A-Level Edexcel Maths Pure Binomial Expansion: The functions f and g are define

The functions f and g are defined by $f: x \mapsto 2x + \ln 2, \quad x \in \mathbb{R},$ g: x \mapsto e^{x^2}, \quad x \in \mathbb{R}.$ (a) Prove that the composite function gf is gf: x \mapsto 4e^{x^2}, \quad x \in \mathbb{R}.$ (b) Sketch the curve with equation $y = gf(x)$, and show the coordinates of the point where the curve cuts the y-axis - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 5

Question 1

$The-functions-f-and-g-are-defined-by--$f:-x-\mapsto-2x-+-\ln-2,-\quad-x-\in-\mathbb{R},$--g:-x-\mapsto-e^{x^2},-\quad-x-\in-\mathbb{R}.$--(a)-Prove-that-the-composite-function-gf-is--gf:-x-\mapsto-4e^{x^2},-\quad-x-\in-\mathbb{R}.$--(b)-Sketch-the-curve-with-equation-$y-=-gf(x)$,-and-show-the-coordinates-of-the-point-where-the-curve-cuts-the-y-axis-Edexcel-A-Level Maths Pure-Question 1-2018-Paper 5.png$

The functions f and g are defined by $f: x \mapsto 2x + \ln 2, \quad x \in \mathbb{R},$ g: x \mapsto e^{x^2}, \quad x \in \mathbb{R}.$ (a) Prove that the composit... show full transcript

Worked Solution & Example Answer:The functions f and g are defined by $f: x \mapsto 2x + \ln 2, \quad x \in \mathbb{R},$ g: x \mapsto e^{x^2}, \quad x \in \mathbb{R}.$ (a) Prove that the composite function gf is gf: x \mapsto 4e^{x^2}, \quad x \in \mathbb{R}.$ (b) Sketch the curve with equation $y = gf(x)$, and show the coordinates of the point where the curve cuts the y-axis - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 5

Step 1

Prove that the composite function gf is

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Answer

To find the composite function gf, we compute:

Start with g(x): $g(x) = e^{x^2}$
Substitute f(x) into g: $gf(x) = g(f(x)) = g(2x + \ln 2) = e^{(2x + \ln 2)^2}$
Expanding the square: $(2x + \ln 2)^2 = 4x^2 + 4x \ln 2 + (\ln 2)^2$
Therefore: $gf(x) = e^{4x^2 + 4x \ln 2 + (\ln 2)^2}$
Simplifying further: $gf(x) = \frac{e^{(\ln 2)^2}}{e^{-4x^2 - 4x \ln 2}} = 4e^{x^2}$

Hence, the composite function is $gf(x) = 4 e^{x^2}, \quad x \in \mathbb{R}.$

Step 2

Sketch the curve with equation $y = gf(x)$, and show the coordinates of the point where the curve cuts the y-axis.

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Answer

To find where the curve cuts the y-axis, set x = 0:

$y = gf(0) = 4e^{0^2} = 4e^0 = 4.$

The coordinates of the point where the curve cuts the y-axis are (0, 4).

Sketch:

The function is an upward-opening curve.
Starts at (0, 4), increases rapidly.

Step 3

Write down the range of gf.

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Answer

The function $gf(x) = 4e^{x^2}$ is always positive for all real x, since $e^{x^2} > 0$ . Therefore, the range of gf is:

$\text{Range of } gf: (0, \infty).$

Step 4

Find the value of x for which $\frac{d}{dx}[gf(x)] = 3$.

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Answer

To find ( \frac{d}{dx}[gf(x)] ), we first differentiate:

Using the chain rule: $\frac{d}{dx}[gf(x)] = \frac{d}{dx}[4e^{x^2}] = 4 \cdot \frac{d}{dx}[e^{x^2}] = 4 \cdot e^{x^2} \cdot 2x = 8xe^{x^2}.$
Set this equal to 3: $8xe^{x^2} = 3.$
Solving for x: $x = \frac{3}{8e^{x^2}}.$

Using numerical methods, one can approximate that: $x \approx -0.418$ .

Prove that the composite function gf is

Sketch the curve with equation $y = gf(x)$, and show the coordinates of the point where the curve cuts the y-axis.

Write down the range of gf.

Find the value of x for which $\frac{d}{dx}[gf(x)] = 3$.

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