Cambridge A Level Mathematics 9709 — 2023 May/June Paper 2 · Variant 1

1 Use logarithms to solve the equation 12x Give your answer correct to 3 significant figures. = 32x+1. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Apply logarithms to both sides and apply power law correctly at least once *M1 OE with x not in a power. Obtain ln12 (2 1)ln3   x x A1 Attempt solution of linear equation DM1 Obtain 3.82 A1 Do not condone incorrect use of logarithms or greater accuracy. 4

More questions on Logarithmic and exponential functions

2 3 ln x 2 A curve has equation y . + 1 2x = + Find the equation of the tangent to the curve at the point 1, 2 . Give your answer in the form 3 ax by c 0, where a, b and c are integers. [5] + + = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Use quotient rule (or equivalent) to find first derivative M1 Obtain 3 2 (1 2 ) (2 3ln )2 (1 2 )     x x x x A1 OE Substitute 1  x and obtain 5 9 A1 Attempt equation of tangent through 2 3 (1, ) with their numerical gradient M1 Must have made an attempt at differentiation. Obtain 5 9 1 0   x y or equivalent of required form A1 5

More questions on Differentiation

a 3 It is given that 3e2x dx 12, where a is a positive constant. Ó 0 −1 = (a) Show that a 1 ln 9 23a . [4] 2 = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Integrate to obtain the form 2 1 2 e  x k k x 1 2 0 k k . Obtain correct 2 3 2 e  x x A1 Use limits correctly and attempt rearrangement at least as far as 2 e ...  a DM1 For DM1, must be equated to 12 and simplified using a correct method. Do not condone verification. Confirm given result 1 2 2 3 ln(9 )   a a with sufficient detail A1 AG 4 3(b) Use iteration process correctly at least once M1 Need to see 1.13434 and 1.13895 . Obtain final answer 1.139 A1 Final answer needed to exactly 4 sf. Show sufficient iterations to 6 sf to justify answer or show a sign change in interval [1.1385, 1.1395] A1 3

More questions on Integration

5 y M x O The diagram shows the curve with parametric equations x 4e2t, y cos 2t, = = 5e−t for The curve has a maximum point M. −14π ≤t ≤14π. dy (a) Find an expression for in terms of t. [3] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the coordinates of M, giving each coordinate correct to 3 significant figures. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Use product rule to find d d y t Obtain d 5e cos2 10e sin2 d     t t y t t t A1 or (unsimplified) equivalent (do not condone poor use of brackets. Obtain 2 d 5e cos2 10e sin 2 d 8e      t t t y t t x A1 OE following their expression for d d y t . 3 5(b) Equate d d y x to zero and simplify at least as far as tan 2 ...  t M1* now condoning any error with d d x t . Obtain 1 2 tan 2  t A1 Obtain 0.231...  t A1 allow 0.232 t  . Substitute negative value of t in expressions for x and y DM1 Obtain 2.52  x and 5.64  y A1 or greater accuracy. 5

More questions on Differentiation

1 @ A 3π 1 3 16 Show that 4 cos2 2x dx 3 [7] 4 Ô 1 + cos2x = + 6π −1. 4π ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 Express 2 4cos 2x in the form 1 2 cos4  k x k M1 where 1 2 0 k k . Obtain correct 2cos4 2  x A1 Allow unsimplified. State or imply 2 2 1 sec cos  x x B1 Maybe implied by integration. Integrate to obtain 3 4 sin 4 tan   k x k x x *M1 where 3 4 0  k k . Obtain correctly 1 2 sin 4 2 tan   x x x A1 Use limits correctly with correct values of 4 3 sin π and 1 3 tan π indicated DM1 Confirm given result 3 1 4 6 3 π 1   with sufficient detail A1 AG 7

More questions on Integration

7 (a) Express 7 cos 24 sin in the form R cos , where R 0 and Give the value of correct to12+decimal1 places. 1 −! > 0Å < ! < 90Å. [3] ! ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Solve the equation 7 cos 24 sin 18 for [4] 1 + 1 = 0Å < 1 < 360Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) As varies, the greatest possible value of " 150 7 cos 1 24 sin 1 50 2" + 2" + is denoted by V. Find the value of V and determine the smallest positive value of (in degrees) for which the value of V occurs. " [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) State 25 R Use appropriate trigonometry to find  M1 Allow if found in radians . Obtain 73.74  A1 or greater accuracy. 3 Question Answer Marks Guidance 7(b) Use correct method to find one value of  M1 Obtain 29.8 (or 117.7) A1 or greater accuracy. Use correct method to find second value of  between 0 and 360 M1 Obtain 117.7 (or 29.8) A1 or greater accuracy; and no others between 0 and 360. 4 7(c) State or imply expression is 1 2 150 25cos( 73.74) 50   B1 FT following their R and . Obtain 6  V B1 Attempt complete method to find positive value from 1 2 cos( 73.74) 1   M1 for their . Obtain 507.5 A1 or greater accuracy. 4

More questions on Trigonometry