66Q1) Obviously there are two of them
Q2) f(x)=1/2 is the only function.
Q3) the given equation defines a circle x2+y2 = 1, so there are two functions: when you take the upper branch and when you take the lower branch.
Among various properties of continous, we have if f is continous function on [a,b] and f(a) f(b) < 0, then there exists a point c in (a,b) such that f(x) = 0 equivalently if f is continous on [a,b] and x belongs to R is such that f(a) < x < f(b) then there is c belonging to (a,b) such that x=f(c). It follows frm the above result that the image of a closed interval under a continous function is a closed interval.
Ques1) The number of continous functions on R which satisfy (f(x))2 = x2 for all x belonging to R is
(a)1 (b)2 (c)4 (d)8
Ques2) Suppose that f(1/2) = 1 and f is cont on [0,1] assuming only rational value in the entire interval. The number of such functions is
(a) infinite (b) 2 (c) 4 (d) 1
Ques3) Let f be a continous function on [-1,1] satisfying (f(x))2 + x2 =1 for all x belonging to [-1,1]
The number of such functions is
(a) 2 (b)1 (c)4 (d) infinitely wrong
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13 Answers
66
3Q3) sir f(x) is depend on the variable x,but u have taken it as y which is another variable.
66yes..sorry..i made a mistake in the second one..i was in a bit of hurry..its f(x)=1
3oops sry sir,i messed up in q3).Sir can u pls explain a bit more in Q2) i cant understand.
66@msp
There is something called the intermediate value theorem which comes into picture here.
Let f be continuous on [a,b]. Let m be the minimum and M be the maximum attained by this function on [a,b]. Then for any ξ satisfying m ≤ ξ ≤ M, there exists at least one c in [a,b] such that f(c)=ξ.
Basically, what it means is that any continuous function will take all real values between its minimum and maximum values.
In this case, the important statement is that the function takes on only rational values. If the function were not constant, then between its minimum and maximum values, there would be a lot of irrational numbers and being continuous, the function must attain them. This however does not happen. Accordingly, the function must be continuous. And since f(1/2)=1, the only possibility is that f(x)=1 for all x in [0,1]
1i don't kn but i got 4 fns for q1....
1.x
2.-x
3.|x|
4.-|x|
means as far as mcqs are concerned...these matter!!!
1to add to Kayamant sir's explanation
see if the function is to be continues the LHL and the RHL got to point at that point and the moment it happens the function got to have a large no of irrational values
1actually to look at the concept lets look at the nnumber line once more....
u might recollect the concept taught in 9th:there are infinite irrational no between 2 rational...so obvious;y their density on the number scale would be larger.....so the image can be somthing like this..........so here if the limit exits the fuction would immediate step on a irrational no. which is not allowed
