Mathematics T (MS1)

SCHEME MID TERM P1 2020 Mark Remarks
B1
No Solutions
1(a) (i) The range of g is { y : y ϵ R, y > 0} M1

y = f −1(x) A1
(ii) 3y − 5 = x
OR
f −1 : x → 1 (x + 5) M1
3
A1
OR
g( y) = e−3y = x
g −1 : x → − 1 ln x

3
y

y=f-1(x)

-5 0 1 x Both
y= g-1(x) D1 graphs

1(b) By comparison Correct
2p + q =2 ………..(1)
2pq + p2 = -15…..(2) M1
p = 3, q = -4 A1 Both

5x2 – 31 = A(x + 3)(x - 4) +B(x - 4) +C(x + 3)2 correct
M1

5x2 − 31 =4− 2 +1

x3 + 2x 2 − 15x − 36 x + 3 (x + 3) 2 x − 4 A1

Q1 Total 8

2(a)  2 1 3 4 − 5 3  A1
PQ = 1 2 3 − 5 4 3 
 2 2 1 2 2 − 3
9 0 0
= 0 9 0
 0 0 9

PQ = 9I

P −1 = 1 Q
9

= 1  4 −5 3  M1
9 −5 4 3 A1
 2 2 − 3 B1
B1
 4 − 5 1 
9 9 3 M1
 5 4 1 A1
=  − 9 9 3 A1

2 2 − 1 
 
 9 9 3

2(b) 6x + 3 y + 9z = 94.50

3x + 6 y + 9z = 99.00

6x + 6 y + 3z = 79.50

 6 3 9  x   94.50
 3 6 9  y  =  99.00
 6 6 3 z   79.50

 2 1 3 x   94.50
3 1 2 3 y  =  99.00

 2 2 1 z   79.50

 x   94.50
3P y  =  99.00

 z   79.50

 x  = P −1  3 1.5 0 
 y  3 3.0 0 

 z   26.50

= 1  31.50
9 Q 33.00

 26.50

= 1  4 −5 3  31.50
9 −5 4 3  33.00
 2 2 − 3 26.50

 4.50
=  6.00

 5.50

x = RM 4.50 y = RM6.00 z = RM5.50

3.
y

( ( 2, 2 )

0 x
Q( 2, 2 )

a) Let point of intersection at x-axis be (4, 0)
Gradient of PT = Gradient of PQ

2 −0 = 2 −2 M1
2−4 2− 2
A1
2 = 2( − ) M1
2−4 ( − )( + )
M1
( + ) = 2 − 4
= −4 A1 With

b) Midpoint of PQ = ( 2+ 2 , 2 +2 ) conclusion

22 B1

M = ( 2+ 2 , + ) All correct

2

Let = 2+ 2 and = +
2

= 1 [( + )2 − 2 ]

2

= 1 [ 2 − 2(−4)]

2

2 = 2 + 8

2 = 2( − 4)
Hence, the locus of M is a parabola.

c) 2 = 4 (1) ( − 4)

2
= 1
2

Vertex is (4, 0)

Focus is (4 1 , 0)
2

Axis of symmetry is = 0

3
y

2 = 2( − 4)

•• x D1
0 (4, 0) Shape-
Open to the
4. (a) y = 4x2 − 24x +11 = 4[x2 − 6x + 11]
4 D1 right
With label
= 4[x2 − 6x + (−3)2 − (−3)2 + 11]
4 M1

= 4[(x − 3)2 − 25] A1
4 A1

= 4(x − 3)2 − 25 M1

Turning point = (3, -25) A1
A1
(b) y = 4x2 − 24x +11 = 4(x − 3)2 − 25
(x − 3)2 = 1
4( y + 25)
x − 3 = 1 y + 25
2
f −1(x) = 3 − 1 x + 25
2

Domain of f −1(x) is Df −1 = x  −9

5.  − 1 2 1   − 1 2 1   − 5 1 9 (Process of
A2 =  − 3 1 4  − 3 1 4 =  0 −1 9 1multiplication
 0 1 2  0 1 2  − 3 3 8 must be seen
−5 1 9 −35 19 18 1at least once,
A2B = ( 0 −1 9) (−27 −13 45) = if not -1
−3 3 8 −3 12 5 mark)
121 0 0
( 0 121 0 ) 1
0 0 121

1 0 0
= 1210 1 0

0 0 1

A2B = 121 I 1

A-1AAB = A-1 121 I 1

I AB = A-1 121 I 1

A-1 = 1 AB 1
121
1 −1 2 1 −35 19 18 1 (allow
= (−3 1 4) (−27 −13 45) premultiply
121 0 1 2 −3 12 with his A-
5 1)
1 −22 −33 77
= ( 66 −22 11) 1
121 −33 11 55
1
2 37
− 11 − 11 11
6 21
= 11 − 11 11

315
(− 11 11 11)

−1 2 1 8

(−3 1 4) ( ) = (15)

0 1 2 4

2 3 7
− 11 − 11
6 11 8
( ) = 2 1 (15)
11 − 11 11
3 5 4
1
(− 11 11 11)

−3
=( 2 )

1

x = -3, y = 2 and z =1

6 (a) 5x2 + 4 y2 − 40x + 8y + 64 = 0

5x2 − 40x + 4 y2 + 8 y + 64 = 0 B1
5(x2 − 8x) + 4( y2 + 2 y) + 64 = 0
5[x2 − 8x + (−4)2 − (−4)2 ] + 4[( y2 + 2 y + (1)2 − (1)2 ] + 64 = 0 M1
5[(x − 4)2 −16] + 4[( y +1)2 −1] + 64 = 0 A1

5(x − 4)2 − 80 + 4( y +1)2 − 4 + 64 = 0
5(x − 4)2 + 4( y +1)2 = 20
(x − 4)2 + ( y +1)2 = 1
45

(b) center = (4, -1) A1(both)
a = 2, b =  5, c2 = b2 − a2 = 5 − 4 =1 A1(both)
Foci = (4, -2) and (4, 0)
Vertices = (4, −1 5)

7 B1
M1
Solution:
(a) f (x) = 10cos2 x − 8sin x cos x + 4sin2 x A1

= 4sin2 x + 4 cos2 x + 6 cos2 x − 4sin 2x
= 4 + 3(1+ cos 2x) − 4sin 2x
= 7 + 3cos 2x − 4sin 2x

f (x) = c + r cos(2x +  )

r cos(2x +  ) = 3cos 2x − 4sin 2x

r[cos 2x cos − sin 2x sin  ] = 3cos 2x − 4sin 2x

r cos = 3, r sin = 4

r2 = 32 + 42  r = 5, B1
B1
sin = 4  tan = 4   = tan−1( 4)   = 53.1 (Both)
cos 3 3 3 A1
B1
f (x) = 7 + 3cos 2x − 4sin 2x = 7 + 5cos(2x + 53.1) A1
−1  cos(2x + 53.1)  1 A1

− 5  5cos(2x + 53.1)  5 B1
7 − 5  7 + 5cos(2x + 53.1)  7 + 5 B1
A1
2  7 + 5cos(2x + 53.1)  12
B1
Greatest and least values of f (x) are 12 and 2 respectively.
B1
7 15
(b) f (x) = 7 + 5cos(2x + 53.1) = 2
A1
 f (x) = cos(2x + 53.1) = −1
2x + 53.1 = 180, 540

x = 63.5, 243.5

(c) 2  f (x)  9  2  7 + 5cos(2x + 53.1)  9
22

 −1  cos(2x + 53.1)  − 1
2

cos(2x + 53.1) = −1, cos(2x + 53.1) = − 1
2

2x + 53.1 = 180, 2x + 53.1 = 120, 240

x = 63.5, x = 33.5, 93.5

x = x : 33.5  x  93.5

8

(a) at P : x = p + 1 , y = p − 1
pp

dx =1− 1 , dy =1+ 1 M1
dp p2 dp p2 (both)
A1
= p2 −1, = p2 +1
p2 p2 M1

m= dy = p2 +1 p2 = p2 +1 M1
dx p2 p2 −1 p2 −1
A1
Equation of the tangent at P : Sub mark Total

y − y1 = m(x − x1) mark

y −  p− 1  = p2 +1  −  p+ 1 
 p  p2 −1 x  
    p 

y −  p 2− 1  = p2 + 1  x −  p2 +1
 p  p2 − 1   
    p 

py − ( p2 −1) = p2 + 1  px − ( p 2 + 1) 
p2 − 1

p( p2 −1) y − ( p2 −1)2 = ( p2 +1) px − ( p2 +1)2

( p2 +1) px − p( p2 −1) y = ( p2 +1)2 − ( p2 −1)2

( p2 +1) px − ( p2 −1) py = ( p2 +1+ p2 −1)( p2 +1− p2 +1)

( p2 +1) px − ( p2 −1) py = 4 p2

( p2 +1)x − ( p2 −1) y = 4 p

No

8

(b) ( p2 +1)x − ( p2 −1) y = 4 p

A : when y = x, ( p2 +1)x − ( p2 −1)x = 4 p B1
p2x + x − p2x + x = 4p
2x = 4p A1
x = 2p (A & B
Correct)
A = (2 p, 2 p) M1

B : when y = -x, ( p2 +1)x − ( p2 −1)(−x) = 4 p A1
p2x + x + p2x − x = 4p A1

2p2x = 4p Sub mark Total
mark
2 p( px − 2) = 0
B1
 2 p  0, px − 2 = 0  x = 2
p

B =  2 ,− 2 
 p p 
 

0 2p 2 0
Area of triangle OAB = 1 p

2 0 2p − 2 0
p

= 1 2 p  − 2  − 2 p  2
2  p   
   p 

= 1 −4 − 4
2

= 4 unit2

Hence, the area of triangle OAB is independent of p.

No x = t +1, y = t −1
8 tt

(c)

x2 =  t + 1 2 , y2 =  t − 1 2
 t   t 

x2 = t2 + 1 + 2, y2 = t2 + 1 − 2
t2 t2

x2 − 2 = t2 + 1 , y2 + 2 = t2 + 1
t2 t2

x2 − 2 = y2 + 2

x2 − y2 = 4 A1

x2 − y2 =1
44

It is a hyperbola.
y

y = -x y=x

2 D1
(shape)
-2 0 2 x 15
-2 D1
(points)

D1
(all
Correct)