200701 Umaths Tma3 Ansguide

1 WUC 112 University Mathematics (January 2007) TMA3 – Answer Guide Question 1 A =

2

h2 3 2   h  = 4  3  3

3 2 x = 4

2 dA h = dh 3 dA = dh h = 2 3

(2 3 ) = 4

2 3

Question 2 (a)

f (x) =

x 2 − 2x 1 2 x − 2x f ' (x) = 2 x −1 = x 2 − 2x

(

( b ) g(x) = g ' (x) = =

) ( 2 x − 2) −

1 2

x x +1 2

x2 + 1 − 2x2

(x

2

)

+1

2

1 − x2

(x

2

)

+1

2

(c)

h (x) = x2 sin 2x h ' (x) = 2x sin 2x + 2x2 cos 2x = 2x ( sin 2x + x cos 2x )

(d)

u(x) =

sin x 1 − cos x

u ' (x) = = =

Question 3

cos x (1 − cos x ) − sin 2 x

(1 − cos x ) 2

cos x − cos 2 x − sin 2 x

(1 − cos x ) 2

cos x − 1

(1 − cos x )

2

=

1 cos x − 1

2 (a)

( i ) x2y + xy2 = 2x Differentiate with respect to x dy dy 2xy + x2 + y2 + 2xy = 2 dx dx dy ( x2 + 2xy ) = 2 – 2xy – y2 dx 2 − 2 xy − y 2 dy = x 2 + 2 xy dx ( ii )

1 1 + =1 x y Differentiate with respect to x , 1 1 dy − 2 − 2 =0 x y dx dy y2 =− 2 dx x

(b)

x2 + y2 = 16 dy =0 dx dy x =− dx y dy At ( 2, 1 ) , = −2 dx 2x + 2y

Equation of tangent to the circle at ( 2, 1 ) is y −1 = − 2 ⇒ y = - 2x + 5 x−2 (c)

( i ) xy = 5 y+x

dy = 0 dx dy y = − dx x

( ii ) x2 - y2 = 4 dy 2x - 2y = 0 dx x dy = y dx The two curves intersect at right angles. Question 4 (a)

f (x) = x3 – x - 2 f (1) = - 2 < 0

and f (2) = 4 > 0.

3 Therefore, there is one real solution in the interval ( 1, 2 ) f ' (x) = 3x2 -1 3

xn+1 = xn -

xn − xn − 2 2

3xn − 1

x1 = 1.5, x2 = 1.5217391, x3 = 1.5213798, x4 = 1.5213797, x5 = 1.5213797. (b)

x (t) = 2 t 3 ( i ) Average velocity =

distance travelled time taken

2 ( 4 )3 = m s −1 4 = 32 m s-1. ( ii ) The instantaneous velocity, At t = 3,

dx = 6 t2 dt

dx = 6 ( 3 )2 = 54 m s-1 dt

d 2x ( iii ) Acceleration = 2 = 12 t dt 2 d x At t = 3 , = 12 ( 3 ) = 36 m s -2 . dt 2 Question 5 (a)

(b)

x x 2 x3 + + 4 2 3 1 + x + x2 f'(x)= 4 f '' ( x ) = 1 + 2x f(x)=

x3 + 2x − 1 x x (3 x 2 + 2 ) − x 3 − 2 x + 1 2x3 + 1 f'(x)= = x2 x2 f(x)=

f '' ( x ) = (c)

6 x 2 ( x 2 ) − 2 x(2 x 3 + 1) 2 x 4 − 2 x 2 ( x 3 − 1) = = x4 x4 x3

f ( x ) = ( x – 1 ) ( x + 1 ) ( x2 + 1 )

4 = ( x2 – 1 ) ( x 2 + 1 ) = ( x4 – 1 ) f ' ( x ) = 4x 3 f '' ( x ) = 12x2

Question 6 ( a ) C (100) = 2000 + 100 (100) – 0.1 (10000) = 11000 ( b ) C' (x) = 100 – 0.2 x C' (100) = 100 – 0.2 (100) = 80 ( c ) The cost of producing one more computer after the first 100 have been made = C (101) – C (100) = 2000 + 100 (101) – 0.1 (101)2 – 11 000 = 11 079.9 – 11 000 = 79.9 ≈ 80

Question 7 (a)

lim x →0

cos x − 1 x

We can write

cos x − 1 cos x − 1 cos x + 1 = × x x cos x + 1 cos 2 x − 1 = x (1 + cos x) − sin 2 x − sin x sin x = × = x(1 + cos x) x 1 + cos x

Since lim x →0

sin x − sin x sin x 0 = − lim = − 1 and lim = =0 x →0 1 + cos x x → 0 x x 2

Hence, lim x →0

cos x − 1 =0 x

Alternatively, students may apply L’ Hospital rule to find the limit. lim x →0

(b)

cos x − 1 − sin x = lim =0 x →0 x 1 cos ( x + h) − cos x d ( cos x) = lim h → 0 dx h

5 cos x cos h − sin x sinh − cos x h →0 h cos h − 1 sin h     − sin x  lim = cos x lim   h   h →0 h   h →0 = − sin x = lim

(c)

d d  cos x  − sin 2 x − cos 2 x  = ( cot x ) = = –1 – cot2 x dx dx  sin x  sin 2 x = – (1 + cot2 x) = – cosec2 x

Question 8 (a) The length of a rectangle is x and the perimeter is 24. The width must be

24 − 2 x = 12 − x 2

Area , A(x) = x ( 12 – x ) = 12x – x2 dA = 12 – 2x dx Given

dx dA = 2, find the value of x when starts to become negative. dt dt dA dA dx = x = (12 – 2x) 2 = 24 – 4x dt dx dt 24 – 4x < 0 x > 6. The value of x is more than 6 cm when the area of the rectangle starts to decrease.

(b)

f ( x ) =1 = lim+ f ( x ) = a + b f (x) is continuous at x = 1 if xlim →1− x →1 Thus, a + b = 1 f (x) is differentiable at x = 1 f (1 + h) − f (1) f (1 + h) − f (1) if lim− = lim+ . h →0 h →0 h h f (1 + h) − f (1) (1 + h)3 − 1 = a. lim− = 3 and lim+ h →0 h →0 h h Therefore, we obtain a = 3 and b = -2

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