CAIE A-Level Mathematics AS 2.6 Numerical Solution of Equations Questions

Practise locating roots by sign changes or graph intersections, rearranging equations for fixed-point iteration and showing enough rounded iterates to justify the requested accuracy.

Exam points

evaluate f at both interval endpoints and use opposite signs to establish a root
rearrange the equation as x = F(x) without changing the required fixed point
iterate xₙ₊₁ = F(xₙ), record values at the stated precision and justify final rounding

Covered syllabus points

2.6.1—Root location 2.6.2—Iterative approximations 2.6.3—Iteration formulae

[Maximum number: 5]

Figure for Question 1 — CAIE A-Level Mathematics AS

The diagram shows the curve with equation $y=\frac{\ln (2 x+1)}{x+3}$ . The curve has a maximum point M.

(a)

Show by calculation that the x-coordinate of M lies between 2.5 and 3.0 .

[ 2 ]

Consider sign of $x-\frac{x+3}{\ln (2 x+1)}+0.5$ or equivalent for 2.5 and 3.0
Obtain $-0.07(0.0696 \ldots)$ and $0.4(0.4166 \ldots)$ and justify
conclusion
Answer given - necessary detail needed.

2

Alternative Method for Question 6(c)

Consider the values of $\mathrm{f}(x)=\frac{x+3}{\ln (2 x+1)}-0.5$ and obtain
$f(2.5)=2.57(2.5696 \ldots)$ and $f(3)=2.58(2.58339 \ldots)$
Conclude f(2.5)<3 and f(3)>2.5 so root lies in given interval
Answer given - necessary detail needed.

(b)

Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

[ 3 ]

[Maximum number: 3]

Figure for Question 2 — CAIE A-Level Mathematics AS

The diagram shows the curve with equation $y=\frac{2 \ln x}{3 x+1}$ . The curve crosses the x-axis at the point A and has a maximum point B. The shaded region is bounded by the curve and the lines x=3 and y=0.

Show by calculation that the x-coordinate of B lies between 3.0 and 3.1.

Equate numerator of first derivative to zero
May be implied.

Consider sign of $\frac{2}{x}(3 x+1)-6 \ln x$ for 3.0 and 3.1
OE

Obtain $0.074 \ldots$ and $-0.14 \ldots$ or equivalents and justify conclusion
AG - necessary detail needed.
0.00075 and -0.001275 .

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2