The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

STEP maths I, II, III 2007 solutions

Scroll to see replies

Reply 20
Avatar for JeremyC
JeremyC
8
eponymous
thats it!
thank you so much!!
i really can't get latex right :frown:

No worries. :smile:

The only reason the latex didn't work as you wanted it to was because of one small mistake: instead of using a } to close a curly bracket you used a ), which as you saw really confused it.

For instance, something as simple as [noparse]
Unparseable latex formula:

\frac{1}{2)

[/noparse] will produce
Unparseable latex formula:

\frac{1}{2)

.

Apart from that your latex was fine (I used a lot more latex tags, so that the ordinary English wouldn't be in italics [any letters within latex tags are considered variables, and so are italicised], and used \arctan instead of arc\tan, but that only made aesthetic differences - with the exception of your tiny error your code would have worked). :smile:

How about trying 2 (ii) now? :smile:
Reply 21
Avatar for eponymous
eponymous
2
fatuous_philomath
No worries. :smile:

The only reason the latex didn't work as you wanted it to was because of one small mistake: instead of using a } to close a curly bracket you used a ), which as you saw really confused it.

For instance, something as simple as [noparse]
Unparseable latex formula:

\frac{1}{2)

[/noparse] will produce
Unparseable latex formula:

\frac{1}{2)

.

Apart from that your latex was fine (I used a lot more latex tags, so that the ordinary English wouldn't be in italics [any letters within latex tags are considered variables, and so are italicised], and used \arctan instead of arc\tan, but that only made aesthetic differences - with the exception of your tiny error your code would have worked). :smile:

How about trying 2 (ii) now? :smile:


wow, i wonder what it would do if i tried to divide by zero...since ")" was a potentially dangerous error lol.

yeah, im currently doing (ii) now, i might have to post it up later since im going out soon.. :smile: (is my part (i) correct ?)
*reminds ukgea he's done III/8 too :redface:*
Ugh. I don't feel particularly inspired by any of the STEP I questions.

II/5:

(i)

f(x)=x+31−3x\displaystyle f(x) = \frac{x+\sqrt{3}}{1-\sqrt{3} x}

f2(x)=(x+31−3x+3)/(1−3x+31−3x)\displaystyle f^2(x) = \left( \frac{x+\sqrt{3}}{1-\sqrt{3} x} + \sqrt{3} \right) / \left( 1 - \sqrt{3} \frac{x+\sqrt{3}}{1-\sqrt{3} x} \right)

=(x+3+3−3x1−3x)/(1−3x−3x−31−3x)\displaystyle = \left( \frac{x+\sqrt{3}+\sqrt{3} -3x}{1-\sqrt{3} x}\right) / \left(\frac{1-\sqrt{3} x - \sqrt{3} x-3}{1-\sqrt{3} x} \right)

=x+3+3−3x1−3x−3x−3\displaystyle = \frac{x+\sqrt{3}+\sqrt{3} -3x}{1-\sqrt{3} x - \sqrt{3} x-3}

=23−2x−2−23x=x−31+3x.\displaystyle = \frac{2\sqrt{3} -2x}{-2-2\sqrt{3} x} = \frac{x - \sqrt{3}}{1 + \sqrt{3} x} .

f3(x)=(x−31+3x+3)/(1−3x−31+3x)\displaystyle f^3(x) = \left( \frac{x-\sqrt{3}}{1+\sqrt{3} x} + \sqrt{3} \right) / \left( 1 - \sqrt{3} \frac{x-\sqrt{3}}{1+\sqrt{3} x} \right)

=(x−3+3+3x1+3x+3)/(1+3x−3x+31+3x)\displaystyle = \left( \frac{x-\sqrt{3}+\sqrt{3} + 3x}{1+\sqrt{3} x} + \sqrt{3} \right) / \left(\frac{1+\sqrt{3} x-\sqrt{3} x+3}{1+\sqrt{3} x} \right)

=x−3+3+3x1+3x−3x+3\displaystyle = \frac{x-\sqrt{3}+\sqrt{3} + 3x}{1+\sqrt{3} x-\sqrt{3} x+3}

=4x4=x.\displaystyle = \frac{4x}{4} = x.

∴x=f3(x)=f3(f3(x))=f3(f3(f3(x)))=…\therefore x = f^3(x) = f^3(f^3(x)) = f^3(f^3(f^3(x))) = \dots

∴x=f3(x)=f6(x)=f9(x)=⋯=f3n(x)=…\therefore x = f^3(x) = f^6(x) = f^9(x) = \dots = f^{3n} (x) = \dots (or induction if you're bored)

∴x=f2007(x)\therefore x = f^{2007}(x) as 3 divides 2007.

(ii)

tan is periodic: since tan⁡(a+nπ)=tan⁡a\tan (a+n\pi) = \tan a, it suffices to prove the result for n = 0, 1, 2 (and then, again, be as fussy as you like about the induction, which I won't).

tan⁡(θ+13⋅0π)=tan⁡θ=x=f0(x)\tan (\theta + \frac{1}{3}\cdot 0\pi ) = \tan \theta = x = f^0(x) as required.

tan⁡(θ+13π)=tan⁡θ+tan⁡13π1−tan⁡θtan⁡13π=x+31−3x=f(x)\displaystyle \tan (\theta + \frac{1}{3} \pi ) = \frac{\tan \theta + \tan \tfrac{1}{3} \pi}{1 - \tan \theta \tan \frac{1}{3} \pi} = \frac{x + \sqrt{3}}{1 - \sqrt{3} x} = f(x) as required.

tan⁡(θ+23π)=tan⁡θ+tan⁡23π1−tan⁡θtan⁡23π=x−31+3x=f2(x)\displaystyle \tan (\theta + \frac{2}{3} \pi ) = \frac{\tan \theta + \tan \tfrac{2}{3} \pi}{1 - \tan \theta \tan \frac{2}{3} \pi} = \frac{x - \sqrt{3}}{1+\sqrt{3} x} = f^2(x) as required.

(Cue tedious inductive step, or rather obvious conclusion as fn(x)=fn+3(x)f^n(x) = f^{n+3}(x) and tan⁡(θ+13nπ)=tan⁡(θ+13(n+3)π).\tan (\theta + \frac{1}{3} n\pi ) = \tan (\theta + \frac{1}{3} (n+3)\pi ).)

(iii)

Bit of guesswork involved, but several hints give it away; |t| < 1, the use of tan in part (ii), and the form of the expression suggest an expression in sin or cos. I chose sin:

Let t=sin⁡ϕ,t = \sin \phi , so that 1−t2=cos⁡ϕ.\sqrt{1 - t^2} = \cos\phi .

Note also that 32=cos⁡π6,12=sin⁡π6.\frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} , \quad \frac{1}{2} = \sin \frac{\pi}{6} .

Then the expression for g can be rewritten as: g(t)=sin⁡ϕcos⁡π6+cos⁡ϕsin⁡π6g(t) = \sin\phi\cos\frac{\pi}{6} + \cos\phi\sin\frac{\pi}{6}

g(t)=sin⁡(ϕ+π6)g(t) = \sin (\phi +\frac{\pi}{6} )

And then another small inductive step... let s=sin⁡(θ),θ=ϕ+π6,s = \sin (\theta ), \quad \theta = \phi + \frac{\pi}{6}, so that s=g(t).s = g(t) . Then:

g(s)=sin⁡(θ+π6)g(s) = \sin (\theta + \frac{\pi}{6} )

g(g(t))=sin⁡((ϕ+π6)+π6)g(g(t)) = \sin ((\phi + \frac{\pi}{6} ) + \frac{\pi}{6} )

g2(t)=sin⁡(ϕ+π3)g^2(t) = \sin (\phi + \frac{\pi}{3} )

etc.

Hence gn(t)=sin⁡(ϕ+nπ6).g^n(t) = \sin (\phi + \frac{n\pi}{6} ).
Reply 24
Avatar for JeremyC
JeremyC
8
eponymous

yeah, im currently doing (ii) now, i might have to post it up later since im going out soon.. :smile: (is my part (i) correct ?)

Looks fine to me. :biggrin:

(You might possibly comment that you've deduced that A+B=π4A+B = \frac{\pi}{4} because you know A and B are acute, but I doubt you'd lose more than a mark for omitting it.)
Reply 25
Avatar for coffeym
coffeym
10
STEP I Q6

(i) We are given that x2−y2=(x−y)3 x^2-y^2=(x-y)^3 (*), and hence (x+y)(x−y)=(x−y)3(x+y)(x-y)=(x-y)^3.

∴x+y=d2⇒y=d2−x\therefore x+y=d^2\qquad \Rightarrow y=d^2-x

Now substituting back into (*):
x2−d4−x2+2xd2=d3x^2-d^4-x^2+2xd^2=d^3

⇒2xd2=d3+d4\Rightarrow 2xd^2=d^3+d^4

⇒x=d+d22\Rightarrow x=\frac{d+d^2}{2} (**)

Now from above, y=d2−xy=d^2-x, and so after substitution of our result (**):

y=d2−d2y=\frac{d^2-d}{2} as required

Now for the next part let m=x2m=x^2 and n=y2n=y^2, thus

m=(d+d2)24m=\frac{(d+d^2)^2}{4} and n=(d2−d)24n=\frac{(d^2-d)^2}{4}

Now choosing d=6d=6, we find that suitable integers mm and nn satisying the required conditions are:

Unparseable latex formula:

\framebox{m=441, n=225}



(ii) This time note that x3−y3=(x−y)(x2+xy+y2)x^3-y^3=(x-y)(x^2+xy+y^2), and hence that x2+y2+xy=d3x^2+y^2+xy=d^3 from the identity given in the question

Hence (x−y)2+3xy=d3(x-y)^2+3xy=d^3

⇒3xy=d3−d2\Rightarrow 3xy=d^3-d^2 as required.

From this result it is clear that y=d3−d23xy=\frac{d^3-d^2}{3x} and so, substituting into x−y=dx-y=d we get:

x−d3−d23x=dx-\frac{d^3-d^2}{3x}=d

⇒3x2−3dx+(d2−d3)=0\Rightarrow 3x^2-3dx+(d^2-d^3)=0 provided x≠0x\neq 0

Now applying the quadratic formula,

x=3d±12d3−3d26x=\frac{3d\pm \sqrt{12d^3-3d^2}}{6}

⇒x=d2±d212d−39\Rightarrow x=\frac{d}{2} \pm \frac{d}{2}\sqrt{\frac{12d-3}{9}}

⇒x=d2±d24d−13\Rightarrow x=\frac{d}{2} \pm \frac{d}{2}\sqrt{\frac{4d-1}{3}}

⇒2x=d±d4d−13\Rightarrow 2x=d \pm d\sqrt{\frac{4d-1}{3}} as required

Now letting m=xm=x and n=yn=y and d=7d=7, we find that two integers satisfying our conditions are

Unparseable latex formula:

\framebox{m=14,n=7}



This completes the question
Reply 26
Avatar for ukgea
ukgea
6
STEP 2 Q7

Since

ddx2sin⁡x=−sin⁡x\displaystyle \frac{d}{dx^2} \sin x = -\sin x

, and −sin⁡x<0-\sin x < 0 in 0<x<π0 < x < \pi, sin⁡x\sin x is concave in this interval.

Similarly,

ddx2ln⁡x=−1x2\displaystyle \frac{d}{dx^2} \ln x = - \frac{1}{x^2},

which is negative for x>0x > 0, (and indeed negative for x<0 x < 0 as well, the only problem being ln⁡x\ln x is not defined there), and thus ln⁡x\ln x is concave in x>0x > 0.

(i) First we note that A+B+C=πA + B + C = \pi since they are angles in a triangle.

Applying Jensen's inequality with n=3n = 3, x1=A,x2=B,x3=3x_1 = A, x_2 = B, x_3 = 3 and f(x)=sin⁡(x)f(x) = \sin(x) yields

13(sin⁡A+sin⁡B+sin⁡C)≤sin⁡A+B+C3\displaystyle \frac{1}{3}\left(\sin A + \sin B + \sin C\right) \leq \sin \frac{A + B + C}{3}.

But since A+B+CA + B + C, the RHS equals sin⁡(π/3)=3/2\sin (\pi / 3) = \sqrt 3 / 2, and thus we have

13(sin⁡A+sin⁡B+sin⁡C)≤32\displaystyle \frac{1}{3}\left(\sin A + \sin B + \sin C\right) \leq \frac{\sqrt{3}}{2}

from which the desired result immediately follows.

(Note that since A,B,CA, B, C are angles in a triangle, 0<A,B,C<π0 < A, B, C < \pi and so our usage of Jensen's inequality was valid.)

(ii)

Setting, in Jensen's inequality, x1=t1,x2=t2⋯xn=tnx_1 = t_1, x_2 = t_2 \cdots x_n = t_n and f(x)=ln⁡xf(x) = \ln x, we get

1n(ln⁡t1+ln⁡t2+⋯+ln⁡tn)≤ln⁡t1+t2+⋯+tnn\displaystyle \frac{1}{n}\left(\ln t_1 + \ln t_2 + \cdots + \ln t_n\right) \leq \ln \frac{t_1 + t_2 + \cdots + t_n}{n}

Exponentiating both sides (note that since exe^x is an increasing function, this preserves the inequality)

Unparseable latex formula:

\displaystyle \left(t_1t_2\cdots t_n\right)^{1/n} \leq \frac{t_1 + t_2 + \cdots + \t_n}{n}

(*)

as required.

a) Setting, in (*), n=4n = 4, t1=x4,t2=y4,t3=z4,t4=16t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 16, we get

16x4y4z44≤x4+y4+z4+164\displaystyle \sqrt[4]{16x^4y^4z^4} \leq \frac{x^4 + y^4 + z^4 + 16}{4}

from which follows (note that we in this step use x,y,z>0x, y, z > 0

2xyz≤x4+y4+z4+164\displaystyle 2xyz \leq \frac{x^4 + y^4 + z^4 + 16}{4}

8xyz≤x4+y4+z4+168xyz \leq x^4 + y^4 + z^4 + 16

as required.

(Again, the usage of Jensen's inequality was valid, since x4,y4,z4>0x^4, y^4, z^4 > 0 which was a requirement for the function to be concave)

b) Let g(x,y,z)=x5+y5+z5−5xyzg(x, y, z) = x^5 + y^5 + z^5 - 5xyz.

Set, in (*), n=5n = 5, t1=x5,t2=y5,t3=z5,t4=1,t5=1t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1, t_5 = 1, and we get

x5y5z55≤x5+y5+z5+1+15\displaystyle \sqrt[5]{x^5y^5z^5} \leq \frac{x^5 + y^5 + z^5 + 1 + 1}{5}

xyz≤x5+y5+z5+25\displaystyle xyz \leq \frac{x^5 + y^5 + z^5 + 2}{5}
t
5xyz≤x5+y5+z5+2\displaystyle 5xyz \leq x^5 + y^5 + z^5 + 2

x5+y5+z5−5xyz≥−2\displaystyle x^5 + y^5 + z^5 - 5xyz \geq -2

(This inequality is valid whenever x,y,zx, y, z are positive, because then x5,y5,z5x^5, y^5, z^5 are positive as well, which was a requirement for (*) to be valid)

Now, this means

g(x,y,z)≥−2g(x, y, z) \geq -2.

But g(x,y,z)g(x, y, z) attains this value at (x,y,z)=(1,1,1)(x, y, z) = (1, 1, 1), which means that this is the function's minimum. (By definition, since we have just proven that for all (x,y,z)(x, y, z), where x,y,z>0x, y, z > 0, we have g(1,1,1)≥g(x,y,z)g(1, 1, 1) \geq g(x, y, z))

and so the desired minimum value for the expression is -2.
Reply 27
Avatar for coffeym
coffeym
10
STEP I Q9

Diagrams clearly help in this question, they should be easy to draw and I assume in this solution you have drawn them.

I will call the first situation described in the question Situation α\alpha, and the second situtation Situation β\beta

Situation α\alpha

Call the normal reaction R1R_1, and the friction up the plane F1F_1.

Resolving perpendicular to the plane we get
R1=Xsin⁡θ+Wcos⁡θR_1=X\sin{\theta}+W\cos{\theta}, and hence that

F1=μ(Xsin⁡θ+Wcos⁡θ)F_1=\mu(X\sin{\theta}+W\cos{\theta})

Resolving parallel to the plane we have
F1+Xcos⁡θ=Wsin⁡θF_1+X\cos{\theta}=W\sin{\theta}

⇒μ(Xsin⁡θ+Wcos⁡θ)=Wsin⁡θ−Xcos⁡θ\Rightarrow \mu(X\sin{\theta}+W\cos{\theta})=W\sin{\theta}-X\cos{\theta} (*)

Situation β\beta

Call the normal reaction here R2R_2 and the friction down the plane F2F_2

Resolving perpendicular to the plane we have:

R2=kXsin⁡θ+Wcos⁡θR_2=kX\sin{\theta}+W\cos{\theta}

⇒F2=μ(kXsin⁡θ+Wcos⁡θ)\Rightarrow F_2=\mu(kX\sin{\theta}+W\cos{\theta})

Resolving parrallel to the plane, we find that

F2+Wsin⁡θ=kXcos⁡θF_2+W\sin{\theta}=kX\cos{\theta}

⇒μ(kXsin⁡θ+Wcos⁡θ)=kXcos⁡θ−Wsin⁡θ\Rightarrow \mu(kX\sin{\theta}+W\cos{\theta})=kX\cos{\theta}-W\sin{\theta} (**)


Combining (*) and (**) we have the two simultaneous equations:

μ(Xsin⁡θ+Wcos⁡θ)=Wsin⁡θ−Xcos⁡θ\mu(X\sin{\theta}+W\cos{\theta})=W\sin{\theta}-X\cos{\theta} (*)
μ(kXsin⁡θ+Wcos⁡θ)=kXcos⁡θ−Wsin⁡θ\mu(kX\sin{\theta}+W\cos{\theta})=kX\cos{\theta}-W\sin{\theta} (**)

Rearranging these equations we have:

X(μsin⁡θ+cos⁡θ)=W(sin⁡θ−μcos⁡θ)X(\mu\sin{\theta}+\cos{\theta})=W(\sin{\theta}-\mu\cos{\theta}) (*)
X(μksin⁡θ−kcos⁡θ)=−W(sin⁡θ+μcos⁡θ)X(\mu k \sin{\theta}-k\cos{\theta})=-W(\sin{\theta}+\mu \cos{\theta}) (**)

So from (*):

W=X(μsin⁡θ+cos⁡θ)(sin⁡θ−μcos⁡θ)\displaystyle W=\frac{X(\mu \sin{\theta}+\cos{\theta})}{(\sin{\theta}-\mu\cos{\theta})}

Substituting this result into (**) and cancelling X on both sides we find that:

(μksin⁡θ−kcos⁡θ)=(μsin⁡θ+cos⁡θ)(sin⁡θ+μcos⁡θ)μcos⁡θ−sin⁡θ(\mu k \sin{\theta}-k\cos{\theta})=\frac{(\mu\sin{\theta}+\cos{\theta})(\sin{\theta}+\mu\cos{\theta})}{\mu \cos{\theta}-\sin{\theta}}


After mutiplying out a few brackets (which is just bookwork) we come to the result:

k[(μ2+1)cos⁡θsin⁡θ]=μ+(μ2+1)cos⁡θsin⁡θ[br]k[(\mu^2+1)\cos{\theta}\sin{\theta}]=\mu+(\mu^2+1)\cos{\theta}\sin{\theta}[br]


⇒(k−1)(μ2+1)cos⁡θsin⁡θ=μ(k+1)[br]\Rightarrow (k-1)(\mu^2+1)\cos{\theta}\sin{\theta}=\mu(k+1)[br] which is the required result

For the final part note that
0≤cos⁡θsin⁡θ≤12[br]0\leq \cos{\theta}\sin{\theta} \leq \frac12[br] in the range of sensible values for θ\theta : 0≤θ≤90o0 \leq \theta \leq 90^{o} (easily provable if necessary using calculus)

Hence we can say from the result we just proved that:

(k−1)(1+μ2)≥2μ(k+1)[br](k-1)(1+\mu^2)\geq 2\mu(k+1)[br]

⇒k(1+μ2−2μ)≥2μ+1+μ2[br]\Rightarrow k(1+\mu^2-2\mu)\geq2\mu+1+\mu^2[br]

⇒k≥(μ+1)2(1−μ)2[br]\displaystyle \Rightarrow k\geq \frac{(\mu+1)^2}{(1-\mu)^2}[br] as required

This completes the question
Reply 28
Avatar for Square
Square
13
Doing I/10
STEP I, Q11: (on the grounds that anything that looks this horrible can't be that bad...)

Let vv be the velocity of the particle as it leaves the tube. The height of the tube is Lsin⁡30o=L/2L \sin 30^o = L/2, and so conservation of energy gives us v2=u2−gLv^2 = u^2 - gL.

Then if xx is the horizontal distance travelled and yy the height above the ground we have:

x=vtcos⁡30o,y=L/2+vtsin⁡30o−12gt2x = vt \cos 30^o, y = L/2 + vt \sin 30^o -\frac{1}{2}gt^2

Then vtsin⁡30o=xtan⁡30o=x/3vt \sin 30^o = x \tan 30^o = x/\sqrt{3}, t2=x2v2cos230o=4x23v2t^2 = \frac{x^2}{v^2 cos^2 30^o} = \frac{4x^2}{3v^2}. So we end up with:

2y=L+2x/3−4x2g3v2\displaystyle 2y = L + 2x/\sqrt{3} - \frac{4x^2g}{3v^2}.

Now we know y = 0 when x=D, so:

0=L+2D/3−4D2g3v2\displaystyle 0 = L + 2D/\sqrt{3} - \frac{4D^2g}{3v^2} so

0=3Lv2+23Dv2−4D2g0 = 3Lv^2 + 2\sqrt{3}Dv^2-4D^2g. Substitute in v2=u2−gLv^2=u^2-gL and we get

4D2g−23D(u2−gL)+3L(u2−gL)=04D^2g -2\sqrt{3}D(u^2-gL)+3L(u^2-gL) = 0 as required.

For the next part, it's just implicit differentiation:

8DgdDdL=3u2−6gL+23dDdL(u2−gL)−23Dg\displaystyle 8Dg\frac{dD}{dL} = 3u^2-6gL+2\sqrt{3}\frac{dD}{dL}(u^2-gL)-2\sqrt{3}Dg, so

[8Dg−23(u2−gL)]dDdL=3(u2−2gL)−23Dg\displaystyle \left[8Dg - 2\sqrt{3}(u^2-gL)\right]\frac{dD}{dL} = 3(u^2-2gL)-2\sqrt{3}Dg and so

dDdL=3(u2−2gL)−23Dg8Dg−23(u2−gL)\displaystyle \frac{dD}{dL} = \frac{3(u^2-2gL)-2\sqrt{3}Dg}{8Dg - 2\sqrt{3}(u^2-gL)}

Now when 2D=L32D = L\sqrt{3}, we get

dDdL=3(u2−2gL)−3gL43gL−23(u2−gL)=3(u2−3gL)23(2gL−u2+gL)=−32\displaystyle \frac{dD}{dL} = \frac{3(u^2-2gL)-3gL}{4\sqrt{3}gL - 2\sqrt{3}(u^2-gL)} = \frac{3(u^2-3gL)}{2\sqrt{3}(2gL-u^2+gL)} = -\frac{\sqrt{3}}{2}

But R=D+L32R = D + \frac{L \sqrt{3}}{2}, so dRdL=dDdL+32\displaystyle \frac{dR}{dL} = \frac{dD}{dL} + \frac{\sqrt{3}}{2} and so dRdL=0\displaystyle \frac{dR}{dL} = 0 as required.

Finally we need to find the range when 2D=L32D = L\sqrt{3}. Substitute into our quadratic for D and we get:

3gL2−3(u2−gL)L3−3L(u2−gL)=03gL^2-\sqrt{3}(u^2-gL)L\sqrt{3}-3L(u^2-gL) = 0

3gL2−3u2L+3gL2−3u2L+3gL2=03gL^2-3u^2L+3gL^2-3u^2L+3gL^2 = 0

3L2g−2u2L=03L^2g-2u^2L = 0, so L(3gL−2u2)=0L(3gL-2u^2) = 0 We clearly want L≠0L\neq 0, so we end up with L=2u23gL = \frac{2u^2}{3g}. Then D=L32D=\frac{L\sqrt{3}}{2}.

Finally, R=D+L32=L3=2u2g3R = D + \frac{L\sqrt{3}}{2} = L \sqrt{3} = \frac{2u^2}{g\sqrt{3}}.
Reply 30
Avatar for Square
Square
13
Doing I/10

i) Taking Saxon Army as Origin and j as the direction travelled by saxon horseman:

h1=Norman Rider[br][br]h2=Saxon Rider\text{h1}=\text{Norman Rider}[br][br]\text{h2}=\text{Saxon Rider}



vh1=−yj[br][br]rh1=∫vh1dt[br][br]rh1=(−yt+C)j[br][br]at t=0 r=d  ⟹  C=d[br][br]  ⟹  rh1=(−yt+d)j\text{v}_{\text{h1}}=-\text{y}\bold{j}[br][br]\text{r}_{\text{h1}}=\int \text{v}_{\text{h1}} \text{dt}[br][br]\text{r}_{\text{h1}}=(-\text{yt}+\text{C})\bold{j}[br][br]\text{at t=0 r=d} \implies \text{C}=\text{d}[br][br]\implies \text{r}_{\text{h1}}=(-\text{yt}+\text{d})\bold{j}



vh2=−xj[br][br]rh2=∫vh2dt[br][br]rh2=(xt+C)j\text{v}_{\text{h2}}=-\text{x}\bold{j}[br][br]\text{r}_{\text{h2}}=\int \text{v}_{\text{h2}} \text{dt}[br][br]\text{r}_{\text{h2}}=(\text{xt}+\text{C})\bold{j}


at t=0 r=0  ⟹  C=0[br][br]  ⟹  rh2=xtj\text{at t=0 r=0} \implies \text{C}=\text{0}[br][br]\implies \text{r}_{\text{h2}}=\text{xt}\bold{j}

First encounter: Implies that: rh1=rh2\text{r}_{\text{h1}}=\text{r}_{\text{h2}} For a given value of t:

−yt+d=xt[br][br]yt+xt=d[br][br]t=dy+x-\text{yt}+\text{d}=\text{xt}[br][br]\text{yt}+\text{xt}=\text{d}[br][br]t=\frac{d}{y+x}

Distance from Saxon army at this time:

x(dy+x)=xdy+x\text{x}(\frac{d}{y+x})=\frac{xd}{y+x} as required.

Gah my latex skills suck, rest of the question on the way.
Reply 31
Avatar for ukgea
ukgea
6
STEP I Q4:

x3−3xbc+b3+c3x^3 - 3xbc + b^3 + c^3

=x3+b3+c3−x2b+x2b−x2c+x2c−b2x+b2x−b2c+b2c−c2x+c2x−c2b+c2b−3xbc = x^3 + b^3 + c^3 - x^2 b + x^2 b - x^2 c + x^2 c - b^2 x + b^2 x - b^2c + b^2c - c^2 x + c^2x - c^2 b + c^2 b - 3xbc

=(x+b+c)(x2+b2+c2−xb−bc−cx) = (x + b + c)(x^2 + b^2 + c^2 - xb - bc - cx)

where thus

Q(x)=x2+b2+c2−xb−bc−cxQ(x) = x^2 + b^2 + c^2 - xb - bc -cx

Then,

2Q(x)=2x2+2b2+2c2−2xb−2bc−cx2Q(x) = 2x^2 + 2b^2 + 2c^2 - 2xb - 2bc - cx

=x2−2xb+b2+b2−2bc+c2+c2−2cx+x2= x^2 - 2xb + b^2 + b^2 - 2bc + c^2 + c^2 - 2cx + x^2

=(x−b)2+(b−c)2+(c−x)2= (x-b)^2 + (b - c)^2 + (c - x)^2

which is the sum of three squares, as required.


We are given that

ak2+bk+c=0ak^2 + bk + c = 0 (1)
bk2+ck+a=0bk^2 + ck + a = 0 (2)

Multiplying (1) by b, and (2) by a we get

abk2+b2k+bc=0abk^2 + b^2k+bc = 0
abk2+ack+a2=0abk^2 + ack+a^2 = 0

and subtracting the former from the latter:

(ac−b2)k+(a2−bc)=0(ac - b^2)k + (a^2 - bc) = 0

(ac−b2)k=bc−a2(ac - b^2)k = bc - a^2 (3)

as required.

Similarly, if we instead multiply (1) by c, and (2) by b:

ack2+bck+c2=0ack^2 + bck + c^2 = 0
b2k2+bck+ab=0b^2k^2 + bck + ab = 0

and subtracting the latter from the former, and rearranging:

(ac−b2)k2=ab−c2(ac - b^2) k^2 = ab - c^2 (4).


Now, squareing (3), and multiplying (4) by (ac - b^2), we get

(ac−b2)2k2=(bc−a2)2(ac - b^2)^2 k^2 = (bc - a^2)^2
(ac−b2)2k2=(ab−c2)(ac−b2)(ac - b^2)^2 k^2 = (ab -c^2)(ac - b^2)

and it is clear that

(ac−b2)(ab−c2)=(bc−a2)2(ac - b^2)(ab - c^2) = (bc -a^2)^2

as required.


Expanding this out, we get

a2bc−ac3−ab3+b2c2=b2c2−2a2bc+a4 a^2bc - ac^3 - ab^3 + b^2 c^2 = b^2 c^2 - 2a^2bc + a^4

0=a4+ab3+ac3−3a2bc0 = a^4 + ab^3 + ac^3 - 3a^2bc

and dividing by a (we are given that a is non-zero)

0=a3+b3+c3−3abc0 = a^3 + b^3 + c^3 - 3abc

as required.

Now, this means (from the factorisation performed in the very first part) that

0=12(a+b+c)((a−b)2+(b−c)2+(c−a)2)\displaystyle 0 = \frac{1}{2}(a + b + c)\left((a-b)^2 + (b - c)^2 + (c -a)^2\right)

i.e. either

a+b+c=0a + b + c = 0 (5)

or

(a−b)2+(b−c)2+(c−a)2=0\displaystyle (a-b)^2 + (b - c)^2 + (c-a)^2 = 0 (6)

If (5) holds, then x=1x = 1 is a root to both (1) and (2). If k≠1k \neq 1, this would mean that the two equations have two common roots, from which it follows that they are equal up to a scalar factor. Assume this factor is λ\lambda. Then

λa=b\lambda a = b
λb=c\lambda b = c
λc=a\lambda c = a

Multiplying these three equations together, and we have λ3=1\lambda^3 = 1 and thus

λ=1\lambda = 1

i.e. the (1) and (2) are identical.

If instead (6):

Since all squares are non-negative, it follows that we must have

(a−b)2=(b−c)2=(c−a)2=0(a - b)^2 = (b - c)^2 = (c - a)^2 = 0

(If any of the squares were larger than zero, then the whole sum must be larger than or equal to that square (since the other two squares are nonnegative), and thus the whole sum must be larger than zero, contradiction)

and thus

a=b,b=c,c=aa = b, b = c, c = a

and (1) and (2) are, again, identical.
Haven't spotted anyone doing III/3 yet, so I'll do that.

(i)

F1=1,F2=1,F3=2,F4=3,F5=5,F6=8,F7=13,F8=21.F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21. Fibonacci, yay!

(ii)

F2k+3F2k+1−F2k+22F_{2k+3}F_{2k+1} - F_{2k+2}^2

=(F2k+2+F2k+1)F2k+1−F2k+22 = (F_{2k+2}+F_{2k+1})F_{2k+1} - F_{2k+2}^2

=F2k+2F2k+1+F2k+12−F2k+22 = F_{2k+2}F_{2k+1} +F_{2k+1}^2 - F_{2k+2}^2

=F2k+2F2k+1+F2k+12−F2k+2(F2k+1F2k) = F_{2k+2}F_{2k+1} +F_{2k+1}^2 - F_{2k+2}(F_{2k+1}F_{2k})

=F2k+2F2k+1+F2k+12−F2k+2F2k+1−F2k+2F2k = F_{2k+2}F_{2k+1} +F_{2k+1}^2 - F_{2k+2}F_{2k+1} - F_{2k+2}F_{2k}

=−F2k+2F2k+F2k+12. = - F_{2k+2}F_{2k}+F_{2k+1}^2.

(iii)

P(n):F2n+1F2n−1−F2n2=1P(n): \quad F_{2n+1}F_{2n-1} - F_{2n}^2 = 1

Choosing n=2 gives:

LHS=F5F3−F42=5⋅2−32=10−9=1=RHS.\text{LHS} = F_5F_3 - F_4^2 = 5\cdot 2 - 3^2 = 10 - 9 = 1 = \text{RHS} .

Assume P(k):

P(k)⇒F2k+1F2k−1−F2k2=1P(k) \Rightarrow F_{2k+1}F_{2k-1} - F_{2k}^2 = 1

⇒F2k+1(F2k+1−F2k)−F2k2=1\Rightarrow F_{2k+1}(F_{2k+1}-F_{2k}) - F_{2k}^2 = 1

⇒F2k+1(F2k+1−F2k)−F2k(F2k+2−F2k+1)=1\Rightarrow F_{2k+1}(F_{2k+1}-F_{2k}) - F_{2k}(F_{2k+2}-F_{2k+1}) = 1

⇒F2k+12−F2k+1F2k−F2kF2k+2+F2kF2k+1=1\Rightarrow F_{2k+1}^2 - F_{2k+1}F_{2k} - F_{2k}F_{2k+2} + F_{2k}F_{2k+1} = 1

⇒F2k+12−F2kF2k+2=1\Rightarrow F_{2k+1}^2 - F_{2k}F_{2k+2} = 1

⇒F2k+1(F2k+3−F2k+2)−(F2k+2−F2k+1)F2k+2=1\Rightarrow F_{2k+1}(F_{2k+3}-F_{2k+2}) - (F_{2k+2}-F_{2k+1})F_{2k+2} = 1

⇒F2k+1F2k+3−F2k+1F2k+2−F2k+22+F2k+1F2k+2=1\Rightarrow F_{2k+1}F_{2k+3} - F_{2k+1}F_{2k+2} - F_{2k+2}^2+F_{2k+1}F_{2k+2} = 1

⇒F2k+1F2k+3−F2k+22=1\Rightarrow F_{2k+1}F_{2k+3} - F_{2k+2}^2 = 1

⇒F2(k+1)+1F2(k+1)−1−F2(k+1)2=1⇒P(k+1).\Rightarrow F_{2(k+1)+1}F_{2(k+1)-1} - F_{2(k+1)}^2 = 1 \Rightarrow P(k+1).

P(k)⇒P(k+1)P(k) \Rightarrow P(k+1) which completes the induction.

Then F2n2+1F2n+1=F2n−1\displaystyle \frac{F_{2n}^2 + 1}{F_{2n+1}} = F_{2n-1} which is an integer.

(iv)

F2n−12+1F_{2n-1}^2 + 1

=(F2n+1−F2n)2+1= (F_{2n+1} - F_{2n})^2 + 1

=(F2n+1−F2n)2+(F2n+1F2n−1−F2n2)= (F_{2n+1} - F_{2n})^2 + (F_{2n+1}F_{2n-1} - F_{2n}^2)

=F2n+12−2F2n+1F2n+F2n2+F2n+1F2n−1−F2n2= F_{2n+1}^2 - 2F_{2n+1}F_{2n} + F_{2n}^2 + F_{2n+1}F_{2n-1} - F_{2n}^2

=F2n+1(F2n+1−2F2n+F2n−1)= F_{2n+1}(F_{2n+1} - 2F_{2n} + F_{2n-1}) as required.
Reply 33
Avatar for eponymous
eponymous
2
Step I Q2 part (ii)

dont fail me latex....

[br]arctan⁡1r+arctan⁡ss+t=π4[br][br]\arctan\frac{1}{r} + \arctan\frac{s}{s+t} = \frac{\pi}{4}[br]
so from part (i);
[br]1r+ss+t1−sr(s+t)=1[br][br]\frac{\frac{1}{r}+\frac{s}{s+t}}{1-\frac{s}{r(s+t)}}= 1[br]

rearranging we get: rt=2s+t

so

r=2st\frac{2s}{t} + 1

for R to be an integer 2st \frac{2s}{t} must be an integer.

now since t and s have HCF 1, t must be equal to 1 or 2 (for 2st\frac{2s}{t}
to be an integer.

so since t= 1 or 2, substituting into: r=2st\frac{2s}{t} + 1

we get:

r= s+1 or 2s+1
Reply 34
Avatar for coffeym
coffeym
10
STEP II Q14 (the one that, had I done it RIGHT, could have got me the 1 I needed!)

(i) The sketch should be straightforward: a curve (in the shape of ln⁡x\ln{x}) between x=1 and x=k, a straight horizontal line up until x=2k, and then a negative gradient straight line which reaches the x-axis at x=4k

(ii) Noticing the less than or equal to signs, we can say that

ln⁡k=a−2bk\ln{k}=a-2bk (by substituting x=2k into the pdf) and also

a−4bk=0a-4bk=0

Solving these simultaneously yields:

a=2ln⁡k,b=ln⁡k2ka=2\ln{k}, b=\frac{\ln{k}}{2k} as required for the first part.

Now the total area under the pdf must be equal to 1, and hence

Unparseable latex formula:

\displaystyle \int^{4k}_1{\ln{x}\,dx+k\ln{k}+k\ln{k}=1

(note the areas of two of the sections are simply rectangles and triangles and so integration is overkill)

Evaluating the integral and gathering like terms, we find

3kln⁡k−k=0⇒k(3ln⁡k−1)=03k\ln{k}-k=0 \Rightarrow k(3\ln{k}-1)=0

Now as k≠0k \neq 0 we can say that k=e13k=e^{\frac13}, and by the relations we derived earlier,

a=23 ,b=16e1/3 ,k=e1/3a=\frac23\,,b=\frac{1}{6e^{1/3}}\,,k=e^{1/3}

(iii) Using as estimate for e1/3e^{1/3} close to 1, we can see that the median lies in the range k≤x≤2kk \leq x \leq 2k

Let us call the median γ\gamma, then using the idea of rectangles and triangles under our pdf as before we have:

Unparseable latex formula:

\displaystyle \int^{e^{1/3}}_1{\ln{x}\,dx+\frac13(\gamma-e^{1/3})=\frac13(2e^{1/3}-\gamma)+\frac13 e^{1/3}

by the definition of the median that:

∫−∞γf(x) dx=∫γ∞f(x) dx\displaystyle \int^{\gamma}_{-\infty}{f(x)}\,dx=\int_{\gamma}^{\infty}{f(x)}\,dx

Evaluating the integral and tidying up, we see

23γ=2e1/3−1\frac23 \gamma=2e^{1/3}-1

⇒γ=32(2e1/3−1)\Rightarrow \gamma= \frac32(2e^{1/3}-1) as required

This completes the question (why oh why didn't I get this in the first place...)
Reply 35
Avatar for Dystopia
Dystopia
10
STEP I, Q7.
i)
p1=(1+2λ)i+2λj+(2−3λ)kp_{1} = (1 + 2\lambda)\textbf{i} + 2\lambda \textbf{j} + (2-3\lambda)\textbf{k}

p2=(4+μ)i+(2μ−2)λj+(9−2μ)kp_{2} = (4 + \mu)\textbf{i} + (2\mu - 2)\lambda \textbf{j} + (9 - 2\mu)\textbf{k}

P1P2=p2−p1=(3+μ−2λ)i+(−2+2μ−2λ)j+(7−2μ+3λ)kP_{1}P_{2} = p_{2} - p_{1} = (3 + \mu - 2\lambda)\textbf{i} + (-2 + 2\mu - 2\lambda)\textbf{j} + (7 - 2\mu + 3\lambda)\textbf{k}

D2=(3+μ−2λ)2+(−2+2μ−2λ)2+(7−2μ+3λ)2D^{2} = (3 + \mu - 2\lambda)^{2} + (-2 + 2\mu - 2\lambda)^{2} + (7 - 2\mu + 3\lambda)^{2}
=(3μ−4λ−5)2+(λ−1)2+36 = (3\mu - 4\lambda - 5)^{2} + (\lambda - 1)^{2} + 36

Therefore the minimum value of D occurs when:
λ−1=0\lambda - 1 = 0

3μ−4λ−5=03\mu - 4\lambda - 5 = 0

Which gives λ=1,  μ=3,  D2=36,  Dmin=6\lambda = 1, \; \mu = 3, \; D^{2} = 36, \; D_{min} = 6

p1=3i+2j−kp_{1} = 3\textbf{i} + 2\textbf{j} - \textbf{k}

p2=7i+4j+3kp_{2} = 7\textbf{i} + 4\textbf{j} + 3\textbf{k}

ii)
In the exam I would probably have 'cheated' and used the scalar triple product, but since this is STEP I I'll use the intended method.

p1=2i+(3+α)j+5kp_{1} = 2\textbf{i} + (3 + \alpha)\textbf{j} + 5\textbf{k}
p2=(3+4kβ)i+(3+(1−k)β)j+6kβkp_{2} = (3 + 4k\beta)\textbf{i} + (3 + (1-k)\beta)\textbf{j} + 6k\beta\textbf{k}

Using the same method as above, you get:
D2=(1+4kβ)2+((1−k)β−α)2+(7+3kβ)2D^{2} = (1 + 4k\beta)^{2} + ((1-k)\beta - \alpha)^{2} + (7 + 3k\beta)^{2}

For the next bit I needed to play around with it a bit. It's fairly similar to completing the square.
D2=(α−(1−k)β)2+25(kβ+1)2+25D^{2} = (\alpha - (1-k)\beta)^{2} + 25(k\beta + 1)^{2} + 25

Minimum when:
kβ+1=0k\beta + 1 = 0

α−(1−k)β=0\alpha - (1-k)\beta = 0

Dmin=5D_{min} = 5 when α=1−1k,  β=−1k\alpha = 1 - \frac{1}{k}, \; \beta = -\frac{1}{k}

When k=0, the lines are parallel; this can easily be seen as their direction vectors become identical. In this case, they are at a constant distance of 50\sqrt{50}.
Ive got three out of four parts of STEP III Question 4 completed. I'm working on the last part now.
Finally finished III/3 (post #33 above). Brain stopped working for a couple of hours there.
STEP III, Q10: Draw the obligatory diagram, and we see that in the (x,y) coord frame, acceleration due to gravity is (−gsin⁡ϕ,−gcos⁡ϕ)(-g \sin \phi, -g \cos \phi).

Thus the equations for x and y are:

x=Vtcos⁡θ−12gt2sin⁡ϕ,y=Vtsin⁡θ−12gt2cos⁡ϕx = Vt \cos \theta - \frac{1}{2}gt^2 \sin \phi, \quad y = Vt \sin \theta - \frac{1}{2}gt^2 \cos \phi

When the particle hits the ground, y=0 (and t≠0t\neq 0), so Vsin⁡θ=12gtcos⁡ϕV \sin \theta = \frac{1}{2}gt \cos \phi. (1)

The rebound only affects the y component of the velocity, so if it is to retrace its path, we must have x˙=0\dot{x}= 0 and so Vcos⁡θ=gtsin⁡ϕV \cos \theta = gt \sin \phi. (2)

Divide (2) by (1) to get cot⁡θ=2tan⁡ϕ\cot \theta = 2 \tan \phi and so 2tan⁡θtan⁡ϕ=12 \tan \theta \tan \phi = 1 (3).

Rewrite (2) to get t=Vcos⁡θgsin⁡ϕt=\frac{V \cos \theta}{g \sin \phi} at time of landing. Substitute into the equation for x to get the maximum range:

R=V2cos⁡2θgsin⁡ϕ−g2V2cos⁡2θg2sin⁡2ϕsin⁡ϕ=V2cos⁡2θ2gsin⁡ϕ\displaystyle R = \frac{V^2 \cos^2 \theta}{g \sin \phi} - \frac{g}{2} \frac{V^2 \cos^2 \theta}{g^2 \sin^2 \phi} \sin \phi = \frac{V^2 \cos^2 \theta}{2g \sin \phi} as required.

So then 2V2gR=2V2g2gsin⁡ϕV2cos⁡2θ\displaystyle \frac{2V^2}{gR} = \frac{2V^2}{g} \frac{2g \sin \phi}{V^2 \cos^2 \theta}

=4sinϕsec⁡2θ= 4 sin \phi \sec^2 \theta

=sin⁡ϕ(4+4tan⁡2θ)= \sin \phi (4 + 4 \tan^2 \theta).

=sin⁡ϕ(4+cot⁡2ϕ)=\sin\phi(4+\cot^2 \phi) (Using (3) to tell us 2tan⁡θ=cot⁡ϕ2\tan \theta = \cot \phi).

=sin⁡ϕ(3+cosec2ϕ)=3sin⁡ϕ+cosec ϕ=\sin \phi (3+\textrm{cosec}^2 \phi) = 3 \sin \phi + \textrm{cosec } \phi as required.

Now 3sin⁡ϕ+cosec ϕ=3(3sin⁡ϕ+13sin⁡ϕ)\displaystyle 3 \sin \phi + \textrm{cosec } \phi = \sqrt{3} \left(\sqrt{3}\sin \phi + \frac{1}{\sqrt{3}\sin \phi}\right)

=3(2+(3sin⁡ϕ−13sin⁡ϕ)2)\displaystyle = \sqrt{3}\left(2+\left(\sqrt{ \sqrt{3}\sin \phi} - \sqrt{\frac{1}{\sqrt{3}}\sin \phi}\right)^2\right)

≥23\geq 2\sqrt{3}, with equality when sin⁡ϕ=13\sin \phi = \frac{1}{\sqrt{3}}.

Thus 2V2gR≥23\frac{2V^2}{gR} \geq 2\sqrt{3} with equality when sin⁡ϕ=13\sin \phi = \frac{1}{\sqrt{3}}.

Then R≤V2g3R \leq \frac{V^2}{g\sqrt{3}} with equality when sin⁡ϕ=13\sin \phi = \frac{1}{\sqrt{3}}.
(edited 13 years ago)
(Ukgea: you've put your solution to 1/4 as II/4 in your first post.)

II/4:

αsin⁡(A−B)+βcos⁡(A+B)=γsin⁡(A+B)\alpha \sin(A-B) + \beta\cos(A+B) = \gamma\sin(A+B)

⇒α(sin⁡Acos⁡B−sin⁡Bcos⁡A)+β(cos⁡Acos⁡B−sin⁡Asin⁡B)=γ(sin⁡Acos⁡B+sin⁡Bcos⁡A)\Rightarrow \alpha(\sin A\cos B - \sin B\cos A) + \beta(\cos A\cos B - \sin A\sin B) = \gamma(\sin A\cos B + \sin B\cos A)

⇒α(tan⁡A−tan⁡B)+β(1−tan⁡Atan⁡B)=γ(tan⁡A+tan⁡B)\Rightarrow \alpha(\tan A - \tan B) + \beta(1 - \tan A\tan B) = \gamma(\tan A + \tan B)

⇒tan⁡Atan⁡B+γ−αβtan⁡A+γ+αβtan⁡B−1=0\Rightarrow \tan A\tan B + \frac{\gamma -\alpha}{\beta} \tan A + \frac{\gamma +\alpha}{\beta} \tan B - 1 = 0

⇒(tan⁡A−m)(tan⁡B−n)=0\Rightarrow (\tan A-m)(\tan B-n) = 0

where

m=−(γ+αβ),n=−(γ−αβ)\displaystyle m = -\left( \frac{\gamma +\alpha}{\beta} \right) , \quad n = -\left( \frac{\gamma -\alpha}{\beta} \right)

and

mn=−1⇔(γ+αβ)(γ−αβ)=−1⇔α2=β2+γ2.mn = -1 \Leftrightarrow \left( \frac{\gamma +\alpha}{\beta} \right)\left( \frac{\gamma -\alpha}{\beta} \right) = -1 \Leftrightarrow \alpha^2 = \beta^2 + \gamma^2 .

(i)

α=2,β=3,γ=1⇒m=−3,n=1/3 \alpha=2,\quad \beta=\sqrt{3} ,\quad \gamma=1 \Rightarrow m = -\sqrt{3},\quad n=1/\sqrt{3}

A=x,B=14πA = x, B = \frac{1}{4}\pi

∴Given equation⇔(tan⁡x+3)(tan⁡14π−13)=0\therefore \text{Given equation} \Leftrightarrow (\tan x+\sqrt{3})(\tan \frac{1}{4}\pi-\frac{1}{\sqrt{3}} ) = 0

⇒tan⁡x=−3⇒x=2π3,    5π3.\Rightarrow \tan x = -\sqrt{3} \Rightarrow x = \frac{2\pi}{3} ,\;\; \frac{5\pi}{3} .

(ii)

m=−3,n=1/3,A=x,B=16πm = -\sqrt{3} ,\quad n = 1/\sqrt{3} ,\quad A = x ,\quad B = \frac{1}{6}\pi

∴Given equation⇔(tan⁡x+3)(tan⁡16π−13)=0\therefore \text{Given equation} \Leftrightarrow (\tan x+\sqrt{3})(\tan \frac{1}{6}\pi-\frac{1}{\sqrt{3}} ) = 0

But tan⁡16π=13\tan \frac{1}{6}\pi=\frac{1}{\sqrt{3}} and so the equation is an identity.

(iii)

m=−3,n=1/3,A=2x+16π,B=x−16πm = -\sqrt{3} ,\quad n = 1/\sqrt{3} ,\quad A = 2x + \frac{1}{6}\pi ,\quad B = x - \frac{1}{6}\pi

∴Given equation⇔(tan⁡A+3)(tan⁡B−13)=0\therefore \text{Given equation} \Leftrightarrow (\tan A+\sqrt{3})(\tan B-\frac{1}{\sqrt{3}} ) = 0

⇒A=2π3,    5π3,    8π3,    11π3,B=π6,    7π6\Rightarrow A = \frac{2\pi}{3} ,\;\; \frac{5\pi}{3} ,\;\; \frac{8\pi}{3} ,\;\; \frac{11\pi}{3} ,\quad B = \frac{\pi}{6} ,\;\; \frac{7\pi}{6}

⇒x=π4,    π3,    3π4,    5π4,    4π3,    7π4.\Rightarrow x = \frac{\pi}{4} ,\;\; \frac{\pi}{3} ,\;\; \frac{3\pi}{4} ,\;\; \frac{5\pi}{4} ,\;\; \frac{4\pi}{3} ,\;\; \frac{7\pi}{4} .

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you