Found the method to solve inequalities at
inequalities-trick-91482.html very useful. Applying it to the questinon above
x^3 - 4x^5 <0
x^3(1 - 4x^2) <0
x3(1−2x)(1+2x)<0
This gives 3 roots - 0 (by equating x3=0), x=1/2 (by equating 1-2x=0) and x=-1/2 (by equating 1+2x=0)
On a number line, we have 4 regions
-------- -1/2 -------- 0 --------- 1/2 -------
I used -1, -1/3, 1/3 and 1 as data sets for each region and put them in x3(1−2x)(1+2x) eq.
For x=-1, x3(1−2x)(1+2x) is a +ve expression (-1*3*-1=3). So function is +vw for x< -1/2 --- range1
For x=-1/3, x3(1−2x)(1+2x) is a -ve expression (-1/27*5/3*1/3). So fn is -ve for 1/2<=x<0 ------ range 2
For x=1/3, x3(1−2x)(1+2x) is +ve. So fn is +ve for 0<=x<1/2------ range 3
For x=1, x3(1−2x)(1+2x) is -ve. So fn is -ve for 1/2<=x------ range 4
The original expression (condition) is x^3 - 4x^5 <0. So we are interested in -ve function only which are given by ranges 2 and 4 only. Thus answer is 1/2<=x<0 and 1/2<=x which is same as choice C (–½ < x < 0 or ½ < x)