InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Let `A={x: 0 le x lt pi//2} and f:R to A` be an onto function given by `f(x)=tan^(-1)(x^(2)+x+lambda)`, where `lambda` is a constant. Then,A. `lambda gt 0`B. `lambda ge 1//4`C. `lambda lt 1//4`D. `0 le lambda le 1` |
|
Answer» Correct Answer - B Since, `f:R to A ` is an onto functions. Therefore Range of f=A. `0 le f(x) le (pi)/(2)" " "for all "x in R` `Rightarrow 0 le tan^(-1) (x^(2)+x+lambda) le pi//2" " "for all "x in R` `Rightarrow 0 le x^(2)+x+lambda le oo " " "for all "x in R` `Rightarrow x^(2)+x+lambda ge 0" " "for all "x in R` `Rightarrow 1-4lambda le 0 Rightarrow lambda ge (1)/(4)" " "for all "x in R` |
|