Home / IBDP Maths AA: Topic : SL 3.8: Solving trigonometric equations: IB style Questions SL Paper 2

IBDP Maths AA: Topic : SL 3.8: Solving trigonometric equations: IB style Questions SL Paper 2

Question

Let \(f(x) = 3\sin x + 4\cos x\) , for \( – 2\pi  \le x \le 2\pi \) .

a.Sketch the graph of f .[3]

b.Write down

(i)     the amplitude;

(ii)    the period;

(iii)   the x-intercept that lies between \( – \frac{\pi }{2}\) and 0.[3]

c.Hence write \(f(x)\) in the form \(p\sin (qx + r)\) .[3]

d.Write down one value of x such that \(f'(x) = 0\) .[2]

e.Write down the two values of k for which the equation \(f(x) = k\) has exactly two solutions.[2]

f.Let \(g(x) = \ln (x + 1)\) , for \(0 \le x \le \pi \) . There is a value of x, between \(0\) and \(1\), for which the gradient of f is equal to the gradient of g. Find this value of x.[5]

Answer/Explanation


     A1A1A1     N3

Note: Award A1 for approximately sinusoidal shape, A1 for end points approximately correct \(( – 2\pi {\text{, }}4)\) \((2\pi {\text{, }}4)\), A1 for approximately correct position of graph, (y-intercept \((0{\text{, }}4)\), maximum to right of y-axis).

[3 marks]

a.

(i) 5     A1     N1

(ii) \(2\pi \)  (6.28)     A1     N1

(iii) \( – 0.927\)     A1     N1

[3 marks]

b.

\(f(x) = 5\sin (x + 0.927)\) (accept \(p = 5\) , \(q = 1\) , \(r = 0.927\) )     A1A1A1     N3

[3 marks]

c.

evidence of correct approach     (M1)

e.g. max/min, sketch of \(f'(x)\) indicating roots


one 3 s.f. value which rounds to one of \( – 5.6\), \( – 2.5\), \(0.64\), \(3.8\)     A1     N2

[2 marks]

d.

\(k = – 5\) , \(k = 5\)     A1A1     N2

[2 marks]

e.

METHOD 1

graphical approach (but must involve derivative functions)     M1

e.g.


each curve     A1A1

\(x = 0.511\)     A2     N2

METHOD 2

\(g'(x) = \frac{1}{{x + 1}}\)     A1

\(f'(x) = 3\cos x – 4\sin x\)     \((5\cos (x + 0.927))\)     A1

evidence of attempt to solve \(g'(x) = f'(x)\)     M1

\(x = 0.511\)     A2     N2

[5 marks]

f.

Question

A Ferris wheel with diameter \(122\) metres rotates clockwise at a constant speed. The wheel completes \(2.4\) rotations every hour. The bottom of the wheel is \(13\) metres above the ground.


A seat starts at the bottom of the wheel.

After t minutes, the height \(h\) metres above the ground of the seat is given by\[h = 74 + a\cos bt {\rm{  .}}\]

Find the maximum height above the ground of the seat.

[2]
a.

(i)     Show that the period of \(h\) is \(25\) minutes.

(ii)     Write down the exact value of \(b\) .

 

[2]
b.

(b)     (i)     Show that the period of \(h\) is \(25\) minutes.

  (ii)     Write down the exact value of \(b\) .

(c)     Find the value of \(a\) .

(d)     Sketch the graph of \(h\) , for \(0 \le t \le 50\) .

[9]
bcd.

Find the value of \(a\) .

[3]
c.

Sketch the graph of \(h\) , for \(0 \le t \le 50\) .

[4]
d.

In one rotation of the wheel, find the probability that a randomly selected seat is at least \(105\) metres above the ground.

[5]
e.
Answer/Explanation

Markscheme

valid approach     (M1)

eg   \(13 + {\rm{diameter}}\) , \(13 + 122\)

maximum height \( = 135\) (m)     A1 N2

[2 marks]

a.

(i)     period \( = \frac{{60}}{{2.4}}\)     A1

period \( = 25\) minutes     AG     N0

(ii)     \(b = \frac{{2\pi }}{{25}}\) \(( = 0.08\pi )\)     A1     N1

[2 marks]

b.

(a)     (i)     period \( = \frac{{60}}{{2.4}}\)     A1

period \( = 25\) minutes     AG     N0

(ii)     \(b = \frac{{2\pi }}{{25}}\) \(( = 0.08\pi )\)     A1     N1

[2 marks]

 

(b)     METHOD 1

valid approach     (M1)

eg   \({\rm{max}} – 74\) , \(\left| a \right| = \frac{{135 – 13}}{2}\) , \(74 – 13\)

\(\left| a \right| = 61\) (accept \(a = 61\) )     (A1)

\(a = – 61\)     A1     N2

METHOD 2

attempt to substitute valid point into equation for h     (M1)

eg   \(135 = 74 + a\cos \left( {\frac{{2\pi  \times 12.5}}{{25}}} \right)\)

correct equation     (A1)

eg   \(135 = 74 + a\cos (\pi )\) , \(13 = 74 + a\)

\(a = – 61\)     A1     N2

[3 marks]

 

(c)

     A1A1A1A1      N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

  A1 for approximately correct sinusoidal shape with \(2\) cycles.

  Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

 

Total [9 marks]

bcd.

METHOD 1

valid approach     (M1)

eg   \({\rm{max}} – 74\) , \(\left| a \right| = \frac{{135 – 13}}{2}\) , \(74 – 13\)

\(\left| a \right| = 61\) (accept \(a = 61\) )     (A1)

\(a = – 61\)     A1     N2

METHOD 2

attempt to substitute valid point into equation for h     (M1)

eg   \(135 = 74 + a\cos \left( {\frac{{2\pi  \times 12.5}}{{25}}} \right)\)

correct equation     (A1)

eg   \(135 = 74 + a\cos (\pi )\) , \(13 = 74 + a\)

\(a = – 61\)     A1     N2

[3 marks]

c.

     A1A1A1A1      N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

  A1 for approximately correct sinusoidal shape with \(2\) cycles.

  Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

 

d.

setting up inequality (accept equation)     (M1)

eg   \(h > 105\) , \(105 = 74 + a\cos bt\) , sketch of graph with line \(y = 105\)

any two correct values for t (seen anywhere)     A1A1

eg   \(t = 8.371 \ldots \) , \(t = 16.628 \ldots \) , \(t = 33.371 \ldots \) , \(t = 41.628 \ldots \)

valid approach     M1

eg   \(\frac{{16.628 – 8.371}}{{25}}\) , \(\frac{{{t_1} – {t_2}}}{{25}}\) , \(\frac{{2 \times 8.257}}{{50}}\) , \(\frac{{2(12.5 – 8.371)}}{{25}}\)

\(p = 0.330\)     A1     N2

[5 marks]

e.
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