JEE Challenger
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Properties of Differentiable Composite and Inverse Functions

Consider differentiable functions f:R→Rf : \mathbb{R} \to \mathbb{R}, g:R→Rg : \mathbb{R} \to \mathbb{R}, and h:R→Rh : \mathbb{R} \to \mathbb{R} such that f(x)=x3+3x+2f(x) = x^3 + 3x + 2, g(f(x))=xg(f(x)) = x, and h(g(g(x)))=xh(g(g(x))) = x for all x∈Rx \in \mathbb{R}. Then

Options

A

g′(2)=115g'(2) = \frac{1}{15}

B

h′(1)=666h'(1) = 666

Correct
C

h(0)=16h(0) = 16

Correct
D

h(g(3))=36h(g(3)) = 36

Step-by-Step Solution

To determine the correct options, let us analyze the given functions and their relationships.

We are given f(x)=x3+3x+2f(x) = x^3 + 3x + 2. The derivative of f(x)f(x) is: f′(x)=3x2+3f'(x) = 3x^2 + 3

Since f′(x)≥3>0f'(x) \ge 3 > 0 for all x∈Rx \in \mathbb{R}, f(x)f(x) is strictly increasing and bijective from R→R\mathbb{R} \to \mathbb{R}.

Given g(f(x))=xg(f(x)) = x for all x∈Rx \in \mathbb{R}, g(x)g(x) is the inverse function of f(x)f(x), i.e., g=f−1g = f^{-1}. Thus, we also have: f(g(x))=xfor all x∈Rf(g(x)) = x \quad \text{for all } x \in \mathbb{R}

Next, we are given h(g(g(x)))=xh(g(g(x))) = x for all x∈Rx \in \mathbb{R}. Let y=g(g(x))y = g(g(x)). Then h(y)=xh(y) = x. Applying ff to both sides of y=g(g(x))y = g(g(x)): f(y)=f(g(g(x)))=g(x)f(y) = f(g(g(x))) = g(x) Applying ff again: f(f(y))=f(g(x))=xf(f(y)) = f(g(x)) = x Since h(y)=xh(y) = x, we get: h(y)=f(f(y))  ⟹  h(x)=f(f(x))for all x∈Rh(y) = f(f(y)) \quad \implies \quad h(x) = f(f(x)) \quad \text{for all } x \in \mathbb{R}


Checking Option (A):

We differentiate g(f(x))=xg(f(x)) = x with respect to xx using the chain rule: g′(f(x))⋅f′(x)=1  ⟹  g′(f(x))=1f′(x)g'(f(x)) \cdot f'(x) = 1 \implies g'(f(x)) = \frac{1}{f'(x)}

To find g′(2)g'(2), we set f(x)=2f(x) = 2: x3+3x+2=2  ⟹  x(x2+3)=0  ⟹  x=0x^3 + 3x + 2 = 2 \implies x(x^2 + 3) = 0 \implies x = 0

So, f(0)=2f(0) = 2. Thus: g′(2)=g′(f(0))=1f′(0)g'(2) = g'(f(0)) = \frac{1}{f'(0)}

Since f′(0)=3(0)2+3=3f'(0) = 3(0)^2 + 3 = 3: g′(2)=13g'(2) = \frac{1}{3}

Hence, Option (A) is incorrect.


Checking Option (B):

Using h(x)=f(f(x))h(x) = f(f(x)), we differentiate with respect to xx: h′(x)=f′(f(x))⋅f′(x)h'(x) = f'(f(x)) \cdot f'(x)

For x=1x = 1: f(1)=13+3(1)+2=6f(1) = 1^3 + 3(1) + 2 = 6 f′(1)=3(1)2+3=6f'(1) = 3(1)^2 + 3 = 6 f′(f(1))=f′(6)=3(6)2+3=108+3=111f'(f(1)) = f'(6) = 3(6)^2 + 3 = 108 + 3 = 111

Therefore: h′(1)=f′(f(1))⋅f′(1)=111⋅6=666h'(1) = f'(f(1)) \cdot f'(1) = 111 \cdot 6 = 666

Hence, Option (B) is correct.


Checking Option (C):

Using h(x)=f(f(x))h(x) = f(f(x)) for x=0x = 0: f(0)=03+3(0)+2=2f(0) = 0^3 + 3(0) + 2 = 2 h(0)=f(f(0))=f(2)h(0) = f(f(0)) = f(2) f(2)=23+3(2)+2=8+6+2=16f(2) = 2^3 + 3(2) + 2 = 8 + 6 + 2 = 16

Thus, h(0)=16h(0) = 16.

Hence, Option (C) is correct.


Checking Option (D):

Using h(x)=f(f(x))h(x) = f(f(x)), we evaluate h(g(3))h(g(3)): h(g(3))=f(f(g(3)))h(g(3)) = f(f(g(3)))

Since f(g(x))=xf(g(x)) = x for all x∈Rx \in \mathbb{R}, we have f(g(3))=3f(g(3)) = 3. Therefore: h(g(3))=f(3)=33+3(3)+2=27+9+2=38h(g(3)) = f(3) = 3^3 + 3(3) + 2 = 27 + 9 + 2 = 38

Hence, Option (D) is incorrect.


Conclusion:

The correct options are B and C.