Question
Consider the function \( f(x) = e^{\cos 2x} \), where \( -\pi \leq x \leq \frac{5\pi}{4} \).
(a) Find the coordinates of the points on the curve \( y = f(x) \) where the gradient is zero. [5]
(b) Using the second derivative at each point found in part (a), show that the curve \( y = f(x) \) has two local maximum points and one local minimum point. [4]
(c) Sketch the curve of \( y = f(x) \) for \( 0 \leq x \leq \pi \), taking into consideration the relative values of the second derivative found in part (b). [3]
(d) (i) Find the Maclaurin series for \( \cos 2x \), up to and including the term in \( x^4 \).
(ii) Hence, find the Maclaurin series for \( e^{\cos 2x – 1} \), up to and including the term in \( x^4 \).
(iii) Hence, write down the Maclaurin series for \( f(x) \), up to and including the term in \( x^4 \). [6]
(e) Use the first two non-zero terms in the Maclaurin series for \( f(x) \) to show that \( \int_{0}^{\frac{1}{10}} e^{\cos 2x} dx \approx \frac{149e}{1500} \). [3]
▶️Answer/Explanation
Detail Solution
(a)
Explanation: Step 1: Find the derivative of the function \( f(x) = e^{\cos 2x} \).
To find the points where the gradient is zero, we first need to calculate the derivative \( f'(x) \) using the chain rule.
The derivative is given by:
$$ f'(x) = e^{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2 e^{\cos 2x} \sin 2x $$
Step 2: Set the derivative equal to zero to find critical points.
We need to solve the equation:
$$ -2 e^{\cos 2x} \sin 2x = 0 $$
Since \( e^{\cos 2x} \) is never zero, we can simplify this to:
\( \sin 2x = 0 \).
Step 3: Find the values of \( x \) where \( \sin 2x = 0 \).
The general solutions for \( \sin 2x = 0 \) are:
$ 2x = n\pi $ for $n \in \mathbb{Z} $
Thus, $$ x = \frac{n\pi}{2} $$
Step 4: Determine the integer values of \( n \) that fit within the interval \( -\pi \leq x \leq \frac{5\pi}{4} \).
Calculating the bounds for \( n \):
For \( x = -\pi \):
$ -\pi = \frac{n\pi}{2} $ gives $ n = -2 $
For \( x = \frac{5\pi}{4} \):
$ \frac{5\pi}{4} = \frac{n\pi}{2} $ gives $ n = \frac{5}{2} $ which is not an integer. Thus, we consider $ n = -2, -1, 0, 1, 2 $
Calculating the corresponding \( x \) values:
– For \( n = -2 \): \( x = -\pi \)
– For \( n = -1 \): \( x = -\frac{\pi}{2} \)
– For \( n = 0 \): \( x = 0 \)
– For \( n = 1 \): \( x = \frac{\pi}{2} \)
– For \( n = 2 \): \( x = \pi \)
Step 5: Evaluate \( f(x) \) at these critical points to find the coordinates.
– For \( x = -\pi \):
$$ f(-\pi) = e^{\cos(-2\pi)} = e^{1} = e $$
Coordinates: \( (-\pi, e) \).
– For \( x = -\frac{\pi}{2} \):
$$ f(-\frac{\pi}{2}) = e^{\cos(-\pi)} = e^{-1} = \frac{1}{e} $$
Coordinates: \( (-\frac{\pi}{2}, \frac{1}{e}) \).
– For \( x = 0 \):
$$ f(0) = e^{\cos(0)} = e^{1} = e $$
Coordinates: \( (0, e) \).
– For \( x = \frac{\pi}{2} \):
$$ f(\frac{\pi}{2}) = e^{\cos(\pi)} = e^{-1} = \frac{1}{e} $$
Coordinates: \( (\frac{\pi}{2}, \frac{1}{e}) \).
– For \( x = \pi \):
$$ f(\pi) = e^{\cos(2\pi)} = e^{1} = e $$
Coordinates: \( (\pi, e) \).
Step 6: Summarize the coordinates of the points where the gradient is zero.
The coordinates are:
1. \( (-\pi, e) \)
2. \( (-\frac{\pi}{2}, \frac{1}{e}) \)
3. \( (0, e) \)
4. \( (\frac{\pi}{2}, \frac{1}{e}) \)
5. \( (\pi, e) \)
(b)
To analyze the function \( f(x) = e^{\cos 2x} \) for local maxima and minima, we first need to find the critical points by determining the first derivative and setting it to zero. Then we will use the second derivative test to classify these critical points.
Step 1: Find the first derivative \( f'(x) \).
Using the chain rule, we have:
\[
f'(x) = e^{\cos 2x} \cdot (-\sin 2x) \cdot 2 = -2 e^{\cos 2x} \sin 2x
\]
Step 2: Set the first derivative to zero to find critical points.
Setting \( f'(x) = 0 \):
\[
-2 e^{\cos 2x} \sin 2x = 0
\]
Since \( e^{\cos 2x} \) is never zero, we have:
\[
\sin 2x = 0
\]
This gives us the solutions:
\[
2x = n\pi \quad \Rightarrow \quad x = \frac{n\pi}{2}
\]
where \( n \) is an integer. We need to find the values of \( n \) such that \( -\pi \leq x \leq \frac{5\pi}{4} \).
Calculating for \( n = -2, -1, 0, 1, 2, 3 \):
– For \( n = -2: x = -\pi \)
– For \( n = -1: x = -\frac{\pi}{2} \)
– For \( n = 0: x = 0 \)
– For \( n = 1: x = \frac{\pi}{2} \)
– For \( n = 2: x = \pi \)
– For \( n = 3: x = \frac{3\pi}{2} \) (not in the interval)
– For \( n = 4: x = 2\pi \) (not in the interval)
Thus, the critical points are \( x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \).
Step 3: Find the second derivative \( f”(x) \).
Using the product rule and chain rule:
\[
f”(x) = \frac{d}{dx}(-2 e^{\cos 2x} \sin 2x)
\]
Applying the product rule:
\[
f”(x) = -2 \left( e^{\cos 2x} \cdot (-\sin 2x) \cdot 2 \cdot \sin 2x + e^{\cos 2x} \cdot \cos 2x \cdot 2 \right)
\]
Simplifying:
\[
f”(x) = 4 e^{\cos 2x} \sin^2 2x – 4 e^{\cos 2x} \cos 2x
\]
Factoring out \( 4 e^{\cos 2x} \):
\[
f”(x) = 4 e^{\cos 2x} (\sin^2 2x – \cos 2x)
\]
Step 4: Evaluate the second derivative at the critical points.
1. For \( x = -\pi \):
\[
f”(-\pi) = 4 e^{\cos(-2\pi)} (\sin^2(-2\pi) – \cos(-2\pi)) = 4 e^{1} (0 – 1) = -4e \quad \text{(local maximum)}
\]
2. For \( x = -\frac{\pi}{2} \):
\[
f”(-\frac{\pi}{2}) = 4 e^{\cos(-\pi)} (\sin^2(-\pi) – \cos(-\pi)) = 4 e^{-1} (0 + 1) = 4e^{-1} \quad \text{(local minimum)}
\]
3. For \( x = 0 \):
\[
f”(0) = 4 e^{\cos(0)} (\sin^2(0) – \cos(0)) = 4 e^{1} (0 – 1) = -4e \quad \text{(local maximum)}
\]
4. For \( x = \frac{\pi}{2} \):
\[
f”(\frac{\pi}{2}) = 4 e^{\cos(\pi)} (\sin^2(\pi) – \cos(\pi)) = 4 e^{-1} (0 + 1) = 4e^{-1} \quad \text{(local minimum)}
\]
5. For \( x = \pi \):
\[
f”(\pi) = 4 e^{\cos(2\pi)} (\sin^2(2\pi) – \cos(2\pi)) = 4 e^{1} (0 – 1) = -4e \quad \text{(local maximum)}
\]
Step 5: Conclusion on local maxima and minima.
The function \( f(x) \) has two local maximum points at \( x = -\pi \) and \( x = 0 \), and one local minimum point at \( x = -\frac{\pi}{2} \).
(c)
Sketch the curve of $y = f(x)$ for $ 0 \leq x \leq \pi$
We know that:
- At $x = 0$, $f(0) = e^{\cos(0)} = e^{1} = e$
- At $x = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = e^{\cos(\pi)} = e^{-1}$
- The function decreases from $e$ at $ x = 0$ to $e^{-1}$ at $x = \frac{\pi}{2}$
The local maximum at $x = 0$ is $e$, and the local minimum at $x = \frac{\pi}{2}$ is $e^{-1}$
The sketch should reflect these values and the behavior of the function, showing the local maximum and minimum clearly.
for $ 0\leq x \leq \pi $
for large interval
(d)
(i)
Step 1: Recall the Maclaurin series expansion formula.
The Maclaurin series for a function \( f(x) \) is given by:
\[
f(x) = f(0) + f'(0)x + \frac{f”(0)}{2!}x^2 + \frac{f”'(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \ldots
\]
Step 2: Calculate the derivatives of \( \cos 2x \).
We need to find \( f(0) \), \( f'(0) \), \( f”(0) \), \( f”'(0) \), and \( f^{(4)}(0) \) for \( f(x) = \cos 2x \).
1. \( f(0) = \cos(2 \cdot 0) = \cos(0) = 1 \)
2. \( f'(x) = -2\sin(2x) \)
\( f'(0) = -2\sin(0) = 0 \)
3. \( f”(x) = -4\cos(2x) \)
\( f”(0) = -4\cos(0) = -4 \)
4. \( f”'(x) = 8\sin(2x) \)
\( f”'(0) = 8\sin(0) = 0 \)
5. \( f^{(4)}(x) = 16\cos(2x) \)
\( f^{(4)}(0) = 16\cos(0) = 16 \)
Step 3: Substitute the derivatives into the Maclaurin series formula.
Now we substitute the values we calculated into the Maclaurin series expansion:
\[
\cos 2x = 1 + 0 \cdot x + \frac{-4}{2!}x^2 + 0 \cdot x^3 + \frac{16}{4!}x^4 + \ldots
\]
Step 4: Simplify the series up to \( x^4 \).
Calculating the coefficients:
\[
\frac{-4}{2!} = \frac{-4}{2} = -2
\]
\[
\frac{16}{4!} = \frac{16}{24} = \frac{2}{3}
\]
Thus, the Maclaurin series for \( \cos 2x \) up to and including the term in \( x^4 \) is:
\[
\cos 2x \approx 1 – 2x^2 + \frac{2}{3}x^4
\]
(ii)
Step 1: Recall the Maclaurin series expansion for \( e^x \).
The Maclaurin series for \( e^x \) is given by:
\[
e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots
\]
Step 2: Substitute \( \cos 2x – 1 \) into the series.
We need to find the Maclaurin series for \( e^{\cos 2x – 1} \). First, we will find the series for \( \cos 2x \).
Step 3: Recall the Maclaurin series expansion for \( \cos x \).
The Maclaurin series for \( \cos x \) is given by:
\[
\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \ldots
\]
Step 4: Substitute \( 2x \) into the series for \( \cos x \).
\[
\cos 2x = 1 – \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} = 1 – \frac{4x^2}{2} + \frac{16x^4}{24} = 1 – 2x^2 + \frac{2}{3}x^4
\]
Step 5: Calculate \( \cos 2x – 1 \).
\[
\cos 2x – 1 = (1 – 2x^2 + \frac{2}{3}x^4) – 1 = -2x^2 + \frac{2}{3}x^4
\]
Step 6: Substitute \( \cos 2x – 1 \) into the series for \( e^x \).
Now we substitute \( -2x^2 + \frac{2}{3}x^4 \) into the series for \( e^x \):
\[
e^{\cos 2x – 1} = e^{-2x^2 + \frac{2}{3}x^4}
\]
Step 7: Use the series for \( e^x \) with \( x = -2x^2 + \frac{2}{3}x^4 \).
We will only consider terms up to \( x^4 \):
\[
e^{-2x^2 + \frac{2}{3}x^4} = 1 + (-2x^2 + \frac{2}{3}x^4) + \frac{(-2x^2 + \frac{2}{3}x^4)^2}{2!}
\]
Step 8: Calculate \( \frac{(-2x^2 + \frac{2}{3}x^4)^2}{2!} \).
\[
(-2x^2 + \frac{2}{3}x^4)^2 = 4x^4 – 2 \cdot 2x^2 \cdot \frac{2}{3}x^4 + \left(\frac{2}{3}x^4\right)^2 = 4x^4 – \frac{8}{3}x^6 + \frac{4}{9}x^8
\]
Since we only need terms up to \( x^4 \), we can ignore higher order terms:
\[
\frac{4x^4}{2} = 2x^4
\]
Step 9: Combine all terms.
Thus, we have:
\[
e^{\cos 2x – 1} = 1 – 2x^2 + \frac{2}{3}x^4 + 2x^4 = 1 – 2x^2 + \left(\frac{2}{3} + 2\right)x^4 = 1 – 2x^2 + \frac{8}{3}x^4
\]
Step 10: Write the final Maclaurin series up to \( x^4 \).
The Maclaurin series for \( e^{\cos 2x – 1} \) up to and including the term in \( x^4 \) is:
\[
1 – 2x^2 + \frac{8}{3}x^4
\]
(iii)
Explanation: Step 1: Identify the function and its derivatives.
The function given is \( f(x) = e^{\cos 2x} \). To find the Maclaurin series, we need to calculate the derivatives of \( f(x) \) at \( x = 0 \).
Step 2: Calculate \( f(0) \).
\[
f(0) = e^{\cos(2 \cdot 0)} = e^{\cos(0)} = e^1 = e.
\]
Step 3: Calculate the first derivative \( f'(x) \).
Using the chain rule, we have:
\[
f'(x) = e^{\cos(2x)} \cdot (-\sin(2x)) \cdot 2 = -2\sin(2x)e^{\cos(2x)}.
\]
Now, evaluate at \( x = 0 \):
\[
f'(0) = -2\sin(2 \cdot 0)e^{\cos(0)} = -2 \cdot 0 \cdot e = 0.
\]
Step 4: Calculate the second derivative \( f”(x) \).
Using the product rule:
\[
f”(x) = -2\left(2\cos(2x)e^{\cos(2x)} – 2\sin(2x)(-\sin(2x)e^{\cos(2x)})\right).
\]
This simplifies to:
\[
f”(x) = -4\cos(2x)e^{\cos(2x)} + 4\sin^2(2x)e^{\cos(2x)}.
\]
Evaluating at \( x = 0 \):
\[
f”(0) = -4\cos(0)e^{\cos(0)} + 4\sin^2(0)e^{\cos(0)} = -4 \cdot 1 \cdot e + 0 = -4e.
\]
Step 5: Calculate the third derivative \( f”'(x) \).
Differentiating \( f”(x) \) requires applying the product rule and chain rule again. After simplification, we find:
\[
f”'(0) = 0.
\]
Step 6: Calculate the fourth derivative \( f^{(4)}(x) \).
Continuing from the previous derivatives, we find:
$$ f^{(4)} (x) =e^{cos(2x)} [48cos(4x) +16 cos(2x)-48 sin^2(2x) cos(2x)] +e^{cos(2x)} (-sin(2x) (2)[12sin(4x) -8 sin^3(2x) +8sin(2x)]$$
\[
f^{(4)}(0) = 64 e.
\]
Step 7: Construct the Maclaurin series.
The Maclaurin series for \( f(x) \) up to \( x^4 \) is given by:
\[
f(x) = f(0) + f'(0)x + \frac{f”(0)}{2!}x^2 + \frac{f”'(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \ldots
\]
Substituting the values we calculated:
\[
f(x) = e + 0 \cdot x + \frac{-4e}{2}x^2 + 0 \cdot x^3 + \frac{64e}{24}x^4.
\]
This simplifies to:
\[
f(x) = e – 2ex^2 + \frac{8e}{3}x^4.
\]
Step 8: Final expression for the Maclaurin series.
Thus, the Maclaurin series for \( f(x) \) up to and including the term in \( x^4 \) is:
\[
f(x) = e – 2ex^2 + \frac{8e}{3}x^4.
\]
(e)
Step 1: Find the Maclaurin series for \( f(x) = e^{\cos 2x} \).
The Maclaurin series for \( e^u \) is given by \( e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots \). Here, we will substitute \( u = \cos 2x \).
Step 2: Calculate \( \cos 2x \) using its Maclaurin series.
The Maclaurin series for \( \cos x \) is \( \cos x = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \ldots \). Thus, for \( \cos 2x \), we have:
\[
\cos 2x = 1 – \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} – \ldots = 1 – 2x^2 + \frac{4x^4}{24} – \ldots = 1 – 2x^2 + \frac{x^4}{6} – \ldots
\]
Step 3: Substitute \( \cos 2x \) into the Maclaurin series for \( e^{\cos 2x} \).
Using the first two non-zero terms of \( \cos 2x \):
\[
f(x) = e^{\cos 2x} \approx e^{1 – 2x^2} \approx e \cdot e^{-2x^2}
\]
Now, using the series expansion for \( e^{-2x^2} \):
\[
e^{-2x^2} = 1 – 2x^2 + \frac{(-2x^2)^2}{2!} + \ldots = 1 – 2x^2 + 2x^4 + \ldots
\]
So,
\[
f(x) \approx e \left(1 – 2x^2 + 2x^4\right)
\]
Step 4: Integrate the approximate function from \( 0 \) to \( \frac{1}{10} \).
We will integrate the first two non-zero terms:
\[
\int_{0}^{\frac{1}{10}} f(x) \, dx \approx \int_{0}^{\frac{1}{10}} e \left(1 – 2x^2\right) \, dx
\]
Calculating the integral:
\[
\int_{0}^{\frac{1}{10}} e \, dx = e \left[x\right]_{0}^{\frac{1}{10}} = e \cdot \frac{1}{10}
\]
\[
\int_{0}^{\frac{1}{10}} -2ex^2 \, dx = -2e \left[\frac{x^3}{3}\right]_{0}^{\frac{1}{10}} = -2e \cdot \frac{1}{3} \cdot \left(\frac{1}{10}\right)^3 = -\frac{2e}{3000}
\]
Step 5: Combine the results of the integrals.
\[
\int_{0}^{\frac{1}{10}} f(x) \, dx \approx e \cdot \frac{1}{10} – \frac{2e}{3000}
\]
Finding a common denominator:
\[
\frac{300e}{3000} – \frac{2e}{3000} = \frac{298e}{3000}
\]
Step 6: Approximate the integral.
Now we need to check if this is approximately \( \frac{149e}{1500} \):
\[
\frac{298e}{3000} = \frac{149e}{1500}
\]
This shows that:
Thus, the approximation holds.
————Markscheme—————–
(a) \( f'(x) = -2\sin 2x e^{\cos 2x} = 0 \)
\( \sin 2x = 0 \)
\( x = 0, \frac{\pi}{2}, \pi \)
Coordinates: \( (0, e) \), \( \left( \frac{\pi}{2}, \frac{1}{e} \right) \), \( (\pi, e) \)
(b) \( f”(x) = -4\cos 2x e^{\cos 2x} + 4\sin^2 2x e^{\cos 2x} \)
At \( x = 0 \) and \( x = \pi \), \( f”(x) < 0 \) (local maxima)
At \( x = \frac{\pi}{2} \), \( f”(x) > 0 \) (local minimum)
(c) Sketch should show maxima at \( x = 0 \) and \( x = \pi \), and a minimum at \( x = \frac{\pi}{2} \).
(d) (i) \( \cos 2x = 1 – 2x^2 + \frac{2x^4}{3} \)
(ii) \( e^{\cos 2x – 1} = 1 – 2x^2 + \frac{8x^4}{3} \)
(iii) \( f(x) = e \left( 1 – 2x^2 + \frac{8x^4}{3} \right) \)
(e) Using the first two non-zero terms:
\( \int_{0}^{\frac{1}{10}} e^{\cos 2x} dx \approx \int_{0}^{\frac{1}{10}} e(1 – 2x^2) dx = \frac{149e}{1500} \)
Question
The function f is defined by f (x) = ex sinx , where x ∈ R.
(a) Find the Maclaurin series for f (x) up to and including the x3 term.
(b) Hence, find an approximate value for \(\int_{0}^{1}e^{x^{2}} sin(x^{2})dx.\)
The function g is defined by g (x) = ex cos x , where x ∈ R.
(c) (i) Show that g(x) satisfies the equation g ″(x) = 2(g′(x) – g(x)).
(ii) Hence, deduce that g(4) (x) = 2(g″′(x) – g ″(x)) .
(d) Using the result from part (c), find the Maclaurin series for g(x) up to and including the x4 term.
(e) Hence, or otherwise, determine the value of \(\lim_{x\rightarrow 0}\frac{e^{x}cos x – 1 – x}{x^{3}}.\)
▶️Answer/Explanation
Ans:
(a) METHOD 1
recognition of both known series
attempt to multiply the two series up to and including x3 term
Note: Condone absence of limits up to this stage.
Note: Accept working with each side separately to obtain -2ex sin x .
Note: Accept working with each side separately to obtain -4ex cos x .
Note: Do not award any marks for approaches that do not use the part (c) result.
Note: Condone the omission of +… in their working.
