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C3 Past Paper Questions

Hey, could anyone give me a hand with some of these past paper questions? The exam is coming up soon and with some questions I can't even remember the methods of doing them and others I just simply can't get anywhere. Thanks in advance.

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June 07
1) Find the exact solutions to the equations:
(a) ln x + ln 3 = ln 6
(b) e^x + 3e^-x = 4

3) A curve C has equation y = (x^2)(e^x)
(b) Find the turning points
(d) Determine the nature of each turning point of the curve C.

5) Functions f and g are defined by:
f:-> ln(2x-1), where x>0.5
g:-> 2/(x-3), where x doesn't equal 3

(a) Find exact value of fg(4)
(b) Find inverse function of f^-1 (x), stating its domain
---
Jan 08
1) Given that:
(2x^4 - 3x^2 + x + 1) / (x^2 - 1) = (ax^2 + bx + c) + (dx+e/x^2-1)
Find values of constants, a, b, c, d and e. Sorry this question is difficult to write out kinda...

2) A curve C has equation:
y= e^2x tan x , where x cannot equal (2n+1)pi/2

(a) Show the turning points on C occur where tan x = -1
(b) Find an equation of the tangent to C at point where x = 0

7) Curve C has equation:
y= 3sin2x + 4cos2x, -Pi < x < Pi The point A (0,4) lies on C.

(a) Find an equation of the normal to curve C at A.

---

Again, some of it was kinda hard to write out but I'd appreciate all the help I can get. :redface:

Reply 1

viksta1000

15

Original post by Wolfiexe

Hey, could anyone give me a hand with some of these past paper questions? The exam is coming up soon and with some questions I can't even remember the methods of doing them and others I just simply can't get anywhere. Thanks in advance.

---
June 07
1) Find the exact solutions to the equations:
(a) ln x + ln 3 = ln 6
(b) e^x + 3e^-x = 4

(a) ln x + ln 3 = ln 6

take ln3 from ln6 so you get: lnx = ln6-ln3 which is the same as ln(6/3) = ln2
so you have lnx = ln2 - so ln's cancel so you have x=2

(b) e^x + 3e^-x = 4

take logs of both sides so you have: xlne + -x ln3e = ln4 (remeber logx^n is the same as nlogx which is why the x's can come to the front

lne = 0 so it can cancel, so now you have

x - x(ln3e) = ln4 --> ln3e = ln3 + lne so again lne cancels

so now you have x-xln3 = ln4

take out factor of x
x(1-ln3) = ln 4

divide ln4 by (1-ln3) and you have x

remember that questions wants the 'exact value' so leave the answer with logs and e's

and btw, I only *think* this is how you do them, im just trying lol

Reply 2

Luke0011

11

Original post by Wolfiexe

Hey, could anyone give me a hand with some of these past paper questions? The exam is coming up soon and with some questions I can't even remember the methods of doing them and others I just simply can't get anywhere. Thanks in advance.

---
June 07
1) Find the exact solutions to the equations:
(a) ln x + ln 3 = ln 6
(b) e^x + 3e^-x = 4

3) A curve C has equation y = (x^2)(e^x)
(b) Find the turning points
(d) Determine the nature of each turning point of the curve C.

5) Functions f and g are defined by:
f:-> ln(2x-1), where x>0.5
g:-> 2/(x-3), where x doesn't equal 3

(a) Find exact value of fg(4)
(b) Find inverse function of f^-1 (x), stating its domain
---
Jan 08
1) Given that:
(2x^4 - 3x^2 + x + 1) / (x^2 - 1) = (ax^2 + bx + c) + (dx+e/x^2-1)
Find values of constants, a, b, c, d and e. Sorry this question is difficult to write out kinda...

2) A curve C has equation:
y= e^2x tan x , where x cannot equal (2n+1)pi/2

(a) Show the turning points on C occur where tan x = -1
(b) Find an equation of the tangent to C at point where x = 0

7) Curve C has equation:
y= 3sin2x + 4cos2x, -Pi < x < Pi The point A (0,4) lies on C.

(a) Find an equation of the normal to curve C at A.

---

Again, some of it was kinda hard to write out but I'd appreciate all the help I can get. :redface:

I'll try to help, by giving some hints!

June 07

1) a) Put ln(x) as the subject, and mess around with the other side into one ln(a/b). Then do e^ of both sides.

b) i did this by rearranging it into a quadratic and solving it that way.

3) b) do dy/dx (product rule) and find stationary points, obv.
d) then d2y/dx^2 to find out the nature.

5)a) work out fg(x) and then sub in x=4, simplify and leave answer in ln.
b) y=f(x)
make x the subject of the equation
then change x for the y's, and make that f^-1(x).
work domain out from that.

Jan 08

1) i'm guessing you split into two quotients and then just simplify until you get what it asks for.

2) a) dy/dx using product rule and make dy/dx=0 and work that out.
b) then work out gradient at x=0 and use y=mx+c to get the tangent.

7)
a) dy/dx, then sub in value of x
then find negative reciprocal for the gradient at that value
and use y=mx+c again.

hope i helped a bit, didnt really know what you wanted as you didnt show working.

Reply 3

ztibor

10

Original post by Wolfiexe

Hey, could anyone give me a hand with some of these past paper questions? The exam is coming up soon and with some questions I can't even remember the methods of doing them and others I just simply can't get anywhere. Thanks in advance.

---
June 07
1) Find the exact solutions to the equations:
(a) ln x + ln 3 = ln 6
(b) e^x + 3e^-x = 4

3) A curve C has equation y = (x^2)(e^x)
(b) Find the turning points
(d) Determine the nature of each turning point of the curve C.

5) Functions f and g are defined by:
f:-> ln(2x-1), where x>0.5
g:-> 2/(x-3), where x doesn't equal 3

(a) Find exact value of fg(4)
(b) Find inverse function of f^-1 (x), stating its domain
---
Jan 08
1) Given that:
(2x^4 - 3x^2 + x + 1) / (x^2 - 1) = (ax^2 + bx + c) + (dx+e/x^2-1)
Find values of constants, a, b, c, d and e. Sorry this question is difficult to write out kinda...

2) A curve C has equation:
y= e^2x tan x , where x cannot equal (2n+1)pi/2

(a) Show the turning points on C occur where tan x = -1
(b) Find an equation of the tangent to C at point where x = 0

7) Curve C has equation:
y= 3sin2x + 4cos2x, -Pi < x < Pi The point A (0,4) lies on C.

(a) Find an equation of the normal to curve C at A.

---

Again, some of it was kinda hard to write out but I'd appreciate all the help I can get. :redface:

For JUn 7
1a) Use the logarithmic identities and that the logaritmic funtion is strictly monotone

lna+lnb=ln(ab)

1b) Use the

e^x=t

substitution and you will get a quadratic equation
3a
Find where is the derivative zero

y'=2xe^x+x^2e^x=xe^x(x+2)=0

3b)If Derivative is + the function is monotone increasing, and decreasing
if the derivative is negative. Where the derivative change the sing and is zero
there is a maximum or minimum.
5a)
THe composite funtion

f[g(x)]=ln(\frac{4}{x-3}-1)

Substitute 4 into x
5b)x=ln(2y-1) Solve to y

JAN8
1)

\displaystyle \frac{2x^4-3x^2+x+1}{x^2-1}=ax^2+bx+c+\frac{dx+e}{x^2-1}

At he left side divide the numerator by the denominator. It will be

2x^2-1+\frac{x}{x^2-1}

Comparing the two sides you will get the values of parameters
2a)
THE derivative of function is zero

2e^{2x}tanx+\frac{1}{cos^2x}=0

take out e^(2x) as factor and solve the equation
Note

\frac{1}{cos^2x}=1+tan^2x

2b)
find the coords of C (x0=0 -> y0)
Write down the derivative (y') and calculate y'(0)=m
the equation:

y-y_0=m(x-x_0)

7a)
Similarly as abobe but the equation

y-y_0=-\frac{1}{m}(x-x_0)

Reply 4

gasmask

June 07
1) Find the exact solutions to the equations:
(a) ln x + ln 3 = ln 6

ln(3x) = ln6
3x = 6

(b) e^x + 3e^-x = 4

e^x + 3/(e^x) = 4
(e^x)^2 + 3 = 4e^x
then substitute in y = e^x
y^2 - 4y + 3 = 0

3) A curve C has equation y = (x^2)(e^x)
(b) Find the turning points

dy/dx = (x^2)(e^x) + 2x(e^x) = 0
xe^x(x +2) = 0
x = 0 or -2, y = 0 or 4e^-2

(d) Determine the nature of each turning point of the curve C.

d^2y/dx^2 = (x^2)(e^x) + 2x(e^x) + 2x(e^x) + 2e^x
=e^x(x^2 + 4x +2)
sub in x = 0 and -2 to determine maximum or minimum

5) Functions f and g are defined by:
f:-> ln(2x-1), where x>0.5
g:-> 2/(x-3), where x doesn't equal 3

(a) Find exact value of fg(4)

g(4) = 2/1 = 2
f(g(4)) = f(2) = ln3

(b) Find inverse function of f^-1 (x), stating its domain

First, the range of f(x) is f(x) belongs to set of real numbers (the double R thing), so that's the domain of f^-1(x)
y = ln(2x - 1)
e^y = 2x - 1
2x = e^y + 1
x = 0.5(e^y + 1)
f^-1(x) = 0.5(e^x + 1)

---
Jan 08
1) Given that:
(2x^4 - 3x^2 + x + 1) / (x^2 - 1) = (ax^2 + bx + c) + (dx+e/x^2-1)
Find values of constants, a, b, c, d and e. Sorry this question is difficult to write out kinda...

Multiply out by (x^2 - 1), and equate coefficients.

2) A curve C has equation:
y= e^2x tan x , where x cannot equal (2n+1)pi/2

(a) Show the turning points on C occur where tan x = -1

dy/dx = -2e^2x = 0

but I don't think this can be right - as N -> 0, lnN -> ?

(b) Find an equation of the tangent to C at point where x = 0

7) Curve C has equation:
y= 3sin2x + 4cos2x, -Pi < x < Pi The point A (0,4) lies on C.

(a) Find an equation of the normal to curve C at A.

dy/dx = 6cos2x - 8sin2x = 6cos0 - 8sin0 = 6 - 0 = 6

therefore, gradient of the normal is -1/6 = y-4/x...
-x = 6y - 24
6y + x = 24

---

Pretty sure this is right, but it's been a while since I did any calculus :P

Reply 5

Wolfiexe

OP

2

Wow, thanks alot everyone. :biggrin:

I think this will help alot, I'll be able to give these a good try now. :smile:

lne=1

Original post by viksta1000

(a) ln x + ln 3 = ln 6

take ln3 from ln6 so you get: lnx = ln6-ln3 which is the same as ln(6/3) = ln2
so you have lnx = ln2 - so ln's cancel so you have x=2

(b) e^x + 3e^-x = 4

take logs of both sides so you have: xlne + -x ln3e = ln4 (remeber logx^n is the same as nlogx which is why the x's can come to the front

lne = 0 so it can cancel, so now you have

x - x(ln3e) = ln4 --> ln3e = ln3 + lne so again lne cancels

so now you have x-xln3 = ln4

take out factor of x
x(1-ln3) = ln 4

divide ln4 by (1-ln3) and you have x

remember that questions wants the 'exact value' so leave the answer with logs and e's

and btw, I only *think* this is how you do them, im just trying lol

lne=1

C3 Past Paper Questions

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