Flex your maths muscles! (1 Viewer)

[LoneShadow]

[Uni people, please let HSC students do the questions]
[I have put the question in a pdf file. I'll type up the naswers as well sometime. Note that the order of questions have changed]
Harder quaestiopns for Maths Ext 2 students:
Some of these questions may prove a bit too difficult.
Your welcome to add your own problems from different areas of HSC Ext 2 Maths.

(2) Show that ∫₀¹{(1-x²)ⁿ dx} = (2²ⁿ(n!)²)/(2n+1)! [Solved by Yip]

(3) Evaluate ∫dx/(x⁷ - x) (Solved by Yip using partial fractions; & also done by icycloud using a nicer method)

(4) [Easier] (Solved by Yip)
(a) Prove that if ƒ is a continuous function, then ∫₀^aƒ(x)dx = ∫₀^aƒ(a - x)dx
(b) Use part (a) to show that ∫₀^π/2sinⁿx dx/(sinⁿx + cosⁿx) = π/4

(5) Show ∫₀^π(xdx)/[1+ cos²x] = π²/(2√2) [Added by Yip. Solved by icycloud]

(6) Evaluate: (Hint: Use substitution) [Solved by Yip]
(a) ∫dx/(√x - ³√x)
(b) ∫dx/(³√x + ⁴√x)

(7) Determine the type of curve represented by the equation x²/k + y²/(k-16) =1 in each of the following cases: [Solved by Yip]
(a) k>16;
(b) 0< k <16; and
(c) k<0.
(d) show that the curves in parts (a) and (b) have the same foci, no matter what the value of k is.

Apparently thse questions are beyond the scope of HSC. So Ignor them, unless you want to and can do them.
(1) Use Integration by parts to show that, for all x>0, 0<&int;₀^&infin;{(sin(t)dt)/ln(1+x+t)}<2/ln(1+x) [Seems this question is a bit too hard]

(8) Evaluate lim_x&rarr;0(e^x - 1 - x)/x² [Solved by Yip using L'Hospital's Rule. It can be done through Taylor Series as well]

(9) Find the first three nonzero terms in the Maclaurin series for:
(a) e^xsinx
(b) tanx

(10) Evaluate &int;e^-x² dx as an infinite series [i.e. Integrate the Taylor/Maclaurin series for &fnof;(x) = e^-x²]

(11) Find the Maclaurin series for the function &fnof;(x) = 1/&radic;(4 - x)

(12) Application of Taylor Polynomials to Physics:
[Difficult]
In Einstein's theory of special relativity the mass of an object moving with velocity v is m = m_o/&radic;(1 - v²/c²) where m_o is the mass of the object when at rest and c is the speed of light. The Kinetic energy of the object is the difference between its total energy and its energy at rest: K = mc² - m_oc²

Show that when v is very small compared with c, this expression for K agrees with classical Newtonian physics: K = &frac12;m_ov²

(13) [Difficult]A paper drinking cup filled with water has the shape of a cone with height h and semivertical angle &theta;. A ball is placed carefully in the cup, thereby displacing some of the water and making it overflow. What is the radius of the ball that causes the greatest volume of water to spill out of the cup.
(17) [Difficult, unless you cheat] A curve called the folium of Descartes is defined by the parametric equations:
x = 3t/(1+t³); y = 3t²/(1+t³)
(a) Show that the curve is symetric with respect to the line y = x. Where does the curve intersect this line?
(b) Find the points on the curve where the tangent lines are horizontal or vertical.
(c) Show that the line y = -x - 1 is an asymptote. [correct name for it is slant asymptote]
(d) Sketch the curve
(e) Find the Cartesian equation of this curve.
(f) Find the area enclosed by the loop of this curve.

(14) Area and Volumes: [Solved by Yip. Haven't checked the answer. Do the checking for yourself. I'm too sleepy]
(a) Find area enclosed between x = y² -4y and x = 2y - y²
(b) Find the volume of folid obtained by rotating the region in (a) about x-axis.
(c) Show that the volume of a segment of height h of a sphere of radius r is: V = &pi;h²(3r - h)/3

(15)[Easier] Find derivative of the function. [Solved by pLuvia]
(a) g(x) = cos^-1(3-2x)
(b) y = &radic;tan^-1x
(c) y = sin^-1(2x + 1)
(d) y = cos^-1(e^2x)

(16) Evaluate∫₀^π dx/(2-cosx)² Added by the grooving V. aka acmilan

(18) Find the equation for the ellipse that shares a vertex and a focus with the parabola x² + y = 100 and that has its other focus at the origin.

(19) The point P(a*cos&theta;, b*sin&theta; ) lies on the ellipse with equation: x²/a² + y²/b² = 1, with a>b.
The foci of the ellipse are S and S' and M, M' are the feet of the perpendiculars from P onto the directrices corresponding to S and S'.
The normal to the ellipse at P meets the major axis of the ellipse at H.
(a) Prove that SP + S'P = 2a [Done by STx]
(b) Show that the coordinates of H are ([(a² - b²)cos&theta;]/a, 0)[Done by STx]
(c) Show that HS/HS' = |1 -e*cos&theta;|/(1 + ecos&theta; ) = PS/PS'
(d) Show PH bisects angle SPS'

(20) [Solved by pLuvia. Efficient method]The point P(a*cos&theta;, b*sin&theta; ) is any point on the ellipse: x²/a² + y²/b² = 1, with focus S.
The point M is the midpoint of the interval SP.
Show that as P moves on the ellipse, M lies on another ellipse whose centre is midway between the origin O and the focus S.

(21) P(cp, c/p) and Q(cq, c/q) lie on hyperbola xy = c². Tangents TP and TQ contact the hyperbola at P and Q. The point M is the midpoint of the chord PQ.
(a) Find the coordinates of T and M.
(b) KMLT is a rectangle whose sides are parallel to the x and y axes. Show that K and L are points on the hyperbola.

(22) If M = &int;_ln2^ln4{e^tdt/(e^t + e^-t)} and N = &int;_ln2^ln4{e^-tdt/(e^t + e^-t)}, find the exact values of:
(a) M + N
(b) M - N
(c) M
(d) N

(23) [Scary looking but very simple. Solved by Riviet]Find &int;(1 + tan²&theta; )e^tan&theta;d&theta;

(24)
(a) Show that, if I_n = &int;₀ⁱ(1+x²)ⁿdx, then I_n = {2n/(2n+1)}I_n-1
(b) Hence find &int;₀ⁱ(1 + x²)³dx

(25) [Easy]Show that, if 0< x < &pi;/2, then sin(5a + 3)x/sin3x - sin(5a - 3)x/sin3x = 2cos(5ax)

(26) [Added by STx]

(27) [Added by STx]
Let the points A₁,A₂,...,A_n represent the nth roots of unity, w₁,w₂,...,w_n, and suppose P represents any complex number z such that |z|=1.

(i) Prove that w₁,w₂,...,w_n= 0

(ii) Show that PA_i² = (z-w_i)(z_bar - w-bar_i) for i=1,2,...,n.

(iii) Prove that ∑_i=1ⁿ PA_i²=2n

 

[Yip]

(2) Let In=∫{(1-x^2)^n dx} [0 to 1]

In=x(1-x^2)^n|(0 to 1) +2n∫x^2(1-x^2)^(n-1)dx [0 to 1]
=(0-0) -2n∫(1-x^2-1)(1-x^2)^(n-1)dx
=-2n[In-I(n-1)]
=2n[I(n-1)]-2n(In)

(2n+1)In=2n[I(n-1)]
In=[2n/(2n+1)]I(n-1)
=[2n/(2n+1)][2(n-1)/2[(n-1)+1]]I(n-2)
=(2^n)[(n-1)(n-2)........(2)(1)]/(2n+1)(2n-1)(2n-3).......
Multiplying by [2n(2n-2)(2n-4)......]/[2n(2n-2)(2n-4)......],
In=2^(2n)[n!]^2/(2n+1)!

 

[KeypadSDM]

Look, I know you told me not to do it, but question 3 is too easy for this heading...

Spoiler

Partial fractions yields something quite nice if you can see the symmetry. It's the x-factor in the bottom, everything just falls out.

See, I didn't solve it either.

FOCUS: I swear I had question 4 in an exam, or a past paper. Where'd you see it?

 

[Yip]

q4a) Prove ∫₀^aƒ(x)dx = ∫₀^aƒ(a - x)dx

Let x=a-t
dx=-dt

∫₀^aƒ(x)dx=-∫ƒ(a-t)dt[bottom a top 0]
=∫₀^af(a-t)dt=∫₀^aƒ(a-x)dx

(b)Show ∫₀^π/2sinⁿx dx/(sinⁿx + cosⁿx)=π/4

Let x=π/2-t
dx=-dt
∫₀^π/2sinⁿx dx/(sinⁿx + cosⁿx)=∫₀^π/2sinⁿ(π/2-t) dt/[sinⁿ(π/2-t) + cosⁿ(π/2-t)]
=∫₀^π/2cosⁿt dt/(sinⁿt + cosⁿt)=∫₀^π/2cosⁿx dx/(sinⁿx + cosⁿx)

Hence
2∫₀^π/2sinⁿx dx/(sinⁿx + cosⁿx)=∫₀^π/2sinⁿx dx/(sinⁿx + cosⁿx)+∫₀^π/2cosⁿx dx/(sinⁿx + cosⁿx)=∫₀^π/2dx=π/2
∫₀^π/2sinⁿx dx/(sinⁿx + cosⁿx)=π/4



q3.

Break 1/(x7 - x) into -1/2[[1/x(x^3+1)]-[1/x(x^3-1)]]
From here, use partial fractions

A(x^3+1)+x(Bx^2+Cx+D)=1
A=-1, B=1, C=0, D=0
E(x^3-1)+x(Fx^2+Gx+H)=1
E=-1, F=-1, G=0, H=0

I=-1/2∫[{(1/x)-(x^2/[x^3+1])}-{(-1/x)+(x^2/[x^3-1])}]dx
=-1/2[2log|x|-1/3[log|x^3+1]-1/3[log|x^3-1|]]
=-log|x|+1/6[log|x^6-1|+c

I have a q to add:

Show ∫₀^π(xdx)/[1+ cos²x]=(π^2)/2(root2)


 

[LoneShadow]

Look, I know you told me not to do it, but question 3 is too easy for this heading...

Spoiler

Partial fractions yields something quite nice if you can see the symmetry. It's the x-factor in the bottom, everything just falls out.

See, I didn't solve it either.

FOCUS: I swear I had question 4 in an exam, or a past paper. Where'd you see it?

Yeah I'd done it before in a half yearly exam in yr 12. I found it in one of my Calculus Books.

 

[icycloud]

(3) Evaluate ∫dx/(x^7 - x)

I=∫dx/(x^7 - x)
=∫dx/(x^4 [x^3 - x^-3])

u = x^-3, du=-3x^-4

I = -1/3∫du/(1/u-u)
=-1/3∫udu/(1-u^2)
=1/6∫d(1-u^2)/(1-u^2)
=1/6ln|1-u^2|+C
=1/6ln|1-x^-6|+C

(5) Show ∫0-->π(xdx)/[1+ cos^2x] = π^2/(2√2)

I = ∫{0-->pi} xdx/(1+cos^2x)
= ∫{0-->pi}(pi-x)dx/(1+cos^2(pi-x)) {property proved by Yip above, won't repeat it}
= ∫{0-->pi}(pi-x)dx/(1+cos^2x)

Thus 2I = ∫{0-->pi}(x+pi-x)/(1+cos^2(x)) dx
I=pi/2 ∫{0-->pi} dx/(1+cos^2(x))

Let u=x-pi/2, du=dx

I=pi/2 ∫{-pi/2 --> pi/2} du/(1+sin^2(u)) {after simplifying}
=pi ∫{0-->pi/2} du/(1+sin^2(u)) {even function}
=2pi ∫{0-->pi/2} du/(3-cos(2u)) {apply sin^2(u)=1/2-cos(2u)/2}
=pi ∫{0-->infinity} dt/(1+2t^2) {apply t=tan(u) then simplify}
=pi/sqrt(2) {arctan(sqrt{2t})} | {0-->infinity}
=pi/sqrt(2) {pi/2 - 0}
=pi^2 / (2sqrt(2)), as required

I think there's an easier way for this lol...

 

[Yip]

6.(a) Evaluate ∫dx/(√x - ³√x)

Let x=t^6
dx=6t^5dt

∫dx/(√x - ³√x)=∫6t^5dt/(t^3-t^2)
=6∫(t^3-1+1)dt/(t-1)
=6∫{(t^2+t+1)+[1/(t-1)]}dt
=2t^3+3t^2+6t+6log|t-1|+c
=2√x+3(³√x)+6(⁶√x)+6log|⁶√x-1|+c

(b) Evaluate ∫dx/(³√x + ⁴√x)

Let x=t^12
dx=12t^11dt

∫dx/(³√x + ⁴√x)=∫12t^11dt/(t^4+t^3)
=12∫(t^8-1+1)dt/(t+1)
=12∫{(t^7-t^6+t^5-t^4+t^3-t^2+t-1)+[1/(t+1)]}dt
=(3/2)t^8-(12/7)t^7+2t^6-(12/5)t^5+3t^4-4t^3+6t^2-12t+log|t+1|+c
=(3/2)(x^2/3)-(12/7)(x^7/12)+2(x^1/2)-(12/5)(x^5/12)+3(x^1/3)-4(x^1/4)
+6(x^1/6)-12(x^1/12)+log((x^1/12)+1)+c
7.(a) Ellipse with major axis on x axis
(b) Hyperbola with extreme points on x-axis
(c) x2/k + y2/(k-16) =1
(k-16)x^2+ky^2=k(k-16)
When k<0, LHS<0 since x^2, y^2>0 for all real x, RHS>0, so there is no curve
(d)When k>16,
k-16=k(1-e^2)
e=4/rootk
a=rootk
Foci: (+-4,0)
When 0<k<16,
16-k=k(e^2-1)[since e>0]
e=4/rootk
a=rootk
Foci: (+-4,0)
Thus, (a) and (b) have the same foci regardless of k

8.Evaluate limx→0(ex - 1 - x)/x2
e^0-1-0=0, 0^2=0
Thus since it is in 0/0 form, we can apply l'hospitals rule
differentiating top and bottom,
limx→0(e^x - 1 - x)/x2=limx→0(e^x-1)/2x
Subbing in 0 gives the indeterminate form 0/0 yet again, so repeating,
limx→0(e^x-1)/2x=limx→0(e^x/2)=1/2

 

[LoneShadow]

very nicely done Yip. Could you please place a few brackets for part 6(a) as it mas confuse people between power and root. Also for 6(b) I think you mistyped log(x¹²+1) instead of log(x^1/12+1)

 

[Yip]

9. (a)
e^x=1+x+(x^2)/4+(x^3)/6+(x^4)/24+....... since if u differentiate this series u are returned with the same series
Let f(x)=sinx
f'(x)=cosx
f''(x)=-sinx
f'''(x)=-cosx
f''''(x)=sinx
So it can be seen that further derivatives continue this cycle
f(2kth derivative)(0)=0, f(2k+1)th derivative (0) =(-1)^k
the (2k+1)th term of the sinx series is (-1)^k/(2k+1)!
So the maclaurin series for sinx is
sinx=x-(x^3)/6+(x^5)/120-(x^7)/5040+.......

e^xsinx=(1+x+x^2/4+x^3/6+x^4/24+.......)(x-(x^3)/6+x^5/120-x^7/5040+.......)
=x+x^2+[(1/2)-(1/6)]x^3+.....
=x+x^2+(1/3)x^3

(b) (tanx)(cosx)=sinx
Let tanx=a0+a1x+a2x^2+a3x^3+........
(a0+a1x+a2x^2+a3x^3+........)(1-x^2/2!+x^4/4!-......)=(x-(x^3)/3!+x^5/5!-x^7/7!+.......)[the series for cosx is found by differentiating the sinx series term by term]
Equating coefficients,
a0=0
a1=1
a2=0
a3-a1(1/2!)=-1/6
a3=1/3
a4-a2(1/2!)+a0(1/4!)=0
a4=0
a5+a1(1/4!)-a3(1/2!)=1/5!
a5=2/15
So tanx=x+(1/3)x^3+(2/15)x^5

10.Evaluate ∫e^(-x^2) dx as an infinite series

e^x=Sigma[0->infinity] [x^n/n!]
e^(-x^2)=Sigma[0->Infinity][{(-1)^n}{x^2n}]/n!
∫e^(-x^2) dx=∫Sigma[0->Infinity][{(-1)^n}{x^2n}]/n!
=Sigma[0->Infinity]{(-1)^n}x^(2n+1)/(2n+1)n!

 

[Slidey]

Taylor & MacLaurin series and L'Hospital's rules are not taught in MX2, so giving students questions which are meant to be solved via these things is a bit odd.

 

[LoneShadow]

I thought they taught Taylor Series in High School. I know L'Hospital's Rule is not taught. Yip used that method, but I had Taylor series in mind for doing that question.

Hmmmm... I'll double check the question I post from now on to make sure they are not beyong HSC. Thanks Slide Rule.

[I should blame Yip for the out of scope questions. He found them easy so i kept feeding him harder and harder questions.]

 

[pLuvia]

May I ask what the L'Hospital Rule and Taylor Series is used for?

 

[Yip]

L'Hospitals rule is used in evaluating limits,basically u just differentiate the top and bottom and take the limit of the resulting equation but it can only be used if it is in indeterminate form ie if u sub the limit value in it gives u 0/0, or +-infinity/+-infinity
http://mathworld.wolfram.com/LHospitalsRule.html
Taylor series are series that express functions as series, its useful in cases where u want to integrate a function that is not easaily integratable, in that case u just integrate its taylor series term by term.
http://mathworld.wolfram.com/TaylorSeries.html

 

[IronMaiden]

Wow, you guys are smart.

I suppose you've got 2 years on me though.

 

[pLuvia]

15.(a) g(x) = cos^-1(3-2x)
Let u=3-2x
du/dx=-2
g(x)=cos^-1u
g'(x)=-1/sqrt{1-u²}*du/dx
g'(x)=-1/[sqrt{1-(3-2x)<sup>2[/sup}]*d/dx(3-2x)
=2/[sqrt{1-(3-2x)2[/sup}]

(b) y = √tan-1x
y'=1/2(tan-1)-1/2[1/(1+x2)]
=1/[2[√tan-1x*(1+x2)]]]

(c) y = sin-1(2x + 1)
Let u=2x+1
du/dx=2
y=sin-1u
y'=1/sqrt{1-u2}*du/dx
y'=[sqrt{1-(2x+1)2}]*d/dx(2x+1)
=2/[sqrt{1-(2x+1)2}]

(d) y = cos-1(e2x)
Let u=e2x
du/dx=2e2x
y=cos-1u
y'=-1/sqrt{1-u2}*du/dx
y'=-1/[sqrt{1-e4x}]*d/dx(e2x)
=-2e2x/[sqrt{1-e4x}]

Edit: Enough working</sup>

 

[Yip]

14.(a) Find area enclosed between x = y² -4y and x = 2y - y2
2y^2-6y=0
y=0,3
A=∫₀³(2y-y^2-y^2+4y)dy
=∫₀³(6y-2y^2)dy
=[3y^2-2/3(y^3)]₀³
=27-18=9u^2

(b) Find the volume of the solid obtained by rotating the region in (a) about x-axis.

This is the same as rotating y=x^2-4x and y=2x-x^2 about the y-axis

V=2π∫₀³[x(2x-x^2)-x(x^2-4x)]dx
=2π∫₀³[6x^2-2x^3]dx
=2π[2x^3-1/2(x^4)]₀^3
=2π[54-81/2]=27πu^3

(c) V=π∫_r-h^r[r^2-x^2]dx
=π[(r^2)x-1/3(x^3)]_r-h<sup>r
=π[2/3(r^3)-(r^3-r^2h-(1/3)(r^3-3(r^2)h+3r(h^2)-(h^3))
=π[rh^2-1/3(h^3)]=[πh^2(3r-h)]/3 as required

 

[pLuvia]

Opps my bad, I'll edit it

 

[Riviet]

Also, your final answer to 15 a) should be positive because derivative of any cos^-1 function is negative and the derivative of 3-2x is negative.

 

[LoneShadow]

Yip, here's the diagram for 14(c). I should've had drawn when I posted the question. There are other parts to that sphere question, one of those parts is similar to that on MathWorld. I omitted them.

 

[pLuvia]

Editted all the bugs to that question 15