Toughest Question Solved of Mathemetics

Mathematical solution
Name
Institution

Question one
a) (i) range on the left side (x>0)
(y>-3)
Ii
ln ( y ) −ln ( y+ 3 ) +ln 4=2 ln (x)
Adding ln(y+3) on both sides
lny+ ln 4=2 ln ( x ) + ln( y +3)
From log rule, we obtain
ln ( y .4 ) =2lnx + ln ( y +3)
ln ( y .4 ) =ln ( x2 ) + ln( y +3)
ln ( y .4 ) =ln ( x2 )( y +3)
y .4= ( x2 ) ( y +3)
Making y the subject of the formula,
y= 3 x2
4−x2
b)
dy
dx of x3−2 xy+ y2
x3−2 xy+ y2=1
Both sides of the equation is differentiated with respect to x,
dy
dx ( x¿¿ 3−2 xy + y2)= dy
dx 1 ¿
dy
dx ( x¿¿ 3−2 xy + y2)=3 x2−2 ¿ ¿+2y d
dx y
d
dx ( 1 )=0
3 x2−2 ¿+2y d
dx y=0
The equation is rewritten as

3 x2−2 ¿+2yy’=0
Isolating y’=1
y ’=−3 x2−2 y
2(−x+ y )
y ’=−3 x2−2 y
2(−x+ y )
d
dx ( y ) −3 x2−2 y
2(−x + y )
By substituting values of x and y, 1 and 2
d
dx ( y ) −3 x 12 −2 x 2
2 (−1+2)
=-0.5
-0.5= y−2
x−1
y-2=-1/2x+0.5
y=0.5x-1.5
C (i)
∫
0
1
v3
v 4 +5 dv
1
4 (Ln6-Ln5) which is equals top
0.04558

Substituting u for the term denoted by v4 +5, ∫
5
6
1
4 u dv, Results from taking the
constant out ∫
a
b
f ( x ) dx=a . f ( x ) dx
= 1
4 . (∫
5
6
1
u du)
Applying the use of Domain integral,
∫
a
b
1
u du=LnIuI
= 1
4 . LnIuI ¿the range of 5 ¿ 6
By performing boundary computation, Ln (6)-ln (5)
This therefore results, ¼(Ln (6)-ln (5))
(ii)
∫
2
5
t
t−1 dt
∫
2
5
t
√t−1 dt =20 /3
Ensuring the substitution for the term √t−1 , by a single term u,
∫
1
2
2 ( u2 +1 ) du
=2. ∫
1
2
( u2 ) +∫
1
2
( 1 ) duEnsuring the application of the sum rule.
∫
1
2
( f + g ) dx∫
1
2
( fdx ) +∫
1
2
( gdx ) dx