x  7 3 2 (i ) f '( x )  6 x 2  18 x  24 f "( x)  12 x  18 At Stationary Point dy  6 x 2  1, then we need to integrate to dx find the function of the curve. Remember the derivative if gives you the gradient function. Integrate the gradient function, f '( x)  0. takes us back to the original function. 6 x 2  18 x  24  0 6x 2  6x  24 x   6 x  24   0  1 dx  2 x 3  x  C We can find the value of the constant by using point (x,y) that were given. 6x  x  4  6  x  4  0 y  2 x3  x  C  x  4  6 x  6   0  5  2(2)3  2  C x-4  0 and 6x  6  0 x4 and x  -1  5  16  2  C  5  14  C  5  14  C f ( x)  2 x 3  9 x 2  24 x  7 f ( x )  2 x 3  9 x 2  24 x  7 f (4)  2(4)3  9(4) 2  24(4)  7 f (1)  2(1)3  9(1) 2  24(1)  7 f (4)  105 f (-1)  20 Stationary Points location: (ii )  C  19  f ( x)  2 x 3  x  19, the equation of the curve. Find the area of the finite region by the curve , the x-axis, the line x = 3 and the line x = 4. Integrate the function of curve.  4, 105 and  1, 20  3 f "( x ) = 12 x  18  0 f "(4)  12(4)  18 f "(1)  12(1)  18 f "(4)  30 f "(-1)  - 30 suggest a minimum suggest a maximum  (4) 4 (4) 2   (3) 4 (3) 2     76      57  2 2  2   2  256 16 81 9        76      57  2  2  2 2   44  21 4  xx 2  65  2  dx Area of the definite region under the curve is 65 units 2 . 2 4 3  2 x  dx 2  x4 x2     4 1 4 2  44 42   24 22        4 1 4 1  256 16  16 4       1   4 1  4  48  0  48  (4) 4 (4) 2   (3) 4 (3) 2     19(4)      19(3)  2 2  2   2   128  8  76   40.5  4.5  57  Problem 6(a) x 4  x4 x2  3 2 x  x  19 dx   2  2  19 x  3 4 Nature of Points: Evaluate 2 Problem 7(c) n 1 2 n  1 51  1 Q2    26th term = 71 2 2 Quartile : (i ) median  (i ) P(Math and Biology)  (ii ) P (Biology only)  n  1 51  1   13th term = 56 4 4 3  n  1 3  51  1 Q3    39th term = 79 4 4 Q1  Problem 7(a) 9 60 11 60 Problem 7(b) (a) (b) 6 1  36 6 6 1  36 6 Prepared by: Mr. Leovany López, B.Sc. Mathematics Sacred Heart College, San Ignacio, Cayo, Belize