Given: (x2)+(x+1)2 = 25 Le! x = I + f, where I is integer and f is fractional part function Then, (I+f)2 + (I+f+1)2 = 25 [0 < f < 1] ⇒ (I+1)2 + (l+2)2 = 25 Using definition of least integer function, we get ⇒ I2+2I + 1 + I2 + 4I + 4 = 25 ⇒ 2I2+6I−20 = 0 ⇒ I2+3I−10 = 0 ⇒ I = 2 , - 5 Thus, x = 2 + f or - 5 + f , where 0 < f < 1 ⇒ 2 < 2 + f < 3 or - 5 < - 5 + f < - 4 [Because 0 < f < 1] ... (i) Again, let I = 1 ⇒ x2+(x+1)2 = 25 ⇒ 2x2+2x+1 = 25 ⇒ x2 + x - 12 = 0 ⇒ x = 3 , - 4 ... (ii) From (i) and (ii) , we get x ∊ (- 5 , - 4] ∪ (2 , 3] Hence option 'B' is correct.