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 Vectors in 3 Dimensions: 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 13 - 2022 - Paper 2

Question 13

$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 13-2022-Paper 2.png$

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 13 - 2022 - Paper 2

Step 1

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

96%

114 rated

Answer

To find the derivative of the function, we apply the quotient rule:

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

This simplifies to:

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

Factoring out $e^{3x}$ :

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

Letting $g(x) = \frac{e^{3x}}{(4x^2 + k)^2}$ , we demonstrate that:

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

Step 2

Given that $y = f(x)$ has at least one stationary point, find the range of possible values of k

99%

104 rated

Answer

A stationary point occurs where the derivative is zero:

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

This is a quadratic equation of the form $ax^2 + bx + c = 0$ , and has at least one solution when the discriminant is non-negative:

$D = b^2 - 4ac = (-8)^2 - 4(12)(3k) = 64 - 144k$

Setting the discriminant greater than or equal to zero for real roots:

$64 - 144k \geq 0 \Rightarrow k \leq \frac{64}{144} = \frac{4}{9}$

Since k is a positive constant:

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

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 13 - 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 13 - 2022 - Paper 2

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

Given that $y = f(x)$ has at least one stationary point, find the range of possible values of k

Join the A-Level students using SimpleStudy...