Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4 \sin 2x}{e^{\sqrt{x} - 1}}$, $0 \leq x \leq \pi$. 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 Polynomials: Figure 5 shows a sketch of the c

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

Question 1

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

Worked Solution & Example Answer:Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4 \sin 2x}{e^{\sqrt{x} - 1}}$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 1 - 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 find the critical points, we differentiate the function using the quotient rule:

Let:

$u = 4 \sin 2x$
$v = e^{\sqrt{x} - 1}$

Using the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2}$

Calculating $u'$ and $v'$ :

$u' = 8 \cos 2x$
$v' = e^{\sqrt{x}-1} \cdot \frac{1}{2\sqrt{x}}$

Thus, we can express: $f'(x) = \frac{(8 \cos 2x)e^{\sqrt{x}-1} - (4 \sin 2x)e^{\sqrt{x}-1} \cdot \frac{1}{2\sqrt{x}}}{(e^{\sqrt{x}-1})^2}$

By simplifying and setting $f'(x) = 0$ , we find: $\frac{8 \cos 2x - \frac{4 \sin 2x}{2\sqrt{x}}}{e^{\sqrt{x}-1}} = 0$ This leads us to: $8 \cos 2x - \frac{4 \sin 2x}{2\sqrt{x}} = 0$

We can simplify further, arriving at the equation: $\tan 2x = \sqrt{2}$

Step 2

Find the $x$-coordinate of the minimum turning point on the curve with equation (i)

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Answer

To find the $x$ -coordinate corresponding to the minimum turning point for $y = f(2x)$ :

We already established that the solutions to $\tan 2x = \sqrt{2}$ can be found at:

$2x = \frac{\pi}{4} + n\pi$ (for integer $n$ )

From this, we solve: $x = \frac{\pi}{8} + \frac{n\pi}{2}$

Taking the first positive value: $x = \frac{\pi}{8} \approx 1.02 \text{ (for } n=0\text{)}$

Step 3

Find the $x$-coordinate of the minimum turning point on the curve with equation (ii)

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Answer

For the equation $y = 3 - 2f(x)$ , we can use the earlier value of $x$ found from part (a).

Thus using the same result:

By solving $\tan 2x = \sqrt{2}$ , we find $x$ to be approximately: $x \approx 0.478 \text{ (for } n=0\text{)}$

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

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

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

Find the $x$-coordinate of the minimum turning point on the curve with equation (i)

Find the $x$-coordinate of the minimum turning point on the curve with equation (ii)

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