The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$, where $x eq 0$. The point P on C has x-coordinate 1. (a) Show that the value of $rac{dy}{dx}$ at P is 3. (b) Find an equation of the tangent to C at P. This tangent meets the x-axis at the point $(k, 0)$. (c) Find the value of k.

A-Level Edexcel Maths Pure Simultaneous Equations: The curve C has equation $y =

The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$, where $x eq 0$ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2

Question 8

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The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$, where $x eq 0$. The point P on C has x-coordinate 1. (a) Show that the value of $rac{dy}{dx}$ at P is 3. ... show full transcript

Worked Solution & Example Answer:The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$, where $x eq 0$ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2

Step 1

Show that the value of $rac{dy}{dx}$ at P is 3.

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Answer

To find $rac{dy}{dx}$ , we differentiate the equation of the curve:

$y = 4x^2 + \frac{5 - x}{x} = 4x^2 + \left( 5x^{-1} - 1 \right)$

Using the quotient rule and product rule, we calculate:

$\frac{dy}{dx} = 8x - 5x^{-2}$

At the point P where $x = 1$ , substitute:

$\frac{dy}{dx} = 8(1) - 5(1)^{-2} = 8 - 5 = 3$

Thus, $rac{dy}{dx}$ at P is shown to be 3.

Step 2

Find an equation of the tangent to C at P.

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Answer

The equation of the tangent line at P can be found using the point-slope form:

$y - y_1 = m(x - x_1)$

Where $(x_1, y_1)$ is point P, which we find by substituting $x = 1$ into the curve equation:

$y = 4(1)^2 + \frac{5 - 1}{1} = 4 + 4 = 8$

Thus, point P is (1, 8) and the slope $m = 3$ . The equation then becomes:

$y - 8 = 3(x - 1)$

Expanding this, we have:

$y = 3x + 5$

Step 3

Find the value of k.

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Answer

To find the value of $k$ , we set the tangent line equal to 0 (the x-axis):

$0 = 3x + 5$

Solving for $x$ gives:

$3x = -5 \Rightarrow x = -\frac{5}{3}$

Therefore, at the point where the tangent meets the x-axis, $(k, 0)$ implies:

$k = -\frac{5}{3}$

The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$, where $x eq 0$ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2

Worked Solution & Example Answer:The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$, where $x eq 0$ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2

Show that the value of $ rac{dy}{dx}$ at P is 3.

Find an equation of the tangent to C at P.

Find the value of k.

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Show that the value of $rac{dy}{dx}$ at P is 3.