Mathematics SM025 Smart Revision Slot 1

Question 1 : 2018/2019

Use Newton - Raphson method to solve the

equation ex − x − 2 = 0 correct tofour decimal
places by taking x =1as the first approximat ion.

Jens Martensson 2

f (x) = ex − x − 2 , f '(x) = ex −1

x1 = 1

x2 =1− e1 −1− 2  = 1.16395
 
 e1 −1 

x3 = 1.16395−  e1.16395 −1.16395− 2  = 1.14642
 
 e1.16395 −1 

x4 = 1.14642−  e1.14642 −1.14642− 2 = 1.14619
 
 e1.14642 −1 

x5 =1.14619

Thus , the solution is x = 1.1462

Suggested Answer Jens Martensson

Question 2 : 2018/2019

An ellipse Ax2 + y2 + Bx + Cy +1 = 0 passes

through points (0,1),(1,−1)and (2,1)

(a) Find the equation of the ellipsein the standard
form.Hence,state the centreand vertices of the ellipse.

(b) Find the foci of the ellipse
(c) Sketch the graph of the ellipse

Jens Martensson

At (0,1) 0 +12 + 0 + C(1) +1 = 0 Jens Martensson

C = −2

At (1,−1) A + (−1)2 + B + C(−1) +1 = 0

A + B = −4 - - - -(1)

At (2,1) 4A +12 + B(2) + C(1) +1 = 0

2A + B = 0 - - - -(2)
(2) − (1) A = 4

from (1) B = −8

Suggested Answer

4x2 + y2 −8x − 2y +1= 0

4x2 −8x + y2 − 2y +1= 0 Center (1,1)

 4 (x −1)2 −1 + (y −1)2 −1+1= 0

4(x −1)2 + ( y −1)2 = 4 Vertices (1,1 2) = (1,−1),(1,3)

(x −1)2 + ( y −1)2 = 1

4

c2 = 4 −1 ( ) ( ) ( )Foci 1,1 3 = 1,1− 3 , 1,1+ 3

c= 3

Suggested Answer Jens Martensson

y V (1,3)

( )F 1,1+ 3

C(1,1) x

( )F 1,1− 3

V (1,−1)

Suggested Answer Jens Martensson

Question 3 : 2018/2019

The line L1 and L2 passes through the point R(2,4,−3) and

S(8,−5,9)in the direction of 2 i − 3 j+ 4k and i − 2 j+ 3k, respectively

~~ ~ ~~~

(a) State the equations for line L1 and L2 in the vector form.

Hence, calculate the acute angle between the line L1 and L2 .

(b) Find the equation of plane containing the line L1 and the point (7,−3,5)

in the Cartesian form.

(c) Determine whether the line L2 is parallel to the plane x + 5y + 3z = 5

Jens Martensson

L1 : r1 = 2 i + 4 j− 3 k + t 2 i− 3 j+ 4 k 
L2 :  
~ ~ ~ ~ ~ ~

r2 = 8i− 5 j+ 9k+ s i − 2 j+ 3k 
~ 
~ ~ ~ ~ ~

 2 + 6 +12 

 = cos−1
  22 + (−3)2 + 42  12 + (−2)2 + 32  

( )( ) = cos−1 20 14  = 6.98
29

Suggested Answer Jens Martensson

P (7,−3,5) n = v  AP

 7   2  5  ~~
AP =  − 3 −  4 =  − 7
i jk
 5   3  8  n= 2 −3 4

= 5i−7 j+8k ~

~~~ 5 −7 8

n = 4i+ 4 j+ k

~ ~ ~~

Jens Martensson

r • (4 i + 4 j+ k) = (7 i − 3 j+ 5k) • (4 i + 4 j+ k)

~ ~~ ~ ~~ ~ ~~

4x + 4 y + z = 28 −12 + 5

4x + 4y + z = 21

Jens Martensson

Solution 5 (c)

n = i+5 j+3k v = i− 2 j+3k

~~ ~ ~ ~~ ~ ~

n• v = (i + 5 j+ 3k) • (i − 2 j+ 3k)

~~ ~ ~ ~ ~ ~ ~

n• v = 1−10 + 9

~~

n• v = 0

~~

thus, the line L2 is parallel to the plane.

Jens Martensson

Question 4 : 2018/2019

A discrete random variable X has the probability distribution function
x +1, x = 2,3,4
16

f (x) = kx, x = 6,8

(a) Show that k = 1 0, otherwise

56

(b) Hence, calculate P(3  X  8).

(c) Determine the values of E( X ) and Var( X )

Thus, evaluate Var( 3X −1).

Jens Martensson

(a) f (x) =1 (b) P(3  X  8)

3 + 4 + 5 + 6k + 8k = 1 = P( X = 3) + P( X = 4) + P( X = 6)
16 16 16 =4+5+6

14k = 1 16 16 56
4 = 75 = 0.6696

k = 1 (shown) 112
56

Jens Martensson

E(X ) = xf (x)

E(X ) = 2 3  + 3 4  + 4 5  + 6 6  + 8 8  = 233
16  16  16   56   56  56

E( X 2 ) = 22  3  + 32 4  + 42 5  + 62 6  + 82  8  = 21
16  16  16   56   56 

Var(X ) = 21−  2332 = 3.69
 56 

Var( 3X −1) = 3Var(x) = 3(3.69) = 11.07

Jens Martensson

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