Home
Class 12
MATHS
Let f(x) be defined by f(x) = x- [x], 0!...

Let f(x) be defined by `f(x) = x- [x], 0!=x in R`, where [x] is the greatest integer less than or equal to x then the number of solutions of `f(x) +f(1/x) =1`

Promotional Banner

Similar Questions

Explore conceptually related problems

If f (x) = x - [x], x (ne 0 ) in R, where [x] is the greatest integer less than or equal to x, then the number of solutions of f(x) + f ((1)/(x)) = 1 is :

If f(x)= x-[x],x(phi0)inR , where [x] is greatest integer less than or equal to x, then the number of solution of f(x)+f(1/x)=1 are

If f(x)=|x-1|-[x] , where [x] is the greatest integer less than or equal to x, then

If f(x)=|x-1|-[x] , where [x] is the greatest integer less than or equal to x, then

If f(x)=x-[x], where [x] is the greatest integer less than or equal to x, then f(+1/2) is:

If f(x)=|x-1|-[x] (where [x] is greatest integer less than or equal to x ) then.

If f(x)=|x|+[x] , where [x] is the greatest integer less than or equal to x, the value of f(-2.5)+f(1.5) is

The function f(x)=x-[x]+cos x, where [x]= the greatest integer less than or equal to x is a

Let f(x)=(x-[x])/(1+x-[x]), where [x] denotes the greatest integer less than or equal to x,then the range of f is

Let f(x) = (x^2-9x+20)/(x-[x]) where [x] denotes greatest integer less than or equal to x ), then