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mathnoob · Joined: 06 Apr 2008 Posts: 5

Hey guys, here are some problems in GR9768 practise book. Please help. Thank you!

24. which of the following sets of vectors is a basis for the subspace of Euclidean 4-subspace consisting all vectors that are orthogonal to both (0,1,1,1) and (1,1,1,)
(A) {(0,-1,-1,0)}
(B) {(1,0,0,0),(0,0,0,1)}
(C) {(-2,-1,1,-2),(0,1,-1,0)}
D,E are obvious wrong.

Ans: C.

28. V1 and V2 are 6-dimensional subspaces of a 10-dimensional vector space V. What is the smallest possible dimension of V1 intersect V2?
(A) 0 (B)1 (C)2 (D)4 (E)6

Ans: C (but why is cant be 1? )

34. f is a differentiable function for which both limf(x) and limf’(x) as x->infinity exist and finite. Which must be true?
(A) limf’(x)(x->infinity)=0
(B) limf’’(x)(x->infinity)=0
(C) limf(x) as x->infinity = limf’(x) as x->infinity
D,E are obvious wrong.

Ans:A

47. x, y are uniformly distributed, independent random variables on [0,1], what is the probability that distance between x and y is less than ½?
(A) 1/4 (B)1/3 (C)1/2 (D)2/3 (E)3/4

Ans:E

55. f is twice-differentiable function on set of real numbers, and f(0), f’(0) and f’’(0) are negative. Suppose f’’ has all the following three properties:
I. It is increasing on [0,infinity)
II. It has a unique zero in [0,infinity)
III. It is unbounded on [0,infinity)
Then which of them does f necessarily have?

Ans: II and III only

tankiitr · Joined: 10 Apr 2008 Posts: 2

24)For solution we can see if the vectors are linearly dependent. For example take vectors k such that k1v1+k2v2+......knvn=0 such that k1=k2=....kn. the solution for k1=k2=k3=0 only for answer c.For a better understanding chk this out http://en.wikipedia.org/wiki/Linearly_independent

2 Cool

Take the gcd of 10 and 6....you will see the result as 2.

47) Basically the random variables is the square with [0,1] X [0,1]. xy=0.5 is a curve. The area of the currve after integration is 0.5(1+log2) which is close to 0.75.

lime · Joined: 04 Dec 2007 Posts: 46

tankiitr · Joined: 10 Apr 2008 Posts: 2

Hey lime ur right.....actually
dim(u+v)= dim(u)+dim(v)-dim(u intersection v)

so dim (u intersection b)= 6+6-10 = 2

nardling · Joined: 11 Apr 2008 Posts: 2

I don't know about the log explanation. Let x run from 0 to 1/2. The probability of y being within 1/2 of x is x + 1/2. That's a line. The area under that line from 0 to 1/2 is 3/8. Double 3/8 (for 1/2 to 1) and that's your 3/4 answer. And yes, that's really the way I solved it. No working backwards Smile

diman · Joined: 15 Jul 2008 Posts: 2

don't agree with solution for:
47. x, y are uniformly distributed, independent random variables on [0,1], what is the probability that distance between x and y is less than ½?
(A) 1/4 (B)1/3 (C)1/2 (D)2/3 (E)3/4

XY=0.5 is not a curve for solutions...

We should use square [0; 1] for x and [0; 1] for y.
distance between x and y is less than ½ means
/x-y/=0.5 so x-y=0.5 and x-y=-0.5
graphically: we cut corners of square by lines y=x-0.5 and y=x+0.5
it will show 2 areas on square where /x-y/>0.5.
it will give us an answer 3/4

JcraigMSU · Joined: 06 Aug 2008 Posts: 7

34. If Lim x -->inf, f'(x) != 0, then as x --> inf, f(x) would be changing, so lim x --> inf, f(x) would not exist, which is a contradiction.

55. F"(0) <0 --> for all x in [0,a) for some a where f"(a) = 0, f'(x)<0 --> f(x) decreasing.
f(x) must cross the x axis because f(0) < 0, and at some b: f'(b) = 0, f(x) becomes a strictly increasing function, thereby crossing the x-axis and never decreasing again afterward.
f(x) is unbounded because lim f'(x) as x-->inf is positive