IBDP Maths AA: Topic: SL 2.10: Solving equations: IB style Questions SL Paper 2

Question

The following diagram shows the graphs of \(f(x) = \ln (3x – 2) + 1\) and \(g(x) = – 4\cos (0.5x) + 2\) , for \(1 \le x \le 10\) .


Let A be the area of the region enclosed by the curves of f and g.

(i)     Find an expression for A.

(ii)    Calculate the value of A.

[6]
a(i) and (ii).

(i)     Find \(f'(x)\) .

(ii)    Find \(g'(x)\) .

[4]
b(i) and (ii).

There are two values of x for which the gradient of f is equal to the gradient of g. Find both these values of x.

[4]
c.
Answer/Explanation

Markscheme

(i) intersection points \(x = 3.77\) , \(x = 8.30\) (may be seen as the limits)     (A1)(A1)

approach involving subtraction and integrals     (M1)

fully correct expression     A2

e.g. \(\int_{3.77}^{8.30} {(( – 4\cos (0.5x) + 2) – (\ln (3x – 2) + 1)){\rm{d}}x} \) , \(\int_{3.77}^{8.30} {g(x){\rm{d}}x – } \int_{3.77}^{8.30} {f(x){\rm{d}}x} \)

(ii) \(A = 6.46\)     A1     N1

[6 marks]

a(i) and (ii).

(i) \(f'(x) = \frac{3}{{3x – 2}}\)     A1A1     N2

Note: Award A1 for numerator (3), A1 for denominator (\({3x – 2}\)) , but penalize 1 mark for additional terms.

 

(ii) \(g'(x) = 2\sin (0.5x)\)     A1A1     N2

Note: Award A1 for 2, A1 for \(\sin (0.5x)\) , but penalize 1 mark for additional terms.

[4 marks]

b(i) and (ii).

evidence of using derivatives for gradients     (M1)

correct approach     (A1)

e.g. \(f'(x) = g'(x)\) , points of intersection

\(x = 1.43\) , \(x = 6.10\)     A1A1     N2N2

[4 marks]

c.

Question

Let \(f(x) = 4{\tan ^2}x – 4\sin x\) , \( – \frac{\pi }{3} \le x \le \frac{\pi }{3}\) .

On the grid below, sketch the graph of \(y = f(x)\) .


[3]
a.

Solve the equation \(f(x) = 1\) .

[3]
b.
Answer/Explanation

Markscheme

     A1A1A1     N3

Note: Award A1 for passing through \((0{\text{, }}0)\), A1 for correct shape, A1 for a range of approximately \( – 1\) to 15.

[3 marks]

a.

evidence of attempt to solve \(f(x) = 1\)     (M1)

e.g. line on sketch, using \(\tan x = \frac{{\sin x}}{{\cos x}}\)

\(x = – 0.207\) , \(x = 0.772\)     A1A1     N3

[3 marks]

b.

Question

The following diagram shows two ships A and B. At noon, ship A was 15 km due north of ship B. Ship A was moving south at 15 km h–1 and ship B was moving east at 11 km h–1.


Find the distance between the ships

(i)     at 13:00;

(ii)    at 14:00.

[5]
a(i) and (ii).

Let \(s(t)\) be the distance between the ships t hours after noon, for \(0 \le t \le 4\) .

Show that \(s(t) = \sqrt {346{t^2} – 450t + 225} \) .

[6]
b.

Sketch the graph of \(s(t)\) .

[3]
c.

Due to poor weather, the captain of ship A can only see another ship if they are less than 8 km apart. Explain why the captain cannot see ship B between noon and 16:00.

[3]
d.
Answer/Explanation

Markscheme

(i) evidence of valid approach     (M1)

e.g. ship A where B was, B \(11{\text{ km}}\) away

\({\text{distance}} = 11\)     A1     N2

(ii) evidence of valid approach     (M1)

e.g. new diagram, Pythagoras, vectors

\(s = \sqrt {{{15}^2} + {{22}^2}} \)    (A1)

\(\sqrt {709}  = 26.62705\)

\(s = 26.6\)     A1     N2

Note: Award M0A0A0 for using the formula given in part (b).

[5 marks]

a(i) and (ii).

evidence of valid approach     (M1)

e.g. a table, diagram, formula \(d = r \times t\)

distance ship A travels t hours after noon is \(15(t – 1)\)     (A2) 

distance ship B travels in t hours after noon is \(11t\)     (A1) 

evidence of valid approach    M1

e.g. \(s(t) = \sqrt {{{\left[ {15(t – 1)} \right]}^2} + {{(11t)}^2}} \)

correct simplification     A1

e.g. \(\sqrt {225({t^2} – 2t + 1) + 121{t^2}} \)

\(s(t) = \sqrt {346{t^2} – 450t + 225} \)     AG     N0

[6 marks]

b.


    A1A1A1     N3

Note: Award A1 for shape, A1 for minimum at approximately \((0.7{\text{, }}9)\), A1 for domain.

[3 marks]

c.

evidence of valid approach     (M1)

e.g.  \(s'(t) = 0\) , find minimum of \(s(t)\) , graph, reference to “more than 8 km”

\(\min = 8.870455 \ldots \) (accept 2 or more sf)     A1

since \({s_{\min}} > 8\) , captain cannot see ship B     R1     N0

[3 marks]

d.
Scroll to Top