AP EAMCET202315 May 2023Evening ShiftMathematicsFunctionsActual
If f(a)= | 1-a 1+a | for a -1,1 , then the set of values of all ' a ', for which f ( 2 a 1+a^2 )>0 is
Options
- A(0, )- 1
- B(- , 0)- -1
- C(- , )- -1,1
- D(-1,1)
Correct answer
B. (- , 0)- -1
Step-by-step solution
Given f(a)= | 1-a 1+a |, a -1,1 Now f ( 2 a 1+a^2 )>0 | 1- 2 a 1+a^2 1+ 2 a 1+a^2 |>0 aligned & | 1+a-2 a 1+a+2 a |>1 | (1) (1) |>1 & (1-a)^2-(1+a)^2 (1+a)^2 >0 or (1-a^2 )+ (1+a^2 ) (1+a^2 ) 0 or 1+a^2 0 a & a 1 ) So a (- , 0)- -1