# CIE A level -Pure Mathematics 1 : Topic : 1.2 Function: transformations of the graph : Exam Questions Paper 1

### Question

The diagram shows part of the graph of $$y=a+b\sin x$$. State the values of the constants a and b.

In this case, the graph is shifted upward by 1 unit.The constant a=1 represents a vertical shift of the graph.The amplitude is the half-range of the oscillation, which is 2 in this case.The constant b=2 represents the amplitude of the sine function. So, the equation  y=1+2sinx describes a sine function with an amplitude of 2, centered at y=1.

### Question

(a)The diagram shows part of the graph of $$y=a+b\sin x$$.Find the values of constant a and b.

(b)(i)Show that the equation $$\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta -\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )$$

may be expressed as $$3\cos ^{2}\Theta -2\cos \Theta -1=0$$.

(iii)Hence solve the equation $$\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta-\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )$$ for $$-180^{\circ}\leq \Theta \leq 180^{\circ}$$

(a) The equation of the graph is $$y = -2 + 3 \sin x$$.
Here’s what these values mean for the graph:
The constant $$a = -2$$ represents a vertical shift of the graph. In this case, the graph is shifted downward by 2 units.
The constant $$b = 3$$ represents the amplitude of the sine function. The amplitude is the half-range of the oscillation, which is 3 in this case. This means the graph oscillates between $$-2 + 3 = 1$$ and $$-2 – 3 = -5$$.
So, the equation $$y = -2 + 3 \sin x$$ describes a sine function with an amplitude of 3, centered at $$y = -2$$.
(b)(i) Let’s simplify the given equation:
\begin{align*} & (\sin \Theta + 2 \cos \Theta)(1 + \sin \Theta – \cos \Theta) = \sin \Theta(1 + \cos \Theta) \\ & (\sin \Theta + 2 \cos \Theta)(1 + \sin \Theta – \cos \Theta) – \sin \Theta(1 + \cos \Theta) = 0 \\ & \sin \Theta + 2 \cos \Theta + \sin^2 \Theta – \cos^2 \Theta + 2 \sin \Theta \cos \Theta – \sin \Theta – \cos \Theta = 0 \\ & \sin^2 \Theta – \cos^2 \Theta + 2 \sin \Theta \cos \Theta = 0 \\ & (\sin \Theta + \cos \Theta)(\sin \Theta – \cos \Theta) + 2 \sin \Theta \cos \Theta = 0 \\ & \sin \Theta + \cos \Theta + 2 \sin \Theta \cos \Theta – \sin \Theta \cos \Theta = 0 \\ & \sin \Theta + \cos \Theta + \sin \Theta \cos \Theta = 0 \\ & \cos \Theta(1 + \sin \Theta) + \sin \Theta \cos \Theta = 0 \\ & \cos \Theta(1 + \sin \Theta + \sin \Theta) = 0 \\ & \cos \Theta(1 + 2 \sin \Theta) = 0. \end{align*}
Now, we know that $$\cos \Theta$$ cannot be zero for all values of $$\Theta$$, so we divide both sides by $$\cos \Theta$$:
$$1 + 2 \sin \Theta = 0$$
$$2 \sin \Theta = -1$$
$$\sin \Theta = -\frac{1}{2}$$
$$\sin^2 \Theta = \frac{1}{4}$$
Using the identity $$\sin^2 \Theta + \cos^2 \Theta = 1$$:
$$\cos^2 \Theta = 1 – \frac{1}{4}$$
$$\cos^2 \Theta = \frac{3}{4}$$
$$3 \cos^2 \Theta – 2 \cos \Theta – 1 = 0$$
(ii)$$3 \cos^2 \Theta – 2 \cos \Theta – 1 = 0$$
$$3\cos ^{2}\Theta -3\cos \Theta +\cos \Theta -1=0$$
$$3\cos \Theta (\cos \Theta -1)+1(\cos \Theta -1)=0$$
$$(3\cos \Theta +1)(\cos \Theta -1)=0$$
$$\cos \Theta =-\frac{1}{3}$$ or $$\cos \Theta =1$$
If $$\cos \Theta = 1$$, then $$\Theta = \cos^{-1}(1)$$. However, the inverse cosine of 1 is $$0^\circ$$.
If $$\cos \Theta = -\frac{1}{3}$$, then $$\Theta = \cos^{-1}\left(-\frac{1}{3}\right)$$. This value is approximately $$109.5^\circ$$ or $$-109.5^\circ$$ because the cosine function is negative in the second and third quadrants.
So,
$$\cos \Theta = 1$$ implies $$\Theta = 0^\circ$$.
$$\cos \Theta = -\frac{1}{3}$$ implies $$\Theta \approx 109.5^\circ$$ or $$-109.5^\circ$$.

### Question

The function f is defined by $$f: x\rightarrow 4 sin x – 1$$ for $$\frac{-\Pi}{2}\leq x\leq\frac{\Pi}{2}$$.
(i) State the range of f.
(ii) Find the coordinates of the points at which the curve y = f(x) intersects the coordinate axes.
(iii) Sketch the graph of y = f(x).
(iv) Obtain an expression for $$f^{-1}(x)$$ , stating both the domain and range of $$f^{-1}$$.

(i) State the range of $$f$$:
The function $$f(x) = 4 \sin x – 1$$ is a sine function with an amplitude of 4 and a vertical shift downward by 1 unit. The range of a sine function is $$[-1, 1]$$, so when you multiply it by 4 and shift it downward by 1, the range becomes $$[-5, 3]$$.
So, the range of $$f$$ is $$[-5, 3]$$.
(ii) Find the coordinates of the points at which the curve $$y = f(x)$$ intersects the coordinate axes:
For the $$y$$-axis, where $$x = 0$$, $$f(0) = 4 \sin 0 – 1 = -1$$. So, the point of intersection is $$(0, -1)$$.
For the $$x$$-axis, where $$y = 0$$, solve $$4 \sin x – 1 = 0$$ for $$x$$:
$$4 \sin x = 1$$
$$\sin x = \frac{1}{4}$$
Taking the arcsine (or inverse sine) of both sides:
$$x = \sin^{-1}\left(\frac{1}{4}\right)$$
$$x \approx 0.253 \text{ radians}$$
So, the point of intersection with the $$x$$-axis is approximately $$(0.253, 0)$$.
The correct coordinates are:
For the $$y$$-axis: $$(0, -1)$$
For the $$x$$-axis: $$(0.253, 0)$$
(iii)
The function $$f(x) = 4 \sin x – 1$$ is a sine function with an amplitude of 4, centered at $$x = 0$$, and shifted downward by 1 unit. The graph oscillates between $$-5$$ and $$3$$.
(iv) Obtain an expression for $$f^{-1}(x)$$, stating both the domain and range of $$f^{-1}$$:
To find $$f^{-1}(x)$$, interchange $$x$$ and $$y$$ and solve for $$y$$:
$$x = 4 \sin y – 1$$
$$4 \sin y = x + 1$$
$$\sin y = \frac{x + 1}{4} \] \( y = \sin^{-1}\left(\frac{x + 1}{4}\right)$$
The domain of $$f^{-1}(x)$$ is determined by the domain of $$f(x)$$, which is $$\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$$. The range of $$f^{-1}(x)$$ is determined by the range of $$f(x)$$, which is $$[-5, 3]$$.
So, the expression for $$f^{-1}(x)$$ is:
$$f^{-1}(x) = \sin^{-1}\left(\frac{x + 1}{4}\right)$$
with the domain $$\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$$ and the range $$[-5, 3]$$.

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