harder 3u question (1 Viewer)

abecina · Oct 12, 2016

So i am stuck on this question:

Question 8 ii) "hence deduce..."

Thanks for your help

abecina · Oct 13, 2016

abecina said:
So i am stuck on this question:

Question 8 ii) "hence deduce..."

Thanks for your help

If there isn't a response in the next little while I might have to concede there was a typo. It came from a past paper from the 90's of a reputable school.
Ive tried to guess the typo, but am interested to know if anyone has solved it.

Thanks

Mahan1 · Dec 4, 2016

abecina said:
If there isn't a response in the next little while I might have to concede there was a typo. It came from a past paper from the 90's of a reputable school.
Ive tried to guess the typo, but am interested to know if anyone has solved it.

Thanks

The deducing part:

$\textrm{from a)ii } \ \ b^{2} - a^{2} \leq 2\sqrt{ab}(b-a), c^{2} - b^{2} \leq 2\sqrt{bc}(c-b)$

$\textrm{adding them together} \ c^{2} - a^{2} \leq 2(\sqrt{ab}(b-a) + \sqrt{bc}(c-b) ) = 2\sqrt{b}(\sqrt{a}(b-a) + \sqrt{c}(c-b))$

Lugia101101 · Dec 4, 2016

$\begin{align*}\text{So we're given that }\sqrt{a}(b-a)+\sqrt{c}(c-b)&>\frac{c^2-a^2}{2\sqrt{b}}\\ \frac{a^2}{2\sqrt{b}}-\sqrt{a}(a-b)&>\frac{c^2}{2\sqrt{b}}-\sqrt{c}(c-b)\end{align*}$
$\begin{align*}\text{So let }f(x)&=\frac{x^2}{2\sqrt{b}}-\sqrt{x}(x-b)\\ f^{\prime}(x)&=\frac{x}{\sqrt{b}}-\frac{3}{2}\sqrt{x}+\frac{b}{2 \sqrt{x}}\end{align*}$
$\text{Let }f^{\prime}(x)=0\text{:}$
$\begin{align*}\frac{x}{\sqrt{b}}-\frac{3}{2}\sqrt{x}+\frac{b}{2 \sqrt{x}}&=0\\ \frac{x}{\sqrt{b}}&=\frac{3}{2}\sqrt{x}-\frac{b}{2 \sqrt{x}}\\ 2x\sqrt{x}&=3x\sqrt{b}-b\sqrt{b}\\ 4x^3&=9bx^2-6b^2x+b^3\\ 4x^3-9bx^2+6b^2x-b^3&=0\\ (x-b)(4x^2-5bx+b^2)&=0\\ (x-b)^2(4x-b)&=0\\ \therefore x=b,\frac{b}{4}\end{align*}$
$\begin{align*} f^{\prime}\left(\frac{b}{4}\right)&=\frac{\frac{b}{4}}{\sqrt{b}}-\frac{3}{2}\sqrt{\frac{b}{4}}+\frac{b}{2 \sqrt{\frac{b}{4}}}\\ &=\frac{\sqrt{b}}{4}-\frac{3 \sqrt{b}}{4}+\sqrt{b}\\ &=\frac{\sqrt{b}}{2}\end{align*}$
$\text{So at }x=\frac{b}{4}\text{, the function has a positive gradient, and this stationary point doesn't exist.}$
$\text{As at }x=b\text{ the function has a horizontal point of inflexion, the gradient's sign doesn't change.}$
$\therefore f(x)\text{ is monotonic increasing for }b>0$
$\begin{align*}\text{From the question, we are given that }\frac{a^2}{2\sqrt{b}}-\sqrt{a}(a-b)&>\frac{c^2}{2\sqrt{b}}-\sqrt{c}(c-b)\\ f(a)&>f(c)\end{align*}$
$\text{As }f(x)\text{ is monotonic increasing, this implies that }a>c\text{ as required.}$

abecina · Dec 13, 2016

Lugia101101 said:
$\begin{align*}\text{So we're given that }\sqrt{a}(b-a)+\sqrt{c}(c-b)&>\frac{c^2-a^2}{2\sqrt{b}}\\ \frac{a^2}{2\sqrt{b}}-\sqrt{a}(a-b)&>\frac{c^2}{2\sqrt{b}}-\sqrt{c}(c-b)\end{align*}$
$\begin{align*}\text{So let }f(x)&=\frac{x^2}{2\sqrt{b}}-\sqrt{x}(x-b)\\ f^{\prime}(x)&=\frac{x}{\sqrt{b}}-\frac{3}{2}\sqrt{x}+\frac{b}{2 \sqrt{x}}\end{align*}$
$\text{Let }f^{\prime}(x)=0\text{:}$
$\begin{align*}\frac{x}{\sqrt{b}}-\frac{3}{2}\sqrt{x}+\frac{b}{2 \sqrt{x}}&=0\\ \frac{x}{\sqrt{b}}&=\frac{3}{2}\sqrt{x}-\frac{b}{2 \sqrt{x}}\\ 2x\sqrt{x}&=3x\sqrt{b}-b\sqrt{b}\\ 4x^3&=9bx^2-6b^2x+b^3\\ 4x^3-9bx^2+6b^2x-b^3&=0\\ (x-b)(4x^2-5bx+b^2)&=0\\ (x-b)^2(4x-b)&=0\\ \therefore x=b,\frac{b}{4}\end{align*}$
$\begin{align*} f^{\prime}\left(\frac{b}{4}\right)&=\frac{\frac{b}{4}}{\sqrt{b}}-\frac{3}{2}\sqrt{\frac{b}{4}}+\frac{b}{2 \sqrt{\frac{b}{4}}}\\ &=\frac{\sqrt{b}}{4}-\frac{3 \sqrt{b}}{4}+\sqrt{b}\\ &=\frac{\sqrt{b}}{2}\end{align*}$
$\text{So at }x=\frac{b}{4}\text{, the function has a positive gradient, and this stationary point doesn't exist.}$
$\text{As at }x=b\text{ the function has a horizontal point of inflexion, the gradient's sign doesn't change.}$
$\therefore f(x)\text{ is monotonic increasing for }b>0$
$\begin{align*}\text{From the question, we are given that }\frac{a^2}{2\sqrt{b}}-\sqrt{a}(a-b)&>\frac{c^2}{2\sqrt{b}}-\sqrt{c}(c-b)\\ f(a)&>f(c)\end{align*}$
$\text{As }f(x)\text{ is monotonic increasing, this implies that }a>c\text{ as required.}$

Interesting approach - would you say it used "hence" though

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