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

The functions f and g are defined by f: x ↦ ln(2x−1), x ∈ ℝ, x > 1/2, g: x ↦ 2/(x−3), x ∈ ℝ, x ≠ 3 - Edexcel - A-Level Maths: Pure - Question 5 - 2007 - Paper 5

Question 5

The functions f and g are defined by f: x ↦ ln(2x−1), x ∈ ℝ, x > 1/2, g: x ↦ 2/(x−3), x ∈ ℝ, x ≠ 3. (a) Find the exact value of fg(4). (b) Find the inverse func... show full transcript

Worked Solution & Example Answer:The functions f and g are defined by f: x ↦ ln(2x−1), x ∈ ℝ, x > 1/2, g: x ↦ 2/(x−3), x ∈ ℝ, x ≠ 3 - Edexcel - A-Level Maths: Pure - Question 5 - 2007 - Paper 5

Step 1

Find the exact value of fg(4).

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Answer

To find fg(4), we start by calculating g(4):

$g(4) = \frac{2}{4-3} = 2.$

Next, we find f(g(4)) = f(2):

$f(2) = \ln(2 \cdot 2 - 1) = \ln(3).$

Thus, the exact value of fg(4) is ( \ln(3) ).

Step 2

Find the inverse function f^{-1}(x), stating its domain.

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Answer

To find the inverse function, we start with the equation:

$y = \ln(2x - 1).$

Exponentiating both sides gives:

$e^y = 2x - 1.$

Rearranging this, we find:

$2x = e^y + 1 \implies x = \frac{e^y + 1}{2}.$

Thus, the inverse function is:

$f^{-1}(x) = \frac{e^x + 1}{2}.$

The domain of f^{-1}(x) is all real numbers,( x ∈ ℝ ).

Step 3

Sketch the graph of y = |g(x)|. Indicate clearly the equation of the vertical asymptote and the coordinates of the point at which the graph crosses the y-axis.

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Answer

To sketch the graph of ( y = |g(x)| = \left| \frac{2}{x-3} \right| ), we first identify the vertical asymptote at ( x = 3 ). The graph will approach this line but never touch it. The graph is mirrored in the x-axis for values of ( g(x) < 0 ).

To find where the graph crosses the y-axis, we evaluate ( g(0) ):

$g(0) = \frac{2}{0-3} = -\frac{2}{3}, \text{ so } |g(0)| = \frac{2}{3}.$

Thus, the graph crosses the y-axis at the point (0, ( \frac{2}{3} )).

Step 4

Find the exact values of x for which $ \frac{2}{|x-3|} = 3 $.

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Answer

To solve ( \frac{2}{|x-3|} = 3 ), we first clear the fraction:

$|x - 3| = \frac{2}{3}.$

This gives us two cases to solve:

( x - 3 = \frac{2}{3} )
( x = 3 + \frac{2}{3} = \frac{11}{3} )
( x - 3 = -\frac{2}{3} )
( x = 3 - \frac{2}{3} = \frac{7}{3} )

Thus, the exact values of x are ( \frac{11}{3} ) and ( \frac{7}{3} ).

The functions f and g are defined by f: x ↦ ln(2x−1), x ∈ ℝ, x > 1/2, g: x ↦ 2/(x−3), x ∈ ℝ, x ≠ 3 - Edexcel - A-Level Maths: Pure - Question 5 - 2007 - Paper 5

Worked Solution & Example Answer:The functions f and g are defined by f: x ↦ ln(2x−1), x ∈ ℝ, x > 1/2, g: x ↦ 2/(x−3), x ∈ ℝ, x ≠ 3 - Edexcel - A-Level Maths: Pure - Question 5 - 2007 - Paper 5

Find the exact value of fg(4).

Find the inverse function f^{-1}(x), stating its domain.

Sketch the graph of y = |g(x)|. Indicate clearly the equation of the vertical asymptote and the coordinates of the point at which the graph crosses the y-axis.

Find the exact values of x for which \( \frac{2}{|x-3|} = 3 \).

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