Cambridge A Level Mathematics 9709 — 2007 May/June Paper 2 · Variant 1

1 Solve the inequality |x −3| > |x + 2|. [4]

Mark scheme: 1 EITHER State or imply non-modular inequality (x – 3)2 > (x + 2)2, or corresponding equation M1 Expand and solve a linear inequality, or equivalent M1 1 Obtain critical value 2 A1 1 1 State correct answer x < (allow x Y ) 2 2 A1 OR State a correct linear equation for the critical value, e.g. 3 – x = x + 2, or corresponding correct inequality, e.g. –(x – 3) > (x + 2) M1 Solve the linear equation, or inequality M1 1 Obtain critical value 2 A1 1 State correct answer x < 2 A1 OR Make recognisable sketches of both y = x − 3 and y = x + 2 on a single diagram B1 Obtain a critical value from the intersection of the graphs M1 1 Obtain critical value 2 A1 1 State final answer x < 2 A1 [4] ( )

More questions on Algebra

2 The variables x and y satisfy the relation 3y = 4x+2. (i) By taking logarithms, show that the graph of y against x is a straight line. Find the exact value of the gradient of this line. [3] (ii) Calculate the x-coordinate of the point of intersection of this line with the line y = 2x, giving your answer correct to 2 decimal places. [3]

Mark scheme: 2 (i) State or imply y ln 3 = ( x + 2 ) ln 4 B1 State that this is of the form ay = bx + c and thus a straight line, or equivalent B1 ln 4 State gradient is , or equivalent (allow 1.26) ln 3 B1 [3] (ii) Substitute y = 2x and obtain a linear equation in x M1* Solve for x M1(dep*) Obtain answer 3.42 A1 [3]

More questions on Logarithmic and exponential functions

3 The parametric equations of a curve are x = 3t + ln(t −1), y = t2 + 1, for t > 1. dy (i) Express in terms of t. [3] dx (ii) Find the coordinates of the only point on the curve at which the gradient of the curve is equal to 1. [4]

Mark scheme: 3 (i) dx 1 dy State = 3 + or = 2t dt t − 1 dt B1 dy dy dx Use = ÷ dx dt dt M1 dy 2t (t − 1) Obtain in any correct form, e.g. dx 3t − 2 A1 [3] (ii) Equate derivative to 1 and solve for t M1 1 Obtain roots 2 and A1 2 State or imply that only t = 2 is admissible c.w.o. A1 Obtain coordinates (6, 5) A1 [4] GCE A/AS LEVEL – May/June 2007 9709 02

More questions on Differentiation

4 The polynomial 2x3 −3x2 + ax + b, where a and b are constants, is denoted by p(x). It is given that (x −2) is a factor of p(x), and that when p(x) is divided by (x + 2) the remainder is −20. (i) Find the values of a and b. [5] (ii) When a and b have these values, find the remainder when p(x) is divided by (x2 −4). [3]

Mark scheme: 4 (i) Substitute x = 2, equate to zero, and state a correct equation, e.g. 16 – 12 + 2a + b = 0 B1 Substitute x = –2 and equate to –20 M1 Obtain a correct equation, e.g. –16 – 12 – 2a + b = –20 A1 Solve for a or for b M1 Obtain a = –3 and b = 2 A1 [5] (ii) Attempt division by x2 – 4 reaching a partial quotient of 2x – 3, or a similar stage by inspection B1 Obtain remainder 5x – 10 B1√ + B1√ [3]

More questions on Quadratics

5 (i) By sketching a suitable pair of graphs, show that the equation secx = 3 −x, where x is in radians, has only one root in the interval 0 < x < 12π. [2] (ii) Verify by calculation that this root lies between 1.0 and 1.2. [2] (iii) Show that this root also satisfies the equation 1 x = cos−1 . [1] 3 −x (iv) Use the iterative formula 1 xn+1 = cos−1 , 3 −xn with initial value x1 = 1.1, to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 5 (i) Make recognisable sketch of a relevant graph, e.g. y = secx B1 Sketch an appropriate second graph, e.g. y = 3 –x, correctly and justify the given statement B1 [2] (ii) Consider sign of secx – (3 – x) at x = 1 and x = 1.2, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Show that the given equation is equivalent to secx = 3 –x, or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.04 A1 Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.035, 1.045) B1 [3] 1 1

More questions on Trigonometry

6 (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 13π cos2x dx = 1 + 1 √3. [4] 6π 8 0 (iii) By using an appropriate trigonometrical identity, deduce the exact value of 3π! sin2x dx. 0 [3]

Mark scheme: 1 1 6 (i) State correct expression + cos 2 x , or equivalent B1 [1] 2 2 (ii) Integrate an expression of the form a + b cos 2 x , where ab ≠ 0 , correctly M1 1 1 State correct integral x + sin 2 x , or equivalent A1 2 4 Use correct limits correctly M1 Obtain given answer correctly A1 [4] (iii) Use identity sin 2 x = 1 − cos 2 x and attempt indefinite integration M1  1 1  Obtain integral x −  x − sin 2 x  , or equivalent A1  2 4  1 3 Use limits and obtain answer π − A1 [3] 6 8 [Solutions that use the result of part (ii), score M1A1 for integrating 1 and A1 for the final answer.] GCE A/AS LEVEL – May/June 2007 9709 02

More questions on Trigonometry

7 The diagram shows the part of the curve y = ex cos x for 0 ≤x ≤12π. The curve meets the y-axis at the point A. The point M is a maximum point. (i) Write down the coordinates of A. [1] (ii) Find the x-coordinate of M. [4] (iii) Use the trapezium rule with three intervals to estimate the value of 12π ex cos x dx, 0 giving your answer correct to 2 decimal places. [3] (iv) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii). [1]

Mark scheme: 7 (i) State coordinates (0, 1) for A B1 [1] (ii) Differentiate using the product rule M1* Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1* 1 Obtain x = π or 0.785 (allow 45°) A1 [4] 4 (ii) Show or imply correct ordinates 1, 1.4619…, 1.4248…, 0 B1 1 Use correct formula or equivalent with h = π and four ordinates M1 6 Obtain correct answer 1.77 with no errors seen A1 [3] (iv) Justify statement that the trapezium rule gives and underestimate B1 [1]

More questions on Differentiation