Is 1 < a < 0 ? (1) a^3 > a (2) a^2 > a^3

1) \(a^3\) > a​

Step 1:
Finding critical points \(a^3\) – a >0 -> a (\(a^2\)-1) >0 -> a (a+1)(a-1)>0​
Critical points are a = -1, 0 , 1​
​
Step 2: Validating the regions​ on the number line
i) -∞ to –1: when a=-2, inequlaity is invalid​
ii) -1 to 0; when a=-1/2, inequlaity is valid​
iii) 0 to –1; when a=½, it is invalid​
iv) 1 to ∞, when a =2, it is valid​

Step 3: Solution set is –1<a<0 and a>1​
No concrete answer to the question

2) \(a^2\) > \(a^3\)​

Step 1:
Finding critical points
\(a^3 – a^2\) < 0​
\(a^2\) (a-1) <0​
Critical points are a = 0, 1​

​Step 2:
Validating the regions​
-∞ to 0, when a=-1, valid ​
i) 0 to 1, when a = ½ , valid​
ii)1 to ∞, when a =2, invalid​

Step 3: Solution set -∞ <a<1​
No concrete answer to the question​

Combining statements (1) and (2)​
Common solution set is –1<a<0​
C is correct​