21Sir, iska kya hua??
NO RESPONSE [2]
8 Answers
210<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]
21SO 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)
62x{x} = 1
let [x]=I
(I+f)f=1
f2+If-1=0
2f= -I±√I2+4
now can you nail this one?
21mere 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
21So now do we (rather I ) hav to find da value which satisfiesthis or vo Computer pe chhodna hai?
Is it done?