Comments on: The Infinite Steps To Salvation http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: The Road To Mathematical Reincarnation – Modulus Functions | Sassy O Level Maths Tuition, Questions & Tips from Singapore’s Favourite Private Tutor http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-38344 The Road To Mathematical Reincarnation – Modulus Functions | Sassy O Level Maths Tuition, Questions & Tips from Singapore’s Favourite Private Tutor Mon, 18 Oct 2010 15:23:20 +0000 http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-38344 [...] out this past question to see how this is applied. Else skip to the graphical approach that is more likely to be used in [...] [...] out this past question to see how this is applied. Else skip to the graphical approach that is more likely to be used in [...]

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11802 Miss Loi Sat, 16 Aug 2008 02:01:38 +0000 http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11802 Wah <b>Cendrine</b> such a big reaction - 发生什么事?! <b>Li-sa:</b> Modulus functions are out of HKCEE but in for Singapore's A-Maths 2009. Thanks to you, Miss Loi is now making a phone call at district 0 < x < 2 or 9 < x < 11 and hopes to have escaped her version of <a href="http://en.wikipedia.org/wiki/Groundhog_Day_(film)" rel="nofollow">Groundhog Day</a> :) Wah Cendrine such a big reaction - 发生什么事?!

Li-sa: Modulus functions are out of HKCEE but in for Singapore's A-Maths 2009. Thanks to you, Miss Loi is now making a phone call at district 0 < x < 2 or 9 < x < 11 and hopes to have escaped her version of Groundhog Day :)

]]> By: Li-sa http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11784 Li-sa Fri, 15 Aug 2008 14:04:59 +0000 http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11784 Solving inequalities with absolute values was once in the HKCEE A. Math syllabus but its final appearance was the 2003 paper. Since then it was cut from the syllabus. Solving this took some time; I needed to think it over. (I didn't know it was also called modulus function.) Can you escape from the <a href="http://www.hkedcity.net/library/book/content.phtml?isbn=9789628959891" title="《女媧之門1時空的裂縫》讀了你就會明白" rel="nofollow">recurrence</a> now? Solving inequalities with absolute values was once in the HKCEE A. Math syllabus but its final appearance was the 2003 paper. Since then it was cut from the syllabus. Solving this took some time; I needed to think it over. (I didn't know it was also called modulus function.)
Can you escape from the recurrence now?

]]> By: cendrine http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11754 cendrine Thu, 14 Aug 2008 09:59:29 +0000 http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11754 i hate Maths. i hate Maths.

]]> By: Li-sa http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11751 Li-sa Thu, 14 Aug 2008 08:13:45 +0000 http://www.exampaper.com.sg/questions/a-maths/the-infinite-steps-to-salvation#comment-11751 <div class="highlight"> Miss Loi is now marking your solution while running down those infinite steps ... Instead of risking life and limb and try to solve that scary-looking <em>modulus function</em> directly, it's always a good idea to follow what the question tells you i.e. <em>use</em> the given inequalities to find your final answer (as you've done below). Note that this is another example of a <a href="/questions/a-maths/differentiation-integration-a-hence-question" rel="nofollow">Hence question</a>. </div> Solve inequality 1 and 2: x<sup>2</sup> - 11x + 18 > 0 and x<sup>2</sup> - 11x < 0 (x-9)(x-2)>0 and x(x-11)<0 (x<2 or x>9) and 0<x<11 .'. 0<x<2 or 9<x<11 <div class="highlight"> Factorising x<sup>2</sup> - 11x + 18 quickly via the <a href="/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-finding-your-roots-with-remainder-factor-theorems" rel="nofollow">Trial & Error</a> method (or otherwise), you'll get (x-9)(x-2) > 0. Ensuring first that the <strong>coefficient of <var>x</var><sup>2</sup> is positive</strong> (hence a U-shaped curve), it's clear from the following diagram that <var>x</var>< 2 or <var>x</var> > 9, since we're finding the range of <var>x</var> for the region(s) > 0. <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/08/quadratic-inequality-solution-1.gif" alt="Solution to first (x-9)(x-2) > 0" /> <strong>NOTE:</strong> Whenever you're dealing with > 0 (for a U-shaped curve), please DON'T make this common mistake of e.g. (x-9)(x-2) > 0 ⇒ x > 2, x > 9 <strong>⇒ WRONG!</strong> <hr /> The second case of <var>x</var>(<var>x</var>-11) is more straightforward as we're interested in the single region that is < 0 in diagram below: <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/08/quadratic-inequality-solution-2.gif" alt="Solution to first x(x-11) < 0" /> <hr /> Next we shall first extend a warm welcome to <strong>Modulus Functions</strong> to this year's A-Maths O-Level Syllabus ;) For this question, it's important to note that for an arbitrary constant <var>k</var> ≥ 0: <div class="attention"> |<var>x</var>| < <var>k</var> ⇔ -<var>k</var> < <var>x</var> < <var>k</var> |f(<var>x</var>)| < <var>k</var> ⇔ -<var>k</var> < f(<var>x</var>) < <var>k</var> |<var>x</var>| > <var>k</var> ⇔ <var>x</var> < -<var>k</var> or <var>x</var> > <var>k</var> |f(<var>x</var>)| > <var>k</var> ⇔ f(<var>x</var>) < -<var>k</var> or f(<var>x</var>) > <var>k</var> *Doesn't this follow the same pattern as our quadratic inequalities diagram above? :) ** Can <var>k</var> ever be < 0 for any modulus expression? </div> </div> From |x<sup>2</sup> - 11x + 9| < 9 we get -9 < x<sup>2</sup> - 11x + 9 < 9 <div class="highlight"> And so we now know that |<var>x</var><sup>2</sup> - 11</var>x</var> + 9| < 9 can be spit into: <var>x</var><sup>2</sup> - 11<var>x</var> + 9 > -9 ⇒ <var>x</var><sup>2</sup> - 11<var>x</var> + 18 > 0 <var>x</var><sup>2</sup> - 11<var>x</var> + 9 < 9 ⇒ <var>x</var><sup>2</sup> - 11<var>x</var> < 0 Which are the two original inequality expressions in our question! Which also means that the solution to |<var>x</var><sup>2</sup> - 11<var>x</var> + 9| < 9 is obtained from the range of <var>x</var> that satisfies <em>both</em> the two original inequalities. </div> i.e. x<sup>2</sup> - 11x < 0 and x<sup>2</sup> - 11x + 18 > 0 Solving this compound inequality gets the same answers above. .'. The required range is 0<x<2 or 9<x<11. <div class="highlight"> Yes this can be seen clearly by representing the solution sets of the two inequalities on a <em>number line</em>: <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/08/quadratic-inequality-number-line-solution.gif" alt="Quadratic inequality number line" /> You'll see that the ranges that satisfies both the equalities are 0 < <var>x</var> < 2 and 9 < <var>x</var> < 11 (i.e. the <strong>red</strong> regions that overlap both solution sets. *<strong>N.B:</strong> Note that that the little cute circles are <em>white</em> ○ in the number line coz it's < / > (exclusive). You should use cute little <em>black</em> circles ☻ instead in situations when there's ≤ / ≥ (inclusive) in your inequality expression. </div>

Miss Loi is now marking your solution while running down those infinite steps ...

Instead of risking life and limb and try to solve that scary-looking modulus function directly, it's always a good idea to follow what the question tells you i.e. use the given inequalities to find your final answer (as you've done below). Note that this is another example of a Hence question.

Solve inequality 1 and 2:
x2 - 11x + 18 > 0 and x2 - 11x < 0
(x-9)(x-2)>0 and x(x-11)<0
(x<2 or x>9) and 0<x<11
.'. 0<x<2 or 9<x<11

Factorising x2 - 11x + 18 quickly via the Trial & Error method (or otherwise), you'll get (x-9)(x-2) > 0.

Ensuring first that the coefficient of x2 is positive (hence a U-shaped curve), it's clear from the following diagram that x< 2 or x > 9, since we're finding the range of x for the region(s) > 0.

Solution to first (x-9)(x-2) > 0

NOTE: Whenever you're dealing with > 0 (for a U-shaped curve), please DON'T make this common mistake of e.g.

(x-9)(x-2) > 0 ⇒ x > 2, x > 9 ⇒ WRONG!

The second case of x(x-11) is more straightforward as we're interested in the single region that is < 0 in diagram below:

Solution to first x(x-11) < 0

Next we shall first extend a warm welcome to Modulus Functions to this year's A-Maths O-Level Syllabus ;) For this question, it's important to note that for an arbitrary constant k ≥ 0:

|x| < k ⇔ -k < x < k
|f(x)| < k ⇔ -k < f(x) < k

|x| > k ⇔ x < -k or x > k
|f(x)| > k ⇔ f(x) < -k or f(x) > k

*Doesn't this follow the same pattern as our quadratic inequalities diagram above? :)
** Can k ever be < 0 for any modulus expression?

From |x2 - 11x + 9| < 9 we get
-9 < x2 - 11x + 9 < 9

And so we now know that |x2 - 11x + 9| < 9 can be spit into:

x2 - 11x + 9 > -9 ⇒ x2 - 11x + 18 > 0
x2 - 11x + 9 < 9 ⇒ x2 - 11x < 0

Which are the two original inequality expressions in our question! Which also means that the solution to |x2 - 11x + 9| < 9 is obtained from the range of x that satisfies both the two original inequalities.

i.e. x2 - 11x < 0 and x2 - 11x + 18 > 0
Solving this compound inequality gets the same answers above.
.'. The required range is 0<x<2 or 9<x<11.

Yes this can be seen clearly by representing the solution sets of the two inequalities on a number line:

Quadratic inequality number line

You'll see that the ranges that satisfies both the equalities are 0 < x < 2 and 9 < x < 11 (i.e. the red regions that overlap both solution sets.

*N.B: Note that that the little cute circles are white ○ in the number line coz it's < / > (exclusive). You should use cute little black circles ☻ instead in situations when there's ≤ / ≥ (inclusive) in your inequality expression.

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