Photo AI
The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2
Question 8
The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$. The point P on C has x-coordinate 1. (a) Show that the value of \( \frac{dy}{dx} \) at P is 3. (b) Find an e... show full transcript
Worked Solution & Example Answer:The curve C has equation $y = 4x^2 + \frac{5 - x}{x}$ - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 2
Step 1
(a) Show that the value of \( \frac{dy}{dx} \) at P is 3.
Answer
To find ( \frac{dy}{dx} ), we first differentiate the equation:
[ y = 4x^2 + \frac{5 - x}{x} = 4x^2 + \frac{5}{x} - 1 ]
Now, applying the differentiation rules, we have:
[ \frac{dy}{dx} = 8x - \frac{5}{x^2} ]
Substituting ( x = 1 ):
[ \frac{dy}{dx} = 8(1) - \frac{5}{(1)^2} = 8 - 5 = 3 ]
Thus, the value of ( \frac{dy}{dx} ) at point P is 3.
Step 2
(b) Find an equation of the tangent to C at P.
Answer
The slope of the tangent at P, where ( \frac{dy}{dx} = 3 ), is 3. The coordinates at P when ( x = 1 ) can be found as:
[ y = 4(1)^2 + \frac{5 - 1}{1} = 4 + 4 = 8 ]
So, point P is (1, 8).
Using the point-slope form of the line:
[ y - y_1 = m(x - x_1) ]
we substitute ( y_1 = 8 ), ( x_1 = 1 ), and ( m = 3 ):
[ y - 8 = 3(x - 1) ]
Simplifying:
[ y = 3x + 5 ]
Thus, the equation of the tangent is ( y = 3x + 5 ).
Step 3