A-Level Edexcel Maths Pure Rational Expressions: Given that $k$ is a negative con

Given that $k$ is a negative constant and that the function $f(x)$ is defined by $$f(x) = 2 - \frac{(x-5k)(x-k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0$$ (a) show that $f(x) = \frac{x+k}{x-2k}$ (b) Hence find $f\prime(x)$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 3

Question 1

$Given-that-$k$-is-a-negative-constant-and-that-the-function-$f(x)$-is-defined-by--$$f(x)-=-2---\frac{(x-5k)(x-k)}{x^2---3kx-+-2k^2},-\quad-x-\geq-0$$--(a)-show-that-$f(x)-=-\frac{x+k}{x-2k}$--(b)-Hence-find-$f\prime(x)$,-giving-your-answer-in-its-simplest-form-Edexcel-A-Level Maths Pure-Question 1-2014-Paper 3.png$

Given that $k$ is a negative constant and that the function $f(x)$ is defined by $$f(x) = 2 - \frac{(x-5k)(x-k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0$$ (a) show that ... show full transcript

Worked Solution & Example Answer:Given that $k$ is a negative constant and that the function $f(x)$ is defined by $$f(x) = 2 - \frac{(x-5k)(x-k)}{x^2 - 3kx + 2k^2}, \quad x \geq 0$$ (a) show that $f(x) = \frac{x+k}{x-2k}$ (b) Hence find $f\prime(x)$, giving your answer in its simplest form - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 3

Step 1

show that $f(x) = \frac{x+k}{x-2k}$

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Answer

To show that ( f(x) = \frac{x+k}{x-2k} ), we will simplify the given expression for ( f(x) ).

Starting with:\n $f(x) = 2 - \frac{(x-5k)(x-k)}{x^2 - 3kx + 2k^2}$

We first factor the denominator: $x^2 - 3kx + 2k^2 = (x-k)(x-2k)$

Thus, we rewrite ( f(x) ) as follows: $f(x) = 2 - \frac{(x-5k)(x-k)}{(x-k)(x-2k)}= 2 - \frac{x-5k}{x-2k}$

Now, let's express '2' in terms of the common denominator: $f(x) = \frac{2(x-2k)}{x-2k} - \frac{x-5k}{x-2k}$

$f(x) = \frac{2x - 4k - (x - 5k)}{x-2k} = \frac{2x - 4k - x + 5k}{x-2k}$

Combining like terms yields: $f(x) = \frac{x + k}{x - 2k}$

This proves that the simplified form is as required.

Step 2

Hence find $f\prime(x)$, giving your answer in its simplest form.

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Answer

Now, we need to find the derivative of ( f(x) ), given by:

$f(x) = \frac{x+k}{x-2k}$

Using the quotient rule: If ( f(x) = \frac{u}{v} ), then ( f\prime(x) = \frac{u\prime v - uv\prime}{v^2} )

Here, ( u = x+k ) and ( v = x-2k
Therefore, ( u\prime = 1 ) and ( v\prime = 1 )

Applying the quotient rule: $f\prime(x) = \frac{(1)(x-2k) - (x+k)(1)}{(x-2k)^2}$

Simplifying gives: $f\prime(x) = \frac{x-2k - x - k}{(x-2k)^2} = \frac{-3k}{(x-2k)^2}$

Thus, ( f\prime(x) = \frac{-3k}{(x-2k)^2} ).

Step 3

State, with a reason, whether $f(x)$ is an increasing or a decreasing function.

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Answer

To determine whether ( f(x) ) is increasing or decreasing, we analyze the sign of ( f\prime(x) ).

We have: $f\prime(x) = \frac{-3k}{(x-2k)^2}$

Since ( k ) is given as a negative constant, ( -3k ) is positive. The denominator ( (x-2k)^2 ) is always positive for all ( x \geq 0 ) because a squared term is non-negative.

Therefore, the entire expression for ( f\prime(x) ) is positive:

Since ( f\prime(x) > 0 ), we conclude that ( f(x) ) is an increasing function on its domain.

show that $f(x) = \frac{x+k}{x-2k}$

Hence find $f\prime(x)$, giving your answer in its simplest form.

State, with a reason, whether $f(x)$ is an increasing or a decreasing function.

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