15 (a) Show that $$ sin x - sin x imes cos 2x \ \approx 2x^3 $$ for small values of x. 15 (b) Hence, show that the area between the graph with equation $$ y = \sqrt{8(sin x - sin x \times cos 2x)} $$ the positive x-axis and the line x = 0.25 can be approximated by $$ \text{Area} = 2^m \times 5^n $$ where m and n are integers to be found. 15 (c) (i) Explain why $$ \int_{6.3}^{6.4} 2x^3 \, dx\ $$ is not a suitable approximation for $$ \int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) \, dx $$ 15 (c) (ii) Explain how $$ \int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) \, dx $$ might be approximated by $$ \int_{a}^{b} 2x^3 \, dx $$ for suitable values of a and b.

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15 (a) Show that $$ sin x - sin x imes cos 2x \ \approx 2x^3 $$ for small values of x - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Question 15

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15 (a) Show that $$ sin x - sin x imes cos 2x \ \approx 2x^3 $$ for small values of x. 15 (b) Hence, show that the area between the graph with equation $$ y = ... show full transcript

Worked Solution & Example Answer:15 (a) Show that $$ sin x - sin x imes cos 2x \ \approx 2x^3 $$ for small values of x - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Step 1

Show that $$ sin x - sin x imes cos 2x \approx 2x^3$$ for small values of x.

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Answer

To demonstrate that ( sin x - sin x \times cos 2x \approx 2x^3 ) for small values of x, we can use the small angle approximation. For small x, we know that:

The Taylor expansion for ( sin x ) is: $sin x \approx x - \frac{x^3}{6} + O(x^5)$
The cosine term can be expanded as: $cos 2x \approx 1 - \frac{(2x)^2}{2} + O(x^4) = 1 - 2x^2 + O(x^4)$
Thus, substituting into ( sin x \times cos 2x ): $sin x \times cos 2x \approx \left( x - \frac{x^3}{6} \right) \left( 1 - 2x^2 \right) \approx x - \frac{x^3}{6} - 2x^3 + O(x^5)$
This reduces to: $sin x - (x - \frac{x^3}{6} - 2x^3) \approx x - \left( x - \frac{x^3}{6} - 2x^3 \right) \approx 2x^3 + O(x^5)$

This validates that ( sin x - sin x \times cos 2x \approx 2x^3 ) for small x.

Step 2

Show that the area between the graph with equation $$ y = \sqrt{8(sin x - sin x \times cos 2x)}$$ the positive x-axis and the line x = 0.25 can be approximated by $$Area = 2^m \times 5^n$$ where m and n are integers to be found.

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Answer

To find the area between the graph and the positive x-axis, we will integrate from x = 0 to x = 0.25:

First, substitute the approximation from part (a): $Area = \int_{0}^{0.25} \sqrt{8(2x^3)} \, dx = \int_{0}^{0.25} \sqrt{16x^3} \, dx = 4 \int_{0}^{0.25} x^{3/2} \, dx$
Integrating: $\int x^{3/2} \, dx = \frac{2}{5} x^{5/2} + C$
Evaluating at the limits: $Area = 4 \left[ \frac{2}{5} x^{5/2} \right]_{0}^{0.25} = 4 \cdot \frac{2}{5} \left( 0.25^{5/2} - 0$

ight) = \frac{8}{5} \left( \frac{1}{4 \sqrt{4}} \right) = \frac{1}{10}

4. Simplifying, we find:

Area \approx 2^m \times 5^n \quad \text{where } m = 0, n = -1

Thus, we conclude with suitable integers m and n.

Step 3

Explain why $$\int_{6.3}^{6.4} 2x^3 \, dx$$ is not a suitable approximation for $$\int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) \, dx$$.

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Answer

The approximation ( \int_{6.3}^{6.4} 2x^3 , dx ) is not suitable because it does not account for the behavior of the function ( sin x - sin x \times cos 2x ) over the interval. The function involves oscillatory behavior due to the sine function, which can vary significantly within the integration limits. In contrast, the polynomial ( 2x^3 ) will vary smoothly but does not capture the necessary variation of the original function, especially where the sine function contributes significant values at certain points.

Step 4

Explain how $$\int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) \, dx$$ may be approximated by $$\int_{a}^{b} 2x^3 \, dx$$ for suitable values of a and b.

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Answer

To approximate ( \int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) , dx ) using ( \int_{a}^{b} 2x^3 , dx ):

We can consider that the behavior of the given function is periodic, and if we evaluate it at different intervals or average its behavior, we can capture its key characteristics.
By selecting suitable constants, particularly through value averaging or selecting representative midpoints that reflect the interval [6.3, 6.4], we can find a noticeable equivalence.
One can use substitutions or linear approximations around the mid-point value which are within an acceptable range for those approximating values of a and b. This effectively allows for the approximation via the integral of a polynomial that captures the average effect over the interval.

15 (a) Show that $$ sin x - sin x imes cos 2x \ \approx 2x^3 $$ for small values of x - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Worked Solution & Example Answer:15 (a) Show that $$ sin x - sin x imes cos 2x \ \approx 2x^3 $$ for small values of x - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Show that $$ sin x - sin x imes cos 2x \approx 2x^3$$ for small values of x.

Show that the area between the graph with equation $$ y = \sqrt{8(sin x - sin x \times cos 2x)}$$ the positive x-axis and the line x = 0.25 can be approximated by $$Area = 2^m \times 5^n$$ where m and n are integers to be found.

Explain why $$\int_{6.3}^{6.4} 2x^3 \, dx$$ is not a suitable approximation for $$\int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) \, dx$$.

Explain how $$\int_{6.3}^{6.4} \left( sin x - sin x \times cos 2x \right) \, dx$$ may be approximated by $$\int_{a}^{b} 2x^3 \, dx$$ for suitable values of a and b.

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