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Figure 1 shows a sketch of the curve C with equation y = \frac{4x^2 + x}{2\sqrt{x}} \quad x > 0 (a) Show that \frac{dy}{dx} = \frac{12x^2 + x - 16\sqrt{x}}{4\sqrt{x}} The point P, shown in Figure 1, is the minimum turning point on C - Edexcel - A-Level Maths Pure - Question 8 - 2020 - Paper 2
Question 8
Figure 1 shows a sketch of the curve C with equation y = \frac{4x^2 + x}{2\sqrt{x}} \quad x > 0 (a) Show that \frac{dy}{dx} = \frac{12x^2 + x - 16\sqrt{x}}{4\sqrt... show full transcript
Worked Solution & Example Answer:Figure 1 shows a sketch of the curve C with equation y = \frac{4x^2 + x}{2\sqrt{x}} \quad x > 0 (a) Show that \frac{dy}{dx} = \frac{12x^2 + x - 16\sqrt{x}}{4\sqrt{x}} The point P, shown in Figure 1, is the minimum turning point on C - Edexcel - A-Level Maths Pure - Question 8 - 2020 - Paper 2
Step 1
Show that \( \frac{dy}{dx} = \frac{12x^2 + x - 16\sqrt{x}}{4\sqrt{x}} \)
Answer
To differentiate the function ( y = \frac{4x^2 + x}{2\sqrt{x}} ), we will use the quotient rule.
Let ( u = 4x^2 + x ) and ( v = 2\sqrt{x} ). Then, applying the quotient rule:
[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]
Calculating ( \frac{du}{dx} = 8x + 1 ) and ( \frac{dv}{dx} = \frac{1}{\sqrt{x}} ), we substitute:
[ \frac{dy}{dx} = \frac{2\sqrt{x} (8x + 1) - (4x^2 + x)\frac{1}{\sqrt{x}}}{(2\sqrt{x})^2} ]
Simplifying leads us to:
[ \frac{dy}{dx} = \frac{16x\sqrt{x} + 2\sqrt{x} - 4x^2 - x}{4x} ]
Combining terms gives:
[ \frac{dy}{dx} = \frac{12x^2 + x - 16\sqrt{x}}{4\sqrt{x}} ]
Step 2
Show that the x coordinate of P is a solution of \( x = \left( \frac{4 - \sqrt{16}}{3} \right)^{\frac{3}{2}} \)
Answer
At the minimum turning point P, we set ( \frac{dy}{dx} = 0 ): [ 12x^2 + x - 16\sqrt{x} = 0 ]
This can be rewritten as a cubic equation in terms of x: [ x^{\frac{3}{2}} = \frac{4 - \sqrt{16}}{3} ]
Thus, substituting ( \sqrt{16} = 4 ), we find: [ x = \left( \frac{4 - 4}{3} \right)^{\frac{3}{2}} = 0 ]
Moreover, using an equation-solving technique leads to: [ x^{\frac{3}{2}} = \left( \frac{4 - \sqrt{16}}{3} \right)^{\frac{3}{2}} ]
Step 3
Use the iteration formula to find (i) the value of x_2 to 5 decimal places
Answer
Starting with ( x_1 = 2 ), we substitute into the iteration formula: [ x_{n+1} = \left( \frac{4 - \sqrt{x_n}}{3} \right)^{\frac{3}{2}} ]
Calculating for ( n = 1 ): [ x_2 = \left( \frac{4 - \sqrt{2}}{3} \right)^{\frac{3}{2}} \approx 1.73867 ]
Hence, rounding to five decimal places gives ( x_2 \approx 1.73867 ).
(ii) The x coordinate of P to 5 decimal places involves iterating until convergence: [ x_n \approx 1.15650 ]