find the non zero real values of x such that x,[x],{x} are in hp
note: a. {x} represents fractional part of x
b.[x] represents greatest integer less than x
see what's the problem in my solution
we have 1/x + 1/{x} = 2 / [x]
====> [x]/x + [x]/{x} = 2
we have [x]= x- {x}
so we have
x/{x} - {x}/x =2
let {x}=y
then x/y - y /x =2 ==> x2-y2=2xy making a quadratic in y we have 0<y<1( y≠0"otherwise the function becomes not defined ") so applying the conditions for the roots lying in the interval (0,1) we have D ≥0 , f(0)>0 , f(1)>0 but f(0)=-x2 that is f(0)≤0 for all values of x so we do not get any solution .... but there are two answers to this question .....i am not getting where am i getting wrong please help me out!!!???
9no surbhi how can u say we have no solution ??
see f(y) = y2+2xy -x2
now we need to find a solution btw 0,1
now parabola is concave upwards and f(0) < 0 so it still can cut the axes btw o and 1 ( ie it can still have one y for a particular x )
but i dont know if the above treatment uve given can obtain value of x as theres now way to ensure y for a particular x is equal to {x}
so think in some other way !!
1i think you are right because if we have one value of y then too we'll have two values of x and the no of answers is 2 only ....so we have someting in our hands so now D>0 and f(1)>0 ..... i dont know how does that help because we simply have that x2 -2x-1<0 and we do not get any perfect value ,we just get a range can you suggest some other method of solving the question ?
9surbhi u r stuck here as ur dealing with x and {x} which are dependent quantities
u shud try dealing with [x] and {x} which are independent
now try doing ( u shud get ans)
i dont want to spoil ur prep by giving ans right away .
if ur stuck (after trying more than 30 min)tell me
derive x= 1/√2 ±1
1okay i got it
we have 1/x +1/{x} = 2/[x]
==> 1/({x}+[x]) =( 2{x}- [x])/({x}[x])
let k={X} and y= [x]
solving the above equation we have ±√2 k= y
so the answers are (1+√2)/√2 and ( 1-√2) /√2
thank you