InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The equation `(x^2+x=1)^2+1=(x^2+x+1)(x^2-x-5)`for `x in (-2,3)`will have number of solutions.`1`b. `2`c. `3`d. 0A. 2B. 4C. 3D. None of these |
|
Answer» Correct Answer - D We have, `(x^(2)+x+1)^(2)+1=(x^(2)+x+1)(x^(2)-x-5)` `implies (x^(2)+x+1)+(1)/((x^(2)+x+1))=x^(2)-x=5` We know that `x^(2)+x+1 gt0` for all `x in R`. Therefore, LHS`=(x^(2)+x+1)+(1)/((x^(2)+x+1)) ge2` But, RHS `=x^(2)-x-5 lt 1` for `x in (-23)`. Hence. the given equation has no solution. |
|