HSC 2015 Maths Marathon (archive) (1 Viewer)

LostInHSC · Dec 15, 2014

Re: HSC 2015 2U Marathon

Fizzy_Cyst said:
Solve for x:

sin^2(x) - 3sin(x)cos(x) + 2cos^2(x) = 0 where 0<=x<=2(Pi)

$$Solve for x:$ \, \, \, \sin^{2}(x) -3\sin(x)\cos(x) +2\cos^{2}x =0 \,\,\,\,\,\, $where 0 \leq x\leq 2\pi \\ \\ $Rewriting:$ \, \, \, \\ sin^{2}x -2\sin(x)\cos(x) +\cos^{2}(x) +\cos(x)\cos(x) -\sin(x)\cos(x) =0 \\ (\sin(x)-\cos(x))^{2} + \cos(x)(\cos(x) - \sin(x)) =0 \\ (\sin(x)-\cos(x))^{2} - \cos(x)(\sin(x) - \cos(x)) =0 \\ \\ $Taking$ $(\sin(x)-\cos(x))$ $out$ $as$ $a$ $common$ $factor:$ \\ (\sin(x)-\cos(x))(\sin(x)-\cos(x) -\cos(x)) =0 \\(\sin(x)-\cos(x))(\sin(x)-2\cos(x))=0 \\ $Now, $ $either$ \sin(x)=\cos(x)\, $or$ \sin(x)=2\cos(x) \,$.$ $However,$ $this$ $can$ $not$ $be$ $true.$ \\ \\ \therefore \sin(x)=\cos(x) \\\therefore \tan(x)=1 $for $ \,0 \leq x\leq 2\pi \\ \therefore x=\frac{\pi}{4} \, $,$ 5\pi/4$

InteGrand · Dec 15, 2014

Re: HSC 2015 2U Marathon

LostInHSC said:
$$Solve for x:$ \, \, \, \sin^{2}(x) -3\sin(x)\cos(x) +2\cos^{2}x =0 \,\,\,\,\,\, $where 0 \leq x\leq 2\pi \\ \\ $Rewriting:$ \, \, \, \\ sin^{2}x -2\sin(x)\cos(x) +\cos^{2}(x) +\cos(x)\cos(x) -\sin(x)\cos(x) =0 \\ (\sin(x)-\cos(x))^{2} + \cos(x)(\cos(x) - \sin(x)) =0 \\ (\sin(x)-\cos(x))^{2} - \cos(x)(\sin(x) - \cos(x)) =0 \\ \\ $Taking$ $(\sin(x)-\cos(x))$ $out$ $as$ $a$ $common$ $factor:$ \\ (\sin(x)-\cos(x))(\sin(x)-\cos(x) -\cos(x)) =0 \\(\sin(x)-\cos(x))(\sin(x)-2\cos(x))=0 \\ $Now, $ $either$ \sin(x)=\cos(x)\, $or$ \sin(x)=2\cos(x) \,$.$ $However,$ $this$ $can$ $not$ $be$ $true.$ \\ \\ \therefore \sin(x)=\cos(x) \\\therefore \tan(x)=1 $for $ \,0 \leq x\leq 2\pi \\ \therefore x=\frac{\pi}{4} \, $,$ 5\pi/4$

Why can't we have $\sin x = 2\cos x$ ?

LostInHSC · Dec 15, 2014

Re: HSC 2015 2U Marathon

InteGrand said:
Why can't we have $\sin x = 2\cos x$ ?

Oops. I was thinking it couldn't equal twice at once.
sinx=2cosx
Tanx=2
x=63.43,243.43
x=1.107 radians, 4.249 radians
Thank you.

InteGrand · Dec 15, 2014

Re: HSC 2015 2U Marathon

LostInHSC said:
Oops. I was thinking it couldn't equal twice at once.
sinx=2cosx
Tanx=2
x=63.43,243.43
x=1.107 radians, 4.249 radians
Thank you.

In the HSC, you'd probably need to give exact values, so $x = \tan ^{-1} 2, \tan ^{-1}2 + \pi$

integral95 · Dec 15, 2014

Re: HSC 2015 2U Marathon

LostInHSC said:
$$Solve for x:$ \, \, \, \sin^{2}(x) -3\sin(x)\cos(x) +2\cos^{2}x =0 \,\,\,\,\,\, $where 0 \leq x\leq 2\pi \\ \\ $Rewriting:$ \, \, \, \\ sin^{2}x -2\sin(x)\cos(x) +\cos^{2}(x) +\cos(x)\cos(x) -\sin(x)\cos(x) =0 \\ (\sin(x)-\cos(x))^{2} + \cos(x)(\cos(x) - \sin(x)) =0 \\ (\sin(x)-\cos(x))^{2} - \cos(x)(\sin(x) - \cos(x)) =0 \\ \\ $Taking$ $(\sin(x)-\cos(x))$ $out$ $as$ $a$ $common$ $factor:$ \\ (\sin(x)-\cos(x))(\sin(x)-\cos(x) -\cos(x)) =0 \\(\sin(x)-\cos(x))(\sin(x)-2\cos(x))=0 \\ $Now, $ $either$ \sin(x)=\cos(x)\, $or$ \sin(x)=2\cos(x) \,$.$ $However,$ $this$ $can$ $not$ $be$ $true.$ \\ \\ \therefore \sin(x)=\cos(x) \\\therefore \tan(x)=1 $for $ \,0 \leq x\leq 2\pi \\ \therefore x=\frac{\pi}{4} \, $,$ 5\pi/4$

Cool method

However I would simply divide both sides by

$\ \ cos^2x$

to get a quadratic in terms of tanx i.e

$sin^{2}(x) -3\sin(x)\cos(x) +2\cos^{2}x =0 \Rightarrow tan^2x -3tanx +2 = 0$

LostInHSC · Dec 15, 2014

Re: HSC 2015 2U Marathon

integral95 said:
Cool method

However I would simply divide both sides by

$\ \ cos^2x$

to get a quadratic in terms of tanx i.e

$sin^{2}(x) -3\sin(x)\cos(x) +2\cos^{2}x =0 \Rightarrow tan^2x -3tanx +2 = 0$

ah i see

$\tan^{2}(x) -3\tan(x) +2 =0 \\ (\tan(x)-2)(\tan(x)-1) =0 \\ \tan(x)=2 \, \, \, \, \, \tan(x) = 1 \\ \\ \therefore x=\frac{\pi}{4} \, ,\tan^{-1}2,\, \frac{5\pi}{4} \, , \tan^{-1}2 +\pi$

leehuan · Jan 12, 2015

Re: HSC 2015 2U Marathon

By considering the graph of $y=\sqrt { 25-{ x }^{ 2 } }$ , evaluate $\int _{ 0 }^{ 5 }{ \sqrt { 25-{ x }^{ 2 } } dx }$

PhysicsMaths · Jan 12, 2015

Re: HSC 2015 2U Marathon

leehuan said:
By considering the graph of $y=\sqrt { 25-{ x }^{ 2 } }$ , evaluate $\int _{ 0 }^{ 5 }{ \sqrt { 25-{ x }^{ 2 } } dx }$

http://imgur.com/UhdTI5U

PhysicsMaths · Jan 12, 2015

Re: HSC 2015 2U Marathon

NEXT QUESTION

Find the derivative of x^3 using first principles

LostInHSC · Jan 12, 2015

Re: HSC 2015 2U Marathon

PhysicsMaths said:
NEXT QUESTION

Find the derivative of x^3 using first principles

$\lim_{h\rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h} \\ \\ = \lim_{h\rightarrow 0}\frac{(x+h)(x^{2}+2hx+h^{2})-x^{3}}{h} \\ \\ = \lim_{h\rightarrow 0}\frac{x^{3}+2x^{2}h+h^{2}x+hx^{2}+2h^{2}x+h^{3}-x^{3}}{h} \\ \\ = \lim_{h\rightarrow 0}\frac{3x^{2}h+3h^{2}x+h^{3}}{h} \\ \\ =\lim_{h\rightarrow 0}3x^{2}+3hx+h^{2} \\ \\ =3x^{2}+3(0)x +(0)^{2} \\ \\ =3x^{2}$

Kaido · Jan 13, 2015

Re: HSC 2015 2U Marathon

Differentiate x^(1/3) by first principles

InteGrand · Jan 13, 2015

Re: HSC 2015 2U Marathon

Let $f(x) = x^{1/3}$ . The inverse function is then $g(x)=x^3$ .

We know from first principles (done in an earlier Q on this page) that $g^\prime(x)=3x^2$ , so the slope of the tangent on the graph of $y=g(x)$ at a point (t, t³) is 3t².

Hence the slope of the tangent on the corresponding point on the inverse function's graph (that is, the point (t³, t) on the graph of y = f(x)) is the reciprocal of this, which is $\frac{1}{3t^2}$ . Replacing t with $t^{\frac{1}{3}}$ (so that t³ becomes $\left (t^{\frac{1}{3}}\right )^3 = t$ ), we find that the slope on the graph of y = f(x) at the point $\left ( t,t^{\frac{1}{3}} \right )$ is $\frac{1}{3\left ( t^\frac{1}{3}\right )^2}=\frac{1}{3}t^{-\frac{2}{3}}$ .

Since the derivative $f^\prime (t)$ gives the slope of the tangent on the graph of y = f(x) at the point $(t,f(t))=\left ( t,t^{\frac{1}{3}} \right )$ , the derivative is $f^\prime (x)=\frac{1}{3}x^{-\frac{2}{3}}$ (replacing t with x).

dan964 · Jan 13, 2015

Re: HSC 2015 2U Marathon

^
here is a proper way to do it by first principles.

I have used Liebnitz notation instead of 'h'
and the limit is Δx→0 not x (that is an error)

Kaido · Jan 13, 2015

Re: HSC 2015 2U Marathon

Interesting solve by integrand. Is that legible in an exam? lol

leehuan · Jan 13, 2015

Re: HSC 2015 2U Marathon

Kaido said:
Interesting solve by integrand. Is that legible in an exam? lol

I don't see why not. The only important thing to notice however is that he used both parametrics and inverse functions, which are parts of the 3U course. (Nonetheless, I haven't heard of 3U knowledge being prohibited in a 2U paper.)

Alternate solution. Same working out, just without the substitution. The difference of two cubes isn't really avoidable.
$y=\sqrt [ 3 ]{ x } \\ \frac { dy }{ dx } =\lim _{ h\rightarrow 0 }{ \frac { \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x } }{ h } } \\ =\lim _{ h\rightarrow 0 }{ \frac { \left( \sqrt [ 3 ]{ x+h } -\sqrt [ 3 ]{ x } \right) \left( { \left( \sqrt [ 3 ]{ x+h } \right) }^{ 2 }+\sqrt [ 3 ]{ x+h } \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 } \right) }{ h\left( { \left( \sqrt [ 3 ]{ x+h } \right) }^{ 2 }+\sqrt [ 3 ]{ x+h } \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 } \right) } }$
$=\lim _{ h\rightarrow 0 }{ \frac { (x+h)-x }{ h\left( { \left( \sqrt [ 3 ]{ x+h } \right) }^{ 2 }+\sqrt [ 3 ]{ x+h } \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 } \right) } } \\ =\lim _{ h\rightarrow 0 }{ \frac { h }{ h\left( { \left( \sqrt [ 3 ]{ x+h } \right) }^{ 2 }+\sqrt [ 3 ]{ x+h } \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 } \right) } }$
$=\lim _{ h\rightarrow 0 }{ \frac { 1 }{ \left( { \left( \sqrt [ 3 ]{ x+h } \right) }^{ 2 }+\sqrt [ 3 ]{ x+h } \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 } \right) } } \\ =\frac { 1 }{ \left( { \left( \sqrt [ 3 ]{ x } \right) }^{ 2 }+\sqrt [ 3 ]{ x } \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 } \right) } \\ =\frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 2 } } }$

FrankXie · Jan 13, 2015

Re: HSC 2015 2U Marathon

Next Question

$The line $ y=2x+1 $ cuts the parabola $ x^2=4y $ at two points P and Q. Without finding the coordinates of P and Q, find the length of chord PQ.$

dan964 · Jan 14, 2015

Re: HSC 2015 2U Marathon

hmm
I would approach this by using the chord of contact formula
and calculate then the angles and lengths necessary to use either sine or cosine rule

InteGrand · Jan 16, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
Next Question

$The line $ y=2x+1 $ cuts the parabola $ x^2=4y $ at two points P and Q. Without finding the coordinates of P and Q, find the length of chord PQ.$

Let P = (x₁, y₁) (so y₁ = 2x₁ + 1), and let Q = (x₂, y₂) (so y₂ = 2x₂ + 1).

Note that the line y = 2x + 1 passes through S(0, 1), the focus of the parabola. The directrix of the parabola is the line y = -1. Let M_P be the foot of the perpendicular from P to the directrix, and let M_Q be the similar point for Q. Let X_P be the foot of the perpendicular from P to the x-axis, and X_Q be the similar point for Q.

Note that by definition of the parabola, PS = PM_P, so PS = PX_P + X_PM_P = y₁ + 1. Similarly, QS = y₂ + 1. Also, PQ = PS + QS.

Now, at the points P and Q:

y = 2x + 1 and $x^2 = 4y \Leftrightarrow y = \frac{x^2}{4}$ , so

$\frac{x^2}{4} = 2x+1$

$\Rightarrow x^2 = 8x+4$

$\Rightarrow x^2 - 8x - 4 = 0$ .

The abscissae of P and Q are the solutions to this quadratic equation, so by sum of roots, x₁ + x₂ = 8.

Therefore,

PQ = PS + QS

= (y₁ + 1) + (y₂ + 1)

= ((2x₁ + 1) + 1) + ((2x₂ + 1) + 1)

= 2(x₁ + x₂) + 4

= 2×8 + 4

= 20.

FrankXie · Jan 21, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
Next Question

$The line $ y=2x+1 $ cuts the parabola $ x^2=4y $ at two points P and Q. Without finding the coordinates of P and Q, find the length of chord PQ.$

My solution applies to more general situations, disregarding the position of the focus, we can find the length of any chord without finding the coordinates of points of intersection.
$$ Let $ P(x_1,y_1) $ and $ Q(x_2,y_2). $ Then $ y_1=2x_1+1, y_2=2x_2+1, $ and $ x_1, x_2 $ are solutions of the quadratic equation $ x^2=4(2x+1), $ or $ x^2-8x-4=0. $ sum of roots $ x_1+x_2=8, $ product of roots $ x_1x_2=-4. $ So $ (x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=80. $ Therefore, $ PQ=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(x_1-x_2)^2+(2x_1+1-2x_2-1)^2}=\sqrt{(x_1-x_2)^2+4(x_1-x_2)^2}=\sqrt{80+320}=20.$

Drsoccerball · Feb 1, 2015

Re: HSC 2015 2U Marathon

New question:
$Find the sum of:$ k^r +k^\frac{rk+1}{k} +k^\frac{rk+2}{k}+...$ for$ k<1. $Express in terms of k and r$

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