IBDP Maths AA: Topic: SL 5.8: Local maximum and minimum points: IB style Questions SL Paper 2

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

    1. Find f (1).. [2]

    2. Solve f (x) = 0. [2]

      The graph of f has a local minimum at point A.

    3. 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.

[3]
a.

Let \(g(x) = f”(x)\) . Explain why the graph of g has no points of inflexion.

[2]
b.
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]

a.

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]

b.

Question

Let \(f(x) =  – 0.5{x^4} + 3{x^2} + 2x\). The following diagram shows part of the graph of \(f\).

M17/5/MATME/SP2/ENG/TZ2/08

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 \(p\).

[2]
a.

Write down the coordinates of A.

[2]
b.i.

Write down the rate of change of \(f\) at A.

[1]
b.ii.

Find the coordinates of B.

[4]
c.i.

Find the the rate of change of \(f\) at B.

[3]
c.ii.

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.

[3]
d.
Answer/Explanation

Markscheme

evidence of valid approach     (M1)

eg\(\,\,\,\,\,\)\(f(x) = 0,{\text{ }}y = 0\)

2.73205

\(p = 2.73\)     A1     N2

[2 marks]

a.

1.87938, 8.11721

\((1.88,{\text{ }}8.12)\)     A2     N2

[2 marks]

b.i.

rate of change is 0 (do not accept decimals)     A1     N1

[1 marks]

b.ii.

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]

c.i.

recognizing rate of change is \(f’\)     (M1)

eg\(\,\,\,\,\,\)\(y’,{\text{ }}f’(1)\)

rate of change is 6     A1     N2

[3 marks]

c.ii.

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]

d.
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