Daily Graph 14-03-09

x{x} = 1

8 Answers

21
tapanmast Vora ·

0<x<1.......... no solution

x<0 jus forget it no +ve * -ve = 1

let x be of the formm I+t (wer I =int, t = deci)

our LHS :

(I+t)*t

==> I.t + t.t

==> now I.t can b fractn or integral..........

if I.t = integer JUS FORGET IT coz t.t can never be integer and hence LHS bcums NON-INTEGRAL

case 2 I.t = fractnal = I' + t' (say)

then LHS : I' + t' + t.t

LHS can b integral only if t.t + t' = 1;
FOFR LHS TO BE 1 I' = 0

and t' + t.t = 1

aage sochne do [12]

21
tapanmast Vora ·

THER4 I.t = t'

x has to b such that I.t<1 --------> @

21
tapanmast Vora ·

SO ALL X of da form ( I + t) SUCH THAT :

I.t<1 (from @)
and I.t + t.t = 1

I wonder how many such nos. will b der (IF ANY)

21
tapanmast Vora ·

Sir, iska kya hua??

NO RESPONSE [2]

62
Lokesh Verma ·

x{x} = 1

let [x]=I

(I+f)f=1

f2+If-1=0

2f= -I±√I2+4

now can you nail this one?

21
tapanmast Vora ·

mere ko yeh to aaya tha naa : and I.t + t.t = 1 but i felt subject bana ke kuch zyada fayda nahi hai [2]

-I - √I^2 + 4

IS RULED OUT COZ THEN RHS < 0

NOW ONLY OPTINO IS : -I + √I^2 + 4

RHS lies b/w (0,2)

which implies

I < √I2 + 4 < 2 + I

21
tapanmast Vora ·

So now do we (rather I ) hav to find da value which satisfiesthis or vo Computer pe chhodna hai?

Is it done?

62
Lokesh Verma ·

yes tapan it is done :)

good work :)

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