# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#10

Question f(x) = x2 + 4kx + (3+11k), where k is a constant. a)   Express f(x) in the form (x + p)2 + q, where p and q are constants to be found in terms of k.  Given that the equation f(x) = 0 has no real roots, b)  find the set of possible values of […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#9

Question a)   Factorise completely x3 – 4x b)  Sketch the curve C with equation y = x3 – 4x showing the coordinates of the points at which the curve meets the x-axis. The point A with x-coordinates -1 and the point B with x-coordinate 3 lie on the curve C. c)   Find an equation of the line which […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#8

Question Figure 1 shows a sketch of part of the curve with equation y = f(x). The curve has a maximum point  (−2, 5) and an asymptote y = 1, as shown in Figure 1. On separate diagrams, sketch the curve with equation a.   y = f(x) + 2 b.   y = 4f(x) c.   y = […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#7

Question Jill gave money to a charity over a 20-year period, from Year 1 to Year 20 inclusive. She gave £150  in Year 1, £160 in Year 2, £170 in Year 3, and so on, so that the amounts of money she gave each  year formed an arithmetic sequence. a.   Find the amount of money she gave […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#6

Question The curve C has equation  , x>0 a.  Find  in its simplest form. b.  Find an equation of the tangent to C at the point where x=2. Solution a.   We are given; We are required to find . Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#5

Question Solve the simultaneous equations Solution We are given simultaneous equations; We rearrange the first equation to find y in terms of x. Substituting this  from first equation in the second equation; We have the algebraic formula; Therefore; Now we have two options. By substituting one-by-one these values of  in first equation, we can find corresponding values of . […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#4

Question  , x>0 Given that y=35 at x=4, find y in terms of x, giving each term in its simplest form. Solution We are given that; We are given coordinates of appoint that is y=35 at x=4 ie  (4,35). We are required to find the equation of y in terms of x. We can find equation of the […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#3

Question The line  has equation 3x+5y-2=0. a.   Find the gradient of line for . The line  is perpendicular to  and passes through the point (3,1). b.   Find the equation of  in the form y=mx+c, where m and c are constants.. Solution a.   We are given equation of the line  as; We are required to find the gradient of the line […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#2

Question a.  Expand and simplify   b.  Express  in the form , where a and b are integers. Solution a.     We are given; Since ; b.     We are given; If we need a rational number in the denominator of a fraction, we need to follow procedure of  “denominator rationalization” as given below. ü If the denominator is of the […]

# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#1

Question Given that  , find . Solution We are given; We are required to find . Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is: Therefore; Rule for differentiation is of  is: Rule for differentiation is of  is: Rule for differentiation is of  is: