Date | May 2009 | Marks available | 9 | Reference code | 09M.1.hl.TZ1.13 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Calculate, Show that, and Hence | Question number | 13 | Adapted from | N/A |
Question
If z is a non-zero complex number, we define L(z) by the equation
L(z)=ln|z|+iarg(z), 0⩽arg(z)<2π.
(a) Show that when z is a positive real number, L(z)=lnz .
(b) Use the equation to calculate
(i) L(−1) ;
(ii) L(1−i) ;
(iii) L(−1+i) .
(c) Hence show that the property L(z1z2)=L(z1)+L(z2) does not hold for all values of z1 and z2 .
Let f be a function with domain R that satisfies the conditions,
f(x+y)=f(x)f(y) , for all x and y and f(0)≠0 .
(a) Show that f(0)=1.
(b) Prove that f(x)≠0 , for all x∈R .
(c) Assuming that f′(x) exists for all x∈R , use the definition of derivative to show that f(x) satisfies the differential equation f′(x)=k f(x) , where k=f′(0) .
(d) Solve the differential equation to find an expression for f(x) .
Markscheme
(a) |z|=z , arg(z)=0 A1A1
so L(z)=lnz AG N0
[2 marks]
(b) (i) L(−1)=ln1+iπ=iπ A1A1 N2
(ii) L(1−i)=ln√2+i7π4 A1A1 N2
(iii) L(−1+i)=ln√2+i3π4 A1 N1
[5 marks]
(c) for comparing the product of two of the above results with the third M1
for stating the result −1+i=−1×(1−i) and L(−1+i)≠L(−1)+L(1−i) R1
hence, the property L(z1z2)=L(z1)+L(z2)
does not hold for all values of z1 and z2 AG N0
[2 marks]
Total [9 marks]
(a) from f(x+y)=f(x)f(y)
for x = y = 0 M1
we have f(0+0)=f(0)f(0)⇔f(0)=(f(0))2 A1
as f(0)≠0, this implies that f(0)=1 R1AG N0
[3 marks]
(b) METHOD 1
from f(x+y)=f(x)f(y)
for y = –x , we have f(x−x)=f(x)f(−x)⇔f(0)=f(x)f(−x) M1A1
as f(0)≠0 this implies that f(x)≠0 R1AG N0
METHOD 2
suppose that, for a value of x, f(x)=0 M1
from f(x+y)=f(x)f(y)
for y=−x, we have f(x−x)=f(x)f(−x)⇔f(0)=f(x)f(−x) A1
substituting f(x) by 0 gives f(0)=0 which contradicts part (a) R1
therefore f(x)≠0 for all x. AG N0
[3 marks]
(c) by the definition of derivative
f′(x)=limh→0(f(x+h)−f(x)h) (M1)
=limh→0(f(x)f(h)−f(x)f(0)h) A1(A1)
=limh→0(f(h)−f(0)h)f(x) A1
=f′(0)f(x)(=k f(x)) AG N0
[4 marks]
(d) ∫f′(x)f(x)dx=∫kdx⇒lnf(x)=kx+C M1A1
lnf(0)=C⇒C=0 A1
f(x)=ekx A1 N1
Note: Award M1A0A0A0 if no arbitrary constant C .
[4 marks]
Total [14 marks]
Examiners report
Part A was answered well by a fair amount of candidates, with some making mistakes in calculating the arguments of complex numbers, as well as careless mistakes in finding the products of complex numbers.
Part B proved demanding for most candidates, particularly parts (c) and (d). A surprising number of candidates did not seem to know what was meant by the ‘definition of derivative’ in part (c) as they attempted to use quotient rule rather than first principles.