or \((V =)\;15(\text{``}{2.5}\text{''})^2 - 4(\text{``}{2.5}\text{''})^3\)
M1
Correct answer only scores full marks (unless from obviously incorrect working) Answer: 31.25
A1
(5)
(5 marks)
Notes
M1: for a correct expression for the volume (condone missing \(V = \ldots\))
M1ft: ft dep on a \(V\) of the form \(ax^3 + bx^2\) where \(a, b \ne 0\)
for a two-term derivative with at least one term correct for their \(V\)
eg \(15(2)x\) or \(30x\) or \(-3(4)x^2\) or \(-12x^2\)
M1ft: dep on previous M mark
for equating their 2-term first derivative to 0 and solving to get a value of \(x\) (the value of \(x\) obtained must be greater than 0 and must not be where their second derivative is equal to 0) or the correct value of \(x\)
M1: dep on M3 for a full and correct substitution of their value of \(x\)
M1: ft dep on M1 for a correct method to solve their 3 term quadratic equation (with at least 2 correct coefficients) using any correct method (if factorising, allow brackets which expanded give 2 out of 3 terms correct) (if using formula allow one sign error and some simplification – allow as far as \(\dfrac{-10 \pm \sqrt{100 - 96}}{24}\)) Derivative must be a 3 term quadratic for this M mark
NB Can be implied by answers of \((x =)-\dfrac{1}{2}\) and \((x =)-\dfrac{1}{3}\)
A1: oe dep on previous M1 Allow –0.33(333) or \(-0.\dot{3}\) for correct \(x\) values
M1: ft dep on previous M1 for substituting at least one \(x\) value into \(y\)
NB Can be implied by one correct value of \(y\)
A1: oe dep on M1 for correct coordinates \((-0.5, -0.25)\), \((-0.33, -0.25(9\ldots))\)
20 The curve with equation \(y = 2x^4 - 64x\) has a minimum point.
Find an equation of the tangent to the curve at the minimum point. Show clear algebraic working.
(4)
Mark scheme
Scheme
Marks
\(4 \times 2x^3\) or \(8x^3\) or \(\pm 64\)
M1
\(8x^3 - 64 = 0\) oe
M1
\(x = \sqrt[3]{\dfrac{64}{8}}\;(= 2)\)
M1
Working required Answer: \(y = -96\)
A1
(4)
(4 marks)
Notes
M1: for differentiating one term correctly
M1: dep on M1 The equation must be in the form \(ax^3 - 64 = 0\) oe where \(a \ne 0\) or \(8x^3 + b = 0\) oe where \(b \ne 0\) where \(a\) and \(b\) are constants
M1: dep on previous M1 for solving for \(x\). The equation must be in the form \(ax^3 - 64 = 0\) oe where \(a \ne 0\) or \(8x^3 + b = 0\) oe where \(b \ne 0\) where \(a\) and \(b\) are constants
A1: oe eg \(y + 96 = 0\) or \(y = 0x - 96\) or \(-y = 96\)
