# Thread: Cambridge Prelim MX1 Textbook Marathon/Q&A

1. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
could you explain how you used the product rule? I don't quite seem to get that. Thanks.
$By the product rule, if we have two functions of x, say u\equiv u(x) and v\equiv v(x), then$

$\frac{\mathrm{d}}{\mathrm{d}x}(uv)=\frac{\mathrm {d}u}{\mathrm{d}x}v+u\frac{\mathrm{d}v}{\mathrm{d} x}.$

$So finding \frac{\mathrm{d}}{\mathrm{d}x}\left(y \frac{\mathrm{d}y}{\mathrm{d}x} \right) just requires us to apply the product rule using u(x)=y and v(x)=\frac{\mathrm{d} y}{\mathrm{d}x}, and hence use \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}y }{\mathrm{d}x} (since u=y) and similarly \frac{\mathrm{d}v}{\mathrm{d}x}=\frac{\mathrm{d}} {\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x} \right) (which is just \frac{\mathrm{d}^2 y}{\mathrm{d}x^2}) in the product rule formula.$

2. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Also, do you know how to answer question 13 from 10D. Thank you.

3. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Also, do you know how to answer question 13 from 10D. Thank you.
What is the question?

4. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Sorry, the question is:

Find positive integers a and b such that

x^2y'' + 2xy' = 12y, where y = x^a + x^-b.

Thank you.

5. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Sorry, the question is:

Find positive integers a and b such that

x^2y'' + 2xy' = 12y, where y = x^a + x^-b.

Thank you.
$y=x^a + x^{-b}, a,b positive integers.$

$So y^\prime = ax^{a-1} -bx^{-b-1}=ax^{a-1} -bx^{-(b+1)} and y''=a(a-1)x^{a-2}+b(b+1)x^{-b-2}=a(a-1)x^{a-2}+b(b+1)x^{-(b+2)}.$

$Since we want x^2y'' + 2xy' = 12y, we will substitute the above expressions into this equation and try and solve for a and b. Upon substitution into this equation,$

$x^2 \left(a(a-1)x^{a-2}+b(b+1)x^{-(b+2)}\right)+2x\left(ax^{a-1} -bx^{-(b+1)}\right)=12\left(x^a + x^{-b}\right).$

$Expanding out and using index laws etc. yields$

$a(a-1)x^a + b(b+1)x^{-b}+2ax^a - 2b x^{-b}=12x^a + 12x^{-b}.$

$Collecting like terms on the left-hand side yields$

$\left( a(a-1)+2a\right)x^a + \left(b(b+1)-2b\right)x^{-b}=12x^a + 12x^{-b}.$

$Equating coefficients of like powers of x on the left- and right-hand sides, we can see we require the following:$

$a(a-1)+2a=12 (A)$

$and$

$b(b+1)-2b=12 (B).$

$Equations (A) and (B) are both quadratic equations that I am sure you can now easily solve to find the suitable a and b, making sure to take the \emph{positive} integer solutions (you should obtain a=3 and b=4).$

6. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thanks for the help with that question. Could you also explain Q 11 b)

Find the nth and (n+1)th derivative of x^n.

I understand that the nth derivative is n(n-1)(n-2) ... x^n-n = n(n-1)(n-2)...1

but am unsure why the (n+1)th derivative is 0. I assume it has something to do with x^n-(n+1). But that equals x^-1.

Thank you a lot for you help.

Cheers.

7. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Thanks for the help with that question. Could you also explain Q 11 b)

Find the nth and (n+1)th derivative of x^n.

I understand that the nth derivative is n(n-1)(n-2) ... x^n-n = n(n-1)(n-2)...1

but am unsure why the (n+1)th derivative is 0. I assume it has something to do with x^n-(n+1). But that equals x^-1.

Thank you a lot for you help.

Cheers.
$The n^{\text{th}} derivative you found there (n!) is a constant (i.e. does not depend on x), so its derivative (which is the (n+1)^{\text{th}} derivative of x^n) is 0, since the derivative of a constant is 0.$

8. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thanks. Also for 12c)

Find the y' and y'' in terms of t:

c) x = 1 - 5t , y =t^3.

y' = -3/5 t^2

then isn't y'' = -6/5t

the answer in the book say y'' = 6/25t

Thanks.

9. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Thanks. Also for 12c)

Find the y' and y'' in terms of t:

c) x = 1 - 5t , y =t^3.

y' = -3/5 t^2

then isn't y'' = -6/5t

the answer in the book say y'' = 6/25t

Thanks.
$By the primes (\prime), I think they mean derivatives with respect to x, rather than t.$

10. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

i still don't understand. I get the same problem in question e)

x = (t -2)^2 , y = 3t

y' = 3/2 ( t -2)^-1

and i get

y'' = -3/2 (t-2)^-2

but the answer for y'' is -3/4 (t-2)^-3

Cheers.

11. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Thanks. Also for 12c)

Find the y' and y'' in terms of t:

c) x = 1 - 5t , y =t^3.

y' = -3/5 t^2

then isn't y'' = -6/5t

the answer in the book say y'' = 6/25t

Thanks.
$So basically, y^\prime \equiv \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y} {\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x} (by the chain rule).$

$But \frac{\mathrm{d}y}{\mathrm{d}t}=3t^2, since y=t^3. Also, \frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\frac{ \mathrm{d}t}{ \mathrm{d}x}}=\frac{1}{-5}=-\frac{1}{5}.$

$Hence y^\prime = 3t^2\times \left(-\frac{1}{5} \right)=-\frac{3}{5}t^2.$

$Now, y''=\frac{\mathrm{d}y^\prime}{\mathrm{d}x}=\frac{ \mathrm{d}y^\prime}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}.$

$But \frac{\mathrm{d}y^\prime}{\mathrm{d}t}=-\frac{6}{5}t, since y^\prime =-\frac{3}{5}t^2. Also, \frac{\mathrm{d}t}{\mathrm{d}x}=-\frac{1}{5} from before.$

$Hence y''=-\frac{6}{5}t\times \left(-\frac{1}{5} \right)=\frac{6}{25}t.$

12. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
i still don't understand. I get the same problem in question e)

x = (t -2)^2 , y = 3t

y' = 3/2 ( t -2)^-1

and i get

y'' = -3/2 (t-2)^-2

but the answer for y'' is -3/4 (t-2)^-3

Cheers.
Try a similar method to the above Q. The key thing to use is that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{ \frac{ \mathrm{d}y}{ \mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}.$

13. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thank you for your help with that question.

For Q5) of 10E

Find the range of value of x for which the curve y = 2x^3 -3x^2 -12x + 8 is:

a) increasing, b)decreasing, c) concave up, d)concave down.

For c) and d), i know you find the y'' and then for c) make it > 0 and d) make it < 0. But am unsure about how to work out increasing and decreasing.

Thank you. Cheers.

14. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Thank you for your help with that question.

For Q5) of 10E

Find the range of value of x for which the curve y = 2x^3 -3x^2 -12x + 8 is:

a) increasing, b)decreasing, c) concave up, d)concave down.

For c) and d), i know you find the y'' and then for c) make it > 0 and d) make it < 0. But am unsure about how to work out increasing and decreasing.

Thank you. Cheers.
$The function is increasing when y^\prime >0 and decreasing when y^\prime < 0.$

15. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For Q7)

Sketch possible graphs of continuous functions with these properties:

a) f(-5) = f(0) = f(5) = 0 , and f'(3) = f'(-3) = 0, and f''(x) > 0 for x < 0, and f''(x) <0 for x > 0

What I don't get is the first bit, is it saying that the x cord. = -5, 0, 5 has y cord = 0 ?

whereas the second bit states that at the point 3 and -3 there is a stationary point.

Cheers.

16. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
For Q7)

Sketch possible graphs of continuous functions with these properties:

a) f(-5) = f(0) = f(5) = 0 , and f'(3) = f'(-3) = 0, and f''(x) > 0 for x < 0, and f''(x) <0 for x > 0

What I don't get is the first bit, is it saying that the x cord. = -5, 0, 5 has y cord = 0 ?

whereas the second bit states that at the point 3 and -3 there is a stationary point.

Cheers.
$Yes, what's wrong with that?$

17. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
For Q7)

Sketch possible graphs of continuous functions with these properties:

a) f(-5) = f(0) = f(5) = 0 , and f'(3) = f'(-3) = 0, and f''(x) > 0 for x < 0, and f''(x) <0 for x > 0

What I don't get is the first bit, is it saying that the x cord. = -5, 0, 5 has y cord = 0 ?

whereas the second bit states that at the point 3 and -3 there is a stationary point.

Cheers.

Yes it does mean that the coordinates are (-5,0), (5,0) and (0,0) < --- The x-intercepts

There are stationary points at x = 3 and x = -3 but since you have been given insufficient information, you can plot these stationary points with any y-coordinate (But the x-coordinates have to stay the same as mentioned above) and it should be enough to get you the mark correct (Make it reasonable though by placing it within the scope of your axis - don't unnecessarily place the y-coordinate at like 1000) . Furthermore, you must be able to plot it with the correct concavity (i.e. Concave up or down).

Also

f''(x) > 0 means that it is a concave up for x < 0

f''(x) < 0 means that it is concave down for x > 0

^ This should give you an idea of the concavity of the stationary points - whether it is minimum or maximum

Therefore:

X-Intercepts:

x = -5

x = 0

x = 5

Minimum Turning Point: x = 3 (Any y-coordinate)

Maximum Turning Point: x = -3 (Any y-coordinate)

You should be able to draw it now, so good luck

18. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Does anyone know how to answer:

Q12 from 10C:

a) Differentiate y = x^1/2 - x^3/2

b) find those values of x for which y' = 0, and hence determine the coordinates of any critical points.

c) Hence sketch a graph of y^2 = x(1-x)^2.

Cheers.

19. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Does anyone know how to answer:

Q12 from 10C:

a) Differentiate y = x^1/2 - x^3/2

b) find those values of x for which y' = 0, and hence determine the coordinates of any critical points.

c) Hence sketch a graph of y^2 = x(1-x)^2.

Cheers.
a) Standard differentiation of functions $y'(x)=\frac{1}{2}x^{-\frac{1}{2}}-\frac{3}{2}x^{\frac{1}{2}}$
b) Set y'(x)=0 and solve for x, will give you critical points for y
c) Since you know the critical points of y, sketch y=f(x) and square the points to give you $y^2$

20. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I am having difficulties doing question 5g from 10F

Without finding inflexions ,sketch the graphs of the function: Indiciate any asymptotes, stationary poitns and intercepts with the axes.

g) y = x^2 + 1/x^2

I don't know how to find the oblique parabola asymptote. Also for some reason I am getting one stationary point instead of the required two.

Cheers.

21. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Put it under a common denominator

y = (x^4 + 1) / x^2

Like you would find the vertical asymptote, take the limit of x approaches infinity of the above equation

In these situations when there's a pronumeral in the denominator you should be expecting an oblique asymptote rather than a straight line

Then divide the numerator and denominator by the HIGHEST POWER OF X IN THE DENOMINATOR

lim
x --> Infinity [(x^4 / x^2) + (1 / x^2)] / (x^2 / x^2)

Simplify this

lim
x --> Infinity [(x^2) + (1 / x^2)] / (1)

Now if you use the limit you'll notice that as x approaches infinity the 1 / x^2 will equal to zero (Sub a huge number into x for this and you will notice it)

Therefore

y = [(x^2) + (0)] / (1)

y = x^2 / 1

y = x^2

22. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Also don't forget the asymptote at x = 0 !!!

As for the stationary points

Stationary points when dy/dx = 0

y' = 2x - (2 / x^3)

0 = 2x - (2 / x^3)

Multiply both sides by x^3

0 = 2x^4 - 2

Factorise

0 = 2(x^4 - 1)

0 = 2(x^2 - 1)(x^2 + 1)

0 = 2(x - 1)(x + 1)(x^2 + 1)

Therefore

x = 1 or x = -1

Continue from here ~

23. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Firstly, notice that the function f, defined as $f(x)=x^2+\frac{1}{x^2}$ is positive for all $x\neq{0}$. Also, note that f(-x) = f(x) which means that f is an even function, so the graph of f will be reflected across the y-axis. You could set the first derivative $f'(x)=2x-\frac{2}{x^3}$ equal to 0, so that $x=\pm{1}$ will give you the stationary points, namely at $(\pm{1}, 2)$. Then sketch the graph accordingly. Also note that as x tends to infinity, your y values will start to tend to $x^2$ and as x tends to 0, your y values will tend to $\frac{1}{x^2}$ so these will essentially become your oblique asymptotes for the graph of f.

24. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Crisium
Put it under a common denominator

y = (x^4 + 1) / x^2

Like you would find the vertical asymptote, take the limit of x approaches infinity of the above equation

In these situations when there's a pronumeral in the denominator you should be expecting an oblique asymptote rather than a straight line

Then divide the numerator and denominator by the HIGHEST POWER OF X IN THE DENOMINATOR

lim
x --> Infinity [(x^4 / x^2) + (1 / x^2)] / (x^2 / x^2)

Simplify this

lim
x --> Infinity [(x^2) + (1 / x^2)] / (1)

Now if you use the limit you'll notice that as x approaches infinity the 1 / x^2 will equal to zero (Sub a huge number into x for this and you will notice it)

Therefore

y = [(x^2) + (0)] / (1)

y = x^2 / 1

y = x^2
Can't you just let x->infinity, making the y values ->x^2, instead of doing all the algebra above

25. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by DatAtarLyfe
Can't you just let x->infinity, making the y values ->x^2, instead of doing all the algebra above
Yes.

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