Edexcel Core Maths C1 January Minutes

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1 Edexcel Core Maths C1 January Minutes

2 (a) Write down the value of 16. Answer: 4 [ Question says write down the answer so no working out is expected. x 1/2 should be recognised as x ] 3 2 (b) Find the value of 16. (1) Answer: 1/64 [ x -n = 1/x n. 16 3/2 = /2 by law of exponents. Seeing 16 1/2 linking with part(a) confirms we are on the right track ] 2. (i) Given that y = 5x 3 + 7x + 3, find (a) d y, dx [ We can differentiate expressions term by term. If we differentiate ax n we get anx n-1 ] 2 d y (b) 2. dx (1) [ Write it down in two steps to show we know that to get the second derivative, we differentiate the first derivative again. That way we have the chance of the mark even if we make a slight mistake] 1 (ii) Find 1 3 x dx. 2 x (4) N23490A 2

3 where C is a constant of integration [ We can integrate term by term. The integral of ax n is ax n+1 /(n+1) ] 3. Given that the equation kx x + k = 0, where k is a positive constant, has equal roots, find the value of k. (4) Suppose the repeated root is a, then the equation is Compare this with the given equation, divided through by k Each power of x must be equal, so and but the question says k is positive, so k=6 Check: put this value of k into the original equation: which has a double root at x=-1, consistent with our value of a [ A quadratic with roots a and b can be factored as (x-a)(x-b)=0. If the roots are equal (sometimes called a double root), this is (x-a)2=0. Saying and means that if a is positive, k is negaitive, and vice versa. The check is not needed, but shows we can do it, and gives us confidence our answer is correct. ] 4. Solve the simultaneous equations x + y = 2 x 2 + 2y = 12. The first equation implies y=2-x. Substitute this for y in the second equation; we will get a quadratic: (6) N23490A 3 Turn over

4 so x=4 and y=2 4 so y= -2, or x=-2 and y=2- -2 so y =4 [Re-arrange the linear equation to express one variable in terms of the other, then substitute this in the quadratic, so we only have one variable. Solve the quadratic, (giving two roots), and substitute into the linear equation. We get two solutions two pairs of x and y ] 5. The rth term of an arithmetic series is (2r 5). (a) Write down the first three terms of this series. -3, -1 and 1 [Write down, so no working. Just use r=1, r=2 and r=3 ] (b) State the value of the common difference. +2 [State means no working. The question really means do you know what common difference means? ] n (c) Show that ( 2r 5) = n(n 4). r 1 The sum of an arithmetic progression = n/2(first term + last) so (1) [This is an a.p. and means sum of. We use the standard formula for the sum of an a.p., substitute in the first and last terms, and re-arrange] N23490A 4

5 6. Figure 1 y O 2 4 x P(3, 2) Figure 1 shows a sketch of the curve with equation y = f(x). The curve crosses the x-axis at the points (2, 0) and (4, 0). The minimum point on the curve is P(3, 2). In separate diagrams sketch the curve with equation (a) y = f(x), [Each point (x,y) on f(x) is replaced by (x,-y) on f(x) a reflection in the x axis ] N23490A 5 Turn over

6 (b) y = f(2x). On each diagram, give the coordinates of the points at which the curve crosses the x-axis, and the coordinates of the image of P under the given transformation. 7. The curve C has equation y = 4x x x, x 0. The point P on C has x-coordinate 1. (a) Show that the value of dy dx at P is 3. (5) At P x=1 so [We re-arrange y to make (5-x)/x easier to differentiate. Then use derivative of ax n is anx n-1. Then substitute the value of x] (b) Find an equation of the tangent to C at P. At P, y=4+4/1 = 8 The tangent is the straight line through (1,8) with slope 3. The equation of this is N23490A 6

7 [We find the co-ordinates of P. The gradient of the tangent is dy/dx. Then use the equation for a straight line through a givemn point with given slope : y-y 1 =m(x-x 1 ) ] This tangent meets the x-axis at the point (k, 0). (c) Find the value of k. At this point, y=0, so 0=3x+5 x=-5/3 so k=-5/3 N23490A 7 Turn over

8 8. Figure 2 y A(1, 7) B(20, 7) D(8, 2) O C( p, q) The points A(1, 7), B(20, 7) and C( p, q) form the vertices of a triangle ABC, as shown in Figure 2. The point D(8, 2) is the mid-point of AC. (a) Find the value of p and the value of q. The x difference from A to D is 7, which is the same from D to C. So p=8+7=15 The y difference from A to D is -5, so q=2-5 = -3 The line l, which passes through D and is perpendicular to AC, intersects AB at E. (b) Find an equation for l, in the form ax + by + c = 0, where a, b and c are integers. Slope of AC = -5/7 So slope of perpendicular =7/5 so equation of AE is (5) [If two lines are perpendicular, their slopes are related by m 1 m 2 =-1 ] (c) Find the exact x-coordinate of E. All points on AE have y=7. So for the point on this line where y=7, so x=, and E is ( N23490A 8

9 9. The gradient of the curve C is given by dy dx = (3x 1) 2. The point P(1, 4) lies on C. (a) Find an equation of the normal to C at P. At C, dy/dx=(3-1) 2 =4 So slope of the normal is -1/4 Equation is (4) (b) Find an equation for the curve C in the form y = f(x). (5) where c= a constant. This goes through (1,4), so so c=3, and f(x)= dy (c) Using = (3x 1) 2, show that there is no point on C at which the tangent is parallel to the dx line y = 1 2x. Slope of this line is -2, but there is no value of x such that (3x 1) 2 could be negative 10. Given that f(x) = x 2 6x + 18, x 0, (a) express f(x) in the form (x a) 2 + b, where a and b are integers. If then N23490A 9

10 So comparing the term in x 2, -6=-2a and so a=3 Comparing the constant term, 18=9+b so b=9 The curve C with equation y = f(x), x 0, meets the y-axis at P and has a minimum point at Q. (b) Sketch the graph of C, showing the coordinates of P and Q. At P, x=0 so y=18 At Q, dy/dx=0, so (4) so x=3 and y= =9 The line y = 41 meets C at the point R. (c) Find the x-coordinate of R, giving your answer in the form p + q 2, where p and q are integers. At R, so (5) N23490A 10

11 END TOTAL FOR PAPER: 75 MARKS N23490A 11