Mathematics Question Thread

[Dragomistress]

Hello!
Many thanks.

[RuiAce]

Hello!
Many thanks.

[Mate2425]

Hi could i please get some help with this question i don't know how to get the five in the final answer.

Find the Exact area enclosed between the curve y =e^2x and the lines y=1 and x=2

Answer: 0.5 (e^4 -5) units squared

Thank you 

[Natasha.97]

Hi could i please get some help with this question i don't know how to get the five in the final answer.

Find the Exact area enclosed between the curve y =e^2x and the lines y=1 and x=2

Answer: 0.5 (e^4 -5) units squared

Thank you 

Hi!





Hope this helps

[mxrylyn]

Hello.

On stationary point. On this worksheet, I am fine doing Q1 I, II, III, and V.

but for Q1 IV , i tried to differentiate Y and ended up with f'(x) = 3x^2 + 6x. and i equated that to 0, but i have no idea what to do with it next.

On Q1 VI, I ended up with X=12 and x=-6, which I know is wrong because the values are in none of the possible answers.

What should I do next for Q1 IV, and how do i know where i am going wrong in Q1 VI?

[Opengangs]

Hello.

On stationary point. On this worksheet, I am fine doing Q1 I, II, III, and V.

but for Q1 IV , i tried to differentiate Y and ended up with f'(x) = 3x^2 + 6x. and i equated that to 0, but i have no idea what to do with it next.

On Q1 VI, I ended up with X=12 and x=-6, which I know is wrong because the values are in none of the possible answers.

What should I do next for Q1 IV, and how do i know where i am going wrong in Q1 VI?

Q1 (iv)


Q1 (vi)


Rinse and repeat from above.

[mxrylyn]

Q1 (iv)


Q1 (vi)


Rinse and repeat from above.

THANK YOU!

It all makes sense now

[mxrylyn]

Hey, not quite sure what method to use to do Q2, I tried to use substitution, but I confused myself and kept going in circles.

[Opengangs]

Hey, not quite sure what method to use to do Q2, I tried to use substitution, but I confused myself and kept going in circles.

Hello, mxrylyn.

If a curve has a stationary point, as in Q1, we need to find when the tangent line is equal to zero. This simply means that we set the derivative to zero.
So, at x = 1, the derivative is equal to zero.



We also know that the cubic curve passes through the point (1, -4). We use these two equations to form two equations which can then be solved simultaneously.


Thus, a is 4 and b is -12.

See if you can do (ii) based on (i).
If you have any more troubles, feel free to reply on this thread.

[Dragomistress]

How do I do f?

[RuiAce]

How do I do f?

\begin{align*}\log_7 14 &= \log_7 2 + \log_7 7\\ &= 0.36 + 1\\ &= 1.36\end{align*}

[swordkillz]

Hey Guys, can someone please help me with this question?

A box of rectangular cross-section sits on a train luggage rack as shown with the point C touching the wall. P, the point in contact with the rack, is the midpoint of AD. If D is 8 cm from the wall and P, the edge of the rack, is 20 cm from the wall, find how far A is from the wall.
Hint: Draw parallels to the wall through A, P and D.

Isn't that a question from our 2017 trial paper?

[RuiAce]

Isn't that a question from our 2017 trial paper?

P.S. It is possible, given the date it was posted

[Dragomistress]

How do I do a?

[brightsky]

Let r be the radius of the circle. By the cosine rule, 8^2 = r^2 + r^2 - 2*r*r*cos(45 degrees). Solving this gives r = 8/sqrt(2-sqrt(2)).

Not too sure what exactly is meant by 'minor segment cut off by the angle' in part b.

If they are asking for area of minor segment cut off by the chord (i.e. the region bounded by the chord and the minor arc intercepted by the chord), then all you need to do is figure out the area of the minor sector and subtract the area of the triangle:

Area of Minor Segment = 45/360*pi*[8/sqrt(2-sqrt(2))]^2 - 1/2*[8/sqrt(2-sqrt(2))][8/sqrt(2-sqrt(2))] sin(45 degrees) = 4.3 units^2 (to 1 decimal place)

If they are asking for area of minor sector cut off by the angle (i.e. the region bounded by the two radii and the minor arc), then all you need to do is find the area of the minor sector by taking the appropriate fraction of entire circle:

Area of Minor Sector = 45/360*pi*[8/sqrt(2-sqrt(2))]^2 = 42.9 units^2 (to 1 decimal place)