HSC 2013 Maths Marathon (archive) (1 Viewer)

HeroicPandas · Oct 7, 2013

Re: HSC 2013 2U Marathon

$$Evaluate $\int_{\frac{-\pi}{4}}^0 \frac{dx}{\sqrt{cosec^2x - 1}}$

mahmoudali · Oct 7, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
$$Evaluate $\int_{\frac{-\pi}{4}}^0 \frac{dx}{\sqrt{cosec^2x - 1}}$

$cosec^2 x -1 = cot^2 x \ \ \ $simple substitution into f(x) $ \ \ \int_{\frac{-\pi}{4}}^0 \ \ \frac{1}{\sqrt{cot^2x}} dx \ \$
$\ \ \int_{\frac{-\pi}{4}}^0 \frac{1}{|cotx|} dx \ \ = \ \ \ int_{\frac{-\pi}{4}}^0 |tanx|\ \ = \ \ $(-ln(cosx)) _\frac{-\pi}{4}^0$$

$|-ln(1)| -|- ln(\frac{1}{\sqrt{2}})| \ \ = \ \ ln(\sqrt{2})$

Menomaths · Oct 7, 2013

Re: HSC 2013 2U Marathon

Code:

\sqrt{x}

$\sqrt{x}$

Sy123 · Oct 7, 2013

Re: HSC 2013 2U Marathon

$Consider the graph$ \ \ y= \frac{1}{x} \ \ $two points$ \ \ A\left(a,\frac{1}{a} \right) \ \ $and$ \ \ B\left(b, \frac{1}{b} \right) \ \ $lie on the graph$

$$i) Show that the straight line$ \ \ AB \ \ $is above the hyperbola$ \ \ \ \fbox{1}$

$$ii) Use the trapezium rule with$ \ \ n$

$function values and part (i) to show that$

$2\int_1^n \frac{1}{x} \ dx + 1 + \frac{1}{n} < 2\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \right) \ \ \ \fbox{3}$

$iii) Hence show that$

$\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > \ln n + \frac{1}{2} + \frac{1}{2n} \ \ \fbox{1}$

$iv) Explain what happens to$

$\sum_{k=1}^{N} \frac{1}{k} \ \ \ \ $as$ \ \ N \to \infty \ \ \fbox{1}$

mahmoudali · Oct 7, 2013

Re: HSC 2013 2U Marathon

too easy sy

integral95 · Oct 7, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$Consider the graph$ \ \ y= \frac{1}{x} \ \ $two points$ \ \ A\left(a,\frac{1}{a} \right) \ \ $and$ \ \ B\left(b, \frac{1}{b} \right) \ \ $lie on the graph$

$$ii) Use the trapezium rule with$ \ \ n$

$function values and part (i) to show that$

$2\int_1^n \frac{1}{x} \ dx + 1 + \frac{1}{n} < 2\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \right) \ \ \ \fbox{3}$

$iii) Hence show that$

$\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > \ln n + \frac{1}{2} + \frac{1}{2n} \ \ \fbox{1}$

for part ii) I got $\int_1^n \frac{1}{x} \ dx + 1 + \frac{1}{n} < 2\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \right) \ \ \ \fbox{3}$

HeroicPandas · Oct 7, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$cosec^2 x -1 = cot^2 x \ \ \ $simple substitution into f(x) $ \ \ \int_{\frac{-\pi}{4}}^0 \ \ \frac{1}{\sqrt{cot^2x}} dx \ \$
$\ \ \int_{\frac{-\pi}{4}}^0 \frac{1}{cotx} dx \ \ = \ \ \int_{\frac{-\pi}{4}}^0 tanx\ \ = \ \ $(-ln(cosx))_\frac{-\pi}{4}^0$$

$-ln(1) - ln(\frac{1}{\sqrt{2}}) \ \ = \ \ ln(\sqrt{2})$

You need to provide more working out when integrating tanx with respect to x

also $\sqrt{cot^2x} = \left | cotx \right |$

integral95 · Oct 7, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
You need to provide more working out when integrating tanx with respect to x

also $\sqrt{cot^2x} = \left | cotx \right |$

it's funny cos he somehow got the right answer with some errors in the working

Sy123 · Oct 7, 2013

Re: HSC 2013 2U Marathon

MethewYan said:
for part ii) I got $\int_1^n \frac{1}{x} \ dx + 1 + \frac{1}{n} < 2\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \right) \ \ \ \fbox{3}$

Hm, was this your expression:

$\int_1^n \frac{1}{x} \ dx< \frac{1}{2}\left(1 + \frac{1}{n} 2\left( \frac{1}{2} + \dots + \frac{1}{n-1} \right) \right)$

??

That is what I got initially

mahmoudali · Oct 8, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
You need to provide more working out when integrating tanx with respect to x

also $\sqrt{cot^2x} = \left | cotx \right |$

i skipped so many steps because i cbfed showing it using this latex bs wastes to much time
and mathewYan can you please tell me where i made mistakes?

Sy123 · Oct 8, 2013

Re: HSC 2013 2U Marathon

$$The line$ \ \ y=mx \ \ $is tangent to the curve$ \ \ y=\ln x$

$Find$ \ \ m \ \ $and hence find the area of the region bounded by the line, the curve and the$ \ \ x$-axis$

HeroicPandas · Oct 8, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
i skipped so many steps because i cbfed showing it using this latex bs wastes to much time
and mathewYan can you please tell me where i made mistakes?

at least in the end of your working, have a quick outline how you integrated tanx with respect to x (unless you memorised it)

AND, have you ever learnt how to integrate something that has an absolute?

$|cotx| = \pm cotx$ (gotta choose plus or minus carefully)

integral95 · Oct 8, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
Hm, was this your expression:

$\int_1^n \frac{1}{x} \ dx< \frac{1}{2}\left(1 + \frac{1}{n} 2\left( \frac{1}{2} + \dots + \frac{1}{n-1} \right) \right)$

??

That is what I got initially

Lol whoops forgot about the 1/2 at the front of the bracket, (sigh I forgot about trapezoidal rule )

Sy123 · Oct 9, 2013

Re: HSC 2013 2U Marathon

$$i) Differentiate$ \ \ x^{a+1} \ln x$

$$ii) Hence find$ \ \ \int_1^e x^{a} \ln x \ dx$

mahmoudali · Oct 9, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$The line$ \ \ y=mx \ \ $is tangent to the curve$ \ \ y=\ln x$

$Find$ \ \ m \ \ $and hence find the area of the region bounded by the line, the curve and the$ \ \ x$-axis$

$\frac{dy}{dx} = \frac{1}{x} \ \ $make the gradient function equal to m$ \ \ \frac{1}{x} = m$

$x= \frac{1}{m} \ \ $substitute into y and you get y=m$ \frac{1}{m} = 1$

$since y= 1 =ln(x) $therefore\ \ point \ \ of \ \tangent$ (1,e)$

$Now the area is found by finding$ \int_0^e \ \ e^y dy - \int_0^1 \frac{1}{e}x \ \ dx = $whatever it gives you$

$fingers crossed$

Sy123 · Oct 9, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$\frac{dy}{dx} = \frac{1}{x} \ \ $make the gradient function equal to$ \ \ m \Rightarrow \ \ \frac{1}{x} = m$

$x= \frac{1}{m} \ \ $substitute into y and you get$ \ y=m \frac{1}{m} = 1$

$$since$ \ \ y= 1 =\ln(x) \ \ $therefore point of tangent$ (1,e)$

$Now the area is found by finding$ \ \ \int_0^e \ \ e^y dy - \int_0^1 \frac{1}{e}x \ \ dx = \ \ $whatever it gives you$

$fingers crossed$

Fixed latex, that is the right idea anyway, if you didn't make any mistakes its right

EDIT: Second integral is wrong
EDIT2: First integral is wrong as well

mahmoudali · Oct 9, 2013

Re: HSC 2013 2U Marathon

why does my writing part leave no spaces

Menomaths · Oct 9, 2013

Re: HSC 2013 2U Marathon

$dollar signs$

superSAIyan2 · Oct 9, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$cosec^2 x -1 = cot^2 x \ \ \ $simple substitution into f(x) $ \ \ \int_{\frac{-\pi}{4}}^0 \ \ \frac{1}{\sqrt{cot^2x}} dx \ \$
$\ \ \int_{\frac{-\pi}{4}}^0 \frac{1}{|cotx|} dx \ \ = \ \ \ int_{\frac{-\pi}{4}}^0 |tanx|\ \ = \ \ $(-ln(cosx)) _\frac{-\pi}{4}^0$$

$|-ln(1)| -|- ln(\frac{1}{\sqrt{2}})| \ \ = \ \ ln(\sqrt{2})$

Although your final answer is correct, your working out suggests that the answer is -ln(root2).

When you square root the cot^2 x it becomes abs (cotx). Now the integral is defined over the interval x = -pi/4 to x = 0. In this region cotx is negative, so abs(Cotx) = -cotx. Therefore you must put a minus sign in front of the integral.

mahmoudali · Oct 9, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$i) Differentiate$ \ \ x^{a+1} \ln x$

$$ii) Hence find$ \ \ \int_1^e x^{a} \ln x \ dx$

i

$x^{a+1}ln(x) \ \ u= x^{a+1}\ \ u'=(a+1) x^a , \ \ v=ln(x) \ \ v'= \frac{1}{x}$

$ln(x)(a+1)x^a \ \ + \ \ \frac{1}{x} x^{(a+1)} = ln(x)(a+1)x^a + x^a$

ii

$\int_1^e \ \ x^{a} \ln x \ dx = \frac{1}{a+1} \big_((x^{(a+1)}\ \ ln(x))- \frac{x^{(a+1)}}{a+1})_1^e$
$By manipulating the equation in part i and shit$
.
$$after substitution$ \frac{1}{1+a} (e^{(a+1)} -\frac{e^{(a+1)}}{a+1} + \frac{1}{a+1})$
.
$\frac{e^{a+1}}{a+1} -\frac{e^{a+1} +1}{(a+1)^2}$
.
$$ after simplifying$ \frac{ae^{(a+1)} +1}{(a+1)^2}$
too much shit to show on latex

HSC 2013 Maths Marathon (archive) (1 Viewer)

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