Question:
A solid metal ornament is in the shape of a right pyramid, with vertex V and square base ABCD. The centre of the base is X. Point V has coordinates (1, 5, 0) and point A has coordinates (-1, 1, 6).
(a) Find AV.
(b) Given that \(A\hat{V}B =40^{0}\), find AB.
The volume of the pyramid is 57.2 cm3, correct to three significant figures.
(c) Find the height of the pyramid, VX.
A second ornament is in the shape of a cuboid with a rectangular base of length 2x cm, width x cm and height y cm. The cuboid has the same volume as the pyramid.
(d) The cuboid has a minimum surface area of S cm2. Find the value of S.
Answer/Explanation
Ans:
(a) attempt to use the distance formula to find AV
(b) METHOD 1
attempt to apply cosine rule OR sine rule to find AB
METHOD 2
Let M be the midpoint of [AB]
attempt to apply right-angled trigonometry on triangle AVM
= 2× 7.48…× sin(200)
= 5.11888…
= 5.12 (cm)
(c) METHOD 1
equating volume of pyramid formula to 57.2
\(\frac{1}{3}\times 5.11….^{2}\times h=57.2\)
h = 6.54886…
h = 6.55 (cm)
METHOD 2
Let M be the midpoint of [AB]
(d) V = x × 2x × y= 57.2
S = 2(2x2+xy+2xy)
Note: Condone use of A.
attempt to substitute \(y = \frac{57.2}{2x^{2}}\) into their expression for surface area
\(\left ( S(x) =\right )4x^{2}+6x\left ( \frac{57.2}{2x^{2}} \right )\)
EITHER
attempt to find minimum turning point on graph of area function
OR
\(\frac{dS}{dx}= 8x – 171.6x^{2}= 0\) OR x = 2.77849….
THEN
92.6401…
minimum surface area = 92.6(cm2)
Question
Let \(f(x)=3x-4^{0.15x^{2}}\) for \( 0\leq x\leq 3\)
(a) Sketch the graph of f on the grid below. [3]
(b) Find the value of x for which \(f{}’\left ( x \right )=0\)
Answer/Explanation
(a)
(b)recognizing that \( f{}’\left ( x \right )=0\) at local maximum
x=2.33084…
x=2.33
Question
Consider the functionf (x) = x2 + x + x + \(\frac{50}{x}\) , x ≠ 0
Find f (1).. [2]
Solve f (x) = 0. [2]
The graph of f has a local minimum at point A.
Find the coordinates of A. [2]
Answer/Explanation
Ans
Question
Let \(f'(x) = – 24{x^3} + 9{x^2} + 3x + 1\) .
There are two points of inflexion on the graph of f . Write down the x-coordinates of these points.
Let \(g(x) = f”(x)\) . Explain why the graph of g has no points of inflexion.
Answer/Explanation
Markscheme
valid approach R1
e.g. \(f”(x) = 0\) , the max and min of \(f’\) gives the points of inflexion on f
\( – 0.114{\text{, }}0.364\) (accept (\( – 0.114{\text{, }}0.811\)) and (\(0.364{\text{, }}2.13)\)) A1A1 N1N1
[3 marks]
METHOD 1
graph of g is a quadratic function R1 N1
a quadratic function does not have any points of inflexion R1 N1
METHOD 2
graph of g is concave down over entire domain R1 N1
therefore no change in concavity R1 N1
METHOD 3
\(g”(x) = – 144\) R1 N1
therefore no points of inflexion as \(g”(x) \ne 0\) R1 N1
[2 marks]
Question
Let \(f(x) = – 0.5{x^4} + 3{x^2} + 2x\). The following diagram shows part of the graph of \(f\).
There are \(x\)-intercepts at \(x = 0\) and at \(x = p\). There is a maximum at A where \(x = a\), and a point of inflexion at B where \(x = b\).
Find the value of \(p\).
Write down the coordinates of A.
Write down the rate of change of \(f\) at A.
Find the coordinates of B.
Find the the rate of change of \(f\) at B.
Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis, the line \(x = b\) and the line \(x = a\). The region \(R\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.
Answer/Explanation
Markscheme
evidence of valid approach (M1)
eg\(\,\,\,\,\,\)\(f(x) = 0,{\text{ }}y = 0\)
2.73205
\(p = 2.73\) A1 N2
[2 marks]
1.87938, 8.11721
\((1.88,{\text{ }}8.12)\) A2 N2
[2 marks]
rate of change is 0 (do not accept decimals) A1 N1
[1 marks]
METHOD 1 (using GDC)
valid approach M1
eg\(\,\,\,\,\,\)\(f’’ = 0\), max/min on \(f’,{\text{ }}x = – 1\)
sketch of either \(f’\) or \(f’’\), with max/min or root (respectively) (A1)
\(x = 1\) A1 N1
Substituting their \(x\) value into \(f\) (M1)
eg\(\,\,\,\,\,\)\(f(1)\)
\(y = 4.5\) A1 N1
METHOD 2 (analytical)
\(f’’ = – 6{x^2} + 6\) A1
setting \(f’’ = 0\) (M1)
\(x = 1\) A1 N1
substituting their \(x\) value into \(f\) (M1)
eg\(\,\,\,\,\,\)\(f(1)\)
\(y = 4.5\) A1 N1
[4 marks]
recognizing rate of change is \(f’\) (M1)
eg\(\,\,\,\,\,\)\(y’,{\text{ }}f’(1)\)
rate of change is 6 A1 N2
[3 marks]
attempt to substitute either limits or the function into formula (M1)
involving \({f^2}\) (accept absence of \(\pi \) and/or \({\text{d}}x\))
eg\(\,\,\,\,\,\)\(\pi \int {{{( – 0.5{x^4} + 3{x^2} + 2x)}^2}{\text{d}}x,{\text{ }}\int_1^{1.88} {{f^2}} } \)
128.890
\({\text{volume}} = 129\) A2 N3
[3 marks]