Practice paper - Difficult, Answers needed. (1 Viewer)

[SoulSearcher]

The Identity:
- int. [f(x)] (a->b), where a is the upper limit and b is the lower limit,
= int. [f(x)] (b->a)

 

[Riviet]

Q5 (a)

(i) Ae^-kt=H+5
dH/dt=-kAe^-kt
=-k(H+5), as required.

(ii) When t=0, H=100,
100=A-5
A=105

(iii) When t=20, H=30,
30=105e^-20k
e^-kt=35/105
=1/3
-20k=ln(1/3)
k=(ln3)/20

When t=30,
H=105e^-(30/20).ln3 - 5
= 15^o (nearest deg.)

(iv) When H=0,
0=105e^-kt - 5
e^-kt=5/105
=1/21
-kt=ln(1/21)
t=(ln21)/[(ln3)/20]
=56 minutes (nearest min.)

(b) (i)
When x=π/4
y₁=sec(π/4)
=sqrt2
y₂=2cos(π/4)
=2/sqrt2
=sqrt2
.'. point of intersection at x=π/4.

(ii) V=π(∫y₁²dx + ∫y₂²dx) {limits for each integral are 0->π/4 and π/4->π/2 respectively}

=π(∫sec²x dx + 4∫cos²x dx) {limits: 0->π/4 and π/4->π/2}

=π[tanx]^π/4₀ + 2π∫(1+cos2x) dx {limits: π/4->π/2}

=π + 2π[x+(sin2x)/2]^π/2_π/4

=π + 2π[(π/2 + 0) - (π/4 + 1/2)]

=π + pi²/2 - π

=π²/2 units³

 

[Riviet]

Q6 (a) solution attached.

 

[Forbidden.]

I'll do the easy ones

Question 1 b)

A (-4,2) B(2,-1) P(x,y) ratio 5:2
since external division, take ratio as -5:2

(x,y) = (mx₁+nx₂) / (m+n) , (my₁+ny₂) / (m+n)
(x,y) = [ (-5)(-4)+(2)(2) ] / -3 , [ (-5)(2)+(2)(-1) ] / -3
(x,y) = (24/-3 , -12/-3)
(x,y) = (-8,4)
P = (-8,4)
......

You are wrong ..

The formula is:

(x,y) = (mx₂+nx₁) / (m+n) , (my₂+ny₁) / (m+n)


Your solution therefore must be:

(x,y) = [ (-5)(2)+(2)(-4) ] / -3 , [ (-5)(-1)+(2)(2) ] / -3
(x,y) = (-18/-3 , 9/-3)
(x,y) = (6,-3)
P = (6,-3)

I'll start us off:

Q3

(c) (i) Let p(x) = x³-15x+4
p(-4)=-64+60+4=0
.'. x=-4 is the other root.

(ii) Let the other two roots be a and 1/a.
Sum of roots = a + 1/a - 4 = 0
a² - 4a + 1 = 0
(a-2)² = 3
a=2+sqrt3
.'. two other roots are x=2+sqrt3 and x=2-sqrt3

1/(2+sqrt3)=(2-sqrt3)/(4-3) by rationalising denominator
=2-sqrt3
.'. 2+sqrt3 and 2-sqrt3 are reciprocals.

I don't know how he got that but that shows how incompetent I am with harder 2 Unit topics that are difficult enough already ...

 

[housemouse]

The paper doesnt look that difficult.

 

[mrzeidan1]

You are wrong ..

The formula is:

(x,y) = (mx₂+nx₁) / (m+n) , (my₂+ny₁) / (m+n)


Your solution therefore must be:

(x,y) = [ (-5)(2)+(2)(-4) ] / -3 , [ (-5)(-1)+(2)(2) ] / -3
(x,y) = (-18/-3 , 9/-3)
(x,y) = (6,-3)
P = (6,-3)

i always manage 2 stuff those questions up

 

[Riviet]

Q4 (a)
(i) Given y=tanx and y=cosx and curves intersect at x=α,
Let y₁ be the common y-coordinate.
Then y₁=tanα=cosα
sinα/cosα = cosα
sinα = cos²α, as required.

(ii) Let m₁ and m₂ be gradients at P of y=tanx and y=cosx respectively.

For y=tanx, dy/dx = sec²x
At x=α, m₁=sec²α

For y=cosx, dy/dx = -sinx
At x=α, m₂=-sinα

.'. m₁.m₂=sec²α.(-sinα)
=-sinα/cos²α
=-sinα/sinα, using result in (i)
=-1

.'. tangents to two graphs are perpendicular.

Q6 (b)
(i) From graph in (a), y=f(x) has an inverse function since for each x-coordinate in the given domain -1 < x < 1, there exists no more than one y-coordinate.

 

[Riviet]

Solution attached.

 

[vafa]

Is there any question left?

 

[Riviet]

Is there any question left?

I think it's only Q7 left, although haque briefly answered some of it before.

 

[Riviet]

Q7 (b)
(iii) We observe that the expression is in the form:
v²=n²(a²-x²) => P is moving in SHM.

(iv) By inspection, maximum displacement (a) = v²/g
.'. displacement midway between N (x=0) and maximum displacement from N (v²/g) = v²/g
Let
.
x be denoted by v.
.'. v²=4w²(V⁴/g² - V⁴/4g²)
=3w²V⁴/g²
.'. v=(sqrt3)wV²/g

Hence speed of P (midway between N and max displacement) is (sqrt3)wV²/g m/s.