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For $$- rac{ heta}{2} \leq y \leq \frac{ heta}{2}$$, sketch the graph of $y = g(y)$ where $$g(x) = \arcsin{x}$$ $$-1 \leq x \leq 1$$ (b) Find the exact value of $x$ for which $$3g(x + 1) + \pi = 0$$ - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 3

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For--$$--rac{-heta}{2}-\leq-y-\leq-\frac{-heta}{2}$$,-sketch-the-graph-of-$y-=-g(y)$-where--$$g(x)-=-\arcsin{x}$$--$$-1-\leq-x-\leq-1$$--(b)-Find-the-exact-value-of-$x$-for-which--$$3g(x-+-1)-+-\pi-=-0$$---Edexcel-A-Level Maths Pure-Question 8-2016-Paper 3.png

For $$- rac{ heta}{2} \leq y \leq \frac{ heta}{2}$$, sketch the graph of $y = g(y)$ where $$g(x) = \arcsin{x}$$ $$-1 \leq x \leq 1$$ (b) Find the exact value of ... show full transcript

Worked Solution & Example Answer:For $$- rac{ heta}{2} \leq y \leq \frac{ heta}{2}$$, sketch the graph of $y = g(y)$ where $$g(x) = \arcsin{x}$$ $$-1 \leq x \leq 1$$ (b) Find the exact value of $x$ for which $$3g(x + 1) + \pi = 0$$ - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 3

Step 1

Sketch the graph of $y = g(y)$ where $g(x) = \arcsin{x}$

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Answer

To sketch the graph of y=g(y)=arcsin⁡xy = g(y) = \arcsin{x}, we must first understand the basic properties of the arcsine function:

  1. The domain of g(x)g(x) is [−1,1][-1, 1] because arcsin is only defined in this interval.
  2. The range of g(x)g(x) is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
  3. The function is increasing in this interval.

The sketch should start from (0,0)(0, 0), moving upwards to (1,π2)(1, \frac{\pi}{2}) and downwards to (−1,−π2)(-1, -\frac{\pi}{2}). Ensure that the curve passes through the origin and has no negative slopes.

Step 2

Find the exact value of $x$ for which $3g(x + 1) + \pi = 0$

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Answer

To find the exact value of xx, start by rearranging the equation:

3g(x+1)+π=03g(x + 1) + \pi = 0 3g(x+1)=−π3g(x + 1) = -\pi g(x+1)=−π3g(x + 1) = -\frac{\pi}{3}

Since g(x)=arcsin⁡xg(x) = \arcsin{x}, we have: arcsin⁡(x+1)=−π3\arcsin{(x + 1)} = -\frac{\pi}{3}

Using the properties of the arcsine function: x+1=sin⁡(−π3)x + 1 = \sin{(-\frac{\pi}{3})}

The value of(-\frac{\pi}{3}) gives us: x+1=−32x + 1 = -\frac{\sqrt{3}}{2}

Solving for xx gives: x=−32−1=−1−32x = -\frac{\sqrt{3}}{2} - 1 = -1 - \frac{\sqrt{3}}{2}

This value represents the point where the initial equation holds true.

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