This

**preview**shows page 1. to view the full**5 pages of the document.**1. (a) Verify whether or not the following statements are correct. No marks for guessing.

(i) [3 marks] If

S, then

}0:)0,{(\}11:),{( 22 >≤≤−∈= xxyxRyx

. )}0,0{(\}11:),{( 22int <<−∈= yxRyxS

NO. }0:)0,{(\}11:),{( 22int ≥<<−∈= xxyxRyxS

(ii) [4 marks] Every continuous function attains its absolute minimum value

RRSf →⊂ 2

:

and its absolute maximum value on the set , where denotes the line segment in

i

i

LS U

∞

=

=

1

i

L

2

R

from the origin to the point )0,0()

1

1,

1

2

i

i

−

( on the circular arc 2

1xy −= .

NO. By the Extreme Value Theorem, continuous function is sure to attain its absolute

minimum value and its absolute maximum value on the set if S is compact. S

Consider the sequence of points =

k

xS

k

k

⊂− )

1

1,

1

2

(, lim so by the S

k

k

∉=

∞→ )1,0(x

Bolzano-Weierstrass S is not compact. Hence the statement is false.

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1.(c) [5 marks] Consider the area A of the parallelogram generated by the vectors

u. Use differentials to answer the following question:

)0,5,1(),0,3,2( −== v

“To which non-zero component of the vectors u, v is the value of the area A

most sensitive?”

(That is a small change of that component causes the biggest change of the value of A)

HINT: Show first that 51

32

−

=

A

=

−

++=×=

2

22

51

32

00|||| vuA|

51

32

|− = 51

32

− since 013

51

32 >=

−

Consider the function yzxw

wz

yx

wzyxA −==),,,(. Then

dA . At the point (xdwydzzdywdx +−−= )5,1,3,2

−

we get dA dwdzdydx 235

+

−+

=

.

Hence a small change of dx causes the largest change in A, and consequently A is most

sensitive to the entry 2 of the vector u.

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