Let f x z x 0 f t dt then f is a continuous function

Let F ( x ) =
Z
x
0
f ( t ) dt , then F is a continuous function. F ( x + y ) =
Z
x + y
0
f ( t ) dt , let
( x + y ) z = t , so t = 0 ⇔ z = 0 , t = x + y ⇔ z = 1. It is enough to consider x > 0, because
f ( x + ( - x )) = f (0) = 0 = f ( x ) + f ( - x ), so f is an odd function of x .
F ( x + y ) =
Z
1
0
f ( xz + yz )( x + y ) dz
F ( x + y )
x + y
=
Z
1
0
f ( xz ) dz +
Z
1
0
f ( yz ) dz
( Let xz = u, yz = v )
=
Z
x
0
f ( u )
du
x
+
Z
y
0
f ( v )
dv
y
=
F ( x )
x
+
F ( y )
y
Thus the function G ( x ) =
F ( x )
x
is continuous and satisfies G ( x + y ) = G ( x ) + G ( y ). So
G ( x ) = xG (1), and F ( x ) = x 2 G (1). Differentiating both sides, f ( x ) = F 0 ( x ) = 2 xG (1), so
f ( x ) = cx , which proves the result.
Question 2(b) Show that the area between the cissoid x = a sin
2
t, y = a
sin
3
t
cos t
and its
asymptote is
3 πa
2
4
.
Solution.
y
2
=
a
2
sin
6
t
cos 2 t
=
x
3
a (1 - sin
2
t )
=
x
3
a - x
.
Thus y 2 ( a - x ) = x 3 , so x = a is an asymp-
tote.
Required Area = 2
Z
a
0
y dx . Sub-
stituting for y and x using the
parametric equations, the area be-
comes 2
Z
π
2
0
a
sin
3
t
cos t
2 a sin t cos t dt =
4 a
2
Z
π
2
0
sin
4
t dt = 4 a
2
3 · 1
4 · 2
π
2
=
3 πa 2
4
as re-
quired.
O A
x = a
δx
y
3

Question 2(c) Show that
ZZ
x
m - 1
y
n - 1
dxdy over the positive quadrant of the ellipse
x 2
a
2
+
y
2
b 2
= 1 is
a
m
b
n
4
Γ(
m
2
)Γ(
n
2
)
Γ(
m
2
+
n
2
+ 1)
.
Solution. Put X = ax, Y = by,
∂ ( x,y )
∂ ( X,Y )
= ab . Thus
I =
ZZ
X ≥ 0 ,Y ≥ 0
X
2
+ Y
2
≤ 1
a
m - 1
X
m - 1
b
n - 1
Y
n - 1
ab dX dY
Put X = r cos θ, Y = r sin θ ⇒
∂ ( X, Y )
∂ ( r, θ )
=
cos θ - r sin θ
sin θ r cos θ
= r . Then
I = a
m
b
n
Z
π
2
0
Z
1
0
r
m + n - 2
sin
m - 1
θ cos
n - 1
θ r dr dθ
=
a
m
b
n
m + n
Z
π
2
0
sin
m - 1
θ cos
n - 1
θ dθ
=
a
m
b
n
m + n
Γ(
n - 1+1
2
)Γ(
m - 1+1
2
)
2Γ(
n - 1
2
+
m - 1
2
+ 1)
=
a
m
b
n
4
m + n
2
Γ(
n
2
)Γ(
m
2
)
Γ(
n
2
+
m
2
)
=
a m b n
4
Γ(
n
2
)Γ(
m
2
)
Γ( n
2
+ m
2
+ 1)
Here we used x Γ( x ) = Γ( x + 1) and
Z
π
2
0
sin
m
θ cos
n
θ dθ =
Γ(
n +1
2
)Γ(
m +1
2
)
2Γ( n
2
+ m
2
+ 1)
Paper II
Question 3(a) Find the shortest distance from the origin to the hyperbola x 2 +8 xy +7 y 2 =
225 , z = 0 .
Solution. We have to minimize f ( x, y ) = x 2 + y 2 subject to the constraint x 2 +8 xy +7 y 2 -
225 = 0. Let F ( x, y ) = x 2 + y 2 + λ ( x 2 +8 xy +7 y 2 - 225) where λ is Lagrange’s undetermined
multiplier. For extreme values
∂F
∂x
= 2 x + 2 λx + 8 λy = 0
∂F
∂y
= 2 y + 8 λx + 14 λy = 0
4

From the second equation we get x =
- y - 7 λy
4 λ
. Substituting the value of x in the first
equation, we get
- (1 + λ )(1 + 7 λ ) y
2 λ
+ 8 λy = 0 or 16 λ 2 - (1 + 7 λ )(1 + λ ) = 0, since y 6 = 0 —
note that y = 0 ⇒ x = 0 which is not possible because x
2
+ 8 xy + y
2
= 225. Thus we get
9 λ
2
- 8 λ - 1 = 0, so λ = 1 , -
1
9
.
We shall now show that λ = 1 is not possible. λ = 1 ⇒ 4 x + 8 y = 0 ⇒ x = - 2 y ⇒
4 y 2 - 16 y 2 + 7 y 2 = 225, which is not possible, thus λ 6 = 1.
λ = -
1
9
⇒ 2 x -
2
9
x -
8
9
y = 0 ⇒ 16 x - 8 y = 0 ⇒ y = 2 x . Thus x 2 + 16 x 2 + 28 y 2 = 225 ⇒
x = ±
√
5 , y = ± 2
√
5. Thus stationary points are (
√
5 , 2
√
5) , ( -
√
5 , - 2
√
5).
F ( x, y ) = x
2
+ y
2
-
1
9
( x
2
+ 8 xy + 7 y
2
- 225)
=
1
9
h
8 x
2
+ 2 y
2
- 8 xy + 225
i
=
1
9
h
2(2 x - y )
2
+ 225
i
which is minimized when 2 x = y . Thus the shortest distance is
√
25 = 5.
Question 3(b) Show that the double integral
ZZ
R
x - y
( x + y )
3
dx dy does not exist over R =
[0 , 1 : 0 , 1] .
Solution. For a fixed x 6 = 0, the function
x - y
( x + y ) 3
is a bounded function of y and
Z
1
0
x - y
( x + y )
3
dy =
Z
1
0
-
x + y
( x + y )
3
+
2 x
( x + y )
3
dy
=
1
x + y
+
2 x
- 2( x + y )
2
1
0
=
1
1 + x
-
x
(1 + x )
2
=
1
(1 + x )
2
Now
Z
1
0
dx
(1 + x )
2
= lim
→ 0 +
Z
1
(1 + x )
- 2
dx = lim
→ 0 +
-
1
2
+
1
1 +
=
1
2
Thus
Z
1
0
Z
1
0
x - y
( x + y ) 3
dy dx =
1
2
.
Similarly, if y 6 = 0,
Z
1
0
x - y
( x + y )
3
dx =
Z
1
0
x + y
( x + y )
3
-
2 y
( x + y )
3
dx
= -
1
x + y
-
2 y
- 2( x + y ) 2
1
0
= -
1
(1 + y ) 2
5

Thus
Z
1
0
Z
1
0
x - y
( x + y )
3
dx dy = -
1
2
, as
Z
1
0
dy
(1 + y )
2
=
1
2
as above.
Thus
Z
1
0
Z
1
0
x - y
( x + y )
3
dy dx 6 =
Z
1
0
Z
1
0
x - y
( x + y )
3
dx dy showing that the double integral
does not exist, because if the double integral exists and the two repeated integrals exist,
these have to be equal. The reason is that the function
x - y
( x + y )
3
is not bounded in the square
R [0 , 1 : 0 , 1].