When this blog was started, Miss Loi had harboured a grandiose vision of the day she clocked her first 100 posts.
will never get to see the likes of this
In her mind she pictured a glorious day of great celebration and festivities, a day of receiving homages from dignitaries around the world, a day of fireworks lighting up the skies and endless processions of performers flooding the streets around Novena!
Hence when a casual glance at the blog stats today revealed that this IS the 100th post, posted on a day less than two weeks away from the O-Levels, anti-climatic was probably too mild a word to describe the magical moment.
With no time to engage street performers or call the fireworks company at such short notice, Miss Loi will just have to make do with a polynomial question (that has frequently appeared in the past three years) to mark this joyous occasion:
The cubic polynomial f(x) is such that the coefficient of x3 is 1 and the roots of f(x) = 0 are -2, 1+√3 and 1-√3.
- Express f(x) as a cubic polynomial in x with integer coefficients.
- Find the remainder when f(x) is divided by x-3.
- Solve the equation f(-x)=0.
If you’ve been following the recent series of questions here, you would’ve noticed the trend: they’re all pretty straightforward once you know how to start. That’s what O-Level maths questions are all about – basically testing your ability to choose the best approach from your collection of approaches.
Given the high standards of Jφss Sticks commentators these days, no relevant equations shall be put up this time to spoil the fun, as a simple hint should be enough for you to yell “Orrrhhh! So simple!”.
And when you eventually solve the question, do sit back, enjoy the imaginary fireworks and wonder how on earth could so many students contrive to get zero marks for this!
13 Comments
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wooo, happy 100th post 😀
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Thanks xinyun. And how did you celebrate yours?
Sorry for calling off the parade and fireworks display ... the police just rejected Miss Loi's public entertainment license application ... sigh ... 🙁
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Congratulation! ( try to imageing the sound of popping champage ) On this milestone! It is ok if not firework or parade! If I am in s'pore I don't mind asking u out for dinner to chelebrate on ur achievement! Have a nice day!
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HAM, you don't mind but wouldn't your gf mind you dining around with other girls in S'pore???
Now that this 100th post milestone had quietly come and gone (sadly without the promised fanfare), time to think of another grand vision for this blog's anniversary. Maybe we'll see a fly-past of aeroplanes by then!
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hmm this seems easy nvm i'll let 123 solve it since he succinctly lamented he din get to solve de last one
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Haha, thnx but.... this question seems abit the complicated @.@
Still thinking though
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Er,,,,,,,,
a) f(x)= (x+2)(x-1)+)√3)(x-1)-√3)
= (x+2)(x-1)^2-√3^2
= (x+2)(x^2-2x+1)-3
= x^3-3x-1
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OOps wrong -.-
Er,,,,,,,,
a) f(x)= (x+2)(x-1)+√3)(x-1)-√3)
= (x+2)(x^2-2x+1-3)
= x^3-6x-2
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OOOOOOOOOOOOOppppps wrong again -.-!!
Er,,,,,,,,
a) f(x)= [x-(-2)][x-(1+√3)][x-(1-√3)]
= (x+2)(x2-2x+1-3)
= x3-6x-4
Ahhh you finally got it right on the nth attempt. Have edited your workings for better readability.
When you see the keyword roots, you'll know that Part 1 tests your understanding of the Factor Theorem: x-a is a factor of f(x) where f(a) = 0. And x=a is a root of the equation. Which you've rightly applied.
Unfortunately some students got too excited and instead formed an arbitrary ax3+bx2+cx+d, whereby they tried to sub in the root values and solve the resultant simultaneous equations. That's mathematical suicide for you.
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f(3)=33-6(3)-4
= 5
Yes Part 2 tests your understanding of the Remainder Theorem i.e. If f(x) is divided by (x-a), the remainder is f(a).
Unfortunately, many Last-Minute Buddha Foot Huggers who quickly flipped through the pages of their textbooks missed this. And proceeded to waste their time in the exam doing long division!
And do note that Part 2 requires your answer from Part 1 to be correct in order for it to be correct. So this is a one die all die question. And you should pay extra attention to these kind of questions when you're checking for careless mistakes during your exam.
Just making a lucky guess for part (iii), is the answer 2, -1+√3, and -1-√3.????
Yes you're lucky. Some day, if you're taking A-Level Math, you'll be taught that f(-x) is actually a reflection of f(x) about the y-axis.
But for now, some students will get stuck in this coz they've never seen it in their textbooks.
Instead of just sitting there and staring blankly at the question, simply sub in -x into the original function to get f(-x), i.e.
f(x) = (x+2)[x-(1+√3)][x-(1-√3)] = 0
f(-x) = [(-x)+2][(-x)-(1+√3)][(-x)-(1-√3)] = 0
(-x+2)(-x-1-√3)(-x-1+√3)=0
(x-2)[x-(-1-√3)][x-(-1+√3)]=0
⇒ x=2, x=-1-√3, x=-1+√3
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Ops, i forgot, Happpy 100th post...hope there will be more 100ths to come ^.^
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hmm should be correct i got de same ans also
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123, Miss Loi has marked your answers. Next time try getting it right on your first attempt, like what you should do in your actual exams.
And thanks for your happy wishes 😀