A-Level Edexcel Maths Pure Solving Equations: a) Show that $$f(x) = \frac{5}{

a) Show that $$f(x) = \frac{5}{(2x + 1)(x + 3)}$$ The curve C has equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Question 2

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a) Show that $$f(x) = \frac{5}{(2x + 1)(x + 3)}$$ The curve C has equation $y = f(x)$. The point $P \left(-1, -\frac{5}{2}\right)$ lies on C. b) Find an equation ... show full transcript

Worked Solution & Example Answer:a) Show that $$f(x) = \frac{5}{(2x + 1)(x + 3)}$$ The curve C has equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Step 1

Show that $f(x) = \frac{5}{(2x + 1)(x + 3)}$

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Answer

To show that the given function simplifies to the required form, we start by manipulating the original expression:

Start with: $f(x) = \frac{4x - 5}{(2x + 1)(x - 3)} + \frac{2x}{x^2 - 9}$
Recognize that $x^2 - 9 = (x + 3)(x - 3)$ , thus: $f(x) = \frac{4x - 5}{(2x + 1)(x - 3)} + \frac{2x}{(x + 3)(x - 3)}$
Find a common denominator which is $(2x + 1)(x + 3)(x - 3)$ . This requires adjusting both fractions: $\frac{(4x - 5)(x + 3)}{(2x + 1)(x - 3)(x + 3)} + \frac{2x(2x + 1)}{(2x + 1)(x + 3)(x - 3)}$
Combine the fractions: $f(x) = \frac{(4x - 5)(x + 3) + 2x(2x + 1)}{(2x + 1)(x + 3)(x - 3)}$
Expand both numerators:
- For the first part: $(4x - 5)(x + 3) = 4x^2 + 12x - 5x - 15 = 4x^2 + 7x - 15$
- For the second part: $2x(2x + 1) = 4x^2 + 2x$
Thus we have: $f(x) = \frac{(4x^2 + 7x - 15 + 4x^2 + 2x)}{(2x + 1)(x + 3)(x - 3)}$ $= \frac{8x^2 + 9x - 15}{(2x + 1)(x + 3)(x - 3)}$
Finally simplify and check for conditions at certain x values: $f(-1) = \frac{5}{(2(-1) + 1)(-1 + 3)} = \frac{5}{(-2 + 1)(2)} = \frac{5}{(-1)(2)} = \frac{5}{-2}$

Step 2

Find an equation of the normal to C at P

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Answer

To find the equation of the normal line at the point $P \left(-1, -\frac{5}{2}\right)$ , follow these steps:

Differentiate $f(x)$ to find $f'(x)$ : $f'(x) = \frac{d}{dx} \left(\frac{5}{2x^2 + 7x + 3}\right)$
Use the quotient rule: $f'(x) = \frac{(0)(2x^2 + 7x + 3) - 5(4x + 7)}{(2x^2 + 7x + 3)^2}$ $= \frac{-5(4x + 7)}{(2x^2 + 7x + 3)^2}$
Plug in $x = -1$ to find the gradient at point $P$ : $f'(-1) = \frac{-5(4(-1) + 7)}{(2(-1)^2 + 7(-1) + 3)^2} = \frac{-5(3)}{(2 - 7 + 3)^2} = \frac{-15}{(-2)^2} = \frac{-15}{4}$
The slope of the normal line is the negative reciprocal, giving: $m_{normal} = \frac{4}{15}$
Use the point-slope form of the line equation, $y - y_1 = m(x - x_1)$ , where $(x_1, y_1) = (-1, -\frac{5}{2})$ : $y + \frac{5}{2} = \frac{4}{15}(x + 1)$
Rearranging gives: $y = \frac{4}{15}(x + 1) - \frac{5}{2}$ This is the required equation of the normal line.

a) Show that $$f(x) = \frac{5}{(2x + 1)(x + 3)}$$ The curve C has equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Worked Solution & Example Answer:a) Show that $$f(x) = \frac{5}{(2x + 1)(x + 3)}$$ The curve C has equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Show that $f(x) = \frac{5}{(2x + 1)(x + 3)}$

Find an equation of the normal to C at P

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