Prelim 2016 Maths Help Thread (6 Viewers)

Unravel · Jun 25, 2016

lmao why does eyeseeyou keep asking people how to state rank in maths lol. Judging by his abilities, shouldn't he be asking how to get at least a band 3?

Green Yoda · Jun 25, 2016

Unravel said:
lmao why does eyeseeyou keep asking people how to state rank in maths lol. Judging by his abilities, shouldn't he be asking how to get at least a band 3?

lmao savage

Green Yoda · Jun 25, 2016

Show that the roots of the equation (3m-5)x^2-(3m^2)x+5m^2=0 for m≠5/3, are rational if m is rational.

davidgoes4wce · Jun 25, 2016

Rathin said:
Show that the roots of the equation (3m-5)x^2-(3m^2)x+5m^2=0 for m≠5/3, are rational if m is rational.

davidgoes4wce · Jun 25, 2016

Denominator term cannot equal to 0

InteGrand · Jun 25, 2016

Rathin said:
Show that the roots of the equation (3m-5)x^2-(3m^2)x+5m^2=0 for m≠5/3, are rational if m is rational.

It's because the discriminant is a perfect square. So if m is rational, the solutions will be rational as the discriminant is a perfect square and the coefficients are all rational, for m ≠ 5/3.

(If m = 5/3, then the given equation is a linear one rather than quadratic, as the quadratic term vanishes.)

eyeseeyou · Jun 25, 2016

How do I differentiate surds?

InteGrand · Jun 25, 2016

eyeseeyou said:
How do I differentiate surds?

$\noindent Note that $\frac{\mathrm{d}}{\mathrm{d}x}\left (x^{p}\right)= px^{p-1}$ for any real $p$. And surds can be put into that form. E.g. $ \frac{ \mathrm{d}}{ \mathrm{d}x} \left( \sqrt{x}\right) = \frac{\mathrm{d}}{\mathrm{d}x} \left( x^{\frac{1}{2}} \right) = \frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.$

Green Yoda · Jun 25, 2016

eyeseeyou said:
How do I differentiate surds?

e.g find y' of √x
= x^(1/2-1)
=1/2*x^-1/2
=1/2sqrt(x)

eyeseeyou · Jun 25, 2016

Rathin said:
e.g find y' of √x
= x^(1/2-1)
=1/2*x^-1/2
=1/2sqrt(x)

Yeah I know that (the basics) but what if it's like y=1+Sqaureroot of (x-3)

Green Yoda · Jun 25, 2016

Its the same logic

eyeseeyou · Jun 25, 2016

Rathin said:
Its the same logic

So you let 1+ sqaureroot of (x-3) be a whole number???

InteGrand · Jun 25, 2016

eyeseeyou said:
Yeah I know that (the basics) but what if it's like y=1+Sqaureroot of (x-3)

$\noindent It'd be $y^\prime = \frac{1}{2\sqrt{x-3}}$ (following from the sum rule, derivative of a constant is 0, chain rule, and derivative of the square root function).$

eyeseeyou · Jun 25, 2016

InteGrand said:
$\noindent It'd be $y^\prime = \frac{1}{2\sqrt{x-3}}$ (following from the sum rule, derivative of a constant is 0, chain rule, and derivative of the square root function).$

So basically eliminate all constants and differentaite the squareroot equation by bringing the 1 to the top and bringing the two in front of the surd?

InteGrand · Jun 25, 2016

eyeseeyou said:
So basically eliminate all constants and differentaite the squareroot equation by bringing the 1 to the top and bringing the two in front of the surd?

$\noindent The additive constants don't affect the derivative because of what I explained above (basically because derivative of $f(x)+C$ is just that of $f(x)$, for any differentiable function $f$, $C$ a constant). Then the derivative of $\sqrt{x-3}$ is just $\frac{1}{2\sqrt{x-3}}$, because $\frac{\mathrm{d}}{\mathrm{d} u }\sqrt{u}=\frac{1}{2\sqrt{u}}$ (as shown in an earlier post). Then let $u=x-3$ and so $\frac{\mathrm{d}u}{\mathrm{d}x}=1$. It follows then from the chain rule that the answer to the question is $\frac{1}{2\sqrt{x-3}}\times 1 = \frac{1}{2\sqrt{x-3}}$. With enough practice, you should be able to do this Q. by inspection.$

eyeseeyou · Jun 25, 2016

Cheers Integrand

eyeseeyou · Jun 25, 2016

1. f(x)=(x-1)^2+3, x>=1

Find f^-1'(x),x>3

2. g^-1'=x^3

Find g^-1'(x)

3. h(x)=x^3-3x, -1<x<1 find h^-1'(0)

InteGrand · Jun 25, 2016

eyeseeyou said:
1. f(x)=(x-1)^2+3, x>=1

Find f^-1'(x),x>3

2. g^-1'=x^3

Find g^-1'(x)

3. h(x)=x^3-3x, -1<x<1 find h^-1'(0)

$\noindent Use the fact (Inverse Function Theorem) that if $b=f(a)$, $g\equiv f^{-1}$, and $f$ is differentiable at $a$ with non-zero derivative, then $g^\prime (b) = \frac{1}{f^\prime (a)}$.$

$\noindent 1) Note $f^\prime (x) = 2(x-1), x >1$. Note the range of $f$ is $\left \{y\in \mathbb{R}:y > 3\right \}$. Then for any $y>3$, we have $g^\prime (y) = \frac{1}{f^\prime (x)}=\frac{1}{2(x-1)}$, where $y=f(x)$. Inverting this, we have $x=f^{-1}(y) = \sqrt{y-3}+1$. So $g^\prime (y) = \frac{1}{2\left[\left(\sqrt{y-3}+1\right)-1\right]}=\frac{1}{2\sqrt{y-3}},y>3$. You can replace all $x$'s in this last equation with $y$ if you want.$

The other questions are quite similar.

$\noindent Note for questions 1) and 2), we could manually invert the functions first and then differentiate them. But for 3) this will be impractical as the function is not easy to invert. So for Q3 we'd want to use the method above $\Big{(}$noting $h(0) = 0$, so take $a = 0$ and $b = 0$. Also $h$ is indeed invertible on $(-1,1)$, since $h^\prime (x) = 3x^2 - 3$, which exists and is non-zero for all $x \in (-1,1)$, which means $h$ is invertible on this interval, due to the Inverse Function Theorem. Also it'll be easier since we don't need a general formula. So the answer would just be $\frac{1}{h^\prime (0)}=\frac{1}{-3}$$\Big{)}$.$

eyeseeyou · Jun 25, 2016

InteGrand said:
$\noindent Use the fact (Inverse Function Theorem) that if $b=f(a)$, $g\equiv f^{-1}$, and $f$ is differentiable at $a$ with non-zero derivative, then $g^\prime (b) = \frac{1}{f^\prime (a)}$.$

$\noindent 1) Note $f^\prime (x) = 2(x-1), x >1$. Note the range of $f$ is $\left \{y\in \mathbb{R}:y > 3\right \}$. Then for any $y>3$, we have $g^\prime (y) = \frac{1}{f^\prime (x)}=\frac{1}{2(x-1)}$, where $y=f(x)$. Inverting this, we have $x=f^{-1}(y) = \sqrt{y-3}+1$. So $g^\prime (y) = \frac{1}{2\left(\sqrt{y-3}+1\right)},y>3$. You can replace all $x$'s in this last equation with $y$ if you want.$

The other questions are quite similar.

$\noindent Note for questions 1) and 2), we could manually invert the functions first and then differentiate them. But for 3) this will be impractical as the function is not easy to invert. So for Q3 we'd want to use the method above (noting $h(0) = 0$, so take $a = 0$ and $b = 0$. Also $h$ is indeed invertible on $(-1,1)$, since $h^\prime (x) = 3x^2 - 3$, which exists and is non-zero for all $x \in (-1,1)$, which means $h$ is invertible on this interval, due to the Inverse Function Theorem.).$

what does the three lines going across mean?

InteGrand · Jun 25, 2016

eyeseeyou said:
what does the three lines going across mean?

Basically 'defined as equal to' here.

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