Question:

Step 1: Differentiate the given functional relationship We are given the relation $p(g(x)) = x$ for all $x \in [0,1]$. We differentiate both sides with respect to $x$ using the chain rule on the left side: $$ \frac{d}{dx} [p(g(x))] = \frac{d}{dx} [x] $$ $$ p'(g(x)) \cdot g'(x) = 1 $$ This allows us to express $g'(x)$ in terms of $p'(g(x))$: $$ g'(x) = \frac{1}{p'(g(x))} $$ Step 2: Establish the domains and ranges of $p(x)$ and $g(x)$ From $p(g(x)) = x$, it is evident that $p$ and $g$ are inverse functions. We are given that $g: [0,1] \to [a,b]$, with specific values $g(0) = a$ and $g(1) = b$. Since $p$ is the inverse of $g$, its domain will be the range of $g$, and its range will be the domain of $g$. Thus, $p: [a,b] \to [0,1]$. Using the given values: For $x=0$, $p(g(0)) = 0 \implies p(a) = 0$. For $x=1$, $p(g(1)) = 1 \implies p(b) = 1$. Step 3: Determine the range of $f'(x)$ We are given that $f'(x)$ is a continuous function which maps from $[0,1]$ to $[p(a), p(b)]$. From Step 2, we found $p(a)=0$ and $p(b)=1$. Therefore, the range of $f'(x)$ is $[0,1]$. This means that for all $x \in [0,1]$, we have: $$ 0 \leq f'(x) \leq 1 $$ Step 4: Characterize $g'(x)$ and its properties Since $p(x)$ is a differentiable function on $[a,b]$, and $p'(g(x)) \cdot g'(x) = 1$, it implies that $p'(g(x)) \neq 0$ for $g'(x)$ to be defined. Also, since $p(g(x))=x$ and $g(x)$ is continuous, $p(x)$ must be strictly monotonic (either increasing or decreasing) on $[a,b]$. Given $p(a)=0$ and $p(b)=1$, $p(x)$ must be strictly increasing, which means $p'(x) > 0$ for $x \in (a,b)$. Consequently, $g'(x) = \frac{1}{p'(g(x))}$ exists and is positive for $x \in (0,1)$. Also, since $p'(x)$ is continuous (as $p'(x) \ne 0$), $g'(x)$ is also continuous on $[0,1]$. Step 5: Verify Options (b) and (c) Let's analyze each option using the derived properties: For Option (b): $f(1) \leq 1 + f(0)$ From Step 3, we know $0 \leq f'(x) \leq 1$ for $x \in [0,1]$. Integrating $f'(x)$ over the interval $[0,1]$: $$ \int_0^1 f'(x) \, dx $$ By the Fundamental Theorem of Calculus: $$ \int_0^1 f'(x) \, dx = f(1) - f(0) $$ Since $f'(x) \leq 1$ for all $x \in [0,1]$, we can integrate this inequality: $$ \int_0^1 f'(x) \, dx \leq \int_0^1 1 \, dx $$ $$ f(1) - f(0) \leq [x]_0^1 $$ $$ f(1) - f(0) \leq 1 - 0 $$ $$ f(1) - f(0) \leq 1 $$ $$ f(1) \leq 1 + f(0) $$ Thus, Option (b) is true. For Option (c): $\dfrac{\displaystyle\int_0^1 f'(x)\, dx}{\displaystyle\int_0^1 g'(x)\, dx} \leq p'(c)$ for some $c \in (a, b)$ From Option (b), the numerator is $\int_0^1 f'(x) \, dx = f(1) - f(0)$. We know $f(1) - f(0) \leq 1$. For the denominator, $\int_0^1 g'(x) \, dx$: Since $g'(x)$ is continuous on $[0,1]$, by the Mean Value Theorem for Integrals, there exists some $x_0 \in (0,1)$ such that: $$ \int_0^1 g'(x) \, dx = g'(x_0) \cdot (1 - 0) = g'(x_0) $$ From Step 1, $g'(x_0) = \frac{1}{p'(g(x_0))}$. Let $c = g(x_0)$. Since $x_0 \in (0,1)$ and $g(0)=a, g(1)=b$ with $g$ being monotonic, it follows that $c \in (a,b)$. So, $\int_0^1 g'(x) \, dx = \frac{1}{p'(c)}$ for some $c \in (a,b)$. Now substitute these into the given inequality: $$ \frac{f(1) - f(0)}{1/p'(c)} \leq p'(c) $$ $$ (f(1) - f(0)) p'(c) \leq p'(c) $$ Since $p'(x) > 0$ for $x \in (a,b)$ (from Step 4), $p'(c) > 0$. We can divide both sides by $p'(c)$ without changing the inequality direction: $$ f(1) - f(0) \leq 1 $$ This is the same inequality derived for Option (b), which we have shown to be true. Thus, Option (c) is true. Step 6: Verify Option (d) For Option (d): There exists $k \in [0,1]$ such that $f'(k) = k$ Consider the function $h(x) = f'(x) - x$. We need to find if there exists $k \in [0,1]$ such that $h(k) = 0$. Since $f'(x)$ is continuous on $[0,1]$ (given) and $x$ is continuous, $h(x)$ is continuous on $[0,1]$. Let's evaluate $h(x)$ at the endpoints of the interval $[0,1]$: At $x=0$: $$ h(0) = f'(0) - 0 = f'(0) $$ From Step 3, we know $0 \leq f'(x) \leq 1$. So, $f'(0) \geq 0$. Thus, $h(0) \geq 0$. At $x=1$: $$ h(1) = f'(1) - 1 $$ From Step 3, we know $f'(1) \leq 1$. Thus, $f'(1) - 1 \leq 0$. So, $h(1) \leq 0$. Since $h(x)$ is continuous on $[0,1]$ and $h(0) \geq 0$ and $h(1) \leq 0$, by the Intermediate Value Theorem, there must exist at least one value $k \in [0,1]$ such that $h(k) = 0$. $$ f'(k) - k = 0 $$ $$ f'(k) = k $$ Thus, Option (d) is true. Final Answer: Options (b), (c), and (d) are true. The final answer is $\boxed{\text{B,C,D}}$.