Question

Let $f: [0,4] \rightarrow \mathbb{R}$ be a differentiable function. Choose the correct option(s) from the following:

• A. For $a, b \in (0,4)$, $(f(4))^2 - (f(0))^2 = 8f'(a)f(b)$
• B. $\int_{0}^{4} f(t) dt = 2 \left[ \alpha f(\alpha^2) + \beta f(\beta^2) \right] \quad \forall 0 < \alpha, \beta < 2$
• C. $\int_{0}^{4} f(t) dt = 3 \left[ \alpha f(\alpha^2) + \beta f(\beta^2) \right] \quad \forall 0 < \alpha, \beta < 2$
• D. $2 \int_{0}^{4} f(t) dt = \left[ \alpha f(\alpha^2) + \beta f(\beta^2) \right] \quad \forall 0 < \alpha, \beta < 2$

Answer 1

Answer: (A)

(a)

Answer 2

Option (a):

By the Mean Value Theorem applied to $g(x) = (f(x))^2$ on the interval $[0,4]$, there exists $c \in (0,4)$ such that:

$g(4) - g(0) = g'(c)(4-0)$

$(f(4))^2 - (f(0))^2 = 4 g'(c)$

Since $g'(x) = 2f(x)f'(x)$, we have:

$(f(4))^2 - (f(0))^2 = 4 [2f(c)f'(c)] = 8f(c)f'(c)$.

Thus, there exists $c \in (0,4)$ such that $(f(4))^2 - (f(0))^2 = 8f(c)f'(c)$. Option (a) states that there exist $a, b \in (0,4)$ such that $(f(4))^2 - (f(0))^2 = 8f'(a)f(b)$. This is equivalent to saying that for any differentiable function $f$ on $[0,4]$, there exist $a, b \in (0,4)$ such that $f(c)f'(c) = f'(a)f(b)$ for some $c \in (0,4)$ given by the MVT.

Let $R_f$ be the range of $f$ on $(0,4)$ and $R_{f'}$ be the range of $f'$ on $(0,4)$. Since $f$ and $f'$ are continuous on $[0,4]$, they are continuous on $(0,4)$. By the Intermediate Value Theorem, $R_f$ and $R_{f'}$ are intervals (or single points).

The range of the function $G(a,b) = f'(a)f(b)$ for $a, b \in (0,4)$ is the set of products $\{xy \mid x \in R_{f'}, y \in R_f\}$. This set is also an interval (or single point).

We have $f'(\xi)f(\xi)$, where $\xi \in (0,4)$. So $f'(\xi) \in R_{f'}$ and $f(\xi) \in R_f$.

The product $f'(\xi)f(\xi)$ is a product of a value from $R_{f'}$ and a value from $R_f$. This product must be in the set $\{xy \mid x \in R_{f'}, y \in R_f\}$, which is the range of $f'(a)f(b)$ for $a, b \in (0,4)$.

Therefore, there exist $a, b \in (0,4)$ such that $f'(a)f(b) = f'(\xi)f(\xi)$.

Options (b), (c), and (d):

These options are false. Consider $f(t) = 1$. Then $\int_0^4 f(t) dt = \int_0^4 1 dt = 4$.

• (b) $4 = 2[\alpha + \beta]$ for all $0 < \alpha, \beta < 2$. This implies $\alpha + \beta = 2$ for all $0 < \alpha, \beta < 2$, which is false (e.g., $\alpha = 0.1, \beta = 0.1$).
• (c) $4 = 3[\alpha + \beta]$ for all $0 < \alpha, \beta < 2$. This implies $\alpha + \beta = \frac{4}{3}$ for all $0 < \alpha, \beta < 2$, which is false.
• (d) $2(4) = \alpha + \beta$ for all $0 < \alpha, \beta < 2$. This implies $\alpha + \beta = 8$ for all $0 < \alpha, \beta < 2$, which is false.

Therefore, only option (a) is correct.