MATHEMATICS T (PAPER 2)
1(i)
lim f ( x ) lim (− 13 x−1) −
1
x →−3− = x →−3− = 3 (-3) – 1 = 0
lim f ( x ) lim √ x+ 3 M1
x →−3+ = x →−3+ = √−3+3 =0
lim f ( x ) lim f ( x )
x →−3− = x →−3+ =0
lim f ( x ) lim f ( x ) A1
x →−3 exists and x →−3 =0
1
−
f(-3) = 3 (-3) – 1 = 0
lim f ( x ) A1
x →−3 = f(-3) = 0
Hence f(x) is continuous at x = -3.
(ii)
2 π
x = sin t , y = sin (t +)
6
dx dy π 1 (Both correct)
=cos t , =cos(t+ ¿ ) ¿
dt dt 6 dy dx
1 (His dt ÷ dt )
π
cos(t + ) dy 1
dy 6
= 1( = seen, can be implied)
dx cos t dx √ 3
π dy 1
When t = , = 1 √3
6 dx √ 3 1 (or ( , ) seen, can be implied)
2 2
π 1 π π √3
x = sin = , y = sin ( + ) = 1 (y – y1 = m(x – x1) )
6 2 6 6 2
y-
√ 3 = 1 ¿)
2 √3 1 6 marks
y=
√ x + √3
3
3 3
3 y 2 1 sin x
M1
M1
M1(sub sinx) M1(simplify)
A1
dy
2y =cos x
dx
d 2 y dy 2
[
2 y 2 +
dx dx ( )]
=−sin x
d2 y dy 2 ( 2 )
2 y 2 +2
dx dx ( )
=− y −1
d2 y dy 2 2
2 y 2 +2
dx dx ( )
+ ( y −1 ) =0(shown )
4 y = 12 – 4x
y = 12 – x3
12 – x3 = 12 – 4x
x3 – 4x = 0
x(x2 – 4) = 0
x = 0, x = -2, x = 2
y = 12, y = 0, y = 4
Points of intersection are (0,12), (-2,20), (2,4). B3
dy
dx = -3x2 M1
2
d y
dx 2 = -6x = 0
M1
x=0
d3 y
dx 3 = -6 ≠ 0
There is a point of inflexion at x = 0. A1
0
M1
2 3 2
π ∫−2 (12−4 x ) −(12−x ) dx
V1 =
0 3 2 6
π ∫−2 24 x +16 x −96 x−x dx
=
2528
=
π [ ]
21
2 M1
π ∫0 (12−x 3 )2 −(12−4 x)2 dx
V2 =
0
π ∫−2 x 6 +96 x−16 x 2 −24 x 3 dx
=
1504
=
π [ ]
21 M1
A1
2528 1504
The volume =
π [ ]
21 +
π [ ]
21
= 192 π
5(a) 1 6x
6 3x 2 1
dx
M1
1
ln 3x 2 1 c
6 A1
(b)
x dx
x 3sin , sin 3cos dx 3cos d
3 , d B1
∫ √ 9−x2 dx=∫ √ 9−(3sinθ)2(3cosθ dθ ) M1
2
= ∫ 9cos θ dθ
∫ 9 (cos 22θ+1 ) dθ 9
∫ 2 (cos2 θ +1)dθ
= =
9 sin 2θ
= 2 ( 2
+θ +c ) A1
9 9
(sin θ cos θ+ θ)+c (θ+ sinθ √ 1−sin2 θ)+ c
= 2 = 2
2
=
9
2 ( x x
sin−1 +
3 3 ( ) √ ( ) )+c
1−
x
3 M1
9 x x
=
2 (
sin−1 +
3 9 ( )√ ) 9−x 2 +c
9 x 1
9 x 2 dx sin 1 x 9 x 2 C
2 3 2 A1
6 dy ( 2 x− y )−1 v−1 B1
Let v=2 x− y; = ¿
dx 2 ( 2 x− y ) +2 2 v+ 2 (Substitute 2x-y)
v=2 x − y
dv dy dy dv
=2− => =2− M1
dx dx dx dx
dy v−1 dv v−1
= => 2− = =>
dx 2 v+2 dx 2 v +2 M1 A1
dv v−1 dv 3 v +5 (Change variable)
=2− => =
dx 2 v +2 dx 2(v +1)
( v +1)
∫ 3 v+ 5 dv=∫ 12 dx
1 2 1 M1
∫ 3 − 3(3 v +5) dv=∫ 2 dx (Integrable form)
1 2 1 A1
[v− ln|3 v+ 5|]= x +c
3 3 2
6 v−4 ln |3 v +5|=9 x+ 18 c
4 ln |3 v +5|=6 v−9 x−18 c ; v=2 x − y
3 9
ln |6 x−3 y +5|= [ x−2 y ] − c
4 2
3 9
( x−2 y)− c
6 x−3 y +5=e 4 2
3 −9 A1
( x−2 y) c
6 x−3 y +5=Ae 4 ; where A=e 2
7(a) y 4 x cos 2 x
dy
4 x 2sin 2 x 4cos 2 x
dx (M1)
4 cos 2 x 2 x sin 2 x
dy
x 4 0 2
4 dx 2 (A1)
y 0 2 x
Equation of tangent is 4 (M1)
y 2 x
4 (A1)
(b)
du
Let u = 4x, dx = 4
dv 1
dx = cos 2x, v = 2 sin2x (M1)
4
4 x cos 2 x dx
A= o
π π
4
= [ 2 x sin 2 x ] - ∫ 2 sin2 xdx
4
0 0 (M1)
π π
4
= 2 - [ −cos 2 x ]0 (M1) (M1)
π
= 2 1 (A1)
4
4 x cos 2 x dx
2
(c) V = o (M1)
4 1
16 x 2 cos 4 x 1 dx
= o 2 (M1)
4
8 x 2 cos 4 x x 2 dx
= o
2 sin 4 x 4 1 4 x
3 4
8 x 4 0 2 x sin 4 x dx 3
4 0 0
= (M1)
cos 4 x 4 4 cos 4 x 3
0 2 2 x 2 dx 8
4 0 0 4 192
= (M1)
sin 4 x 4 4
(1)
= 4 4 0 24 (M1)
2 4
= 4 24
4 2 2 2
1
24 4 4 6 or 1.591 cm3
= (A1)
dv k ( 2 2¿
8. = u −valignl¿ ¿ ¿ )
dt v
∫v 2 2¿
dv=∫ kdt (M1)
(u −valignl¿ ¿ ¿)
1 −2v
- ∫ dv=∫ kdt (M1)
2 ( u2−valignl ¿ ¿ 2 ¿¿ )
1
- ln ( u2−v 2 )=kt+c (A1)
2
1
when v= u, let t=t 1
4
2
1 2 1
2 ( ( ))
− ln u − u =kt 1+c
4
1 15 (M1)
( )
− ln u 2 =kt 1+c.............. (1 )
2 16
1
when v= u, let t=t 2
2 (M1)
1 3
( )
− ln u2 =kt 2 +c ............... ( 2 )
2 4
1 3 2 15 2
(2)−(1) −
2 4 ( 16 )
ln u −ln u =k (t 2 −t 1 )
1 3 16
( )
- ln × =k(t 2 −t 1 )
2 4 15
1 5
(t 2−t 1 )= ln (shown)
2k 4
1 1 1 5
The time taken by particle from speed u to speed u is ln (A1)
4 2 2k 4
Let
dv dx
a is acceleration, a = dt ; v is velocity, v = dt (B1)
By applying chain rule,
dv dv dx dv dv
= × = ×v=v
dt dx dt dx dx (A1)
dv k
v = (u2−v 2 )
dx v
v2
∫ 2 2 dv=∫ kdx (M1)
(u −v )
2
∫ −1+ uu 2−v 2 dv=∫ kdx
[ ] (M1)
∫−1 dv + u2∫ 1u2−v 2 dv=∫ kdx
u2 1
∫−1 dv +2 u ∫ u+v +1u−v dv =∫ kdx (M1)
u
−v + [ ln(u+v )−ln(u−v ) ] =kx+c
2
u u+v
−v + ln
2 u−v( ) =kx +c (A1)
1
when v= u . and let x =x1
4
1
u+ u
1 u
− u+ ln
4 2
1 u 5
( )4
1
u− u
4
=kx 1 +c .
4 2 3 ()
− u+ ln =kx 1 +c .. .. . .. .(3) (M1)
1
when v= u. and let x=x 2
2
1
u+ u
1 u
− u+ ln
2 2
1 u
u− u
2
1
2
( ) =kx 2 +c .
(M1)
− u+ ln ( 3 )=kx 2 +c. . .. ..( 4 )
2 2
(4 )−(3 )
1 u
− u+ ln ( 3 )=kx 2 +c. . .. ..( 4 )
2 2
1 u 5
()
− u+ ln =kx 1 +c . .. . .. ..(3 )
4 2 3
u 1 9 1
x 2− x 1 = ( ln −
k 2 5 4 )
u 9 (A1)
hence the distance is
4k (2 ln −1
5 )
