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Maths Test 005
Maths Test 005
Note
Test 005 covers chapter 4 of the Edexcel Maths AS course.
There is no time limit - the average person should complete the test in an hour and a half.
The test will remain available until midnight on 24 July 2020.
The test total is 120 marks.
- Sketch the following equations and show the points where they cross the coordinate axes (20)
- \( y = (x - 3)(x + 1)(2 - x) \)
- \( y = -x(2x + 1)(3 - x) \)
- \( y = (x - 2)^2(1 - x) \)
- \( y = (x + 5)^3 \)
- \( y = (x - 1)(x^2 + 1) \)
- Sketch each equation below, showing points of intersection on the sketch (20)
- \( y = 12x^3 - 3x \)
- \( y = x^3 + 2x^2 + x \)
- \( y = - (x - \frac{1}{2})^3 \)
- \( y = (x - 10)(x^2 -2x) + 12x \)
- \( y = 3x + 2x^2 - x^3 \)
- Sketch the following equations and show the points where they cross the coordinate axes (20)
- \( y = (x + 2)(x - 1)(2 - x)(x + 1) \)
- \( y = x^2(4x + 1)(3 - x) \)
- \( y = (3 - x)^4 \)
- \( y = (x + 1)(x - 2)(x^2 - 4x + 3) \)
- \( y = (x - 2)^2(3x^2 + 7x - 6) \)
- Sketch the two equations on the same set of axes and estimate the points of intersection (20)
- \( y = x^2(1 - x) \) and \( y = -\frac{2}{x} \)
- \( y = x(x - 4) \) and \( y = (x - 3)^3 \)
- \( y = 4 \) and \( y = x(x - 1)(x + 2)(x + 3)^2 \)
- \( y = x^2(x - 1)(x + 1) \) and \( y = \frac{1}{3}x^3 + 1 \)
- Consider the equations \( f(x) = \frac{4}{x^2}\) and \( g(x) = 3x + 7 \). (20)
- Sketch \( f(x) \) and \( g(x) \) on the same axes.
- Write down the number of real solutions for the equation \( \frac{4}{x^2} = 3x + 7 \).
- Show that the equation can be written as \( (x + 1)(x + 2)(3x - 2) = 0 \).
- Determine the exact coordinates of the points of intersection.
- Given the equations \( f(x) = \frac{1}{2x}\), \( g(x) = (2x)^2(2x - 3) \), and \( 3h(x) = -(x)^2(2 - x)^2 \), sketch (20)
- \( f(-x) \) and \( \frac{1}{2}g(3x) \)
- \( -g(x + 2) \) and \( h(2x) + 1 \)
- \( f(x + 2) + 1 \) and \( h( \frac{x}{4}) - 1 \)
- \( 3g(x + 1) - 2 \) and \( 3f(2 - x) \)
End of Maths Test 005
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