# Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

#### pikachu975

##### I love trials
Moderator

Q 7 (b) EX 9A

I just wondering has there been a mistake in this answer of the 3U Year 11 textbook?

$7 (b) A point P(x,y) moves so that the sum of the squares of its distances from the points A,B and C is 77. Show that the locus is a circle and find its centre and radius.$

$PA^2 = x^2 + (y - 4)^2$
$PB^2 = (x - 1)^2 + (y + 4)^2 and$
$PC^2 = (x - 6)^2 + (y - 3)^2$

ii) ==> PA² + PB² + PC² = 3x² + 3y² - 14x - 6y + 78 = 77

==> 3x² + 3y² - 14x - 6y = -1

==> x² + y² - 14x/3 - 2y = -1/3

Grouping, (x² - 14x/3) + (y² - 2y) = -1/3

==> (x² - 14x/3 + 49/9) + (y² - 2y + 1) = -1/3 + 49/9 + 1

==> (x - 7/3)² + (y - 1)² = 55/9

$This is the form of circle equation (x - h)^2 + (y - k)^2 = r^2,$
$whose center is (h,k) and radius = r$

$Hence the above represents a circle equation,$
$whose center is (\frac{7}{3}, 1) and radius = \frac{(\sqrt{55})}{3} units$

The back of the book has the answer as (2,1) for centre and radius of $\sqrt{5}$
You didn't give us point A B and C

#### jathu123

##### Active Member

Q 7 (b) EX 9A

I just wondering has there been a mistake in this answer of the 3U Year 11 textbook?

$7 (b) A point P(x,y) moves so that the sum of the squares of its distances from the points A,B and C is 77. Show that the locus is a circle and find its centre and radius.$

$PA^2 = x^2 + (y - 4)^2$
$PB^2 = (x - 1)^2 + (y + 4)^2 and$
$PC^2 = (x - 6)^2 + (y - 3)^2$

ii) ==> PA² + PB² + PC² = 3x² + 3y² - 14x - 6y + 78 = 77

==> 3x² + 3y² - 14x - 6y = -1

==> x² + y² - 14x/3 - 2y = -1/3

Grouping, (x² - 14x/3) + (y² - 2y) = -1/3

==> (x² - 14x/3 + 49/9) + (y² - 2y + 1) = -1/3 + 49/9 + 1

==> (x - 7/3)² + (y - 1)² = 55/9

$This is the form of circle equation (x - h)^2 + (y - k)^2 = r^2,$
$whose center is (h,k) and radius = r$

$Hence the above represents a circle equation,$
$whose center is (\frac{7}{3}, 1) and radius = \frac{(\sqrt{55})}{3} units$

The back of the book has the answer as (2,1) for centre and radius of $\sqrt{5}$
You made a small silly mistake, the point B has coordinates (0,-4). So PB^2 would be x^2+(y+4)^2

#### davidgoes4wce

##### Well-Known Member

You didn't give us point A B and C
That was the question: (apologies about the cut off of page)

#### davidgoes4wce

##### Well-Known Member

OK I decided to redo Q 7 a and 7b from EX 9A

$Q 7 (a) EX 9A$

$d^2_{PA}=x^2+(y-4)^2$

$d^2_{PB}=x^2+(y+4)^2$

$d^2_{PC}=(x-6)^2+(y-3)^2$

$Q 7 (b) EX 9A$

$PA^2+PB^2+PC^2=77$

$3x^2+3y^2-12x-6y+77=77$

$3x^2+3y^2-12x-6y=0$

$3(x^2+y^2-4x-2y)=0$

$x^2+y^2-4x-2y=0$

$(x-2)^2+(y-1)^2-\sqrt{5}=0$

$(x-2)^2+(y-1)^2=(\sqrt{5})^2$

$(x-2)^2+(y-1)^2=5$

$\therefore the locus is a circle , with centre (2,1) and radius 5.$

#### pikachu975

##### I love trials
Moderator

That was the question: (apologies about the cut off of page)

#### trea99

##### Member

Hi, need help with Exercise 8H Page 311 Extension Question 27.
If alpha and beta are the roots of x^2 = 5x - 8. Find (a)^1/3 + (b)^1/3 without finding the roots.
Tried cubing (a^1/3 + b^1/3) but got stuck.
Any help would be appreciated.

Also, with Extension question 26 (same page),
If the equations mx^2 + 2x +1 =0 and x^2 + 2x +m = 0 have a common root, find the possible values of m and the value of the common root in each case. I let first equation be p(x) and the second be f(x) and the common root be alpha.
If alpha is the common root p(a) = f(a) = 0 and hence you get,
a^2(m-1) + (1-m) = 0, and that the common root a is x=-1 when m = 1
I had no idea how to get the second common root of the two equations (its x = 1 when m = -3),
So I used product of roots of the above equation a^2(m-1) + (1-m) = 0 as a quadratic is a, and the product of roots is thus -1
Hence I found the other root to be x = -1 and subbing x = -1 into any of the given functions p(x), f(x) and letting it equal 0 I then got m = -3.
Is there a way to find the 2nd m value prior to finding the 2nd common root?
(because this is what the question seems to imply find m then find the common root)
Sorry for long post couldn't find anything on either of the questions
Thanks!

#### Drongoski

##### Well-Known Member

Hi, need help with Exercise 8H Page 311 Extension Question 27.
If alpha and beta are the roots of x^2 = 5x - 8. Find (a)^1/3 + (b)^1/3 without finding the roots.
Tried cubing (a^1/3 + b^1/3) but got stuck.
Any help would be appreciated.

Ok - to save time I will use a and b for alpha and beta.

$\alpha + \beta = -(-5) = 5 and \alpha \beta = 8 \\ \\a + b = 5 and ab = 8\\ \\ (a^ \frac {1}{3} + b^\frac {1}{3})^3 = M^3 = a + b + 3a^\frac {2}{3} b^\frac {1}{3} + 3 a^\frac {1}{3} b^\frac {2}{3} = a + b + 3a^\frac{1}{3} b^\frac {1}{3} (a^\frac {1}{3} + b^\frac {1}{3}) \\ \\ \therefore M^3 = 5 + 3 \times 8^\frac {1}{3}(M) = 5 + 6M \\ \\ \therefore m^3 - 6M - 5 = (M+1)(M^2 - M - 5) = 0 \\ \\ \therefore M = -1 or \frac {1 \pm \sqrt {21}}{2}$

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#### trea99

##### Member

Yes, I got that part lol.
But what would you do to get a^(1/3) + b^(1/3) from a+b = 5 and ab = 8
What I tried was,
(a^(1/3) + b^(1/3))^3 and got a + 3a^(2/3)b^(1/3) + 3a^(1/3)b^(1/3) + b
Tried doing something with the 3a^(2/3)b^(1/3) + 3a^(1/3)b^(2/3) but ended up getting more a^(1/3) + b^(1/3) terms when I tried factorising by taking out powers of ab
Or would cubing a^(1/3) + b^(1/3) be useless for this question?

#### trea99

##### Member

Thank you so much

#### Drongoski

##### Well-Known Member

Thank you so much
You were so close to finishing. Never done this type of question before.

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#### davidgoes4wce

##### Well-Known Member

EX7D Q 9 (b) is there a mistake with the answer?

$Q 9 (a) Find where y=-2x meets y=(x+2)(x-3)$

$(b) Find, to the nearest minute, the angles that the tangents to y=(x+2)(x-3) at these points make with the x-axis. Sketch this situation.$

$Answers for Q9 (b) 71^\circ 34' \ and \ 97^\circ 8'$

#### davidgoes4wce

##### Well-Known Member

One of the angles I believe to be right (the 71 34') and the other I believe to be wrong. Would love it , if somebody could confirm.

Seeing that the two points of intersection are at x=2, -3

I then substituted this back into the derivative which is dy/dx=2x-1=tan $\theta$

#### Drongoski

##### Well-Known Member

The 2 angles in degrees: 71.565 and 98.13

being: inverse tan (3) and 180 - inverse tan (7)

Maybe my calculator is exceptionally accurate.

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#### davidgoes4wce

##### Well-Known Member

I had an attempt at the question and I can't quite get it

$EX7D Q9 (a)$

$-2x=x^2-x-6$

$0=x^2+x-6$

$0=(x+3)(x-2)$

$x=-3,2$

$My thinking for part 9 (b) was just to work out the derivative of y=x^2-x-6 \implies \frac{dy}{dx}=2x-1$

$At x=-3 , \frac{dy}{dx}=2(-3)-1=-7 = tan \theta$

$At x=2, \frac{dy}{dx}=2(2)-1=3 = tan \theta$

$Solving for \theta = 71^\circ 34' and I now have the understanding that we find the other \theta by going = 180^\circ -81^\circ 52'= 98^\circ 8'$

#### davidgoes4wce

##### Well-Known Member

My reasoning as to why I subtract 180 from 81'52' is that the tan theta value is a value of negative 7, which lies somewhere in the 2nd quadrant of the ASTC diagram.

#### lypoon

##### New Member

Anyone know how to do 12B Question 16? thanks.

#### integral95

##### Well-Known Member

Anyone know how to do 12B Question 16? thanks.
Post it up, not everyone has the book.

#### Mahan1

##### Member

Anyone know how to do 12B Question 16? thanks.
I believe that is the question you are referring to:

a)
$ln(x^{2} + 5x) = 2ln(x+1) = ln((x+1)^{2})$
$x^{2} + 5x = (x+1)^{2} = x^{2} + 2x+1$
$3x = 1$ implies
$x = \frac{1}{3}$

b)
$log(7x-12) = 2 log(x) = log(x^{2})$
$x^{2} = 7x -12$ implies
$x^{2} - 7x +12 = (x-3)(x-4) = 0$
therefore: $x = 3 , 4$

#### studyingg

##### New Member

Exercise 6J q11b

Every two hours, half of a particular medical isotope decays. If there was originally 20g, how much remains after a day (to two significant figures)?

Thanks!

#### integral95

##### Well-Known Member
$20*(\frac{1}{2})^{12}$