Inequality problem- 15th Jan'16

Question

!!(log_2 (4y^2 - y - 1))/log_2 (y^2 + 1) > 1!!

Then !!y!! lies in the interval

!!a)" "(-oo, -2/3)!!

!!b)" "(1, oo)!!

!!c)" "(-2/3, 0)!!

!!d)" "!!None

Solution

!!(log_2 (4y^2 - y - 1))/log_2 (y^2 + 1) > 1!!

!!=> 4y^2 - y - 1 > y^2 + 1!!

!!=> 3y^2 - y - 2 > 0!!

!!=> 3y^2 - 3y + 2y - 2 > 0!!

!!=> 3y(y-1) + 2(y-1) >0!!

!!=> (3y + 2)(y - 1) > 0!!

Solving this equation on inequality curve, we will get

!!y in!! !!(-oo, -2/3)!! and !!(1,oo)!!