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6. (a) Given that \[ \frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2} \]\nfind the values of the constants A and B - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 3
Question 7
![6.-(a)-Given-that--\[-\frac{x^4-+-x^3---3x^2-+-7x---6}{x^2-+-x---6}-\equiv-x^2-+-A-+-\frac{B}{x---2}-\]\nfind-the-values-of-the-constants-A-and-B-Edexcel-A-Level Maths Pure-Question 7-2016-Paper 3.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2016_3_1_QP_6_2016 .jpg)
6. (a) Given that \[ \frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2} \]\nfind the values of the constants A and B. (b) Hence or o... show full transcript
Worked Solution & Example Answer:6. (a) Given that \[ \frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2} \]\nfind the values of the constants A and B - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 3
Step 1
Hence or otherwise, using calculus, find an equation of the normal to the curve with equation \(y = f(x)\) at the point where \(x = 3\).
Answer
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Calculate the derivative:
First, we find the derivative, (f'(x)), using the quotient rule: [ f'(x) = \frac{(x^2 + x - 6)(4x^3 + 3x^2 - 6x + 7) - (x^4 + x^3 - 3x^2 + 7x - 6)(2x + 1)}{(x^2 + x - 6)^2} ] Evaluating (f'(3)) gives a slope of the normal line at that point. -
Find coordinates at (x = 3):
Substitute (x = 3) into the function to find (f(3)):
[ f(3) = \frac{3^4 + 3^3 - 3 \cdot 3^2 + 7 \cdot 3 - 6}{3^2 + 3 - 6} = \frac{81 + 27 - 27 + 21 - 6}{9 + 3 - 6} = \frac{96}{6} = 16 ] Thus, the point is ((3, 16)). -
Equation of the normal line:
The slope of the normal is the negative reciprocal of the derivative at that point:
[ \text{slope of normal} = -\frac{1}{f'(3)}. ] Use the point-slope form of the equation: [y - y_1 = m(x - x_1)] where ( (x_1, y_1) = (3, 16) ) and (m) is the slope we calculated. -
Final form:
Rearranging gives the equation of the normal line.