## URGENT--- GRE 9367 Q-42 AND Q-23

Forum for the GRE subject test in mathematics.
lifeisgood88
Posts: 5
Joined: Tue Nov 01, 2011 5:44 am

### URGENT--- GRE 9367 Q-42 AND Q-23

42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?

1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> |f'''(x)|< b for all x in [0,1]

a)1
b)2
c)6
d)12
e)24

23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=-1
3. f(x)>0 for all x in [0,1)

a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3

mhyyh
Posts: 9
Joined: Mon Oct 10, 2011 9:12 pm

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

lifeisgood88 wrote:42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?

1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> |f'''(x)|< b for all x in [0,1]

a)1
b)2
c)6
d)12
e)24

23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=-1
3. f(x)>0 for all x in [0,1)

a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3

23.
according to intermediate value theorem, 1 is obvious.
according to mean value theorem, 2 is obvious.
and the last one is still obvious.....show u a example:y=2x^2-3x+1

42.(I hope you know the language of Wolfram's software--Mathematica...)
Integrate[f'''(x),{x,0,1}]=f''(1)-f''(0), and so do f''(x) and f'(x)
5>Max[f(1)]=Max[Integrate[f'(x),{x,0,1}]+f(0)=Max[Integrate[f''(x),{x,0,1}]+f'(0)+f(0)=Max[Integrate[f'''(x),{x,0,1}]+f''(0)+f'(0)+f(0)

Apparently, Max[f'''(x)]=b -> Max[f''(x)]=bx+2 -> Max[f'(x)]=(b/2)*x^2+2x+1 -> Max[f(x)]= ((b^3)/6)x^3+x^2+x+1
So Max[f(1)]=b/6+1+1+1<5 -> b<=12

I'm not a math major student. If you think my solution is not convinced, you can prove it yourself..=_=iiii

lifeisgood88
Posts: 5
Joined: Tue Nov 01, 2011 5:44 am

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

mhyyh wrote:
lifeisgood88 wrote:42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?

1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> |f'''(x)|< b for all x in [0,1]

a)1
b)2
c)6
d)12
e)24

23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=-1
3. f(x)>0 for all x in [0,1)

a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3

23.
according to intermediate value theorem, 1 is obvious.
according to mean value theorem, 2 is obvious.
and the last one is still obvious.....show u a example:y=2x^2-3x+1

42.(I hope you know the language of Wolfram's software--Mathematica...)
Integrate[f'''(x),{x,0,1}]=f''(1)-f''(0), and so do f''(x) and f'(x)
5>Max[f(1)]=Max[Integrate[f'(x),{x,0,1}]+f(0)=Max[Integrate[f''(x),{x,0,1}]+f'(0)+f(0)=Max[Integrate[f'''(x),{x,0,1}]+f''(0)+f'(0)+f(0)

Apparently, Max[f'''(x)]=b -> Max[f''(x)]=bx+2 -> Max[f'(x)]=(b/2)*x^2+2x+1 -> Max[f(x)]= ((b^3)/6)x^3+x^2+x+1
So Max[f(1)]=b/6+1+1+1<5 -> b<=12

I'm not a math major student. If you think my solution is not convinced, you can prove it yourself..=_=iiii
Thanks a lot... but can you explain me the intermediate value theorem part.... i am bit confused about it... as it states that if f(x) is continuous on [a,b] and k is any number between f(a) and f(b), then there exists at least one no. c such that a<c<b and f(c)=k.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

define g(x)=f(x)-x which is continuous on [0,1]

g(0)=f(0)-0=1>0
g(1)=f(1)-1=0-1=-1<0

By IVT,
There is a x in (0,1) such that g(x)=f(x)-x=0
=>
f(x)=x

Wishing u all the best for your GRE.

lifeisgood88
Posts: 5
Joined: Tue Nov 01, 2011 5:44 am

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

yoyobarn wrote:define g(x)=f(x)-x which is continuous on [0,1]

g(0)=f(0)-0=1>0
g(1)=f(1)-1=0-1=-1<0

By IVT,
There is a x in (0,1) such that g(x)=f(x)-x=0
=>
f(x)=x

Wishing u all the best for your GRE.
Thanks a lot... that cleared my concepts regarding IVT.... just one more favour....
can you solve Q-42 without use of mathematica....... just for curiosity...

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

For 42, the worst case scenario is that your 3rd derivative is constantly equal to B. This is the case for the poly (B/6)x^3+x^2+x+1. (which satisfies the other conditions). Plug in one, so you get 3+B/6 which needs to be less than 5, thus B is less than 12

lifeisgood88
Posts: 5
Joined: Tue Nov 01, 2011 5:44 am

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

blitzer6266 wrote:For 42, the worst case scenario is that your 3rd derivative is constantly equal to B. This is the case for the poly (B/6)x^3+x^2+x+1. (which satisfies the other conditions). Plug in one, so you get 3+B/6 which needs to be less than 5, thus B is less than 12
Thanks again.... your help means a lot...

mhyyh
Posts: 9
Joined: Mon Oct 10, 2011 9:12 pm

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

lifeisgood88 wrote:
mhyyh wrote:
lifeisgood88 wrote:42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?

1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> |f'''(x)|< b for all x in [0,1]

a)1
b)2
c)6
d)12
e)24

23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=-1
3. f(x)>0 for all x in [0,1)

a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3

23.
according to intermediate value theorem, 1 is obvious.
according to mean value theorem, 2 is obvious.
and the last one is still obvious.....show u a example:y=2x^2-3x+1

42.(I hope you know the language of Wolfram's software--Mathematica...)
Integrate[f'''(x),{x,0,1}]=f''(1)-f''(0), and so do f''(x) and f'(x)
5>Max[f(1)]=Max[Integrate[f'(x),{x,0,1}]+f(0)=Max[Integrate[f''(x),{x,0,1}]+f'(0)+f(0)=Max[Integrate[f'''(x),{x,0,1}]+f''(0)+f'(0)+f(0)

Apparently, Max[f'''(x)]=b -> Max[f''(x)]=bx+2 -> Max[f'(x)]=(b/2)*x^2+2x+1 -> Max[f(x)]= ((b^3)/6)x^3+x^2+x+1
So Max[f(1)]=b/6+1+1+1<5 -> b<=12

I'm not a math major student. If you think my solution is not convinced, you can prove it yourself..=_=iiii
Thanks a lot... but can you explain me the intermediate value theorem part.... i am bit confused about it... as it states that if f(x) is continuous on [a,b] and k is any number between f(a) and f(b), then there exists at least one no. c such that a<c<b and f(c)=k.
Well, just wiki it http://en.wikipedia.org/wiki/Intermediate_value_theorem...

f() is a continuous function on [a,b].
for every min{f(a),f(b)}<=y_0<=max{f(a),f(b)}
there exist at least one x from [a,b]
s.t. f(x_0)=y_0..

orangebaby
Posts: 2
Joined: Fri Mar 21, 2014 2:05 am

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

For (I), the intermediate value theorem only works in a closed interval, no?

If I take f(x)=-x^2+1, there is no x in an open interval (0, 1) where f(x)=x.

Last edited by orangebaby on Fri Mar 21, 2014 3:33 am, edited 1 time in total.

Ryker
Posts: 74
Joined: Mon Jul 08, 2013 11:27 pm

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

The answer is right above you, can you not even be bothered by at least looking at the post?

safetybelts
Posts: 18
Joined: Tue Oct 29, 2013 10:29 am

### Re: URGENT--- GRE 9367 Q-42 AND Q-23

orangebaby wrote:For (I), the intermediate value theorem only works in a closed interval, no?

If I take f(x)=-x^2+1, there is no x in an open interval (0, 1) where f(x)=x.

Sure there is. Find the intersection of the curves y = -x^2 + 1 and y = x.

x is approximately .62.

miguel
Posts: 11
Joined: Thu Sep 15, 2011 6:51 pm