VCE Methods Question Thread!

[keltingmeith]

Hey Fellas, i've got a question to ask!

Find the values of a and b such that the graph of y=alog2(x+b) goes through the points (8,10) and (32,14)
Thanks

You'll need to sub in the two points and solve simultaneously. So, we first get the two equations:

10=alog2(8+b).........(1)
14=alog2(32+b).......(2)

Now, if we take (2)/(1), we get:

14/10=alog2(8+b)/alog2(32+b)

The two a's cancel out, so by removing all of the fractions, we get:
7log2(32+b)=5log2(8+b)

Now, try using log laws to solve for just b. From there, it should be easy to find a.

[Stewart98]

Thanks for the insight, however, how did you get it to equal:
                  7log2(32+b)=5log2(8+b)
shouldn't it be like this:
                 7/5= log2(32+b) / log2(28+b)
and also, if i were to simplify it using log laws, wouldn't the b's cancel out also? as division is subtraction ?
Thanks

[keltingmeith]

Thanks for the insight, however, how did you get it to equal:
                  7log2(32+b)=5log2(8+b)
shouldn't it be like this:
                 7/5= log2(32+b) / log2(28+b)

No...? Where'd you get the 28 from?

and also, if i were to simplify it using log laws, wouldn't the b's cancel out also? as division is subtraction ?
Thanks

Not quite - you've got the rule backwards, which is the following:

[eth-dog]

P and Q are the points of intersection of the line y/2 + x/3 = 1 with the x and y axes respectively. The gradient of QR is 1 and the point R has x-coordinate 2a, a > 0
a Find the y coordinate of R in terms of a.
b Find the value of a if the gradient of PR is −2

The answer is probably right in front of me but i can't seem to figure it out. Help would be much appreciated... thanks

[Stewart98]

Oh haha I see where i went wrong, thanks

[exit]

P and Q are the points of intersection of the line y/2 + x/3 = 1 with the x and y axes respectively. The gradient of QR is 1 and the point R has x-coordinate 2a, a > 0
a Find the y coordinate of R in terms of a.
b Find the value of a if the gradient of PR is −2

The answer is probably right in front of me but i can't seem to figure it out. Help would be much appreciated... thanks

a)
y/2+x/3=1 is y=-2x/3+1/2

Q is the y intercept so x=0, sub that into the above equation and you get 1/2 for the y-value so the y-int is (0,1/2) which is on the line QR

QR has a gradient of 1 so if you take into consideration that the equation is y=x+(1/2)

R is x=2a, sub that into the equation for your y-coordinate which is 2a+(1/2)

b) P's y-coordinate is 0 since it intersects with x, so it is (x,0) and sub that into original equation y=-2x/3+1/2

0=-2x/3+1/2, 1/2=2x/3, 3/2=2x, x=3/4 so P is (3/4,0)

R's coordinate is (2a,2a+(1/2))

They are both on the line y=-2x+c as gradient for PR is -2

sub P in for c, 0=-2(3/4)+c, c is 3/2

equation is  y=-2x+3/2

sub R, 2a+1/2=-2(2a)+3/2

equating coefficients, 6a=1

a=1/6

First helping post in this thread so correct me if there's a problem.

[StupidProdigy]

Which of the following functions is strictly increasing on the interval (-infinity, -1]?
A) f(x) = x^2
B) f(x) = x^4
C) f(x) = x^(1/5)
D) f(x) = root(4 - x)
E) f(x) = -x^3

Book says E, but I'm confident it's C

P.S: Why is x^2 strictly increasing for (0,infinity) and not [0,infinity)?

C is correct. strictly increasing excludes 'flat' lines (0 gradient). x^2 has 'flatness', i.e 0 gradient as x=0

[natdogg]

Can someone explain to me why x^2 = 16 gives two solutions,
whereas x = 16^1/2 only gives the positive?

[qazser]

Can someone explain to me why x^2 = 16 gives two solutions,
whereas x = 16^1/2 only gives the positive?

x^2=16

provides +-(sqr root[16])

whereas 16^1/2 is sqr root 16 where a sqr root which must be greater or equal to 0. Note:Not possible(w/o using complex numbers) to sqr root negative numbers

[byCrypt]

Thanks.
Can someone explain how they got the answer? (Answer is C)

To me it looks like the answer is B. Testing different values, B makes more sense than C.

eg. From inspection,

• g(0)=-1 -> f(-1)=-1, hence f(g(0)) should be at -1
• g(1)=0 -> f(0)=-2, hence f(g(1)) should be at -2
• g(2)=1 -> f(1)=-1, hence f(g(2)) should be at -1

The above all seem to match option B.

[eth-dog]

a)
y/2+x/3=1 is y=-2x/3+1/2

Q is the y intercept so x=0, sub that into the above equation and you get 1/2 for the y-value so the y-int is (0,1/2) which is on the line QR

QR has a gradient of 1 so if you take into consideration that the equation is y=x+(1/2)

R is x=2a, sub that into the equation for your y-coordinate which is 2a+(1/2)

b) P's y-coordinate is 0 since it intersects with x, so it is (x,0) and sub that into original equation y=-2x/3+1/2

0=-2x/3+1/2, 1/2=2x/3, 3/2=2x, x=3/4 so P is (3/4,0)

R's coordinate is (2a,2a+(1/2))

They are both on the line y=-2x+c as gradient for PR is -2

sub P in for c, 0=-2(3/4)+c, c is 3/2

equation is  y=-2x+3/2

sub R, 2a+1/2=-2(2a)+3/2

equating coefficients, 6a=1

a=1/6

First helping post in this thread so correct me if there's a problem.

Question a was correct but you made a few errors with question b.
When we calculate the value of a for the line PR, we already know the coordinate of P (3,0) and we know the coordinate of R (2a,a+2). With this information, we can find the gradient and solve for a. We have also been given the gradient of PR which is -2.
(a+2-0)/(2a-3)=-2
a+2/2a-3=-2
a+2=-2*(2a-3)
a+2=-4a+6
rearrange and collect like terms
a+4a=-2+6
5a=4
a=4/5

[Stewart98]

Hey guys, i've got a question... I'll show you what i've worked out so far;
Polonium-210 is a radioactive substance. The decay of polonium-210 is described by the formula M=Mo * e^-k * t , where m is the mass in grams, left after t days, and Mo and k are constants. At t = 0, M = 10g and at t = 140, M = 5g.
a) Find the values of Mo and k.
I understand that at t=140 is the half-life however, i worked it out like this:
Mo = 10g          k = ?
5=10 * e^-k*140
5/10 = e^-k*140
ln(1/2) = ln(e^-k*140)
ln(2^-1) = -k * 140
-k = ln 2^-1/140
however it is not correct. I've probably made a fundamental error which i'm blind to see haha, thanks for the help in advance. 

[keltingmeith]

Hey guys, i've got a question... I'll show you what i've worked out so far;
Polonium-210 is a radioactive substance. The decay of polonium-210 is described by the formula M=Mo * e^-k * t , where m is the mass in grams, left after t days, and Mo and k are constants. At t = 0, M = 10g and at t = 140, M = 5g.
a) Find the values of Mo and k.
I understand that at t=140 is the half-life however, i worked it out like this:
Mo = 10g          k = ?
5=10 * e^-k*140
5/10 = e^-k*140
ln(1/2) = ln(e^-k*140)
ln(2^-1) = -k * 140
-k = ln 2^-1/140
however it is not correct. I've probably made a fundamental error which i'm blind to see haha, thanks for the help in advance. 

Looks fine to me. What are they claiming the answer is?

[Stewart98]

They're claiming the answer to be k = 4.95 *10^-3

[Maz]

hey
can someone please help me with this...
how do you determine if the point of inflection is vertical or horizontal- after using calculus to find stationary points and
nature of the graph?

thanks in advance