Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4\sin 2x}{e^{\sqrt{x}} - 1}, \quad 0 \leq x \leq \pi$. The curve has a maximum turning point at $P$ and a minimum turning point at $Q$ as shown in Figure 5. (a) Show that the $x$ coordinates of point $P$ and point $Q$ are solutions of the equation $$\tan 2x = \sqrt{2}$$. (b) Using your answer to part (a), find the $x$-coordinate of the minimum turning point on the curve with equation (i) $y = f(2x)$. (ii) $y = 3 - 2f(x).$

A-Level Edexcel Maths Pure Applications of Differentiation: Figure 5 shows a sketch of the c

Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4\sin 2x}{e^{\sqrt{x}} - 1}, \quad 0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Question 15

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Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4\sin 2x}{e^{\sqrt{x}} - 1}, \quad 0 \leq x \leq \pi$. The curve has a maximum t... show full transcript

Worked Solution & Example Answer:Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4\sin 2x}{e^{\sqrt{x}} - 1}, \quad 0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Step 1

Show that the $x$ coordinates of point $P$ and point $Q$ are solutions of the equation

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Answer

To show that the $x$ coordinates of points $P$ and $Q$ are solutions of $\tan 2x = \sqrt{2}$ , we start by differentiating the function $f(x)$ using the quotient rule.

Differentiate $f(x)$ : $f'(x) = \frac{(e^{\sqrt{x}})(8\cos2x) - (4\sin2x)(\frac{1}{2\sqrt{x}} e^{\sqrt{x}})}{(e^{\sqrt{x}} - 1)^2}$
Set the derivative $f'(x) = 0$ to find turning points: This gives us: $8\cos2x - \frac{4\sin2x}{2\sqrt{x}} = 0$
Rearranging, we find: $\tan 2x = \sqrt{2}$ confirming that the coordinates correspond to the points where the tangent is equal to $\\sqrt{2}$ .

Step 2

Using your answer to part (a), find the x-coordinate of the minimum turning point on the curve with equation

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Answer

(i) For $y = f(2x)$ , Solve $\tan(4x) = \sqrt{2}$ . This implies: $4x = \frac{\pi}{4} + n\pi , n \in \mathbb{Z}$ The minimum turning point closest to the origin occurs when: $x = \frac{1}{4}\left(\frac{\pi}{4}\right) = 0.785.$

(ii) For $y = 3 - 2f(x)$ , Solve $\tan(2x) = \sqrt{2}$ : $2x = \frac{\pi}{4} + n\pi , n \in \mathbb{Z}$ The smallest positive solution is: $x = \frac{1}{2}\left(\frac{\pi}{4}\right) = 0.392.$

Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4\sin 2x}{e^{\sqrt{x}} - 1}, \quad 0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Worked Solution & Example Answer:Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4\sin 2x}{e^{\sqrt{x}} - 1}, \quad 0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Show that the $x$ coordinates of point $P$ and point $Q$ are solutions of the equation

Using your answer to part (a), find the x-coordinate of the minimum turning point on the curve with equation

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