sir i usually am blank when i see this type of problems(till now i attempted by guessing the possibilites...but now i want some help and to know the method to approach the problems of this type)
1.The number of continuous functions on R which satisy (f(x))2=x2 is _______.
2.Let f be continuous function on [-1,1] satisying (f(x))2+x2=1.then the number of such functions is ______.
sir plz explain the process in as much detail as apossible![1]
1sir usually when we encounter problems where argument of sine function tends to zero we usually use the simplification fo replacing it by x!and it does work well in many problems but in some problems like this...
lim(x→0) {((2+cos x)/(x3 *sinx))-(3/x4)}
here if we replace sin x by x we get different answer than when we directly aplly L-hospital's rule!
so i wanted to know when are we allowed to replace sin x by x and when not!
1ok i got it myself as far as the second problem of limits is concerned....but still i need help with the problem of finding number of possible functions!
49for first post I would use coordinate geometry.. tho I am not quite sure how good and reliable the method is...
1.) y2=x2
Two functions, y=±x
2.) y2+x2=1
Again, 2 functions, two semicircles, one above y-axis and the other below y-axis! :)
1WELL THTS WHEE SLIP OCCURS BY AD-HO REASONING LIKE THAT...N THS Y I ANTED SOMETHING SYSTEMATIC.....(BECAUSE SOMETIMES OUR GUESS MIGHT BE INCOMPLETE[3][3])
GOOD PART IS THAT FOR FIRST QUESTION U ANSWERED WE BOTH GOT SAME ANWER "INITIALLY"=2 ON READING SOLUTION I HAD TO AGREE THAT THERE ARE FOUR SUCH FUNCTIONS!
and having done 66.66% work correct and still not getting the rank u could have got(chem tha na woh toh[5]..ye toh maths hai...anyways no issues[3]] is how painful ask me!!![3][3]
66@kunl, its not that one simply replaces sin x with x. Implicitly one is using the series expansion of sin x which goes like
sin x = x - x33! + x55! + . . .
How many terms you retain depends on other terms. For example, if we take the limit of sin xx as x→0, we see that
sin xx = 1 - x23! + x45! + . . .
So now when x approaches zero, the powers ox x go to zero leaving only 1.
1yes sir thts exactly where i was missing earlier
realzied it a bit late!
thank you
can u plz guide me with the problems sent in first post!
341The first one is quite easy to figure out isnt it:
at any point f(a) = a or f(a)=-a so that f(0)=0
f(x)=x and f(x)=-x are obviously both solutions.
Now, if we have some two points a>0 and b>0 such that f(a)=a and f(b)=-b, then by continuity f(x) = 0 for some x in (a,b) which is a contradiction as we then have 0 \in (a,b). This means if f(a)=a for some a>0, then f(x)=x for all x>0.
Similarly, if f(a)=-a for some a>0, then obviously f(x)=-x for all x>0
By now its clear that we can argue on the same lines for all x<0
So, we have two choices for the two halves giving four possible continuous functions.
f(x)=x, f(x)=-x, f(x)=|x| and f(x)=-|x|.
If the continuity requirement is dropped then of course you have infinitely many functions