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

1 Find the value of the constant c for which the line y = 2x + c is a tangent to the curve y2 = 4x. [4]

Mark scheme: 1 Eliminates x or y completely M1 Aims to make x or y subject + subst y 2 − 2 y + 2c or 4 x 2 + x ( 4c − 4) + c 2 = 0 M1 Correct quadratic – not nec =0 Use of b 2 −ac4 = 0 A1 Used correctly on “quadratic=0” → c = ½ A1 co [or gradients equal 2=1/√x M1A1 [4] → value for x,y and c. M1A1] ∫√

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2 1 The diagram shows the curve y = 3x 4. The shaded region is bounded by the curve, the x-axis and the lines x = 1 and x = 4. Find the volume of the solid obtained when this shaded region is rotated completely about the x-axis, giving your answer in terms of π. [4]

Mark scheme: 2 V = π ∫9√xdx M1 For integral of y² (ignore π here) 3 2 A1 All correct (ignoring π here) 9x = π 3 2 DM1 Correct use of correct limits. [ ] at 4 − [ ] at 1 → 42π A1 co. [4]

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1 −tan2x 3 Prove the identity ≡1 −2 sin2x. [4] 1 + tan2x

Mark scheme: 3 Use of t=s/c M1 tan completely removed → (c²−s²) ÷ (c²+s²) A1 May omit the denominator (=1) Use of c²+s²=1 M1 Whenever used appropriately → (c²−s²) → 1 − 2sin²x A1 ag Beware fortuitous answers [4]

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18 1 4 Find the real roots of the equation + = 4. [4] x4 x2

Mark scheme: 4 × (x4 ) → 4 x 4 −x 2 − 18 = 0 M1 Recognition of quad in x² or 1÷(x²) ( 4 x 2 − 9)( x 2 + 2) = 0 DM1 Solution of quadratic. x = 1.5 or x = − 1.5 A1 A1√ Positive root. For recognition of (−ve) [4] The A1√ assumes no other real answers. 1

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5 In the diagram, OAB is a sector of a circle with centre O and radius 12 cm. The lines AX and BX are tangents to the circle at A and B respectively. Angle AOB = 13π radians. (i) Find the exact length of AX, giving your answer in terms of √3. [2] (ii) Find the area of the shaded region, giving your answer in terms of π and √3. [3]

Mark scheme: 5 (i) tan 16 π =AX÷12 or other valid method M1 Use of trig with tangent in correct ∆ tan 16 π =√3÷3 → AX = 4√3 A1 Co (12 ÷ √3 ok) [2] (ii) area AOC = ½r²θ (= 24π) M1 Correct formula + attempt with radians Area of ∆AOX = ½ × AX×12 M1 Use of ½bh in correct ∆ (once ok) → shaded area = 48√3 − 24π A1 co (144 ÷ √3 ok) [3]

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6 The diagram shows a rectangle ABCD. The point A is (2, 14), B is (−2, 8) and C lies on the x-axis. Find (i) the equation of BC, [4] (ii) the coordinates of C and D. [3]

Mark scheme: 6 (i) m of AB = 1.5 ( or 1½) B1 co anywhere m of BC = −1 ÷ (m of AB) = −⅔ M1 Use of m1m2 = −1 → Eqn y − 8 = − 23 ( x + 2) or 3y+2x=20 M1 A1√ Correct form used – or y = mx + c . co [4] (√ needs both M marks) (ii) Put y = 0 → C (10, 0) B1√ √ in his linear equation. Vector move → D (14, 6) M1A1 completely correct method. co (or sim eqns 3y+2x=46 and 2y=3x−30) [3] GCE A/AS LEVEL – May/June 2007 9709 01 a

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7 The second term of a geometric progression is 3 and the sum to infinity is 12. (i) Find the first term of the progression. [4] An arithmetic progression has the same first and second terms as the geometric progression. (ii) Find the sum of the first 20 terms of the arithmetic progression. [3]

Mark scheme: a 7 (i) ar=3 and = 12 B1 B1 co for each one. 1 −r Solution of sim eqns → a = 6 M1 A1 Needs to eliminate a or r correctly. co [4] (M mark needs a quadratic) (ii) a =6, d = −3 B1√ For d = 3 − his “6”. S20 = 10(12 − 57) M1 Sum formula must be correct and used. → −450 A1 co. [3]

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8 The function f is defined by f(x) = a + b cos 2x, for 0 ≤x ≤π. It is given that f(0) = −1 and f 12π = 7. (i) Find the values of a and b. [3] (ii) Find the x-coordinates of the points where the curve y = f(x) intersects the x-axis. [3] (iii) Sketch the graph of y = f(x). [2]

Mark scheme: 8 (i) f(x) = a + b cos 2 x , → a + b = −1 B1 co and a − b = 7 B1 co Solution → a = 3 and b = −4 B1 co [3] (ii) 3 − 4cos2x = 0 → cos2x = ¾ M1 cos2x subject and finds cos-1 before ÷2 → x = 0.36 and 2.78 A1 A1√ co. √ for π − 1st answer and no other [3] answers in the range.(Degrees max ⅓) (iii) B1 Ignore anything outside 0 to π. Must be 1 oscillation only. B1 Everything ok including curves, not [2] blatant lines and from −1 to 7. − 1  − 2  AB  1  AC  2 

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9 Relative to an origin O, the position vectors of the points A and B are given by 4 3 −−→ −−→ OA = 1 and OB = 2 . −2 −4 −−→ −−→ (i) Given that C is the point such that AC = 2−−→AB, find the unit vector in the direction of OC. [4] 1 −−→ The position vector of the point D is given by OD = 4 , where k is a constant, and it is given that k −−→ OD = m−−→OA + n−−→OB, where m and n are constants. (ii) Find the values of m, n and k. [4]

Mark scheme: 9 (i) AB =  1  and AC =  2  M1 For ±( b−a ) (not b+a)     − 2  − 4   2    OC = OA + AC =  3  A1 Co   −6    2    M1 Division by the modulus 1 Unit vector = 7  3  A1√ √for his OC   −6   [4]  4   3   1        m (ii)  1  + n  2  =  4        −2 −4 k       → 4m + 3n = 1 and m + 2n = 4 M1 Forming 2 simultaneous equations → m = −2 and n = 3 A1 co → k = − 8 M1A1 Equation for k in terms of m and n. co [4] GCE A/AS LEVEL – May/June 2007 9709 01 dy 16 3 /

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8 10 The equation of a curve is y = 2x + x2. dy d2y (i) Obtain expressions for and . [3] dx dx2 (ii) Find the coordinates of the stationary point on the curve and determine the nature of the stationary point. [3] (iii) Show that the normal to the curve at the point (−2, −2) intersects the x-axis at the point (−10, 0). [3] (iv) Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x = 2. [3]

Mark scheme: dy 16 −16/x3.10 (i) = 2 − 3 B1 For dx x d 2 y 48 B1 For “2” and for “0”. 2 = 4 B1√ For d/dx of his −16/x3 providing −ve dx x [3] power differentiated. dy (ii) =0 → x = 2, y = 6. M1 Sets dy/dx to 0 + attempt at x. dx A1 Needs both coordinates. d 2 y 2 is +ve Minimum. A1√ Looks at sign. Correct conclusion for dx [3] his x and his 2nd differential. (iii) x = −2 m = 4 Perp gradient = −¼ M1 Uses m1m2 = −1 with dy/dx. y + 2 = − 14 ( x + 2) DM1 Correct form of equation (not for tan) Sets y to 0 → x = −10 A1 Co nb answer given. [3]  2 8  (iv) Area = x −x  B1 B1 For each term  Evaluated from 1 to 2 → 7 B1 Co. (−7 ⇒ 7 gets B0) [3] 2

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11 6 The diagram shows the graph of y = f(x), where f : x → for x ≥0. 2x + 3 (i) Find an expression, in terms of x, for f′(x) and explain how your answer shows that f is a decreasing function. [3] (ii) Find an expression, in terms of x, for f−1(x) and find the domain of f−1. [4] (iii) Copy the diagram and, on your copy, sketch the graph of y = f−1(x), making clear the relationship between the graphs. [2] The function g is defined by g : x →12x for x ≥0. (iv) Solve the equation fg(x) = 32. [3]

Mark scheme: 11 (i) f´(x) = −6(2x+3)-2 × 2 B1 B1 co.(−ve power ok) B1 for ×2 Always −ve → Decreasing B1√ Answer given. Correct explanation. [3] 6 (ii) y = M1 Reasonable attempt in making x the 2 x + 3 subject (ok to interchange x, y first) M1 Order of operations must be correct ie 1  6  → f -1(x) =  −3  ÷ y, −3 then ÷ 2. 2  x  A1 Correct expression as f–1(x). Gets 2/3 for correct expression with y. Domain of f –1: 0 < x ≤ 2 B1 Could be independent of answer for f -1. [4] Condone < or ≤ (iii) B1 Correct graph for f -1(curve, stops on axis) B1 Makes clear on graph, or in words or [2] by the line y=x marked, the symmetry. 6 (iv) fg(x) = M1 x + 3 = 1.5 → x = 1 M1 g first, then f. Reverse (3÷(2x+3) M0 [or , using f-1, → g(x) = ½, ⇒ x = 1.] A1 Not DM – so can get this if attempt ok [ M1 M1 A1] co [3] DM1 for quadratic. Quadratic must be set to 0. Factors. Attempt at two brackets. Each bracket set to 0 and solved. Formula. Correct formula. Correct use, but allow for numerical slips in b² and −4ac.

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