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

The functions f and g are defined by $f : x \mapsto 7x - 1, \quad x \in \mathbb{R}$ g : x \mapsto \frac{4}{x - 2}, \quad x \neq 2, \ x \in \mathbb{R}$ (a) Solve the equation fg(x) = x (b) Hence, or otherwise, find the largest value of a such that g(a) = f^{-1}(a) - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 3

Question 3

$The-functions-f-and-g-are-defined-by--$f-:-x-\mapsto-7x---1,-\quad-x-\in-\mathbb{R}$--g-:-x-\mapsto-\frac{4}{x---2},-\quad-x-\neq-2,-\-x-\in-\mathbb{R}$--(a)-Solve-the-equation-fg(x)-=-x--(b)-Hence,-or-otherwise,-find-the-largest-value-of-a-such-that-g(a)-=-f^{-1}(a)-Edexcel-A-Level Maths Pure-Question 3-2016-Paper 3.png$

The functions f and g are defined by $f : x \mapsto 7x - 1, \quad x \in \mathbb{R}$ g : x \mapsto \frac{4}{x - 2}, \quad x \neq 2, \ x \in \mathbb{R}$ (a) Solve t... show full transcript

Worked Solution & Example Answer:The functions f and g are defined by $f : x \mapsto 7x - 1, \quad x \in \mathbb{R}$ g : x \mapsto \frac{4}{x - 2}, \quad x \neq 2, \ x \in \mathbb{R}$ (a) Solve the equation fg(x) = x (b) Hence, or otherwise, find the largest value of a such that g(a) = f^{-1}(a) - Edexcel - A-Level Maths Pure - Question 3 - 2016 - Paper 3

Step 1

Solve the equation fg(x) = x

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Answer

To solve the equation $fg(x) = x$ , we first need to find the expression for $fg(x)$ :

Calculate $g(x)$ :
[ g(x) = \frac{4}{x - 2} ]
Then, substitute $g(x)$ into $f(x)$ :
[ f(g(x)) = f\left(\frac{4}{x - 2}\right) = 7\left(\frac{4}{x - 2}\right) - 1 ] [ = \frac{28}{x - 2} - 1 = \frac{28}{x - 2} - \frac{x - 2}{x - 2} = \frac{28 - (x - 2)}{x - 2} = \frac{30 - x}{x - 2} ]
Set this equal to $x$ :
[ \frac{30 - x}{x - 2} = x ]
Cross-multiply to eliminate the fraction:
[ 30 - x = x(x - 2) ] [ 30 - x = x^2 - 2x ]
Rearranging gives us a quadratic equation:
[ x^2 - x - 30 = 0 ]
Factor the quadratic:
[ (x - 6)(x + 5) = 0 ]
Thus, the solutions are:
[ x = 6 \text{ and } x = -5 ]

Step 2

Hence, or otherwise, find the largest value of a such that g(a) = f^{-1}(a)

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Answer

To find the largest value of $a$ such that $g(a) = f^{-1}(a)$ , we first need to determine $f^{-1}(x)$ :

Start with the function $f(x) = 7x - 1$ :
[ y = 7x - 1 ]
Solve for $x$ :
[ x = \frac{y + 1}{7} \to f^{-1}(y) = \frac{y + 1}{7} ]
Now, we set $g(a) = f^{-1}(a)$ :
[ g(a) = \frac{4}{a - 2} ]
[ f^{-1}(a) = \frac{a + 1}{7} ]
Equate the two expressions:
[ \frac{4}{a - 2} = \frac{a + 1}{7} ]
Cross-multiply:
[ 4 \cdot 7 = (a - 2)(a + 1) ]
[ 28 = a^2 - a - 2 ]
Rearranging gives us:
[ a^2 - a - 30 = 0 ]
Factor this quadratic:
[ (a - 6)(a + 5) = 0 ]
Thus, the possible solutions are:
[ a = 6 \text{ and } a = -5 ]

Therefore, the largest value of $a$ such that $g(a) = f^{-1}(a)$ is [ a = 6 ].

Solve the equation fg(x) = x

Hence, or otherwise, find the largest value of a such that g(a) = f^{-1}(a)

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