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