heelllppp pls binomial q (1 Viewer)

yashbb · Sep 22, 2021

idk what i have done wrong, but there are no worked solutions to this problem.

how do i do it.
(srry for blurry image)

Lith_30 · Sep 22, 2021

$\\\text{General Term}\\\\{7 \choose k}\cdot5^{7-k}\cdot(2x^2)^k\\\\={7 \choose k}\cdot5^{7-k}\cdot2^k\cdot{x^{2k}}$

Since x is to the power of 6, k has to equal three

$\\{7 \choose 3}\cdot5^{7-3}\cdot2^3\cdot{x^{2(3)}}\\\\=35\cdot625\cdot8\cdot{x^6}\\=175000x^6$

Therefore the coefficient of $x^6$ is 175000

icycledough · Sep 22, 2021

=)(= · Sep 22, 2021

Lith_30 said:
$\\\text{General Term}\\\\{7 \choose k}\cdot5^{7-k}\cdot(2x^2)^k\\\\={7 \choose k}\cdot5^{7-k}\cdot2^k\cdot{x^{2k}}$

Since x is to the power of 6, k has to equal three

$\\{7 \choose 3}\cdot5^{7-3}\cdot2^3\cdot{x^{2(3)}}\\\\=35\cdot625\cdot8\cdot{x^6}\\=175000x^6$

Therefore the coefficient of $x^6$ is 175000

Can you help with this pls

20210920_135959.jpg

icycledough · Sep 22, 2021

=)(= said:
Can you help with this pls

20210920_135959.jpg

Just to confirm, its number 3 yh? The one with the blue asterisk?

=)(= · Sep 22, 2021

yes please than you

CM_Tutor · Sep 22, 2021

You can also do this with the whole expression in sigma notation. Since $(a+b)^n = \sum_{r=0}^{r=n} \binom{n}{r}a^{n-r}b^r$ , we can say:

$\left(5+2x^2\right)^7 = \sum_{r=0}^{r=7} \binom{7}{r}\times 5^{7-r} \times \left(2x^2\right)^r = \sum_{r=0}^{r=7} \binom{7}{r}\times 5^{7-r} \times 2^r \times x^{2r}$

The only term in $x^6$ in this expansion will occur when $2r = 6 \implies r = 3$ .

So, the required term is:

$\begin{align*} \binom{7}{3}\times 5^{7-3} \times 2^3 \times x^{6} &= \cfrac{7!}{3! \times 4!} \times 5^4 \times 2^3 x^6 \\ &= \cfrac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} \times 5^4 \times 2^3 x^6 \\ &= \cfrac{7 \times 5}{1} \times 5^4 \times 2^3 x^6 \\ &= 2^3.5^5.7x^6 \qquad \text{with the coefficient given as its prime factorisation} \\ \text{So, the required coefficient is}\ 2^3.5^5.7 &= 175\ 000\ \text{confirming the answer given by Lith-30 and icycledough above.} \end{align*}$

yashbb · Sep 22, 2021

THANK YOU EVERYONEEEEEEEE

Lith_30 · Sep 22, 2021

=)(= said:
Can you help with this pls

I've tried to solve it, but I haven't learnt any Ext 2 level projectile motion stuff yet. I'll try to learn it over the next few weeks if possible.

username_2 · Sep 23, 2021

=)(= said:
Can you help with this pls

20210920_135959.jpg

So i remember doing this question and I couldn't figure it out then but I have now. I have put the solution down below. Anything you didn't understand, shoot a reply to this... here goes.

Ok so I would even think that many people will get up to this point. But here's the little trick - the keyword in the question is show:

You don't have to solve the equation to show that k = 0.158, you can just substitute in the value for LHS and RHS and check if this is a valid solution to the equation. (Btw there is no good manual way to solve this equation other than precise graphing but that is impossible in an exam). Hence, is my solution - ie the next steps - below. (This part might not be the best but do give suggestions about any other methods you might know)

Hope that is a bit more helpful!

=)(= · Sep 23, 2021

THANK YOU!!!!

CM_Tutor · Sep 23, 2021

Looking at @username_2's equation 1, and setting $B=4.35\cos{10^\circ}$ , we have:

$\begin{align*} m\left|\cfrac{B}{B-18k}\right|&=\cfrac{k}{m} \\ \cfrac{m^2B}{\left|B-18k\right|}&=k \qquad \qquad \text{as $m,\ B > 0$} \end{align*}$

The solutions are thus the intersection of a diagonal line through the origin $(y = k)$ and the absolute value of a hyperbola that has as its two asymptotes the $k$ -axis and $k=\cfrac{B}{18}$ . There will likely be three solutions, one with $k>\cfrac{B}{18} = 0.2379...$ , and likely two between $0 < k < 0.2379...$ . We know we aren't looking for the solution above 0.2379..., so must be in the region where $B-18k>0$ , and so can solve:

$\begin{align*} \cfrac{m^2B}{\left|B-18k\right|} &= k \\ \cfrac{m^2B}{B-18k} &= k \qquad \text{as $B -18k > 0$} \\ m^2B &= k(B - 18k) \\ m^2B &= kB - 18k^2 \\ 18k^2 - kB + m^2B &= 0 \\ k &= \cfrac{-(-B) \pm \sqrt{(-B)^2 - 4(18)(m^2B)}}{2(18)} \\ k &= \cfrac{B \pm \sqrt{B^2 - 72m^2B}}{36} \\ &= \cfrac{4.2839... \pm \sqrt{4.2839...^2 - 72(0.145)^2 \times 4.2839...}}{36} \\ &= \cfrac{4.2839... \pm 3.4448...}{36} \\ &= 0.2146...\ \text{or}\ 0.023307... \end{align*}$

The other solution, above 0.2379..., is in the region where $B-18k<0$ , and so can solve:

$\begin{align*} \cfrac{m^2B}{\left|B-18k\right|} &= k \\ \cfrac{m^2B}{18k-B} &= k \qquad \text{as $B -18k < 0$} \\ m^2B &= k(18k - B) \\ m^2B &= 18k^2 - kB \\ 18k^2 - kB - m^2B &= 0 \\ k &= \cfrac{-(-B) \pm \sqrt{(-B)^2 - 4(18)(-m^2B)}}{2(18)} \\ k &= \cfrac{B \pm \sqrt{B^2 + 72m^2B}}{36} \\ &= \cfrac{4.2839... \pm \sqrt{4.2839...^2 + 72(0.145)^2 \times 4.2839...}}{36} \\ &= \cfrac{4.2839... \pm 4.9836...}{36} \\ &= 0.2574...\ \text{or}\ -0.01943... \\ &\approx 0.257 \qquad \text{as $k > 0$} \end{align*}$

So, the solution certainly can be found by HSC methods. There are indeed three solutions. And, either I have made an algebraic mistake(s) or there is a flaw in equation 1, as I haven't got $k=0.158$ .

username_2 · Sep 23, 2021

CM_Tutor said:
Looking at @username_2's equation 1, and setting $B=4.35\cos{10^\circ}$ , we have:

$\begin{align*} m\left|\cfrac{B}{B-18k}\right|&=\cfrac{k}{m} \\ \cfrac{m^2B}{\left|B-18k\right|}&=k \qquad \qquad \text{as $m,\ B > 0$} \end{align*}$

The solutions are thus the intersection of a diagonal line through the origin $(y = k)$ and the absolute value of a hyperbola that has as its two asymptotes the $k$ -axis and $k=\cfrac{B}{18}$ . There will likely be three solutions, one with $k>\cfrac{B}{18} = 0.2379...$ , and likely two between $0 < k < 0.2379...$ . We know we aren't looking for the solution above 0.2379..., so must be in the region where $B-18k>0$ , and so can solve:

$\begin{align*} \cfrac{m^2B}{\left|B-18k\right|} &= k \\ \cfrac{m^2B}{B-18k} &= k \qquad \text{as $B -18k > 0$} \\ m^2B &= k(B - 18k) \\ m^2B &= kB - 18k^2 \\ 18k^2 - kB + m^2B &= 0 \\ k &= \cfrac{-(-B) \pm \sqrt{(-B)^2 - 4(18)(m^2B)}}{2(18)} \\ k &= \cfrac{B \pm \sqrt{B^2 - 72m^2B}}{36} \\ &= \cfrac{4.2839... \pm \sqrt{4.2839...^2 - 72(0.145)^2 \times 4.2839...}}{36} \\ &= \cfrac{4.2839... \pm 3.4448...}{36} \\ &= 0.2146...\ \text{or}\ 0.023307... \end{align*}$

The other solution, above 0.2379..., is in the region where $B-18k<0$ , and so can solve:

$\begin{align*} \cfrac{m^2B}{\left|B-18k\right|} &= k \\ \cfrac{m^2B}{18k-B} &= k \qquad \text{as $B -18k < 0$} \\ m^2B &= k(18k - B) \\ m^2B &= 18k^2 - kB \\ 18k^2 - kB - m^2B &= 0 \\ k &= \cfrac{-(-B) \pm \sqrt{(-B)^2 - 4(18)(-m^2B)}}{2(18)} \\ k &= \cfrac{B \pm \sqrt{B^2 + 72m^2B}}{36} \\ &= \cfrac{4.2839... \pm \sqrt{4.2839...^2 + 72(0.145)^2 \times 4.2839...}}{36} \\ &= \cfrac{4.2839... \pm 4.9836...}{36} \\ &= 0.2574...\ \text{or}\ -0.01943... \\ &\approx 0.257 \qquad \text{as $k > 0$} \end{align*}$

So, the solution certainly can be found by HSC methods. There are indeed three solutions. And, either I have made an algebraic mistake(s) or there is a flaw in equation 1, as I haven't got $k=0.158$ .

Hey... um its not m abs(B/(B-18k)) = k/m, its ln(abs(B/(B-18k))) = k/m. Sorry for the terrible handwriting

yashbb · Sep 23, 2021

how did binomial turn into projectile motion

CM_Tutor · Sep 23, 2021

username_2 said:
Hey... um its not m abs(B/(B-18k)) = k/m, its ln(abs(B/(B-18k))) = k/m. Sorry for the terrible handwriting

*That* explains my confusion!

In that case, I agree that there is no reasonable way to solve the equation.

Though, for MX2, if we want to be rigorous, you might look at

$f(k)=\ln\left|\cfrac{B}{B-18k}\right|-\cfrac{k}{m}$

and note that it is defined for $k>0$ and continuous so long as $k\neq \cfrac{B}{18}$ , and then show that $f(0.1576)$ and $f(0.1584)$ have opposite signs to conclude that there must be at least one root on $0.1576 < k < 0.1584$ and thus there must be a root at $k=0.158$ (to 3 significant figures).

heelllppp pls binomial q (1 Viewer)

yashbb

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