1.

Solve the inequation |a^(2x)+a^(x+2)-1|ge1 for all values of a(agt0,a!=1)

Answer»

Solution :Using `a^(x)=t`
the GIVEN inequation can be WRITTEN in the form
`|t^(2)+a^(2)t-1|ge1`…i
`:'agt0` and `a!=1` then `a^(x)gt0`
`:.tgt0` ………ii
Inequation (i) write in the forms
`t^(2)+a^(2)t-1ge1` and `t^(2)+a^(2)t-1le-1`
`:.tle(-a^(2)-sqrt(a^(4)+8)/2,tge(-a^(2)+sqrt((a^(4)+8)))/2`
and `-a^(2)letle0`
But `tgt0` [from Eq (ii)]
`:.tge(-a^(2)+sqrt((a^(4)+8)))/2`
`:.a^(x)ge(-a^(2)+sqrt((a^(4)+8)))/2`
For `0ltalt1`
`xlelog_(a)((-a^(2)+sqrt(a^(4)+8)))/2)`
`:.x epsilon[-oo,log_(a)((-a^(2)+sqrt((a^(4)+8)))/2)]`
and for `agt1,xgelog_(a) ((-a^(2)+sqrt((a^(4)+8)))/2)`
`:.x epsilon(log_(a)((-a^(2)+sqrt((a^(4)+8)))/2),oo)`


Discussion

No Comment Found

Related InterviewSolutions