Practise using graph intersections or sign changes to locate roots and applying convergent fixed-point iterations with enough displayed values to justify the requested decimal…
By sketching a suitable pair of graphs, show that the equation
has only one root in the interval 0<x<21π.
Sketch a relevant graph, e.g. y= 2 x
May be sketching 2 x and -e^-x
For y= 2 x correct shape, asymptote in correct position. Cuts at (0,1)
For -e^x correct shape, cuts at (0,1)
For y= 2 x correct shape, correct max/min, cuts axis at /4
For -e^-x correct shape, cuts at (0,-1)
Complete scale not needed, but should have enough
to imply the key points for both graphs.
Sketch a second relevant graph, e.g. y=-e^x and justify the given statement
Needs to mark intersection with a dot, a cross, or say roots at points of intersection, OE.
Show by calculation that this root lies between 0.9 and 1 .
Calculate the values of a relevant expression or pair of expressions at
x=0.9 and x=1
E.g. aligned & -4.401<-2.460
& -2.403>-2.718aligned or aligned & -1.94<0
& 0.315>0aligned
(or gathered0.179>0
-0.048<0gathered for the reciprocal graphs).
Complete the argument correctly with correct calculated values
Show that if a sequence of values given by the iterative formula
converges, then it converges to the root of the equation in part (a).
State the equation x=1/2 ^-1(-e^-x) and rearrange to obtain 2 x=-e^x
Or rearrange 2 x=-e^x as x=1/2 ^-1(-e^-x) and state x_n+1=1/2 ^-1(-e^-x_n)
Use the iterative formula given in part (c) to calculate x correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
Use the iterative process correctly at least once
Obtain final answer 0.978
Show sufficient iterations to at least 5 d. p. to justify 0.978 to 3 d.p. or show there
is a sign change in the interval (0.9775,0.9785)
E.g.
1, 0.97376, 0.97903,0.97796, 0.97813
0.95, 0.98395, 0.97697, 0.97838, 0.97809
0.9, 0.99475, 0.97480, 0.97882, 0.97800, 0.97817
The constant a is such that ∫0axe21x dx=6.
Verify by calculation that a lies between 2.2 and 2.4.
Calculate the value of a relevant expression or values of a relevant pair of expressions at 2.2 and 2.4. Values to at least 2 sf.
Need all relevant values but only one (pair) needs to be correct to score M1. Complete set of values for their expression. If not comparing with 0, then the comparison must be clear.
Example 12.2<2.33 and 2.4>2.30
Example 2 f(a)=a-2-e^-1/2 a OEf(2.2)=-0.132f(2.4)=0.0988
Example 3 f(a)=2 a e^1/2 a-4 e^1/2 a+4f(2.2)=5.20f(2.4)=6.65
Example 4 f(a)=2 a e^1/2 a-4 e^1/2 a-2f(2.2)=-0.80f(2.4)=0.65
Complete the argument correctly with correct calculated values. Accept truncated
values. Allow work on a smaller interval.
Example 1 A clear explanation is needed.
Example 2 f(2.2)=-0.132 <0f(2.4)=0.0988 >0 OE
Example 3 f(2.2)=5.20 <6f(2.4)=6.65 >6
Example 4 f(2.2)=-0.80 <0f(2.4)=0.65 >0
Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
Use iterative process correctly at least once
Obtain final answer 2.31
Show sufficient iterations to at least 4 dp to justify 2.31 to 2 dp
Marking guidance:
Allow recovery.
Allow truncation.
E.g. 2.3, 2.3166, 2.3140, 2.3144.