Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 3 · Variant 3

1 Sketch the graph of y eax where a is a positive constant. [2] = −1

Mark scheme: 1 Draw curve with increasing gradient existing for negative and positive values of x M1 Draw correct curve passing through the origin A1 [2]

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3 A curve has equation 2 x y −tan 1 tan x. = + 1 Find the equation of the tangent to the curve at the point for which x giving the answer in the form y mx c where c is correct to 3 significant figures. = 40, [6] = +

Mark scheme: 3 Use correct quotient rule or equivalent to find first derivative M1* − (1 + tan x ) sec 2 x − sec 2 x ( 2 − tan x ) Obtain or equivalent A1 1( + tan x ) 2 1 Substitute x = π to find gradient dep M1* 4 3 Obtain − A1 2 1 Form equation of tangent at x = π M1 4 3 Obtain y = − x + .1 68 or equivalent A1 [6] 2 d y y& dy

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8 A plane has equation 4x 39. A straight line is parallel to the vector i 4k and passes through the point A 0, 2,−y +. 5ÏThe= line meets the plane at the point B. −3j + −8 (i) Find the coordinates of B. [3] (ii) Find the acute angle between the line and the plane. [4] (iii) The point C lies on the line and is such that the distance between C and B is twice the distance between A and B. Find the coordinates of each of the possible positions of the point C. [3]

Mark scheme: 8 (i) Express a general point on the line in single component form, e.g. ( λ , 2 − 3λ , − 8 + 4λ ) , substitute in equation of plane and solve for λ M1 Obtain λ = 3 A1 Obtain (,3 − ,7 4) A1 [3] (ii) State or imply normal vector to plane is 4i – j + 5k B1 Carry out process for evaluating scalar product of two relevant vectors M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate sin −1 or cos−1 of the result. M1 Obtain 548.° or 0.956 radians A1 [4] (iii) Either Find at least one position of C by translating by appropriate multiple of direction vector i – 3j + 4k from A or B M1 Obtain ( −,3 11, − 20 ) A1 Obtain (,9 − 25, 28) A1 Or Form quadratic equation in λ by considering BC 2 = 4AB 2 M1 Obtain 26λ 2 − 156λ − 702 = 0 or equivalent and hence λ = − ,3 λ = 9 A1 Obtain ( −,3 11, − 20 ) and (,9 − 25, 28) A1 [3]

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