Q1: Solutions (1 Viewer)

[MarsBarz]

Question 1:

a. Find ∫ [dx/(x²+49)].
∫ [dx/(x²+49)] = (1/7).tan^-1(x/7) + C

b. Sketch the region in the plane defined by y<=|2x+3|.
Sketch y=|2x+3|, and shade under the function.

c. State the domain and range of y=cos^-1(x/4).
Domain: -1<=(x/4)<=1
Domain: -4<=x<=4
Range: 0<=y<=pi

d. Using the substitution u=2x²+1, or otherwise, find ∫ x(2x²+1)^5/4dx.
u=2x²+1
du/dx = 4x
du = 4x.dx
∫ x(2x²+1)^5/4dx = (1/4).∫ 4x(2x²+1)^5/4dx
= (1/4).∫ u^5/4du
= (1/4).(4u^9/4/9) + C
= u^9/4/9 + C
= (2x²+1)^9/4/9 + C

e. The point P(1,4) divides the line segment joining A(-1,8) and B(x,y) internally in the ratio 2:3. Find the coordinate of the point B.
1 = [2x+3(-1)]/[2+3]
5 = 2x-3
2x = 8
x = 4
4 = [2y+3(8)]/[2+3]
20 = 2y+24
2y = -4
y = -2
So the point is B(4,-2).

f. The acute angle between the lines y=3x+5 and y=mx+4 is 45°. Find the two possible values of m.
y = 3x+5
y' = 3
.:. m₁ = 3
y = mx+4
.:. m₂ = m
tan@ = |(m₁-m₂)|/|1+m₁m₂|
tan45° = |(3-m)|/|1+3m|
There are two possibilities:
1 = (3-m)/(1+3m)
1+3m = 3-m
4m = 2
m = 1/2
OR
-1 = (3-m)(1+3m)
-1-3m = 3-m
-2m = 4
m = -2
.:. So m = 1/2 or m = -2

#

 

[richz]

for f. i think ur wrong, -.5 doesnt work

 

[redw0lf]

-2m = 4
m = -1/2

------------
m= -2 not -1/2
so its m = 1/2 or -2

 

[richz]

yep, those are the 2 answers i got

 

[LuKiN]

agreed. for part f i got 1/2 and -2.

the two posibilities are
tan@ = |(m1-m2)|/|1+m1m2|
and
tan@ = |(m2-m1)|/|1+m2m1|

swapping m2 and m1 does make a difference.

the rest seem pretty correct though

 

[MarsBarz]

Ok I'll fix it, that took ages to format.

 

[kyu_chan]

Thanks mate ^-^

You posting solutions to 2 as well?

(Damn!! I didn't get part e!!!)

 

[MarsBarz]

Yeah I'll post question 2 in a bit.

 

[Lil Clutch]

yay i got q1 out. feeling better-jokes! the paper was good, but i always need like 15 minutes extra time-that would work out so well for me! i got to the end of the paper, but i wouldve liked to have gone back to all the questions, or parts more correctly that i circled on the paper-the q's i didnt completely get out and was hoping to go back to! oh well, its done. yay. until i do maths for engineering at uni next year! oh boy, cant wait! (sarcasm)

 

[lourai*87]

dammit for putting things round the wrong way. Solutions were rather unrealisitc..however i couldnt have been bothered trying to fix it...ran out of time as it was. That'll teach me for not remembering formulae, hell that'll teach me for not dropping 3unit