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A curve with equation $y = f(x)$ passes through the point (2, 10) - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 1

Question 9

A curve with equation $y = f(x)$ passes through the point (2, 10). Given that $f'(x) = 3x^2 - 3x + 5$ find the value of $f(0)$. (5)

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (2, 10) - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 1

Step 1

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Answer

To find $f(x)$ , we need to integrate $f'(x)$ .

$f(x) = \int (3x^2 - 3x + 5) \, dx$

Calculating the integral, we have:

$f(x) = x^3 - \frac{3}{2} x^2 + 5x + c$

where $c$ is the constant of integration.

Step 2

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Answer

We know the curve passes through the point (2, 10). This gives us:

$f(2) = 2^3 - \frac{3}{2}(2^2) + 5(2) + c = 10$

Simplifying this:

$10 = 8 - 6 + 10 + c$
\ Which leads to:
$10 = 12 + c$
$c = -2$

Step 3

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101 rated

Answer

Now, substituting $c = -2$ back into the equation for $f(x)$ , we have:

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

To find $f(0)$ :

$f(0) = 0^3 - \frac{3}{2}(0^2) + 5(0) - 2 = -2$

Thus, the value of $f(0)$ is $-2$ .

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