Let $f:[1,2] \to R$ be a differentiable function with $f'(x)$ as a non-decreasing function such that $f(1) = 2$ and $f'(2) \leq 1$, then identify the correct statement(s):
$f(x) \leq x+1 \ \forall \ x \in [1,2]$
$f(x) \geq x+1 \ \forall \ x \in [1,2]$
$f'(2) - f(2) \geq -2$
$\displaystyle\int_1^2 e^{f(x)}\,dx \leq \int_1^2 e^{x^2+1}\,dx$
Step-by-Step Solution
Key Concept: The key idea is to leverage the property that a non-decreasing derivative implies the function is convex. This allows bounding the derivative $f'(x)$ for $x \in [1,2]$ and then applying the Mean Value Theorem to establish inequalities for the function $f(x)$ and its values at specific points.
Step 1: Since $f'(x)$ is non-decreasing and $f'(2) \leq 1$, we have $f'(x) \leq f'(2) \leq 1$ for all $x \in [1,2]$.
Step 2: Apply the Mean Value Theorem. For $x \in [1,2]$: $f(x) - f(1) = f'(c)(x-1)$ for some $c \in (1,x)$. Since $f'(c) \leq 1$, we get $f(x) - 2 \leq x - 1$, i.e., $f(x) \leq x + 1$. So option (a) is correct.
Step 3: Check option (c): $f'(2) - f(2) \geq -2$. From Step 2, $f(2) \leq 3$. Also $f'(2) \leq 1$. We need $f'(2) - f(2) \geq -2$. Since $f'(x)$ is non-decreasing and $f'(2) \leq 1$, by MVT: $f(2) = f(1) + f'(c) \cdot 1 = 2 + f'(c)$ for some $c \in (1,2)$. Since $f'(c) \leq f'(2)$, we have $f(2) \leq 2 + f'(2)$, so $f'(2) - f(2) \geq f'(2) - 2 - f'(2) = -2$. Thus $f'(2) - f(2) \geq -2$. Option (c) is correct.
Step 4: Check option (d): Since $f(x) \leq x+1$ for all $x \in [1,2]$ and $e^t$ is increasing, $e^{f(x)} \leq e^{x+1}$. But we need to compare with $e^{x^2+1}$. Since $x \in [1,2]$, $x^2 \geq x$, so $x^2+1 \geq x+1 \geq f(x)$. Therefore $e^{f(x)} \leq e^{x^2+1}$, giving $\int_1^2 e^{f(x)}\,dx \leq \int_1^2 e^{x^2+1}\,dx$. Option (d) is correct.
Correct Answer: 1, 3, 4