200701 Umaths Tma1 Ansguide

WUC 112 University Mathematics (January 2007) TMA1 – Answer Guide Question 1 (a) Domain: x  4  0 x  4 Range: x  4  0 , 3 x  4  0 D f = [ 4, ) , R f = [0 , ) (b) g ( x)  ( x 2  3 x  1) 2   3 9      x     1 2 4   



2

3 5   x    2 4  5 D g = (  , ) R , R g = (  , ] 4 (c) (i) x  2  x3 (ii)   0  x2  ( x  3)  0 and ( x  2)  0 , thus x  3 and x  2  (2 , )  ( x  3)  0 and ( x  2)  0 , thus x  3 and x  2  (  ,  3] D h = ( ,  3]  (2, ) , R h = [0, 1)  (1, ) Question 2 a. Graph of y = 2x x x

8 6 4 2 y

0 -10

Df = ( , 0)  (0, ) Rf = ( ,  2)  (2, )

-5

-2

0

-4 -6 -8 x

5

10

b.

Graph of y =

x [ x]

Df = ( , 0)  [1, ) Rf = (0, 2)

2 1.5 y

1 0.5 0 -4

-3

-2

-1

0 x

Question 3 ( 2 a Graph of y = 2  x )

( b )

( c )

Graph of y = 2x  x 2

Graph of y =

x2 x2

1

2

3

4

Question 4 (a)

 Graph of y  2 sin ( x  ) 4 Df = (  , ) , Rf = [ 2 , 2]

−π

x

π  sin  x −  4 

0.71

3 − π 4 0

π  2 sin  x −  4 

1.41

0

(b)

π

0 π π π π − − − 2 3 4 6 -0.71 -0.97 -1 -0.97 -0.71

π 6 -0.26

π 4 0

π 3 0.26

π 2 0.71

3 π 4 1

0.71

-1.41 -1.93 -2

-0.52

0

0.52

1.41

2

1.41

−

-1.93 -1.41

π Graph of y = 2 sin[2 ( x − )] 4 Dg = (− ∞ , ∞) , Rg = [− 2 , 2]

x

−π

  π  sin  2  x −   4   

-1

3 − π 4 0

  π  2 sin  2  x −   4   

-2

0

1

π 3 0.5

2

1

−

π 2

−

0

π 6 -0.5

0

-1

−

π 4

π

-1

π 6 -0.5

π 4 0

π 3 0.5

π 2 1

3 π 4 0

-1

-2

1

0

1

2

0

-2

0

−

(c)

Graph of y = sin x + cos x Dh = (− ∞ , ∞) , R h = [− 2 , 2]

x

−π

sin x

0

3 − π 4 -0.71

cos x sin x + cos x

-1 -1

-0.71 -1.41

π 2 -1

π π − 3 4 -0.87 -0.71

π 6 -0.5

0 -1

0.5 0.71 -0.37 0

0.87 0.37

−

−

Question 5 x x+4  x  f  g(x) = f    x+4  x  = 2−   x+4  2( x + 4) − x  =  x+4  

(a) f (x) = 2 – x , g (x) =

−

π

0

π 6 0.5

π 4 0.71

π 3 0.87

π 2 1

3 π 4 0.71

0

1 1

0.87 1.37

0.71 1.41

0.5 1.37

0 1

-0.71 0

-1 -1

0

 2x + 8 − x  =   x+4   x+8  =   x+4 g  f (x) = g (2 – x)  2− x  =   2− x+4  2− x  =   6− x  (b) f (x) = x2 , g (x) = f  g(x) = f =





9− x 9 x

9 x





2

=9–x g  f (x) = g (x2) = 9  x2 Question 6 D f = [ 2,  ) and R f = [ 3,  ) ; D g = [ 1,  ) and R g = [ 0,  ) For f  g to exist, the range of g must be a subset of the domain of f , But, [ 0,  )  [ 2,  ). Thus, f  g does not exist. f  g will only exist if g is in [ 2,  ) , that is ,

x  1  2  x – 1  4 , x  5.

Therefore, f  g will exist if its domain is [ 5,  ). Question 7 (a)

x = r cos θ, y = r sin θ xy=4 (r cos θ) (r sin θ) = 4 r 2 cos θ sin θ = 4

(b)

x = r cos θ, y = r sin θ x2 + y2 = 4 (r cos θ)2 + (r sin θ)2 = 4

r2 (cos2 θ + sin2 θ) = 4 r2 = 4 r =  2 (c)

x = r cos θ, y = r sin θ y2 = 4 x r2 sin2 θ = 4 r cos θ r sin2 θ = 4 cos θ

Question 8 (a)

r = cot θ csc θ  cos    1  r     sin    sin   1   sin         cos    r sin   1   tan      r sin   Substitution tan θ = (y/x), y = r sin θ y 1  x y y2 = x (b)

x = r cos θ, y = r sin θ r cos θ + r sin θ = 1 x+y=1

(c) r2 = – 4 r cos θ Substitution r2 = x2 + y2 and x = r cos θ, x2 + y2 = – 4 x x2 + 4 x + y2 = 0 ( x + 2 )2 – 4 + y2 = 0 ( x + 2 )2 + y2 = 4