3 Unit Maths HSC Exam Revision (3 Viewers)

hscishard · Jul 9, 2010

3log(9/5)

There is a discontinuity at x=4.

random-1006 · Jul 9, 2010

$Find \int \frac{dx}{9x^2 + 24x +16}$ , i like this one

random-1006 · Jul 9, 2010

$\text { Are there any real values of a and b, with a less than b such that} \int_{a}^{b} \frac{dx} {2x -4x^2 -1} > 0$ .

If so, give a pair of values that satisfies the condition. If not, explain why not

I made this one up myself, its amazing the sort of questions you can make up if you really think about the content.

hscishard · Jul 9, 2010

random-1006 said:
$Find \int \frac{dx}{9x^2 + 24x +16}$ , i like this one

-1/3(3x+4) + C

NOOO stupid C

random-1006 · Jul 9, 2010

hscishard said:
-1/3(3x+4)

ohh wheres your C lol

hscishard · Jul 9, 2010

random-1006 said:
$\text { Are there any real values of a and b, with a less than b such that} \int_{a}^{b} \frac{dx} {2x -4x^2 -1} > 0$ .

If so, give a pair of values that satisfies the condition. If not, explain why not

I made this one up myself, its amazing the sort of questions you can make up if you really think about the content.

No, the graph is a negative definite, thus the definite integral is always negative as b is greater than a

Oh sht it's the reciprocal.
Still..the graph is still negative

random-1006 · Jul 9, 2010

hscishard said:
No, the graph is a negative definite, thus the definite integral is always negative as b is greater than a

nice

random-1006 · Jul 9, 2010

hscishard said:
No, the graph is a negative definite, thus the definite integral is always negative as b is greater than a

Oh sht it's the reciprocal

lol, your lucky, it doesnt make a difference, didnt you notice that first time, 1/ negative is still negative

Trebla · Jul 9, 2010

Hopefully part d) will make you think lol.

$a) Prove that $\quad\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left (\dfrac{x+y}{1-xy} \right )\\\\$b) Prove that $ \tan^{-1}(-x)=-\tan^{-1}x\\\\$c) Now consider $\int_{-2}^{3}\dfrac{dx}{1+x^2}\quad$ explain why $ \int_{-2}^{3}\dfrac{dx}{1+x^2} > 0\\\\$d) A student evaluates the integral in c) as follows:$\\\\\int_{-2}^{3}\dfrac{dx}{1+x^2}\\\\=\left [\tan^{-1}x\right]_{-2}^3\\=\tan^{-1}3-\tan^{-1}(-2)\\=\tan^{-1}3+\tan^{-1}2 \quad ($using result in b)$)\\=\tan^{-1}\left (\dfrac{3+2}{1-(3)(2)}\right)\quad ($using result in a$))\\\\=\tan^{-1}(-1)\\\\=-\dfrac{\pi}{4}\\\\\therefore \int_{-2}^{3}\dfrac{dx}{1+x^2} <0\\\\$Why is there a contradiction? Explain the flaw.$

random-1006 · Jul 9, 2010

Trebla said:
Hopefully part d) will make you think lol.

$a) Prove that $\quad\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left (\dfrac{x+y}{1-xy} \right )\\\\$b) Prove that $ \tan^{-1}(-x)=-\tan^{-1}x\\\\$c) Now consider $\int_{-2}^{3}\dfrac{dx}{1+x^2}\quad$ explain why $ \int_{-2}^{3}\dfrac{dx}{1+x^2} > 0\\\\$d) A student evaluates the integral in c) as follows:$\\\\\int_{-2}^{3}\dfrac{dx}{1+x^2}\\\\=\left [\tan^{-1}x\right]_{-2}^3\\=\tan^{-1}3-\tan^{-1}(-2)\\=\tan^{-1}3+\tan^{-1}2 \quad ($using result in b)$)\\=\tan^{-1}\left (\dfrac{3+2}{1-(3)(2)}\right)\quad ($using result in a$))\\\\=\tan^{-1}(-1)\\\\=-\dfrac{\pi}{4}\\\\\therefore \int_{-2}^{3}\dfrac{dx}{1+x^2} <0\\\\$Why is there a contradiction? [B]Explain the flaw[/B].$

Magic

cutemouse · Jul 9, 2010

Trebla said:
Hopefully part d) will make you think lol.

$a) Prove that $\quad\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left (\dfrac{x+y}{1-xy} \right )\\\\$b) Prove that $ \tan^{-1}(-x)=-\tan^{-1}x\\\\$c) Now consider $\int_{-2}^{3}\dfrac{dx}{1+x^2}\quad$ explain why $ \int_{-2}^{3}\dfrac{dx}{1+x^2} > 0\\\\$d) A student evaluates the integral in c) as follows:$\\\\\int_{-2}^{3}\dfrac{dx}{1+x^2}\\\\=\left [\tan^{-1}x\right]_{-2}^3\\=\tan^{-1}3-\tan^{-1}(-2)\\=\tan^{-1}3+\tan^{-1}2 \quad ($using result in b)$)\\=\tan^{-1}\left (\dfrac{3+2}{1-(3)(2)}\right)\quad ($using result in a$))\\\\=\tan^{-1}(-1)\\\\=-\dfrac{\pi}{4}\\\\\therefore \int_{-2}^{3}\dfrac{dx}{1+x^2} <0\\\\$Why is there a contradiction? Explain the flaw.$

For part (d) it's something to do with the fact that you need to do 2pi - pi/4 for some reason... I think ... Will have a look later.

Maybe because in actual fact the result in (a) should be stated as arctana + arctanb=arctan[(a+b)/(1-ab)] +2*pi*k (for appropriate integral value of k) due to some trigonometric property...?

random-1006 · Jul 9, 2010

when we are proving the formula part 1, we are assuming x is acute

A= arctanx, B = arctany
tanA=x, tanB= y
etc.

the above step is assuming acute x, and the limits of integration are 3 and -2 RADIANS ( which is larger than pi/2, ie larger than 90 degrees, not an acute angle.

either that or something about the range of arctan is : -pi/2 < y < pi/2

Trebla · Jul 9, 2010

jm01 said:
Maybe because in actual fact the result in (a) should be stated as arctana + arctanb=arctan[(a+b)/(1-ab)] +2*pi*k (for appropriate integral value of k) due to some trigonometric property...?

Note quite. If you remember the way that the y = tan^-1x was defined, it requires that for each x there must be a unique value of y. The addition of 2kπ seems to imply non-uniqueness.

random-1006 said:
when we are proving the formula part 1, we are assuming x is acute

A= arctanx, B = arctany
tanA=x, tanB= y
etc.

the above step is assuming acute x, and the limits of integration are 3 and -2 RADIANS ( which is larger than pi/2, ie larger than 90 degrees, not an acute angle.

x and y are not "angles". The inverse tangent of them are "angles". So from your labelling, A and B are the angles, not x and y. Even with that correction, tan^-14 and tan^-12 both give acute angles, so that doesn't quite explain the flaw.

cutemouse · Jul 9, 2010

hscishard said:
There is a discontinuity at x=4.

Well technically speaking you should say that there's a gap in the domain. Talking about continuity at a point not in the domain of the function doesn't make sense.

Although at HSC level your reasoning should suffice.

Trebla · Jul 10, 2010

random-1006 said:
something about the range of arctan is : -pi/2 < y < pi/2

You're on the right track with that...

cutemouse · Jul 10, 2010

I know...

$\quad\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left (\dfrac{x+y}{1-xy} \right )\\\\$

Implies that the LHS and RHS must both be within the (-pi/2, pi/2)... if the LHS is outside this range then the equation is not defined/meaningless, which is the case if the LHS = arctan3+arctan2...

Trebla · Jul 10, 2010

Bingo

random-1006 · Jul 12, 2010

Rezen · Jul 12, 2010

random-1006 said:
$Find \ \frac{d}{dx} [\int_{0}^{10} e^{x^2} \ dx]$

isn't it 0?

3 Unit Maths HSC Exam Revision (3 Viewers)

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