maths

[anon_econ]

Suppose n>=9 is an integer. Let u= n^1/2 + n^1/3 + n ^1/4.
Then which of the following relationships between n and u is correct?
A. n=u
B. n>u
C. n<u
D. none of the above.

Also, I have a doubt in dse 2008 question 55. http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/28_june_2008_option_a.pdf
I'm getting the limit to be 0 for all x other than 0, and infinity for x=0. Plz confirm this. And if this is correct then is the answer d? I mean if a sequence has an infinite limit do we say the limit does not exist?

[AJ]

I am getting these answers...

for first question:
u= n^1/2(1+1/n^1/6+1/n^1/4) .... i took n^1/2 common ...

then for values small values like 9 or 10 ... this is ofcourse less than n...

for larger values... the bracketed expression is 1+ something .. its less than 2 always...
so at max.. u = 2.n^1/2.. i.e. (twice of root n)... for all values above 4.. twice of root of a number is less than the no..
so,

B. n>u

(thought after typing this, m feeling m wrong.. :p )

for second question:

What does this notation [0,1/n] ..in base of function g ..means...???
 

[Chocolate Frog]

AJ's solution is acceptable, of course, but a slightly different approach will make our lives even simpler. Notice that both n^(1/6) and n^(1/4) are greater than 1 for n > 1. So, their reciprocals will be less than 1. In symbols, u = n^(1/2) + n^(1/3) + n ^(1/4) < 3n^(1/2). Hence for n >= 9, u < n.

As for the DSE question, fn = n if x belongs to [0,1/n], and 0 otherwise. Limit fn as n tends to infinity exists everywhere except at x = 0. So, you're right, the answer should be (d).

[AJ]

from dskul..2009..

[Chocolate Frog]

Answer to the second question should be (c). It follows directly from the definition of derivative of a function.

[AJ]

I keep on having trouble with derivative.. plss explain this... see...

f'(x)= 2xsin(1/x)-cos(1/x) , if x is not equal to zero
       and 0        , if x is equal to zero...

is this right??

ok, now.. f'(0)=0.. that's fine...

but how do we find the limits of f'(x) when x approaches zero... to check for continuity..

can u please explain me continuity of this sin(1/x) kinda function.. i dont get this....

[ritu]

hi AJ....U CAN DO FIRST QUESTION LIKE THIS....
WE KNOW THAT THE FUNCTION OF "STANDARD" NORMAL DEVIATE ie.Z IS 1/root 2 pie.e^(-x^2/2)...{check
 any normal stats book for this function if u cant read it properly}
now try to imagine a standard normal curve.....area covered btwn (-1.96,1.96) is 95% or 0.95....
this means we are left with 0.05 area of which 0.025 each lies to the left and right of(+- 1.96)
now we want area from 1.96 till infinity.....
integral of {1/ root 2 pie .e^-x^2/2} over (1.96, infinity) =0.025
we can take out constant 1/root2 pie....with the desired integral in brackets....taking constant other side u get answer as 0.025(under root 2 pie)....ie. option d
i kno its quite messy ....sry dnt have access to this graph technology...:P

[anon_econ]

Thanks guys

[Chocolate Frog]

AJ, first, an observation: having confidently arrived at f'(0) = 0, do you need to check for the second part? :)

More seriously, having found out f'(x), when you apply the limit "as x tends to zero" to the expression, you'll find that the limit does not exist. Hence, the discontinuity at x = 0.

Limit of sin(1/x) as x approaches zero does not exist. You can, however, evaluate the limit of xsin(1/x) as x approaches zero by Sandwich theorem.

Ritu, it's not messy at all! I must admit it did not strike me immediately. Great job!

[AJ]

@chocolate frog... thanks..
@ ritu... Neat..!!!