6. (a) Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B. (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 Trigonometric Functions: 6. (a) Given that $$\frac{x^4

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

Question 8

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

6. (a) Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$ find the values of the constants A and B. (b) Hence or o... show full transcript

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

Step 1

Given that $$\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} \equiv x^2 + A + \frac{B}{x - 2}$$, find the values of the constants A and B.

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Answer

To find the constants A and B, we first need to perform polynomial long division of the numerator by the denominator:

Polynomial Long Division:
Divide $x^4 + x^3 - 3x^2 + 7x - 6$ by $x^2 + x - 6$ .
This will yield a quotient and a remainder.

We start by dividing the leading term:
- The leading term $x^4$ divided by the leading term $x^2$ gives us $x^2$ .
  This makes the first part of the quotient.
Multiply and Subtract:
Multiply $x^2$ by $x^2 + x - 6$ , which results in:
$x^4 + x^3 - 6x^2$ .
Now subtract this from the original polynomial: $\left( x^4 + x^3 - 3x^2 + 7x - 6 \right) - \left( x^4 + x^3 - 6x^2 \right) = 3x^2 + 7x - 6$ .
Repeat the Process:
Next, divide $3x^2 + 7x - 6$ by $x^2 + x - 6$ .
The leading term $3x^2$ divided by $x^2$ gives us $3$ .
Final Remainder:
Multiplying $3$ by $x^2 + x - 6$ gives us $3x^2 + 3x - 18$ .
Subtract this from $3x^2 + 7x - 6$ :
$\left( 3x^2 + 7x - 6 \right) - \left( 3x^2 + 3x - 18 \right) = 4x + 12$ .
Setting Up the Equation:
With the polynomial division complete, we have: $\frac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6} = x^2 + 3 + \frac{4x + 12}{x^2 + x - 6}$ .
Compare this with the provided equation:
$x^2 + A + \frac{B}{x - 2}$ .
Identifying Constants:
From the equation, we can infer by identifying that:
- $A = 3$
- The remainder $\frac{4x + 12}{x^2 + x - 6}$ indicates we need to rearrange to obtain a remainder of the form $\frac{B}{x - 2}$ .
  Factor the denominator:
  $x^2 + x - 6 = (x - 2)(x + 3)$ .
Finding B:
Now, express $4x + 12$ in terms of the factors of the denominator and find the value of B. Solving gives us: $B = 18$ .

Therefore, the constants are:
- $A = 3$
- $B = 18$ .

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 the point where $x = 3$ , we follow these steps:

Find f(3):
Substitute $x = 3$ into the original function $f(x)$ to determine the corresponding y-value: $f(3) = \frac{3^4 + 3^3 - 3(3^2) + 7(3) - 6}{3^2 + 3 - 6}$ .
Calculate:

$= \frac{81 + 27 - 27 + 21 - 6}{9 + 3 - 6} = \frac{96}{6} = 16$ . Thus, the point is $(3, 16)$ .
Calculate f'(x):
Differentiate $f(x)$ using the quotient rule:

$f'(x) = \frac{(g(x)h'(x) - h(x)g'(x))}{(h(x))^2}$ ,
where $g(x) = x^4 + x^3 - 3x^2 + 7x - 6$ and $h(x) = x^2 + x - 6$ .
Find $f'(3)$ :

After calculating, we obtain: $f'(3) = 2$ .
Find the Slope of the Normal Line:
The slope of the normal line is the negative reciprocal of the slope of the tangent: $m = -\frac{1}{f'(3)} = -\frac{1}{2}$ .
Equation of the Normal:
Use the point-slope form of the line equation: $y - y_1 = m(x - x_1)$ , where $(x_1, y_1) = (3, 16)$ :

$y - 16 = -\frac{1}{2}(x - 3)$ .
Rearranging gives:

$y = -\frac{1}{2}x + \frac{3}{2} + 16$
$= -\frac{1}{2}x + \frac{35}{2}$ .
Final Answer:
The equation of the normal line is:

$y = -\frac{1}{2}x + \frac{35}{2}$ .

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

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

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

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