A-Level Edexcel Maths Pure Exponential & Logarithms: For the constant $k$, where $k >

For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x ightarrow ext{ln}(x + k), ext{ where } x > -k,$ g: x ightarrow |2x - k|, ext{ where } x ext{ is in } extbf{R} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

Question 7

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For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x ightarrow ext{ln}(x + k), ext{ where } x > -k,$ g: x ightarrow |2x - k|, e... show full transcript

Worked Solution & Example Answer:For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x ightarrow ext{ln}(x + k), ext{ where } x > -k,$ g: x ightarrow |2x - k|, ext{ where } x ext{ is in } extbf{R} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

Step 1

a) On separate axes, sketch the graph of $f$ and the graph of $g$. On each sketch state, in terms of $k$, the coordinates of points where the graph meets the coordinate axes.

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Answer

The graph of $f(x) = ext{ln}(x + k)$ is defined for $x > -k$ . It intersects the $y$ -axis at $(0, ext{ln}(k))$ and approaches $- rac{ ext{ln}(k)}{ ext{e}}$ as $x$ approaches $-k$ from the right. The graph decreases and then increases.

The graph of $g(x) = |2x - k|$ has a vertex at $( rac{k}{2}, 0)$ , hence intersects the $x$ -axis at $( rac{k}{2}, 0)$ and meets the $y$ -axis at $(0, |k|)$ (for $k > 0$ ).

Step 2

b) Write down the range of $f$.

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Answer

The range of $f(x) = ext{ln}(x + k)$ is $(− ext{∞}, ext{∞})$ since the logarithmic function can take any real value.

Step 3

c) Find $f(g(rac{k}{4}))$ in terms of $k$, giving your answer in its simplest form.

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Answer

First, evaluate $g( rac{k}{4}) = |2( rac{k}{4}) - k| = |( rac{k}{2} - k)| = | rac{k}{2} - rac{2k}{2}| = | rac{-k}{2}| = rac{k}{2}.$

Next, substitute into $f$ :

$f(g( rac{k}{4})) = f( rac{k}{2}) = ext{ln}( rac{k}{2} + k) = ext{ln}( rac{3k}{2}).$

Step 4

d) The curve $C$ has equation $y = f(x)$. The tangent to $C$ at the point with $x$-coordinate 3 is parallel to the line with equation $9y = 2x + 1$. Find the value of $k.$

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Answer

The slope of the line $9y = 2x + 1$ is $rac{2}{9}$ , i.e., $m = rac{2}{9}$ . The derivative of $f(x)$ gives the slope of the tangent:

$f'(x) = rac{1}{x + k}.$

Setting $x = 3$ gives

$f'(3) = rac{1}{3 + k}.$

Setting this equal to $rac{2}{9}$ yields:

$rac{1}{3 + k} = rac{2}{9}$

Cross-multiplying gives:

ightarrow 9 = 6 + 2k ightarrow 2k = 3 ightarrow k = rac{3}{2}.$$

For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x ightarrow ext{ln}(x + k), ext{ where } x > -k,$ g: x ightarrow |2x - k|, ext{ where } x ext{ is in } extbf{R} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

Worked Solution & Example Answer:For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x ightarrow ext{ln}(x + k), ext{ where } x > -k,$ g: x ightarrow |2x - k|, ext{ where } x ext{ is in } extbf{R} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 4

a) On separate axes, sketch the graph of $f$ and the graph of $g$. On each sketch state, in terms of $k$, the coordinates of points where the graph meets the coordinate axes.

b) Write down the range of $f$.

c) Find $f(g( rac{k}{4}))$ in terms of $k$, giving your answer in its simplest form.

d) The curve $C$ has equation $y = f(x)$. The tangent to $C$ at the point with $x$-coordinate 3 is parallel to the line with equation $9y = 2x + 1$. Find the value of $k.$

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c) Find $f(g(rac{k}{4}))$ in terms of $k$, giving your answer in its simplest form.