(a) (i) On the axes, sketch the graph of \(y=\sin x\) for \(0^{\circ}\leq x\leq 360^{\circ}.\)
(ii) Describe fully the symmetry of the graph of \(y= \sin x\) for \(0^{\circ}\leq x\leq 360^{\circ}.\)
(b) Solve \(4 \sin x – 1 = 2\) for \(0^{\circ}\leq x\leq 360^{\circ}.\)
(c) (i) Write \(x^{2}+10x+14\) in the form \((x+a)^{2}+b.\)
(ii) On the axes, sketch the graph of \(y=x^{2}+10x+14\), indicating the coordinates of the turning point.
▶️ Answer/Explanation
(a)(i) Sketch a sine curve starting at \((0°,0)\), peaking at \((90°,1)\), crossing \((180°,0)\), troughing at \((270°,-1)\), and ending at \((360°,0)\).
(a)(ii) Ans: Rotational symmetry order 2 about \((180°,0)\).
The graph repeats every \(360°\) and is symmetric about the point \((180°,0)\).
(b) Ans: \(x = 48.6°\) and \(x = 131.4°\).
Solve \(4\sin x – 1 = 2 \Rightarrow \sin x = 0.75\). The solutions in \(0° \leq x \leq 360°\) are \(x = \sin^{-1}(0.75)\) and \(x = 180° – \sin^{-1}(0.75)\).
(c)(i) Ans: \((x + 5)^2 – 11\).
Complete the square: \(x^2 + 10x + 14 = (x^2 + 10x + 25) – 11 = (x + 5)^2 – 11\).
(c)(ii) Sketch a parabola with vertex at \((-5, -11)\), opening upwards.
(a) 
The diagram shows a sketch of the curve \(y = x^{2} + 3x – 4\).
(i) Find the coordinates of the points A, B and C.
(ii) Differentiate \(x^{2} + 3x – 4\).
(iii) Find the equation of the tangent to the curve at the point (2, 6).
(b) 
(i) On the diagram, sketch the graph of \(y = \tan x\) for \(0° \leq x \leq 360°\).
(ii) Solve the equation \(5\tan x = -7\) for \(0° \leq x \leq 360°\).
▶️ Answer/Explanation
(a)(i) Ans: A(–4, 0), B(1, 0), C(0, –4)
To find A and B (x-intercepts), solve \(x^{2} + 3x – 4 = 0\). Factorizing: \((x + 4)(x – 1) = 0\), so \(x = -4\) or \(x = 1\). Thus, A(-4, 0) and B(1, 0).
To find C (y-intercept), substitute \(x = 0\): \(y = -4\). Thus, C(0, -4).
(a)(ii) Ans: \(2x + 3\)
Differentiate \(y = x^{2} + 3x – 4\) with respect to \(x\): \(\frac{dy}{dx} = 2x + 3\).
(a)(iii) Ans: \(y = 7x – 8\)
First, find the gradient at \(x = 2\): \(\frac{dy}{dx} = 2(2) + 3 = 7\).
Using point-slope form: \(y – 6 = 7(x – 2)\), simplifying to \(y = 7x – 8\).
(b)(i) Ans: Correct sketch of \(y = \tan x\)
The graph of \(y = \tan x\) has vertical asymptotes at \(x = 90°\) and \(x = 270°\) and passes through \((0°, 0)\), \((180°, 0)\), and \((360°, 0)\).
(b)(ii) Ans: \(125.5°\) or \(125.53°\) to \(125.54°\) and \(305.5°\) or \(305.53°\) to \(305.54°\)
Solve \(5\tan x = -7\): \(\tan x = -1.4\). The principal solution is \(x = \tan^{-1}(-1.4) ≈ -54.46°\).
In the range \(0° \leq x \leq 360°\), the solutions are \(180° – 54.46° ≈ 125.54°\) and \(360° – 54.46° ≈ 305.54°\).