jnu

[ritu]

1.suppose we know that IaI<3.WHICH OF THE FOLLOWING CONDITIONS  is enough to imply that IbI<5???
a.Ia+bI<8
b.2<Ia-bI<8
c. Ia-bI<=2
d. 3<Ia-bI<=5

i know it has been discussed earlier...bt m still confused btwn option c and d ...pls tell me how to do it...


2.in jnu 2011,questions 24-27 ,,.... a story is given about a person having three stages in his life .....nd so on...
how to go about "QUESTION 27"?????


3.Let P(n,m) be a property about two integers n and m.if we want to disprove the claim that "'for every  integer n ,there exists an integer m such that P(n,m) is true""then we need to prove that:
a.there exists a integers   n, m   such that P(n,m) is false
b.there exists an integer m such that P(n,m) is false for all integers n
c.there exists an integer n such that P(n,m) is false for all integers m
d.foe every integer n ,there exists an integer m such that P(n,m) is false


(i think it should be option a ??????right ya wrong??

[a.m]

answer to 3. is c !

[tim]

Cn u pls confirm the answer to this questn ?
 f(x) = -e^x-2, then the range of f is given by the interval
A. ( - infinity , -2)
B. ( -infinity , + infinity)
C. (-2 , +infinity)
D. (-infinity, + infinity)

I think its C.

[a.m]

@ Tim i think its a. !

how have you reached to the conclusion. please elaborate

[tim]

Hey I found a mistake in the way I was dng it..ya its nt C cn u temme how did u get A ?

[Ankur]

The Range of natural exponential function is (0,+ infinity) , -2 is subtracted from it so the range will also include values greater than -2 ,
so the Range now is (-2, + infinity).

[ritu]

why answer for 3rd is c....pls explain????

[lovekesh]

ritu
It's just a definition of a function. n is domain and m is codomain. All u need is one n which does not belong to any m that is codomain, that means its not a function.

[a.m]

@ Ankur ..

F(x) = -e^x - 2
or y = -e^x - 2
or e^x = -y-2
or log e^x = log(-y-2)
or x = log (-y-2)

so -y-2>0
or -y>2
or y < -2

so i arrived at the range for the function is (-infinity, -2)


Please tell me where am i going wrong.

[tim]

I wud like to point out log fn is never defined for a negative value..

[Ankur]

Your answer is correct but you method is wrong,
sorry I didnt notice the negative sign before e^x.
f(x)=-e^x-2, the min value of e^x is 0 will be the maximum value of -e^x(range of e^x (0,+infinity))
f(x)=-2 at e^x= 0 and hence range of f(x) would be (-infinty,-2)...as all values are negative.
(can also be solved as taking -f(X)=e^x+2..)

[a.m]

yes time ! log is undefined for negative values i understand.
thats why i wrote the total value of -y-2 should be greater than 0.

@ Ankur !! I still am a lil clueless. please elaborate the other method if you can !