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2u Mathematics Marathon v1.0 (1 Viewer)

SoulSearcher

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kathelle said:
Found one, its pretty easy though :<o></o>
<o></o>
i) Show that d (tanθ -θ)=tan<sup>2</sup>θ
....................dθ<o></o>
<o></o>
ii) Hence evaluate
π/4
∫ tan<sup>2</sup>θ . dθ
0
Yeah, someone skips a question. I'll do it anyway.

1) d/dθ (tanθ - θ) = sec2θ - 1
= 1/cos2θ - 1
= (1-cos2θ)/cos2θ
= sin2θ/cos2θ
= tan2θ

ii)
π/4
∫ tan2θ . dθ
0
= [tanθ - θ] (π/4 -> 0)
= (tan (π/4) - π/4) - (0-0)
= 1 - π/4 units2
 

Riviet

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*Looks towards to SoulSearcher* :karate:

Rules are rules. :D
 
Last edited:

drynxz

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SoulSearcher said:
Simple enough:

Show that lim (x->0) (1-cos x)/x2 = 1/2
lol ill give it a shot
lim (x->0) (1-cos x)/x2 = 1/2
=lim (x->0) (2 sin2 x/2 ) / x2
= 2 (sin2 x/2) / x2
= 2 (sin x/2) / (x) x (sin x/2) / (x)
= 2 (sin x/2) / (x/2) x (sin x/2) / (x/2) x 1/4
using the fact lim x->0 (sin x) / x = 1
then 2.1.1.1/4
=1/2
 

ice ken

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a particle moves in a straight line with velocity function v(t)= cos2t cm/sec, for 0<t<pie (both have or equal to) seconds.

a) wen does the particle reverse direction?

easy enough :)
 

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