Author Topic: Mathematics Question Thread (Read 1625954 times) Tweet Share

RuiAce · « **Reply #3225 on:** January 22, 2018, 02:59:31 pm »

Quote from: gumscape on January 22, 2018, 02:47:47 pm

hellooooooo! can anyone give me a general guideline/tips & tricks when it comes to sketching the primitive function? I'm having a hard time with it

Think about how you sketched the derivative first. You converted the gradient of the tangent, into the y-coordinate.

For the primitive, you're doing the exact opposite. So if the curve is above the x-axis, then its primitive will be increasing. If the curve is below the x-axis, then its primitive will be decreasing. (And of course if you have an x-intercept then the primitive has a stationary point).

Remember that in general, a primitive function can be as high or as low as you want it to be (thanks to the +C you get from integrating). So unless they specify something about a point that the primitive passes through, you have flexibility over this.

BridgitteH · « **Reply #3226 on:** January 23, 2018, 06:08:50 pm »

Hi! Could someone please help with this question? I haven't touched maths since the end of last term and forgot how to do it (whoops!)

"The sum of the first 5 terms of an arithmetic series is 100 and the sum of the first 10 terms is 320. Find the first term and the common difference."

RuiAce · « **Reply #3227 on:** January 23, 2018, 06:14:36 pm »

Quote from: BridgitteH on January 23, 2018, 06:08:50 pm

Hi! Could someone please help with this question? I haven't touched maths since the end of last term and forgot how to do it (whoops!)

"The sum of the first 5 terms of an arithmetic series is 100 and the sum of the first 10 terms is 320. Find the first term and the common difference."

$\text{Let the first term and the common difference be}\\ a\text{ and }d\text{ in the usual sense.}$
$\text{Recalling that }S_n = \frac{n}{2} \left[2a + (n-1)d \right]\text{ we are given that}\\ \begin{align*}100&= \frac{5}{2} (2a + 4d)\\ 40&=2a+4d\tag{1}\\ 320 &= \frac{10}{2}(2a + 9d)\\ 64&=2a+9d\tag{2}\end{align*}$
$\text{This is now a pair of simultaneous equations.}\\ (2)-(1)\text{ gives }5d = 24\implies d=\frac{24}{5}.$
The rest should be easy

BridgitteH · « **Reply #3228 on:** January 23, 2018, 06:23:26 pm »

Quote from: RuiAce on January 23, 2018, 06:14:36 pm

$\text{Let the first term and the common difference be}\\ a\text{ and }d\text{ in the usual sense.}$
$\text{Recalling that }S_n = \frac{n}{2} \left[2a + (n-1)d \right]\text{ we are given that}\\ \begin{align*}100&= \frac{5}{2} (2a + 4d)\\ 40&=2a+4d\tag{1}\\ 320 &= \frac{10}{2}(2a + 9d)\\ 64&=2a+9d\tag{2}\end{align*}$
$\text{This is now a pair of simultaneous equations.}\\ (2)-(1)\text{ gives }5d = 24\implies d=\frac{24}{5}.$
The rest should be easy

Thank you! Now I get it

))

angela23 · « **Reply #3229 on:** January 24, 2018, 10:11:32 am »

Hi! Could someone help me with this sequences and series question:

For which angles in the domain 0 ≤ x ≤ 2π does the series 1 + cos²x + cos⁴x +... not converge?

Thanks heaps!

RuiAce · « **Reply #3230 on:** January 24, 2018, 10:14:00 am »

Quote from: angela23 on January 24, 2018, 10:11:32 am

Hi! Could someone help me with this sequences and series question:

For which angles in the domain 0 ≤ x ≤ 2π does the series 1 + cos²x + cos⁴x +... not converge?

Thanks heaps!

$\text{The series is an infinite geometric series}\\ \text{with first term }a=1\text{ and common ratio }r=\cos^2 x$
$\text{The series does }\textbf{not}\text{ converge when }|r| \ge 1.\\ \text{Hence, we are considering }|\cos^2 x| \ge 1\\ \text{for }0\le x \le 2\pi.$
$\text{Since }|x^2| = |x|^2\text{, we then have }|\cos x|^2\ge 1.\\ \text{Noting that absolute values are never negative, we then have }|\cos x| \ge 1$
Edit: Mistake halfway through
____________________________________________________
$\text{Of course, }|\cos x| > 1\text{ never happens.}$
This is essentially because the function $ f(x) = \cos x$ has range $-1 \le y \le 1$, which is the same thing as $ y\le 1$.
$\text{So the equation reduces down to }|\cos x| = 1\\ \text{i.e. }\cos x = \pm 1.\\ \text{From a quick sketch, the solutions are }x=0,\pi,2\pi.$

kauac · « **Reply #3231 on:** January 24, 2018, 10:52:11 am »

Hi...
My holidays have turned my maths brain to mush...

I keep getting stuck on questions like this:
Find the points of inflexion for f(x) = 3x^5 - 10x^3 + 7

I can't quite work out how to manipulate the equation after finding the second derivative... So far I haven't had any luck with factorisation or anything else...

Any help greatly appreciated!

RuiAce · « **Reply #3232 on:** January 24, 2018, 10:57:08 am »

Quote from: kauac on January 24, 2018, 10:52:11 am

Hi...
My holidays have turned my maths brain to mush...

I keep getting stuck on questions like this:
Find the points of inflexion for f(x) = 3x^5 - 10x^3 + 7

I can't quite work out how to manipulate the equation after finding the second derivative... So far I haven't had any luck with factorisation or anything else...

Any help greatly appreciated!

$\text{Since the second derivative is }f^{\prime\prime}(x) = 60x^3-60x,\\ \text{possible points of inflexion occur when }\boxed{60x^3-60x=0}.$
$\text{To factorise this,}\\ \begin{align*}60x^3-60x&=0\\ 60x(x^2-1)&=0\\ 60x(x-1)(x+1)&=0\end{align*}\\ \text{Hence }x=-1,0,1.$
Testing a bit to both sides we see that all of them are points of inflexion.
$\text{So noting that }f(1)=0, f(0)=7\text{ and }f(-1)=14,\\ \text{our points of inflexion are at }(-1,14),(0,7)\text{ and }(1,0)$

kauac · « **Reply #3233 on:** January 24, 2018, 11:02:29 am »

Quote from: RuiAce on January 24, 2018, 10:57:08 am

$\text{Since the second derivative is }f^{\prime\prime}(x) = 60x^3-60x,\\ \text{possible points of inflexion occur when }\boxed{60x^3-60x=0}.$
$\text{To factorise this,}\\ \begin{align*}60x^3-60x&=0\\ 60x(x^2-1)&=0\\ 60x(x-1)(x+1)&=0\end{align*}\\ \text{Hence }x=-1,0,1.$
Testing a bit to both sides we see that all of them are points of inflexion.
$\text{So noting that }f(1)=0, f(0)=7\text{ and }f(-1)=14,\\ \text{our points of inflexion are at }(-1,14),(0,7)\text{ and }(1,0)$

Thanks so much!
I think I was on the right track with the factorisation, but now I can understand how to approach these kinds of questions. Thanks again.

vic321 · « **Reply #3234 on:** January 26, 2018, 06:36:13 pm »

How do you do this
Simplify:
sin(pi -x)

Thanks

RuiAce · « **Reply #3235 on:** January 26, 2018, 06:38:51 pm »

Quote from: vic321 on January 26, 2018, 06:36:13 pm

How do you do this
Simplify:
sin(pi -x)

Thanks

This is just $ \sin x $. Recall your ASTC identities from prelim: $ \sin (180^\circ - \theta) = \sin \theta$.

vic321 · « **Reply #3236 on:** January 26, 2018, 06:40:54 pm »

Quote from: RuiAce on January 26, 2018, 06:38:51 pm

This is just $ \sin x $. Recall your ASTC identities from prelim: $ \sin (180^\circ - \theta) = \sin \theta$.

ohh ok, thanks

Calley123 · « **Reply #3237 on:** January 28, 2018, 06:49:59 pm »

Hey,
I keep getting the wrong answer for this question !
Find the area enclosed by the curve y=x+e^-x, the x-axis and the lines x=o and y=2, correct to 2 decimal places

Thanks

RuiAce · « **Reply #3238 on:** January 28, 2018, 07:11:36 pm »

Quote from: Calley123 on January 28, 2018, 06:49:59 pm

Hey,
I keep getting the wrong answer for this question !
Find the area enclosed by the curve y=x+e^-x, the x-axis and the lines x=o and y=2, correct to 2 decimal places

Thanks

This question lacks information. Ignoring that the sketch of $ y = x + e^{-x} $ is certainly not 2U material, there are two regions that could be implied by this information. One in the first quadrant, and one in the second quadrant.

Calley123 · « **Reply #3239 on:** January 28, 2018, 08:02:16 pm »

Quote from: RuiAce on January 28, 2018, 07:11:36 pm

This question lacks information. Ignoring that the sketch of $ y = x + e^{-x} $ is certainly not 2U material, there are two regions that could be implied by this information. One in the first quadrant, and one in the second quadrant.

This question is from the log and Exponential Functions, a two unit topic but I think its for 3U students...
Sorry

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