Assignment-1
ANU, EMET7001, Semester-1 2023
Dr Reza Hajargasht
1- Consider the function
(i)
f ( x) = ln ( x 2 + 1)1/2 + x 
(10 marks)
Show that this function is increasing.
Since ln is an increasing function, it is enough to focus only on (1 + x 2 )1/2 + x which is
obviously increasing for x  0 . For x  0 , note that multiplying and dividing by
1
,
(1 + x 2 )1/2 − x gives us ( x 2 + 1)1/ 2 + x =
2 1/ 2
(1 + x ) − x
the nonnegative quantity (1 + x 2 )1/2 − x is obviously decreasing in x  0 which is
equivalent to ( x 2 + 1)1/2 + x being increasing.
(ii)
Show that it is an odd function.
We need to show that f (− x) = − f ( x)
f (− x) = ln ((− x) 2 + 1)1/2 − x)  = ln ( x 2 + 1)1/2 − x)  multiply and divide by ( x 2 + 1)1/2 + x
 (( x 2 + 1)1/2 − x)(( x 2 + 1)1/2 + x) 


1
 f (− x) = ln 
=
ln



2
1/2
( x 2 + 1)1/2 + x
 ( x + 1) + x 


using the fact that ln(1/ a) = − ln(a)  f (− x) = − ln ( x 2 + 1)1/2 + x)  = − f ( x)
(iii)
Find the domain of the function.
The domain is R since ( ( x 2 + 1)1/2 + x  0 for all x  R and ln is defined over positive
values.
(iv)
Find the inverse of the function.
x = ln ( y 2 + 1)1/2 + y ) 
 ( y 2 + 1)1/2 + y = e x
( y 2 + 1)1/2 = (e x − y )
y 2 + 1 = (e x − y ) 2 = e2 x − 2e x y + y 2
y=
(v)
e2 x − 1
e2 x − 1
−1

f
(
x
)
=
2e x
2e x
Find the range of the function.
The range of this function should be R as well. Since ( x 2 + 1)1/2 + x takes all value
between zero and infinity then its log takes all values between − and + .
2- Knowing
(i)
(ii)
(5 marks)
sin( A + B ) = sin( A) cos( B ) + cos( A) sin( B )
sin( A − B) = sin( A) cos( B ) − cos( A) sin( B )
Prove using induction that
cos( x) + cos(3x) + cos(5 x) + .... + cos[(2n + 3) x] =
sin[2(n + 2) x]
2sin( x)
Proof
(i)
(ii)
(iii)
sin[2(−1 + 2) x] sin(2 x) 2sin( x) cos( x)
=
=
= cos( x)
2sin( x)
2sin( x)
2sin( x)
Assume this is true for k > -1 i.e.
sin[2(k + 2) x]
cos( x) + cos(3x) + cos(5 x) + .... + cos[(2k + 3) x] =
2sin( x)
Show the formula is true for (k+1) i.e.
For n = −1  S (−1) = cos( x) &
cos( x) + cos(3x) + cos(5 x) + .... + cos[(2(k + 1) + 3) x] =
Note that S (k + 1) =
sin[2((k + 1) + 2) x] sin[2(k + 3) x]
=
2sin( x)
2sin( x)
sin[(2k + 4) x]
sin[(2k + 4) x] + 2sin( x) cos[(2k + 5) x]
+ cos[(2k + 5) x] =
2sin( x)
2sin( x)
by adding (i) and (ii) we can see that 2sin( x) cos[(2k + 5) x] = sin[(2k + 6) x] − sin[(2 k + 4) x]
S (k + 1) =
sin[(2k + 4) x] + sin[(2k + 6) x] − sin[(2k + 4) x] sin[2( k + 3) x]
=
2sin( x)
2sin( x)
I have started my induction from -1, it is possible to start from 1 (or 0) as well where the
starting point would be cos( x) + cos(3 x) + cos(5 x) .
(10 marks)
A = {1, 2,3, 4,5, 6, 7,8,9,10,11,12,13,14,15,16,17,18,19, 20}
B = {2,,5,8,11,14,17, 20,....}
C = {3,5, 7,9,11,13,15,17,19, 21,....}
A  B  C = {5,11,17}
( A  B) \ C = {2,5,8,11,14,17, 20} \{3,5, 7,9,11,13,15,17,19, 21,....} = {2,8,14, 20}
( A  C ) \ B = {3,5, 7,9,11,13,15,17,19} \{2,5,8,11,14,17, 20} = {3, 7,9,13,15,19}
For any set E and D, using Venn diagram or otherwise, find
(d) D \ ( E \ D)
(e) D \ ( D \ E )
(d )Note that A \ B = A  B
D \ ( E \ D) = D  ( E  D)
= D  ( E UD) Due to De'Morgan laws
=(D  E )U ( D  D) = D since (D  E )  D
(e)Note that A \ B = A  B
D \ ( D \ E ) = D  ( D  E )
= D  ( DUE ) Due to De'Morgan laws
=(D  D)U ( D  E ) = D  E since (D  D) = 