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7. (a) Show that f(x)= \frac{4x-5}{(2x+1)(x-3)} \quad ; \quad x \neq \pm 3, \; x \neq \frac{1}{2} The curve C has equation y=f(x) - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3
Question 8
7. (a) Show that f(x)= \frac{4x-5}{(2x+1)(x-3)} \quad ; \quad x \neq \pm 3, \; x \neq \frac{1}{2} The curve C has equation y=f(x). The point P \left( -1, -\frac{... show full transcript
Worked Solution & Example Answer:7. (a) Show that f(x)= \frac{4x-5}{(2x+1)(x-3)} \quad ; \quad x \neq \pm 3, \; x \neq \frac{1}{2} The curve C has equation y=f(x) - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3
Step 1
Show that f(x)=\frac{5}{(2x+1)(x+3)}
Answer
To show that [ f(x) = \frac{5}{(2x+1)(x+3)} ], we simplify the expression for ( f(x) ).
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Start with the given function: [ f(x) = \frac{4x-5}{(2x+1)(x-3)} + \frac{2x}{x^2-9} ]
Note that ( x^2-9 = (x+3)(x-3) ). -
Rewrite the second term: [ f(x) = \frac{4x-5}{(2x+1)(x-3)} + \frac{2x}{(x+3)(x-3)} ]
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Find a common denominator, which is ( (2x+1)(x+3)(x-3) ), and combine the fractions: [ f(x) = \frac{(4x-5)(x+3) + 2x(2x+1)}{(2x+1)(x+3)(x-3)} ]
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Expand both numerators: [ (4x-5)(x+3) = 4x^2 + 12x - 5x - 15 = 4x^2 + 7x - 15 ] [ 2x(2x+1) = 4x^2 + 2x ]
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Combine these results: [ 4x^2 + 7x - 15 + 4x^2 + 2x = 8x^2 + 9x - 15 ]
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Thus we have: [ f(x) = \frac{8x^2 + 9x - 15}{(2x+1)(x+3)(x-3)} ]
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Now factor the numerator, observing that ( 8x^2 + 9x - 15 = (2x+1)(x+3) ): [ f(x) = \frac{(2x+1)(x+3)}{(2x+1)(x-3)(x+3)} = \frac{5}{(2x+1)(x+3)} \text{ as required.} ]
Step 2
Find an equation of the normal to C at P.
Answer
To find the equation of the normal to the curve C at the point P \left( -1, -\frac{5}{2} \right):
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Start by determining ( f(-1) ) to verify the point P lies on C: [ f(-1) = \frac{5}{(2(-1)+1)(-1+3)} = \frac{5}{(-2)(2)} = -\frac{5}{2} ]
This confirms P is on C. -
Next, compute the derivative ( f'(x) ): Since ( f(x) = \frac{5}{(2x+1)(x+3)} ): [ f'(x) = \frac{-5(2)}{(2x+1)^2 (x+3)^2} ]
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Evaluate the derivative at ( x=-1 ): [ f'(-1) = \frac{-5(2)}{(2(-1)+1)^2(-1+3)^2} = \frac{-10}{(-1)^2(2)^2} = \frac{-10}{4} = -\frac{5}{2} ]
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The gradient of the normal is the negative reciprocal of the gradient of the curve: [ m_{normal} = -\frac{1}{f'(-1)} = \frac{2}{5} ]
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Use point-slope form to write the equation of the normal: [ y - (-\frac{5}{2}) = \frac{2}{5}(x - (-1)) ]
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Simplifying yields: [ y + \frac{5}{2} = \frac{2}{5}(x + 1) ]
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Finally, express this in standard linear form: [ y = \frac{2}{5}(x + 1) - \frac{5}{2} ] Alternatively, this can also be expressed as: [ y + \frac{5}{2} = \frac{2}{5}(x + 1) \text{ or } y = \frac{2}{5}x + \frac{2}{5} - \frac{5}{2} ]