2024 JC 2 H2 Mathematics (9758)

Non-Weighted Assessment 1

Date: 18 January 2023 Duration: 50 mins

Name: CG: Mark: / 27

Instructions to students: Answer all questions.

1 The equations of the line l1 and the plane  are given by

y −4 z −5
l1 : x − 2 = = ,
2 −2
 : x + 2 y + 3z = 24 .
(a) Show that l1 and  intersect at the point A with position vector 3i + 6j + 3k. . [2]

(b) Find the acute angle between l1 and  . [2]

The point B has position vector −i + j + 6k.

(c) Find the shortest distance from B to  . [2]

(d) The points C and D are on l1 such that they are a distance of 5 3 away from B. Find the
position vectors OC and OD . [4]
(e) Find the exact area of the triangle OAB. [3]

1 a
The line l2 has equation r =  0  +   b  ,  
 
, where a and b are constants.
 2  −1
   
(f) Find a condition on a and b for l2 and  to intersect at exactly one point. [1]

[Solution]
1(a)  2 1 1
   
l1 : r =  4  +   2  ,   ----(1);  : r  2  = 24 ---(2)
5  −2   3
     
 2+  1
   
Subst (1) into (2):  4 + 2   2  = 24
 5 − 2   3
   
( 2 +  ) + 2 ( 4 + 2 ) + 3(5 − 2 ) = 24   =1
 2  1   3
     
Subst  = 1 into (1), OA =  4  +  2  =  6 
 5   −2   3 
     
(b) Let  be the acute angle between l1 and  . d n



Method 1 Method 2
1 1  1  1
      
 2  2  2   2
 −2   3  −2   3 
    1    1
sin  = = cos  = =
9 ( ) 14 3 14 9 14 ( ) 3 14
 = 5.1 (1 d.p.) or 0.0892 rad  = 84.889
 = 90 −  = 5.1 (1 d.p.)

(c) Let N be the foot of perpendicular from B to 

AB n B
BN = (length of projection of AB on n) n
n
 −4   1  A
1   
= −5   2 
2 
1 +2 +3    
2 2

 3   3
5
= or 0.802 units (3 s.f.)
14
(d)  2 1
   
Let OR =  4  +   2  be the position vector of a point on l1 for some   .
5  −2 
   
B
 2 1
   
l1 : r =  4  +   2  ,   .
5  −2 
    d
l1
 2   1    −1  3 +   C D
       
BR =  4  +   2   −  1  =  3 + 2 
 5   −2    6   −1 − 2 
     
 BR = ( 3 +  )2 + ( 3 + 2 )2 + ( −1 − 2 )2 =5 3

9 + 6 +  2 + 9 + 12 + 4 2 + 1 + 4 + 4 2 = 75
9 2 + 22 − 56 = 0
(  + 4 )( 9 − 14 ) = 0
14
 = −4 or  =
9

 2   1   −2 
     
When  = −4, OC =  4  − 4  2  =  −4 
 5   −2   13 
     
 2 1  32 
14   14   1  
When  = , OD =  4  +  2  =  64 
9   9   9 
5  −2   17 
(e) Area of triangle OAB
 3   −1
1 1   
= OA  OB =  6    1 
2 2   
 3  6 
 33   11   11   −1
1   3    
=  −21 = 2  −7  Check:  −7   1  = −11 − 7 + 18 = 0
2  9   3
     36
  
3
= 121 + 49 + 9
2
3
= 179 units 2
2
(f) For l2 and  1 to intersect,
l2 is not parallel to 
d n0
 a  1
  
 b   2  0
 −1  3 
  
a + 2b  3
 1 2x
Integrate by parts twice to show that  e sin x dx = e (2sin x − cos x) + c .
2x
2 (a) [3]
5
(b) The diagram below shows the shaded region R that is bounded by part of the curve

x−
y=e 2
sin x , the line 4x =  ( y + 1) and the x-axis. The line 4x =  ( y + 1) cuts the
   
curve and the x-axis at  , 1 and  , 0  respectively.
2  4 
y

R
x

Show that the volume generated when region R is rotated 2 radians about the x-axis is
( )
given by A B + e− + C 2 , where A, B and C are exact constants to be determined. [4]
[Solution]
2(a) 1 2x 1 2x
 e sin x dx = 2 e sin x − 2  e cos x dx
2x

1 1 1 1 
= e 2 x sin x −  e 2 x cos x +  e 2 x sin x dx 
2 2 2 2 
1 1 1
= e 2 x sin x − e 2 x cos x −  e 2 x sin x dx + c
2 4 4
5 2x 1 1
4  e sin x dx = e2 x sin x − e2 x cos x + c
2 4
2 2x 1 2x
 e sin x dx = 5 e sin x − 5 e cos x + c
2x '

Alternatively,

e
2x
sin x dx = e2 x ( − cos x ) −  ( − cos x ) 2e 2 x dx
= −e2 x cos x + 2 e2 x cos x dx

= −e2 x cos x + 2 e2 x ( sin x ) −  ( sin x ) 2e 2 x dx 

= −e2 x cos x + 2e2 x sin x − 4 e 2 x sin x dx + c

5 e2 x sin x dx = −e2 x cos x + 2e2 x sin x + c
2 2x 1
e sin x dx = e sin x − e2 x cos x + c'
2x

5 5
(b) Volume
2
 x−

 1  
=   2  e 2 sin x  dx −  (1) 2  
0
  3 4

1 2
=   2 e 2 x − sin x dx − 
0 12

 1 2
=
e   0
2
e 2 x sin x dx −
12


 1 2 1
=   e (2sin x − cos x)  −  2 using (i)
2x

e 5  0 12
      1 2
= − − −  − 12 
0
e (2sin cos ) e (2sin 0 cos0)
5e  
2 2 

=  ( 2e + 1) −  2
1
5e 12

= ( 2 + e − ) −  2
1
5 12
3 A curve C has parametric equations
 
x = 2sin t +1, y = 2cos3t + 4sin t , − . t 
3 3
In the diagram below, the line L is the tangent to C at the point (1, 2) and L cuts the line x = 2
at the point (2, 4).

y
L

Show that the area bounded by C, L and the line x = 2 is given by


a −  4 cos 3t cos t + 8sin t cos t dt ,
0

where a and  are constants to be determined in exact form.

Hence find the exact value of this area. [6]
[Solution]
3 When x = 1, 2sin t +1 =1, sin t = 0, t = 0
1 
When x = 2, 2sin t + 1 =2, sin t = , t =
2 6
dx
x = 2sin t + 1, = 2cos t
dt

Area
2
= Area of trapezium – 1
y dx

1
= ( 2 + 4 ) (1) −  6
(2cos3t + 4sin t )(2cos t ) dt
2 0


= 3− 0
6
(4cos3t cos t + 8sin t cos t ) dt (shown)

Use factor formula:


= 3− 0
6
(2cos 4t + 2cos 2t + 4sin 2t ) dt cos A cos B =
1
2
( cos ( A + B ) + cos ( A − B ) )
Use double angle formula:
1
sin A cos A = sin 2 A
2
 
 sin 4t 6
= 3−  + sin 2t − 2 cos 2t 
 2 0
 2 
 sin 3   
= 3−  + sin − 2 cos − (−2) 
 2 3 3 
 
3 3 
= 3−  + 1
 4 
3 3
= 2− units 2
4