Practise Graphs of sin, cos and tan. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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Question 2Medium7 marks
The diagram shows the curve with equation \(y = a\cos(x - \alpha) + b\), \(0 \le x \le 360^\circ\), where \(a\), \(b\) and \(\alpha\) are constants, \(a \gt 0\) and \(0 \lt \alpha \lt 90^\circ\).
The curve has a maximum point at \((20^\circ, 1)\) and a minimum point at \((200^\circ, -7)\).
[object Object]
(a) Find the values of \(a\), \(b\) and \(\alpha\).[3]
(b) Find the \(y\)-coordinate of the point where the curve meets the \(y\)-axis, giving your answer to 3 significant figures.[1]
(c) Find the values of \(x\), for \(0 \le x \le 360^\circ\), for which \(y = -3\).[3]
Show the answer and mark scheme
(a) Answer: \(a = 4, \ b = -3, \ \alpha = 20^\circ\)
- B1 for \(a = 4\) (half the difference between the maximum and minimum values)
- B1 for \(b = -3\) (the mean of the maximum and minimum values)
- B1 for \(\alpha = 20^\circ\)
Worked solution: \(a = \frac{1 - (-7)}{2} = 4\), \(b = \frac{1 + (-7)}{2} = -3\). The maximum of \(\cos(x - \alpha)\) occurs when \(x = \alpha\), so \(\alpha = 20^\circ\).
(b) Answer: \(0.759\)
- B1ft for awrt \(0.759\) (\(4\cos(-20^\circ) - 3\))
Worked solution: \(y = 4\cos(0 - 20^\circ) - 3 = 0.759\)
(c) Answer: \(x = 110^\circ, \ 290^\circ\) °
- M1 for \(\cos(x - 20^\circ) = 0\)
- A1 for \(x = 110^\circ\)
- A1 for \(x = 290^\circ\) and no other values in the range
Worked solution: \(4\cos(x - 20^\circ) - 3 = -3 \Rightarrow \cos(x - 20^\circ) = 0\), so \(x - 20^\circ = 90^\circ, \ 270^\circ\) (for \(-20^\circ \le x - 20^\circ \le 340^\circ\)), giving \(x = 110^\circ, \ 290^\circ\).
Question 3Hard6 marks
The function \(\mathrm{f}\) is defined by \[\mathrm{f}(x) = 5\sin(x - 60^\circ) + 2, \quad 0 \le x \le 270^\circ.\]
(a) Sketch the graph of \(y = \mathrm{f}(x)\) for \(0 \le x \le 270^\circ\), showing the coordinates of any turning points and of the end points of the graph.[3]
(b) Find the set of values of \(k\) for which the equation \(\mathrm{f}(x) = k\) has exactly two solutions in the interval \(0 \le x \le 270^\circ\).[3]
Show the answer and mark scheme
(a) Answer: A sine-shaped curve with a maximum at \((150^\circ, 7)\); end points \((0, 2 - \frac{5\sqrt{3}}{2})\) and \((270^\circ, -\frac{1}{2})\).
- B1 for a sine-shaped curve over the correct interval
- B1 for the turning point correct: a maximum at \((150^\circ, 7)\)
- B1 for the end points \((0, 2 - \frac{5\sqrt{3}}{2})\) and \((270^\circ, -\frac{1}{2})\)
Worked solution: The graph is \(y = \sin x\) translated 60° to the right, stretched vertically by scale factor 5 and translated 2 units up. Over the interval it has a maximum at \((150^\circ, 7)\); it starts at \((0, 2 - \frac{5\sqrt{3}}{2})\) and ends at \((270^\circ, -\frac{1}{2})\).
(b) Answer: \(-\frac{1}{2} \le k \lt 7\)
- M1 for using the critical values \(2 - \frac{5\sqrt{3}}{2}, \ -\frac{1}{2}, \ 7\) (turning values and end values)
- A1 for one correct piece of the set
- A1 for \(-\frac{1}{2} \le k \lt 7\) with correct strict and non-strict inequalities
Worked solution: Counting the intersections of horizontal lines \(y = k\) with the graph: \(k = 2 - \frac{5\sqrt{3}}{2}\): 1; \(2 - \frac{5\sqrt{3}}{2} \lt k \lt -\frac{1}{2}\): 1; \(k = -\frac{1}{2}\): 2; \(-\frac{1}{2} \lt k \lt 7\): 2; \(k = 7\): 1.
So exactly two solutions when \(-\frac{1}{2} \le k \lt 7\).