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

## Need your help on GR9768 24,28,34,47,55

### Replies

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

28) 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.

28) 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.

### And for the probability....

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

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

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