A curve with equation y = f(x) passes through the point (4, 9). Given that f'(x) = \frac{3 \sqrt{x}}{2} - \frac{9}{4 \sqrt{x}} + 2, \quad x > 0 a) find f(x), giving each term in its simplest form. Point P lies on the curve. The normal to the curve at P is parallel to the line 2y + x = 0 b) Find the x coordinate of P.

A-Level Edexcel Maths Pure Integration: A curve with equation y = f(x) p

A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 1

Question 1

A curve with equation y = f(x) passes through the point (4, 9). Given that f'(x) = \frac{3 \sqrt{x}}{2} - \frac{9}{4 \sqrt{x}} + 2, \quad x > 0 a) find f(x), givi... show full transcript

Worked Solution & Example Answer:A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 1

Step 1

find f(x), giving each term in its simplest form.

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Answer

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

Start with the given derivative: $f'(x) = \frac{3 \sqrt{x}}{2} - \frac{9}{4 \sqrt{x}} + 2$
Rewrite the terms with simpler powers: $f'(x) = \frac{3}{2} x^{1/2} - \frac{9}{4} x^{-1/2} + 2$
Now, integrate each term:
- The integral of (x^{n}) is (\frac{x^{n+1}}{n+1}):
- For the first term: $\int \frac{3}{2} x^{1/2} \ dx = \frac{3}{2} \cdot \frac{x^{3/2}}{3/2} = x^{3/2}$
- For the second term: $\int -\frac{9}{4} x^{-1/2} \ dx = -\frac{9}{4} \cdot 2x^{1/2} = -\frac{9}{2} x^{1/2}$
- For the constant term: $\int 2 \ dx = 2x$
Combine the results: $f(x) = x^{3/2} - \frac{9}{2} x^{1/2} + 2x + C$
To find the constant C, use the point (4, 9): $9 = f(4) = 4^{3/2} - \frac{9}{2} \cdot 4^{1/2} + 2 \cdot 4 + C$
- Calculate each term:
  - (4^{3/2} = 8)
  - (4^{1/2} = 2)
- Substitute: $9 = 8 - \frac{9}{2} \cdot 2 + 8 + C$
- This simplifies to: $9 = 8 - 9 + 8 + C$ $C = 9 - 7 = 2$ Hence, $f(x) = x^{3/2} - \frac{9}{2} x^{1/2} + 2x + 2$ .

Step 2

Find the x coordinate of P.

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Answer

To find the x-coordinate of point P, we first need to determine the slope of the line defined by 2y + x = 0:

Rearranging the line equation to y form gives: $y = -\frac{1}{2}x$
- Therefore, the slope (m) of this line is -\frac{1}{2}.
The normal line at point P is parallel to this line, thus it also has a slope of -\frac{1}{2}.
The derivative at point P, f'(x_P), must equal the negative reciprocal of this slope: $f'(x_P) = 2$
Set the derivative equal and solve for x: $\frac{3 \sqrt{x_P}}{2} - \frac{9}{4 \sqrt{x_P}} + 2 = 2$
Simplifying this:
- Subtract 2 from both sides: $\frac{3 \sqrt{x_P}}{2} - \frac{9}{4 \sqrt{x_P}} = 0$
- Therefore: $\frac{3 \sqrt{x_P}}{2} = \frac{9}{4 \sqrt{x_P}}$
Cross-multiplying yields: $3 \sqrt{x_P} \cdot 4 \sqrt{x_P} = 18$ $12x_P = 18$
- Solving for x_P gives: $x_P = \frac{18}{12} = \frac{3}{2}$

Thus, the x coordinate of point P is (x_P = \frac{3}{2}).

A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 1

Worked Solution & Example Answer:A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 1

find f(x), giving each term in its simplest form.

Find the x coordinate of P.

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