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The diagram below shows the graph of the function $f : x \mapsto \sin 2x$ - Leaving Cert Mathematics - Question 5 - 2014

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The diagram below shows the graph of the function $f : x \mapsto \sin 2x$. The line $2y = 1$ is also shown. (a) On the same diagram above, sketch the graphs of $g ... show full transcript

Worked Solution & Example Answer:The diagram below shows the graph of the function $f : x \mapsto \sin 2x$ - Leaving Cert Mathematics - Question 5 - 2014

Step 1

On the same diagram above, sketch the graphs of $g : x \mapsto \sin x$ and $h : x \mapsto 3\sin 2x$. Indicate clearly which is $g$ and which is $h$.

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Answer

To sketch the graphs of the functions:

  1. Graph of g:xsinxg : x \mapsto \sin x:

    • This is a standard sine wave oscillating between -1 and 1 with a period of 2π2\pi. It crosses the x-axis at integer multiples of π\pi.
  2. Graph of h:x3sin2xh : x \mapsto 3\sin 2x:

    • This function oscillates between -3 and 3 due to the amplitude of 3, and it has a period of π\pi because the frequency is doubled (2 in front of x). It also crosses the x-axis at x=nπ2x = \frac{n\pi}{2} where nn is any integer.
  3. Indicate the functions:

    • Use different colors for clarity: let gg be in green and hh in red. Clearly label both graphs on the diagram.

Step 2

Find the co-ordinates of the point $P$ in the diagram.

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Answer

To find the coordinates of point PP which is the intersection of the line y=12y = \frac{1}{2} and the curve y=sin2xy = \sin 2x:

  1. Setting up the equation:

    • Set sin2x=12\sin 2x = \frac{1}{2}.
  2. Solving for xx:

    • The general solution for sinθ=12\sin \theta = \frac{1}{2} is θ=π6+2nπ\theta = \frac{\pi}{6} + 2n\pi or θ=5π6+2nπ\theta = \frac{5\pi}{6} + 2n\pi, where nZn \in \mathbb{Z}.
    • For θ=2x\theta = 2x, we have:
      1. 2x=π6+2nπ2x = \frac{\pi}{6} + 2n\pi
        • This gives x=π12+nπx = \frac{\pi}{12} + n\pi
      2. 2x=5π6+2nπ2x = \frac{5\pi}{6} + 2n\pi
        • This gives x=5π12+nπx = \frac{5\pi}{12} + n\pi
  3. Finding suitable xx values:

    • Since we are considering xx in the range [0,2π]\left[0, 2\pi\right], the valid values from these solutions are:
      • x=π12,5π12,13π12,17π12x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}.
  4. Identifying the y-coordinate of point PP:

    • The y-coordinate is clearly 12\frac{1}{2} since point PP lies on the line y=12y = \frac{1}{2}.

Thus, the coordinates of point PP are:

o (17π12,12)\left(\frac{17\pi}{12}, \frac{1}{2}\right).

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