This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Hey bilgate, good to see you again. Here are the solutions for the questions from the image: Question No. 3 (i) Given the functions f: x 2x + 3 and g: x x - 1. So, f(x) = 2x + 3 and g(x) = x - 1. (a) Find f(-2) Step 1: Substitute x = -2 into the function f(x). f(-2) = 2(-2) + 3 f(-2) = -4 + 3 f(-2) = -1 The value of f(-2) is -1. (b) Find g^-1(x) Step 1: Let y = g(x), so y = x - 1. Step 2: Swap x and y to find the inverse relation. x = y - 1 Step 3: Solve for y to express g^-1(x). y = x + 1 The inverse function g^-1(x) is x + 1. (c) Find fg(x) Step 1: Substitute g(x) into f(x). fg(x) = f(g(x)) = f(x - 1) Step 2: Replace x in the expression for f(x) with (x - 1). fg(x) = 2(x - 1) + 3 fg(x) = 2x - 2 + 3 fg(x) = 2x + 1 The composite function fg(x) is 2x + 1. (d) Find the value of x, for which f(x) = g(x) Step 1: Set the expressions for f(x) and g(x) equal to each other. 2x + 3 = x - 1 Step 2: Solve the equation for x. 2x - x = -1 - 3 x = -4 The value of x is -4. (ii) Using only a pencil, ruler and a pair of compasses, (a) Draw the line PQ = 6cm Draw a straight line segment and mark two points P and Q on it such that the distance between them is 6 cm. (b) Construct the perpendicular bisector of PQ to meet PQ at M 1. Place the compass needle at P and open it to a radius greater than half the length of PQ. Draw arcs above and below the line segment PQ. 2. Without changing the compass radius, place the needle at Q and draw arcs above and below PQ, intersecting the first set of arcs. 3. Draw a straight line connecting the two points where the arcs intersect. This line is the perpendicular bisector of PQ. 4. Mark the point where the perpendicular bisector intersects PQ as M. (c) Draw the line PS = 5cm, where S is a point on the perpendicular bisector 1. Place the compass needle at P and open it to a radius of 5 cm. 2. Draw an arc that intersects the perpendicular bisector (constructed in part b). 3. Mark the point of intersection on the perpendicular bisector as S. 4. Draw a straight line segment connecting P and S. The length of PS will be 5 cm. --- Question No. 4 (i) Given the function f(x) = x^2 - 3x - 4, for -2 x 5: (a) Copy and complete the following table Step 1: Calculate f(x) for each given value of x using the formula f(x) = x^2 - 3x - 4. • For x = -2: f(-2) = (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 • For x = -1: f(-1) = (-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0 • For x = 0: f(0) = (0)^2 - 3(0) - 4 = 0 - 0 - 4 = -4 • For x = 1: f(1) = (1)^2 - 3(1) - 4 = 1 - 3 - 4 = -6 • For x = 2: f(2) = (2)^2 - 3(2) - 4 = 4 - 6 - 4 = -6 • For x = 3: f(3) = (3)^2 - 3(3) - 4 = 9 - 9 - 4 = -4 • For x = 4: f(4) = (4)^2 - 3(4) - 4 = 16 - 12 - 4 = 0 • For x = 5: f(5) = (5)^2 - 3(5) - 4 = 25 - 15 - 4 = 6 Step 2: Complete the table with the calculated values. | x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---|---|---| | y = f(x) | 6 | 0 | -4 | -6 | -6 | -4 | 0 | 6 | (b) Find the minimum value of y = f(x) and the corresponding value of x Step 1: The minimum value of a quadratic function ax^2 + bx + c occurs at x = -(b)/(2a). For f(x) = x^2 - 3x - 4, we have a=1 and b=-3. x = -(-3)/(2(1)) = (3)/(2) = 1.5 Step 2: Substitute x = 1.5 into f(x) to find the minimum value. f(1.5) = (1.5)^2 - 3(1.5) - 4 f(1.5) = 2.25 - 4.5 - 4 f(1.5) = -6.25 The minimum value of y = f(x) is -6.25 at x = 1.5. (c) Solve the equation f(x) = 0 Step 1: Set the function f(x) equal to 0. x^2 - 3x - 4 = 0 Step 2: Factor the quadratic equation. (x - 4)(x + 1) = 0 Step 3: Solve for x. x - 4 = 0 or x + 1 = 0 x = 4 or x = -1 The solutions for f(x) = 0 are x = -1 or x = 4. (ii) Given a 2 × 2 matrix, P defined by P = 3 & 2k \\ k & 6 , where k is a constant. (a) Find P when k = 2 Step 1: Substitute k = 2 into the matrix P. P = 3 & 2(2) \\ 2 & 6 P = 3 & 4 \\ 2 & 6 The matrix P when k=2 is 3 & 4 \\ 2 & 6 . (b) Hence or otherwise, find Adj. P, the adjoint of P. Step 1: For a 2 × 2 matrix A = a & b \\ c & d , its adjoint is given by Adj A = d & -b \\ -c & a . Step 2