For the constant $k$, where $k > 1$, the functions $f$ and $g$ are defined by $f: x o ext{ln}(x + k), ext{ for } x > -k,$ g: x o |2x - k|, ext{ for } x ext{ in } ext{R}. (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 $fg(rac{k}{4})$ 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 $9y = 2x + 1$. (e) Find the value of $k.$

A-Level Edexcel Maths Pure Rational Expressions: For the constant $k$, where $k >

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

Question 1

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

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

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

Step 1

Sketch the graph of $f$ and the graph of $g$

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Answer

To sketch the graph of $f$ , we first note the asymptote at $x = -k$ . The graph is defined for $x > -k$ and approaches negative infinity as x approaches -k from the right. The y-intercept occurs at $x = 0$ , giving the point $(0, ext{ln}(k))$ .

For function $g$ , the absolute value function gives a V shape. The vertex occurs at $x = rac{k}{2}$ , where $g( rac{k}{2}) = 0$ . It intersects the x-axis at the points where $2x - k = 0$ , which means $x = rac{k}{2}$ . The graph intercepts the y-axis at $g(0) = | -k | = k$ . Thus:

The graph of $f$ crosses the axes at the points: $(0, ext{ln}(k))$ and $( -k, - ext{infinity})$
The graph of $g$ crosses the axes at the points: $( rac{k}{2}, 0)$ and $(0, k)$ .

Step 2

Write down the range of $f$

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Answer

The range of the function $f(x) = ext{ln}(x + k)$ is $(- ext{infinity}, + ext{infinity})$ since the natural logarithm function can take on all real values as $x$ goes from $-k$ to $+ ext{infinity}$ .

Step 3

Find $fg(rac{k}{4})$

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Answer

First, find the value of $g( rac{k}{4})$ :

$gigg( rac{k}{4}igg) = |2 imes rac{k}{4} - k| = | rac{k}{2} - k| = | - rac{k}{2}| = rac{k}{2}.$

Now, we find $figg(gigg( rac{k}{4}igg)igg)$ :

$figg( rac{k}{2}igg) = ext{ln}igg( rac{k}{2} + kigg) = ext{ln}igg( rac{3k}{2}igg).$

Thus, $fgigg( rac{k}{4}igg) = ext{ln}igg( rac{3k}{2}igg).$

Step 4

Find the value of $k$

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Answer

To find the value of $k$ , we equate the gradient of the tangent:

The slope of the line $9y = 2x + 1$ is $rac{2}{9}$ . At $x = 3$ , we find:

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

Evaluating at $x = 3$ , $f'(3) = rac{1}{3 + k}.$

Setting these equal gives:

Cross-multiplying yields:

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

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

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

Sketch the graph of $f$ and the graph of $g$

Write down the range of $f$

Find $fg( rac{k}{4})$

Find the value of $k$

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Find $fg(rac{k}{4})$