Cambridge A Level Mathematics 9709 — 2011 May/June Paper 3 · Variant 3

giving your answer correct to 3 significant figures.1 Use logarithms to solve the equation 52x−1 = 2(3x), [4]

Mark scheme: 1 Use law for the logarithm of a product, power or quotient M1* Obtain a correct linear equation, e.g. (2 x − 1) ln 5 = ln 2 + x ln 3 A1 Solve a linear equation for x M1(dep*) Obtain answer x = 1.09 A1 [4] x x  25  [SR: Reduce equation to the form a = b M1*, obtain   = 10 Al, use correct method to  3  calculate value of x M1(dep*), obtain answer 1.09 A1.]

More questions on Logarithmic and exponential functions

ln x 2 The curve y has one stationary point. Find the x-coordinate of this point. [4] = x3

Mark scheme: 2 Use correct quotient or product rule M1 3 ln x 1 Obtain correct derivative in any form, e.g. − + A1 x 4 x 4 Equate derivative to zero and solve for x an equation of the form ln x = a , where a > 0 M1 1 Obtain answer exp( ), or 1.40, from correct work A1 [4] 3 1 1 ( ) − x − x ∫

More questions on Differentiation

1 2x 23 Show that 4e −1 [5] dx = −2. ã 0 (1 −x)e−1

Mark scheme: x x , or equivalent M13 Attempt integration by parts and reach k (1 − x )e 2 ± k ∫ e 2 d 1 1 − x − x x , or equivalent A1 Obtain − 2(1 − x )e 2 − 2 ∫ e 2 d 1 1 − x − x Integrate and obtain − 2(1 − x )e 2 + 4 e 2 , or equivalent A1 Use limits x = 0 and x = 1, having integrated twice M1 Obtain the given answer correctly A1 [5]

More questions on Integration

4 (i) Show that the equation k tan(60◦+ θ) + tan(60◦−θ) = can be written in the form (2 √3)(1 + tan2θ) = k(1 −3 tan2θ). [4] (ii) Hence solve the equation 3√3, tan(60◦+ θ) + tan(60◦−θ) = giving all solutions in the interval [3] 0◦≤θ ≤180◦.

Mark scheme: 4 (i) Use tan(A ± B) formula correctly at least once and obtain an equation in tanθ M1 Obtain a correct horizontal equation in any form A1 Use tan60° = 3 throughout M1 Obtain the given equation correctly A1 [4] 1 (ii) Set k = 3 3 and obtain tan2θ = B1 11 Obtain answer 16.8° B1√ Obtain answer 163.2° B1√ [3] [Ignore answers outside the given interval. Treat answers in radians (0.293 and 2.85) as a misread.] GCE AS/A LEVEL – May/June 2011 9709 33 1

More questions on Trigonometry

5 The polynomial ax3 bx2 5x where a and b are constants, is denoted by It is given that + + −2, p(x). is a factor of and that when is divided by the remainder is 12. (2x −1) p(x) p(x) (x −2) (i) Find the values of a and b. [5] (ii) When a and b have these values, find the quadratic factor of [2] p(x).

Mark scheme: 1 5 (i) Substitute x = and equate to zero, or divide, and obtain a correct equation, e.g. 2 1 1 5 a + b + − 2 = 0 B1 8 4 2 Substitute x = 2 and equate result to 12, or divide and equate constant remainder to 12 M1 Obtain a correct equation, e.g. 8a + 4b + 10 – 2 = 12 A1 Solve for a or for b M1 Obtain a = 2 and b = –3 A1 [5] 1 2 (ii) Attempt division by 2x – 1 reaching a partial quotient ax + kx M1 2 Obtain quadratic factor x2 – x + 2 A1 [2] [The M1 is earned if inspection has an unknown factor Ax2 + Bx + 2 and an equation in A 1 2 and/or B, or an unknown factor of ax + Bx + C and an equation in B and/or C.] 2

More questions on Quadratics

6 (i) By sketching a suitable pair of graphs, show that the equation cotx 1 x2, = + 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 0.5 and 0.8. [2] (iii) Use the iterative formula xn+1 = tan−1 1 1 x2n + to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Make recognisable sketch of a relevant graph over the given range B1 Sketch the other relevant graph and justify the given statement B1 [2] 2 x = 0.5 and x = 0.8, or equivalent M1 (ii) Consider the sign of cot x − (1 + x ) at Complete the argument with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once with 5.0 ≤ x n ≤ 8.0 M1 Obtain final answer 0.62 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.615, 0.625) A1 [3]

More questions on Trigonometry

7 (i) Find the roots of the equation 4 0, ß2 + (2p3)ß + = giving your answers in the form x iy, where x and y are real. [2] + (ii) State the modulus and argument of each root. [3] (iii) Showing all your working, verify that each root also satisfies the equation ß6 = −64. [3]

Mark scheme: 7 (i) Use the quadratic formula, completing the square, or the substitution z = x + iy to find a root and use i2 = –1 M1 Obtain final answers − 3 ± i , or equivalent A1 [2] (ii) State that the modulus of both roots is 2 B1√ 5 State that the argument of − 3 + i is 150° or π (2.62) radians B1√ 6 5 State that the argument of − 3 − i is –150° (or 210°) or – π (–2.62) radians or 6 7 π (3.67) radians B1√ [3] 6 (iii) Carry out an attempt to find the sixth power of a root M1 Verify that one of the roots satisfies z6 = –64 A1 Verify that the other root satisfies the equation A1 [3] GCE AS/A LEVEL – May/June 2011 9709 33

More questions on Complex numbers

8 y M 1 x O 2p The diagram shows the curve y 5 sin3x cos2x for 0 2π, and its maximum point M. = ≤x ≤1 (i) Find the x-coordinate of M. [5] (ii) Using the substitution u cos x, find by integration the area of the shaded region bounded by the = curve and the x-axis. [5]

Mark scheme: 8 (i) Use product and chain rule M1 Obtain correct derivative in any form, e.g. 15 sin 2 x cos 3 x − 10 sin 4 x cos x A1 Equate derivative to zero and obtain a relevant equation in one trigonometric function M1 Obtain 2 tan 2 x = 3 , 5 cos 2 x = 2 , or 5 sin 2 x = 3 A1 Obtain answer x = 0.886 radians A1 [5] du (ii) State or imply d u = − sin x d x , or = − sin x , or equivalent B1 dx Express integral in terms of u and du M1 Obtain ± 5(u 2 − u 4 ) ∫ d u , or equivalent A1 1 Integrate and use limits u = 1 and u = 0 (or x = 0 and x = π ) M1 2 2 Obtain answer , or equivalent, with no errors seen A1 [5] 3 dx ( )( )

More questions on Differentiation

9 In a chemical reaction, a compound X is formed from two compounds Y and Z. The masses in grams of X, Y and Z present at time t seconds after the start of the reaction are x, 10 and 20 −x −x respectively. At any time the rate of formation of X is proportional to the product of the masses of Y dx and Z present at the time. When t 0, x 0 and 2. = = dt = (i) Show that x and t satisfy the differential equation dx dt = 0.01(10 −x)(20 −x). [1] (ii) Solve this differential equation and obtain an expression for x in terms of t. [9] (iii) State what happens to the value of x when t becomes large. [1]

Mark scheme: dx 9 (i) State or imply = k (10 − x )(20 − x ) and show k = 0.01 B1 [1] dt (ii) Separate variables correctly and attempt integration of at least one side M1 1 A B Carry out an attempt to find A and B such that ≡ + , or (10 − x )(20 − x ) 10 − x 20 − x equivalent M1 1 1 Obtain A = and B = − , or equivalent A1 10 10 1 1 Integrate and obtain − ln (10 − x ) + ln (20 − x ) , or equivalent A1√ 10 10 Integrate and obtain term 0.01t, or equivalent A1 Evaluate a constant, or use limits t = 0, x = 0, in a solution containing terms of the form a ln (10 − x ) , b ln (20 − x ) and ct M1 1 1 1 Obtain answer in any form, e.g. − ln (10 − x ) + ln (20 − x ) = .001t + ln 2 A1√ 10 10 10 Use laws of logarithms to correctly remove logarithms M1 Rearrange and obtain x = 20(exp (1.0t ) − 1) / (2 exp (1.0t ) − )1 , or equivalent A1 [9] (iii) State that x approaches 10 B1 [1] GCE AS/A LEVEL – May/June 2011 9709 33

More questions on Differential equations

10 With respect to the origin O, the lines l and m have vector equations r 2i k and = + + λ(i −j + 2k) r 2j 6k 2j respectively. = + + µ(i + −2k) (i) Prove that l and m do not intersect. [4] (ii) Calculate the acute angle between the directions of l and m. [3] (iii) Find the equation of the plane which is parallel to l and contains m, giving your answer in the form ax by d. [5] + + cß =

Mark scheme: 10 (i) EITHER: Express general point of l or m in component form, e.g. (2 + λ, –λ, 1 + 2λ) or (µ, 2 + 2µ, 6 – 2µ) B1 Equate at least two pairs of components and solve for λ or for µ M1 1 Obtain correct answer for λ or µ (possible answers for λ are –2, , 7 and for 4 1 1 µ are 0, 2 , − 4 ) A1 4 2 Verify that all three component equations are not satisfied A1 OR: State a relevant scalar triple product, e.g. (2i – 2j – 5k) . ((i – j + 2k) × (i + 2j – 2k)) B1 Attempt to use the correct method of evaluation M1 Obtain at least two correct simplified terms of the three terms of the expansion of the triple product or of the corresponding determinant, e.g. –4, –8, –15 A1 Obtain correct non-zero value, e.g. –27, and state that the lines do not intersect A1 [4] (ii) Carry out the correct process for evaluating scalar product of direction vectors for l and m M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain answer 47.1° or 0.822 radians A1 [3] (iii) EITHER: Use scalar product to obtain a – b + 2c = 0 B1 Obtain a + 2b – 2c = 0, or equivalent, from a scalar product, or by subtracting two point equations obtained from points on m, and solve for one ratio, e.g. a : b M1* Obtain a : b : c = –2 : 4 : 3, or equivalent A1 Substitute coordinates of a point on m and values for a, b and c in general equation and evaluate d M1(dep*) Obtain answer –2x + 4y + 3z = 26, or equivalent A1 OR1: Attempt to calculate vector product of direction vectors of l and m M1* Obtain two correct components A1 Obtain –2i + 4j + 3k, or equivalent A1 Form a plane equation and use coordinates of a relevant point to evaluate d M1(dep*) Obtain answer –2x + 4y + 3z = 26, or equivalent A1 OR2 : Form a two-parameter plane equation using relevant vectors M1* State a correct equation e.g. r = 2j + 6k + s(i – j + 2k) + t(i + 2j – 2k) A1 State three correct equations in x, y, z, s and t A1 Eliminate s and t M1(dep*) Obtain answer –2x + 4y + 3z = 26, or equivalent A1 [5]

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