The inequality|x||x−1|<1is equivalent to12<x(x−1)2<x2|x−1|2<|x|21<x2(x−1)21<|x|2|x−1|2|x|2>|x−1|2−x<−12−x>−12

The inequality|x||x−1|<1is equivalent to12<x(x−1)2<x2|x−1|2<|x|21<x2(x−1)21<|x|2|x−1|2|x|2>|x−1|2−x<−12−x>−12

Solve for  x  :  2[x2 + 1x2] − 9[x + 1x] + 14  =  0  :Select an answerA 12, 1, 2 B 2, 4, 13 C 13, 4, 1 DNone of these

Let y=loge(1−x21+x2),−1<x<1. Then at x=12, the value of 225(y′−y′′) is equal to

If X is a random variable such that E(X)= 3 and E(X2)=13, use Chebychev's inequality to determine a lower bound for P(-2<X<8).

rite down the Lagrangian function for the following maximization problem.max x + ln y s.t. x2 + y2 ≤ 1, and y ≥ 12