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Date May 2019 Marks available 4 Reference code 19M.1.SL.TZ2.S_9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Find Question number S_9 Adapted from N/A

Question

Let θ be an obtuse angle such that  sin θ = 3 5 .

Let  f ( x ) = e x sin x 3 x 4 .

Find the value of tan θ .

[4]
a.

Line L passes through the origin and has a gradient of tan θ . Find the equation of L .

[2]
b.

The following diagram shows the graph of f  for 0 ≤ x ≤ 3. Line M is a tangent to the graph of f at point P.

Given that M is parallel to L , find the x -coordinate of P.

[4]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

evidence of valid approach       (M1)

eg   sketch of triangle with sides 3 and 5,  co s 2 θ = 1 si n 2 θ

correct working       (A1)

eg  missing side is 4 (may be seen in sketch),  cos θ = 4 5 ,   cos θ = 4 5

tan θ = 3 4        A2 N4

[4 marks]

a.

correct substitution of either gradient or origin into equation of line        (A1)

(do not accept y = m x + b )

eg    y = x tan θ ,    y 0 = m ( x 0 ) ,    y = m x

y = 3 4 x      A2 N4

Note: Award A1A0 for  L = 3 4 x .

[2 marks]

b.

valid approach to equate their gradients       (M1)

eg    f = tan θ ,    f = 3 4 e x cos x + e x sin x 3 4 = 3 4 ,    e x ( cos x + sin x ) 3 4 = 3 4

correct equation without  e x         (A1)

eg    sin x = cos x ,   cos x + sin x = 0 ,   sin x cos x = 1

correct working       (A1)

eg    tan θ = 1 ,   x = 135

x = 3 π 4 (do not accept  135 )       A1 N1

Note: Do not award the final A1 if additional answers are given.

[4 marks]

 

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
d.

Syllabus sections

Topic 5 —Calculus » SL 5.4—Tangents and normal
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