The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Maclaurin Question (FP3)

Hey, can someone help me with this FP3 question?

It's question 6(d)(i) on here:

http://filestore.aqa.org.uk/subjects/AQA-MFP3-QP-JAN13.PDF

So far I have

ln(e3xcosx)=3x−x22−x412ln(e^{3x}cosx) = 3x - \frac{x^{2}}{2} - \frac{x^{4}}{12}

ln(1+px)=px−p2x22ln(1+px) = px - \frac{p^{2}x^{2}}{2}

and I worked towards

lim⁡x→0(3−x2−x312−p+p2x2)\lim_{x \to 0} (3-\frac{x}{2} - \frac{x^{3}}{12} - p + \frac{p^{2}x}{2})

How do I know when it exists?

Thanks!
Original post by Cubic
Hey, can someone help me with this FP3 question?

It's question 6(d)(i) on here:

http://filestore.aqa.org.uk/subjects/AQA-MFP3-QP-JAN13.PDF

So far I have

ln(e3xcosx)=3x−x22−x412ln(e^{3x}cosx) = 3x - \frac{x^{2}}{2} - \frac{x^{4}}{12}

ln(1+px)=px−p2x22ln(1+px) = px - \frac{p^{2}x^{2}}{2}

and I worked towards

lim⁡x→0(3−x2−x312−p+p2x2)\lim_{x \to 0} (3-\frac{x}{2} - \frac{x^{3}}{12} - p + \frac{p^{2}x}{2})

How do I know when it exists?

Thanks!


I think you've made a mistake. You should have:

lim⁡x→0(3−px+p2−12−x212.....)\lim_{x \to 0} (\frac{3-p}{x}+\frac{p^2-1}{2} - \frac{x^{2}}{12} .....)

Now note that the first one you will have it tend to infinity as x tends to zero. what value of p will stop that?
(edited 9 years ago)
Reply 2
Avatar for Cubic
Cubic
OP
8
Original post by Slowbro93
I think you've made a mistake. You should have:

lim⁡x→0(3−px+p2−12−x212.....)\lim_{x \to 0} (\frac{3-p}{x}+\frac{p^2-1}{2} - \frac{x^{2}}{12} .....)

Now note that the first one you will have it tend to infinity as x tends to zero. what value of p will stop that?




Ah, I divided by x instead of x squared.

So it's 3?
Original post by Cubic
Ah, I divided by x instead of x squared.

So it's 3?


Indeed :h:
Reply 4
Avatar for Cubic
Cubic
OP
8
Original post by Slowbro93
Indeed :h:


Thanks! :smile:
Original post by Cubic
...
Original post by Slowbro93
...


Did you get 4 for the final part of the question?
Original post by Khallil
Did you get 4 for the final part of the question?


I don't know, I assisted at the point where the OP was stuck :tongue:
Original post by Slowbro93
...


Does this look OK?

φ =  lim⁡x→0[3−px+p2−12−112x2]=p=3 lim⁡x→0[3−3x+32−12−112x2+⋯ ] ≈  lim⁡x→0[4−112x2] ≈ 4\begin{aligned} \varphi & \ = \ \ \lim_{x \to 0} \left[ \dfrac{3-p}{x} + \dfrac{p^2 - 1}{2} - \dfrac{1}{12}x^2 \right] \\ & \overset{p=3}= \ \displaystyle \lim_{x \to 0} \left[ \dfrac{3-3}{x} + \dfrac{3^2 - 1}{2} - \dfrac{1}{12}x^2 + \cdots \right] \\ & \ \approx \ \ \lim_{x \to 0} \left[ 4 - \dfrac{1}{12} x^2 \right] \\ & \ \approx \ 4 \end{aligned}
(edited 9 years ago)
Original post by Khallil
Does this look OK?

φ =  lim⁡x→0[3−px+p2−12−112x2]=p=3 lim⁡x→0[3−3x+32−12−112x2+⋯ ] ≈  lim⁡x→0[4−112x2] ≈ 4\begin{aligned} \varphi & \ = \ \ \lim_{x \to 0} \left[ \dfrac{3-p}{x} + \dfrac{p^2 - 1}{2} - \dfrac{1}{12}x^2 \right] \\ & \overset{p=3}= \ \displaystyle \lim_{x \to 0} \left[ \dfrac{3-3}{x} + \dfrac{3^2 - 1}{2} - \dfrac{1}{12}x^2 + \cdots \right] \\ & \ \approx \ \ \lim_{x \to 0} \left[ 4 - \dfrac{1}{12} x^2 \right] \\ & \ \approx \ 4 \end{aligned}


Seems right :h:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you