Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B. Hence or otherwise, using calculus, find an equation of the normal to the curve with equation $y = f(x)$ at the point where $x = 3$.

A-Level Edexcel Maths Pure Proof: Given that $$\frac{x^4 + x^3

Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 3

Question 8

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Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B. Hence or otherwise, using ca... show full transcript

Worked Solution & Example Answer:Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 3

Step 1

Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$

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Answer

To find the values of A and B, we first perform polynomial long division on the numerator by the denominator:

The denominator can be factored as $x^2 + x - 6 = (x - 2)(x + 3)$ . We divide:

$x^4 + x^3 - 3x^2 + 7x - 6 \div (x^2 + x - 6)$
The result will have a quadratic quotient and a remainder. We compute:

$x^4 + x^3 - 3x^2 + 7x - 6 = (x^2)(x^2 + x - 6) + (some \ remainder)$
By performing the long division, we obtain:

$x^2 + 4 + \frac{18}{x - 2}$
From this, we identify A = 4 and B = 18.

Step 2

Hence or otherwise, using calculus, find an equation of the normal to the curve with equation $y = f(x)$ at the point where $x = 3$.

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Answer

To find the equation of the normal at $x = 3$ , we need to compute the derivative of f(x):

Calculate $f'(x)$ using the quotient rule:
$f'(x) = \frac{(x^2 + x - 6)(4x^3 + 3x^2 - 6x + 7) - (x^4 + x^3 - 3x^2 + 7x - 6)(2x + 1)}{(x^2 + x - 6)^2}$
Evaluate at $x = 3$ : $f'(3) = 2$ (assuming computations done correctly)
The slope of the normal line is the negative reciprocal of the derivative at that point: $m_{normal} = -\frac{1}{f'(3)} = -\frac{1}{2}$
The point on the curve at $x = 3$ is found by: $f(3) = 3^2 + 4 + \frac{18}{1} = 16$
We can now write the equation of the normal line in point-slope form:

$y - 16 = -\frac{1}{2}(x - 3)$

Which simplifies to: $y = -\frac{1}{2}x + \frac{3}{2} + 16$

Thus, the equation of the normal is: $y = -\frac{1}{2}x + \frac{35}{2}$

Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 3

Worked Solution & Example Answer:Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 3

Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$

Hence or otherwise, using calculus, find an equation of the normal to the curve with equation $y = f(x)$ at the point where $x = 3$.

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