The correct options are
A f′(x)=0 for some x∈(0,1)
D f(x) is monotonically increasing in (0,23)∪(1,∞)
Here, we can see that f(x)≠f(−x)
So, clearly f(x) is not an even function.
Consider f(x)=x4(1−x)2, x∈[0,1]
∴f(x) being a polynomial, is continuous and differentiable also.
f(0)=0=f(1)
So, Rolle's theorem is applicable here.
f′(x)=4x3+6x5−10x4
So, there exists at least one x∈(0,1) such that f′(x)=0
f′(x)=4x3+6x5−10x4
=2x3(2+3x2−5x)
=2x3(x−1)(3x−2)
f′(x)>0 in (0,23)∪(1,∞)
So, f(x) is monotonically increasing in (0,23)∪(1,∞)
f′′(x)=0
⇒12x2+30x4−40x3=0
⇒2x2(6+15x2−20x)=0
⇒x=0 or 6+15x2−20x=0
15x2−20x+6=0
D=400−360=40>0
So, 15x2−20x+6=0 has two distinct real roots.
Hence, f′′(x)=0 has three distinct solutions.