To determine the correct options, let us analyze the given functions and their relationships.
We are given f(x)=x3+3x+2.
The derivative of f(x) is:
f′(x)=3x2+3
Since f′(x)≥3>0 for all x∈R, f(x) is strictly increasing and bijective from R→R.
Given g(f(x))=x for all x∈R, g(x) is the inverse function of f(x), i.e., g=f−1. Thus, we also have:
f(g(x))=xfor all x∈R
Next, we are given h(g(g(x)))=x for all x∈R.
Let y=g(g(x)). Then h(y)=x.
Applying f to both sides of y=g(g(x)):
f(y)=f(g(g(x)))=g(x)
Applying f again:
f(f(y))=f(g(x))=x
Since h(y)=x, we get:
h(y)=f(f(y))⟹h(x)=f(f(x))for all x∈R
Checking Option (A):
We differentiate g(f(x))=x with respect to x using the chain rule:
g′(f(x))⋅f′(x)=1⟹g′(f(x))=f′(x)1
To find g′(2), we set f(x)=2:
x3+3x+2=2⟹x(x2+3)=0⟹x=0
So, f(0)=2. Thus:
g′(2)=g′(f(0))=f′(0)1
Since f′(0)=3(0)2+3=3:
g′(2)=31
Hence, Option (A) is incorrect.
Checking Option (B):
Using h(x)=f(f(x)), we differentiate with respect to x:
h′(x)=f′(f(x))⋅f′(x)
For x=1:
f(1)=13+3(1)+2=6
f′(1)=3(1)2+3=6
f′(f(1))=f′(6)=3(6)2+3=108+3=111
Therefore:
h′(1)=f′(f(1))⋅f′(1)=111⋅6=666
Hence, Option (B) is correct.
Checking Option (C):
Using h(x)=f(f(x)) for x=0:
f(0)=03+3(0)+2=2
h(0)=f(f(0))=f(2)
f(2)=23+3(2)+2=8+6+2=16
Thus, h(0)=16.
Hence, Option (C) is correct.
Checking Option (D):
Using h(x)=f(f(x)), we evaluate h(g(3)):
h(g(3))=f(f(g(3)))
Since f(g(x))=x for all x∈R, we have f(g(3))=3. Therefore:
h(g(3))=f(3)=33+3(3)+2=27+9+2=38
Hence, Option (D) is incorrect.
Conclusion:
The correct options are B and C.