The function f is defined by $$f(x) = \frac{e^{3x}}{4x^2 + k}$$ where k is a positive constant. (a) Show that $$f'(x) = (12x^2 - 8x + 3k)g(x)$$ where g(x) is a function to be found. (3) Given that the curve with equation $y = f(x)$ has at least one stationary point, (b) find the range of possible values of k. (3)

A-Level Edexcel Maths Pure Differential Equations: The function f is defined by $$

The function f is defined by $$f(x) = \frac{e^{3x}}{4x^2 + k}$$ where k is a positive constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Question 14

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The function f is defined by $$f(x) = \frac{e^{3x}}{4x^2 + k}$$ where k is a positive constant. (a) Show that $$f'(x) = (12x^2 - 8x + 3k)g(x)$$ where g(x) is a ... show full transcript

Worked Solution & Example Answer:The function f is defined by $$f(x) = \frac{e^{3x}}{4x^2 + k}$$ where k is a positive constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Step 1

Show that $f'(x) = (12x^2 - 8x + 3k)g(x)$

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Answer

To find the derivative $f'(x)$ , we apply the quotient rule, which states that if $f(x) = \frac{u}{v}$ , then

$f'(x) = \frac{u'v - uv'}{v^2}$

Here, let:

$u = e^{3x}$ and $u' = 3e^{3x}$
$v = 4x^2 + k$ and $v' = 8x$

Applying the quotient rule:

$f'(x) = \frac{(3e^{3x})(4x^2 + k) - (e^{3x})(8x)}{(4x^2 + k)^2}$

Simplifying the numerator:

$= e^{3x} \frac{(12x^2 + 3k) - 8x}{(4x^2 + k)^2}$

This can be factored and rearranged to yield:

$f'(x) = \frac{e^{3x}(12x^2 - 8x + 3k)}{(4x^2 + k)^2}$

Letting $g(x) = \frac{e^{3x}}{(4x^2 + k)^2}$ , we have shown that the form holds.

Thus,

$f'(x) = (12x^2 - 8x + 3k)g(x)$

Step 2

find the range of possible values of k

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Answer

For the curve $y = f(x)$ to have at least one stationary point, the derivative $f'(x)$ must equal zero at some value of $x$ :

$12x^2 - 8x + 3k = 0$

This is a quadratic equation in $x$ . For this equation to have at least one real solution, the discriminant ( $b^2 - 4ac$ ) must be non-negative:

$(-8)^2 - 4 \cdot 12 \cdot 3k \geq 0$

Calculating the discriminant gives:

$64 - 144k \geq 0$

Simplifying this inequality results in:

$64 \geq 144k$ $k \leq \frac{64}{144} = \frac{4}{9}$

Since $k$ is defined as a positive constant, we also have:

$0 < k \leq \frac{4}{9}$

Therefore, the range of possible values for k is:

$k \in (0, \frac{4}{9}]$

The function f is defined by $$f(x) = \frac{e^{3x}}{4x^2 + k}$$ where k is a positive constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Worked Solution & Example Answer:The function f is defined by $$f(x) = \frac{e^{3x}}{4x^2 + k}$$ where k is a positive constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

Show that $f'(x) = (12x^2 - 8x + 3k)g(x)$

find the range of possible values of k

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