For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x ightarrow ext{ln}(x + k), ext{ for } x > -k,$ g: x ightarrow |2x - k|, ext{ for } x ext{ in } ext{R.}$ (a) On separate axes, sketch the graph of $f$ and the graph of $g$ - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 4

The function $g(x) = |2x - k|$ has the following features:

• This function is in a V-shape with its vertex at x = rac{k}{2}.
• The $y$-intercept occurs when $x = 0$, giving the point $(0, |2(0) - k|) = (0, k)$ if $k > 0$.
• The graph intersects the $x$-axis at x = rac{k}{2}, leading us to the point igg(rac{k}{2}, 0igg).

Where it meets the axes:

• $y$-axis: $(0, k)$
• $x$-axis: igg(rac{k}{2}, 0igg)

The range of the function $f(x) = ext{ln}(x + k)$ is $(- ext{infinity}, ext{infinity})$ since logarithmic functions can take any real value as output, particularly going to $- ext{infinity}$ as $x$ approaches $-k$.

To find figg(rac{k}{4}igg), substitute rac{k}{4} into the function:

figg(rac{k}{4}igg) = ext{ln}igg(rac{k}{4} + kigg) = ext{ln}igg(rac{k + 4k}{4}igg) = ext{ln}igg(rac{5k}{4}igg).

To find the value of $k$, we need to determine the slope of the given line $9y = 2x + 1$. Dividing by 9 gives us:

y = rac{2}{9}x + rac{1}{9},

indicating a slope of rac{2}{9}. Next, we calculate the derivative of $f(x)$:

rac{dy}{dx} = rac{1}{x + k}

Setting $x = 3$ provides the derivative value:

rac{dy}{dx}igg|_{x=3} = rac{1}{3 + k}.

We set this equal to rac{2}{9}:

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

Cross-multiplying results in: