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Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = 8 ext{sin} igg( rac{1}{2}x igg) - 3x + 9$ $x > 0$ and $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 8 - 2022 - Paper 2

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Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation-$y-=-f(x)$-where--$f(x)-=-8--ext{sin}-igg(--rac{1}{2}x-igg)---3x-+-9$---$x->-0$--and-$x$-is-measured-in-radians-Edexcel-A-Level Maths Pure-Question 8-2022-Paper 2.png

Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = 8 ext{sin} igg( rac{1}{2}x igg) - 3x + 9$ $x > 0$ and $x$ is measured in ... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = 8 ext{sin} igg( rac{1}{2}x igg) - 3x + 9$ $x > 0$ and $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 8 - 2022 - Paper 2

Step 1

a) find the $x$ coordinate of $P$, giving your answer to 3 significant figures.

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Answer

To find the xx coordinate of point PP, we first need to determine the derivative of the function, which gives us the slope of the curve at any point:

f'(x) = 4 ext{cos} igg( rac{1}{2} x igg) - 3

To find the local maximum point, we set the derivative equal to zero:

4 ext{cos} igg( rac{1}{2} x igg) - 3 = 0

This leads to:

ext{cos} igg( rac{1}{2} x igg) = rac{3}{4}

Now we take the inverse cosine:

rac{1}{2} x = ext{cos}^{-1} igg( rac{3}{4} igg)

Hence:

x = 2 ext{cos}^{-1} igg( rac{3}{4} igg)

Calculating this value yields:

xext(approx)=2imes0.722=1.444x ext{ (approx)} = 2 imes 0.722 = 1.444

Therefore, when rounded to three significant figures:

xP=1.44x_P = 1.44

Step 2

b) explain why $a$ must lie in the interval $[4, 5]$.

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Answer

To explain why aa must lie in the interval [4,5][4, 5], we evaluate the function values provided:

  • f(4)=4.274>0f(4) = 4.274 > 0
  • f(5)=1.212<0f(5) = -1.212 < 0

Since f(4)f(4) is positive and f(5)f(5) is negative, by the Intermediate Value Theorem, there must be at least one root in the interval [4,5][4, 5] because the function changes sign. This confirms that aa, the point where the curve crosses the x-axis, must lie within this interval.

Step 3

c) apply the Newton-Raphson method once to $f(x)$ to obtain a second approximation to $a$.

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Answer

For the Newton-Raphson method, we will use the formula:

x_{n+1} = x_n - rac{f(x_n)}{f'(x_n)}

Given our first approximation x0=5x_0 = 5, we calculate:

  1. Evaluate f(5)f(5): f(5)=1.212f(5) = -1.212

  2. Evaluate f(5)f'(5): f'(5) = 4 ext{cos} igg( rac{1}{2} imes 5 igg) - 3 = 4 ext{cos}(2.5) - 3 Using a calculator, we find: f(5)ext(approx)=4imes(0.801)3agapproximately=6.204f'(5) ext{ (approx)} = 4 imes (-0.801) - 3 ag{approximately} = -6.204

Now, substituting into the Newton-Raphson formula:

x_1 = 5 - rac{-1.212}{-6.204} x1ext(approx)=5+0.195=4.805x_1 ext{ (approx)} = 5 + 0.195 = 4.805

Rounding this to three significant figures, we have:

x1ext(approx)=4.80x_1 ext{ (approx)} = 4.80

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