Easy Question 10 or Easiest? (1 Viewer)

[benji6667]

discuss this years question 10

imo i thought it was the easiest question 10 ever set in a 2 unit paper

 

[gaaay]

10 a was easy

10 b got me good
i couldnt find the area of the upper triangle :S

 

[Alex499]

i couldn't answer any of it
/win

 

[Arithela]

agreed 10b got me as well

 

[dolbinau]

10b got it all but for a, was it a trick? How did we do it?

 

[gloworm14]

i didnt get it LOL
i tried... i tried...

 

[Jaspal]

Yeah 10.b i couldnt do

 

[axlenatore]

there has been easier, i couldnt do 10b (ii) and (iii)

 

[alez]

was pretty good esp compared to the other years
what did u guys get for the last bit?

 

[dolbinau]

My minimum value of x was l/root 2 iirc

The length was a like root something minus soomething times x

 

[betraya]

I got 10a correct and i proved 10bi) and b ii) right. Ran out of time for iii).
I thought I've seen this question somewhere else but worded differently in a past paper. I was prepared to do 10a because it was nearly the same with my school's trial where we had to find the area of logx between 7 and 4 or something like that. So then it was easily done by most of the grade

 

[karnage]

10 a was easy

10 b got me good
i couldnt find the area of the upper triangle :S

Yea i couldnt find the top length of the triangle MP or whatever it was
Spent ages dawdling on it, then i decided to copy the statement and try to backtrack but that didnt go far. But yea, i did part ii and iii

The last bit in Q9 got me (the y value of that beam thingy), and i stuffed up one of my reasons for the last proof in Q8 (thats 1 mark right?) had the right idea though

 

[hjp.]

10 a was easy

10 b got me good
i couldnt find the area of the upper triangle :S

neither.
so i gave up.

 

[m.incognito]

10 was strangely easy this year...

For b you had to sub MP as s(l-x)/x, since MPO and JKO are similar.
Then sub that into the 1/2a.b.sinC, then factor out s and sina.

Differentiate - solve for x then sub value into MP

 

[squeenie]

I was doing just fine until I got to question 10b. That really caught me off guard...

 

[pissedoff]

For 10 a

you find the area of the rectangle

then you can find the area of the unshaded region by making x the subject

so y = ln (x-2) becomes x= e^y +2

then rectangle - unshaded = shaded

for part b...

you needed to prove similarity then equating sides find PM

then you needed to use 1/2 ab sin c

and voila...question 10 is over.

 

[janiceee]

My minimum value of x was l/root 2 iirc

The length was a like root something minus soomething times x

Yup thats what i got for min value too

 

[dolbinau]

how did you do 9a, was it just change to x in terms of y between 0 and ln5?

 

[alez]

awesome i got l/rt2 as well
to find mp, i proved the triangles were similar, so mp/l-x=s/x or something like that

 

[karnage]

For b you had to sub MP as s(l-x)/x, since MPO and JKO are similar.

Damn it, my mind was just relieved of how to find fucking MP . I see what you did there....