Specimen 2017 H2 Mathematics Paper 2 Question 1

Functions

Answers

(i)

f(x)=13cos⁡(x+0.588).{f(x)=\sqrt{13}\cos(x+0.588).}
Rf=[−13,13].{R_f = \left[ -\sqrt{13}, \sqrt{13} \right].}

(ii)

Largest value of b=2.55.{b=2.55.}
g−1(x)=cos⁡−1x13−0.588.{g^{-1}(x) = \cos^{-1} \frac{x}{\sqrt{13}} - 0.588.}

Full solutions

(i)

R=32+22=13  ■α=tan⁡−1(23)=0.588 (3 s.f.)  ■\begin{align*} R &= \sqrt{3^2 + 2^2}\\ &= \sqrt{13} \; \blacksquare \\ \alpha &= \tan^{-1} \left( \frac{2}{3} \right) \\ &= 0.588 \textrm{ (3 s.f.)} \; \blacksquare \end{align*}
From the graph,
Rf=[−13,13]  ■R_f = \left[ -\sqrt{13}, \sqrt{13} \right] \; \blacksquare

(ii)

Since g−1{g^{-1}} exists, g{g} is one-one.

Hence the largest value of b{b} is the x-coordinate{x\textrm{-coordinate}} of the minimum point of the graph.

Largest value of b=2.55  ■{b = 2.55 \; \blacksquare}

y=3cos⁡x−2sin⁡xy=13cos⁡(x+0.588)x=cos⁡−1y13−0.588\begin{gather*} y = 3 \cos x - 2 \sin x \\ y = \sqrt{13} \cos (x + 0.588) \\ x = \cos^{-1} \frac{y}{\sqrt{13}} - 0.588 \end{gather*}
g−1(x)=cos⁡−1x13−0.588  ■g^{-1}(x) = \cos^{-1} \frac{x}{\sqrt{13}} - 0.588 \; \blacksquare
0{0}
x{x}
y{y}
(−π,−3){(-\pi, -3)}
(π,−3){(\pi, -3)}
(−α,13){(-\alpha,\sqrt{13})}
(π−α,−13){(\pi-\alpha,-\sqrt{13})}
y=f(x){y=f(x)}