The correct option is C 0.9 and 1.35
Given, F1 and F2 be the fractional parts of (44−√2017)2017 and (44+√2017)2017 respectively.
Therefore,
(44−√2017)2017=0+F1 [∵√2017=44.91]
and (44+√2017)2017=I1+F2
Using binomial expansion,
(44−√2017)2017=2017C0(44)2017−2017C1(44)2016(√2017)+2017C2(44)2015(√2017)2−⋯−(√2017)2017=0+F1 ⋯(i)
(44+√2017)2017=2017C0(44)2017+2017C1(44)2016(√2017)+2017C2(44)2015(√2017)2+⋯+(√2017)2017=I1+F2 ⋯(ii)
Adding equation (i) and (ii), we get
I1+F1+F2=2[2017C0(44)2017+2017C2(44)2015(√2017)2+⋯+2017C2(44)1(√2017)2016]
⇒I1+F1+F2=I2, where I2 is integer.
∴F1+F2 is an integer.
0<F1<1
0<F2<1
∴0<F1+F2<2
Between 0 and 2, only integer is 1.
∴F1+F2=1 which lies between 0.9 and 1.35