AP EAMCET202023 Sep 2020Morning ShiftMathematicsFunctionsActual
Let R = ( 5 5 + 11 ) 2 n + 1 and f = R - R , where x denotes the greatest integer less than or equal to x , then R f =
Options
- A2 n + 1
- B2 2 n + 1
- C4 n + 1
- D4 2 n + 1
Correct answer
D. 4 2 n + 1
Step-by-step solution
Given, R = 5 5 + 11 2 n + 1 and f = R - R = R If I is the integral part of R , then R = I + f = 5 5 + 11 2 n + 1   … i and 0 < f < 1 Now since 5 5 - 11 = 0 . 18 < 1 Let us assume f 1 = 5 5 - 11 2 n + 1   … ii and 0 < f 1 < 1 equation i - equation ii , we get I + f - f 1 = 5 5 + 11 2 n + 1 - 5 5 - 11 2 n + 1 ⇒ I + f - f 1 = 2 C 1 2 n + 1 5 5 2 n 11 + C 3 2 n + 1 5 5 2 n - 2 11 3 + . . . . . . . . ⇒ I + f - f 1 = Even integer ⇒ f - f 1 must also be an intege