Question

in the diagram below, the graphs of $f(x)=\tan x$ and $g(x)=2\sin 2x$ are drawn for the interval $x\in-180^{\circ};180^{\circ}$. a$(60^{\circ};k)$ and b are two points of intersection of $f$ and $g$. 6.1 write down the period of $g$. (1) 6.2 calculate the: 6.2.1 value of $k$ (1) 6.2.2 coordinates of b (1) 6.3 write down the range of $2g(x)$. (2)

Explanation:

Step1: Find period of $g(x)$

For $y = A\sin(Bx + C)+D$, period $T=\frac{360^{\circ}}{|B|}$. In $g(x)=2\sin2x$, $B = 2$. So $T=\frac{360^{\circ}}{2}=180^{\circ}$.

Step2: Calculate $k$

Substitute $x = 60^{\circ}$ into $g(x)=2\sin2x$. Then $k = 2\sin(2\times60^{\circ})=2\sin120^{\circ}=2\times\frac{\sqrt{3}}{2}=\sqrt{3}$.

Step3: Find coordinates of B

Since $f(x)=\tan x$ and $g(x)=2\sin2x$ are symmetric about the origin in the given interval $x\in[- 180^{\circ};180^{\circ}]$, if $A(60^{\circ},\sqrt{3})$ is a point of intersection, then the other intersection point $B$ has $x=-60^{\circ}$. Substitute $x = - 60^{\circ}$ into $g(x)$: $y=2\sin(2\times(-60^{\circ}))=2\sin(-120^{\circ})=-\sqrt{3}$. So $B(-60^{\circ},-\sqrt{3})$.

Step4: Find range of $2g(x)$

The range of $\sin t$ is $[-1,1]$. For $g(x)=2\sin2x$, the range of $g(x)$ is $[-2,2]$. Then for $y = 2g(x)=4\sin2x$, the range is $[-4,4]$.

Answer:

6.1: $180^{\circ}$
6.2.1: $\sqrt{3}$
6.2.2: $(-60^{\circ},-\sqrt{3})$
6.3: $[-4,4]$