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The curve with equation y = f(x) passes through the point (1, 6) - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 1

Question 9

The curve with equation y = f(x) passes through the point (1, 6). Given that f'(x) = 3 + \frac{5x^2 + 2}{x^3}, x > 0, find f(x) and simplify your answer.

Worked Solution & Example Answer:The curve with equation y = f(x) passes through the point (1, 6) - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 1

Step 1

Find f(x) by integrating f'(x)

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Answer

To find f(x), we integrate f'(x):

$f(x) = \int \left(3 + \frac{5x^2 + 2}{x^3}\right) dx$

This simplifies to:

$f(x) = \int 3 \, dx + \int \left(\frac{5}{x} + \frac{2}{x^3}\right) dx$

Calculating the integrals:

For (3):
(\int 3 , dx = 3x)
For (\frac{5}{x}):
(\int \frac{5}{x} , dx = 5 \ln|x|)
For (\frac{2}{x^3}):
(\int \frac{2}{x^3} , dx = -\frac{1}{x^2})

Putting it all together gives us:

$f(x) = 3x + 5 \ln|x| - \frac{1}{x^2} + C$

Step 2

Apply the initial condition

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Answer

We know that f(1) = 6, thus:

$f(1) = 3(1) + 5 \ln(1) - \frac{1}{1^2} + C = 6$

Substituting (\ln(1) = 0):

$3 - 1 + C = 6$

This simplifies to:

$C = 6 - 2 = 4$

Thus, substituting C back into f(x):

$f(x) = 3x + 5 \ln|x| - \frac{1}{x^2} + 4$

Step 3

Final simplification

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Answer

The answer can be simplified further depending on the context.

Thus, the final result is:

$f(x) = 3x + 5 \ln|x| - \frac{1}{x^2} + 4$

An alternative simplified form:

$f(x) = 3x + 5 \ln x - \frac{1}{x^2} + 4$ for (x > 0).

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