Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 2 · Variant 2

2 (a) Sketch, on the same diagram, the graphs of y x 3 and y 2x . [2] = + = −1 (b) Solve the equation x 3 2x . [3] + = −1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 12y 12y 3 2 5 . Give your answer correct to 3 significant (c) Find the value of y such that 5 + = . × −1. figures. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis *B1 Must be straight lines. Draw (more or less) correct graph of 3 = + y x with smaller gradient together with a V shaped graph DB1 And crossing y-axis above y-intercept of first graph. Intersection in the first quadrant may be implied. 2 Question Answer Marks Guidance 2(b) Solve 3 2 1 + = − x x to obtain 4 = x B1 Attempt solution of linear equation where signs of 2x and xare different M1 Obtain 2 3 = − x A1 Alternative method for question 2(b) State or imply non-modulus equation 2 2 ( 3) (2 1) + = − x x B1 Attempt solution of 3-term quadratic equation obtained from squaring both terms. M1 Must have B1. Obtain 2 3 − and 4 A1 3 2(c) Apply logarithms and use power law for 1 2 5 = y k where 0 > k M1 Using their positive root from part (b). Allow M1 for 5 2log 4 = y . Obtain 1.72 = y A1 AWRT; and no other values. 2

More questions on Quadratics

3 The curve with equation y 5x tan 2x = −2 has exactly one stationary point in the interval 0 1 ≤x < 4π. Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Differentiate to obtain first derivative of form 2 1 2 sec 2 + k k x 1 2 0 ≠ k k . Obtain correct 2 5 4sec 2 − x A1 Equate first derivative of form 2 1 2 sec 2 + k k x to zero and obtain value for 2 cos 2x M1 OE Or value for 2 tan 2xwhich must be > 0. –1 ⩽ cos 2x ⩽ 1. Obtain 2 cos2 5 = x or (decimal) equivalent A1 Or 1 2 tan2 = x Obtain 0.232 = x A1 AWRT Obtain 0.159 = y A1 AWRT 6

More questions on Differentiation

a+14 1 4 Given that dx ln 2, find the value of the positive constant a. [5] 3x = Ôa ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 k x Apply limits and obtain 1 1 3 3 ln( 14) ln ln2 + − = a a or 1 1 3 3 ln(3 42) ln(3 ) ln 2 + − = a a A1 OE Use one relevant logarithm property correctly M1 Apply correct process to obtain equation without logarithms M1 OE. M0 if incorrect logarithm property used. Obtain 14 8 + = a a or equivalent and hence 2 = a A1 5

More questions on Integration

5 A curve has equation x2 4x cos 3y 6. + = Find the exact value of the gradient of the normal to the curve at the point 2, 1 . [6] 12π ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 to obtain 1 2 d cos3 sin3 d + y k y k x y x *M1 1 2 0 ≠ k k Obtain d 4cos3 12 sin3 d − y y x y x A1 Allow unsimplified. Obtain d 2 4cos3 12 sin3 0 d + − = y x y x y x A1 Allow unsimplified. Substitute x- and y-values to find value of first derivative DM1 Apply d 1/ d − y x to find gradient of normal M1 Obtain 3 2 − or 3 2 2 − or exact equivalent A1 6

More questions on Differentiation

6 (a) By sketching a suitable pair of graphs on the same diagram, show that the equation ln x = 2e−x has exactly one root. [2] (b) Verify by calculation that the root lies between 1.5 and 1.6. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Show that if a sequence of values given by the iterative formula xn+1 = e2e−xn converges, then it converges to the root of the equation in part (a). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Use the iterative formula in part (c) to determine the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) Draw correct sketch of ln = y x or 2e− = x y B1 = 2e− = x y must extend into 1st and 2nd quadrants. Draw correct sketch of second curve and indicate one root B1 Point of intersection must be circled/identified or a statement such that ‘there is only one point of intersection so one root only’ or similar. 2 Question Answer Marks Guidance 6(b) Consider sign of ln 2e− − x x , or equivalent, for 1.5 and 1.6 M1 Obtain 0.04... − and 0.06... or equivalents and justify conclusion A1 2 6(c) Replace 1 + nx and nx by x and apply logarithms to confirm result B1 AG Allow if done ‘in reverse’ but nx and 1 + nx need to be seen in the final statement. 1 6(d) Use iteration process correctly at least once M1 Need to see 3 correct values. Obtain final answer 1.54 A1 Answer required to exactly 3sf Show sufficient iterations to 5sf to justify answer or show sign change in interval [1.535, 1.545] A1 3

More questions on Logarithmic and exponential functions