Differential Equations
Properties of differentiable functions and inequalities
GRB_1000_MCQ
Grade Class 12
Question:
Let $f(x): R \to R$ and $g(x): R \to R$ be two differentiable functions, such that $f(x)$; $g(x)$; $x - g'(x)$ and $f'(x) + f(x)g'(x)$ are non-negative for all real $x$. Then:
$g(1) - g(0) \leq k \ \forall \ k \in (5, \infty)$
$g(1) - g(0) \leq k \ \forall \ k \in (0, \infty)$
Maximum value of $\dfrac{f(0)}{f(1)}$ is $e^{1/4}$
Maximum value of $\dfrac{f(0)}{f(1)}$ is $e^{1/2}$
Step-by-Step Solution
Key Concept: The key idea involves two main calculus techniques: using definite integration to establish bounds for functions from inequalities involving their derivatives, and recognizing and manipulating differential inequalities using the concept of an integrating factor to determine the monotonicity of a composite function.
Step 1: Since $x - g'(x) \geq 0$, we have $g'(x) \leq x$ for all real $x$. Integrate both sides from 0 to 1:
$$g(1) - g(0) \leq \int_0^1 x\, dx = \frac{1}{2}$$
So $g(1) - g(0) \leq \frac{1}{2}$, which means $g(1) - g(0) \leq k$ for all $k \in (0, \infty)$ (since $\frac{1}{2} > 0$, option 2 is correct; option 1 requires $k \in (5,\infty)$ which is too restrictive).
Step 2: Since $f'(x) + f(x)g'(x) \geq 0$, multiply through: this means $\frac{d}{dx}[f(x)e^{g(x)}] \geq 0$, so $f(x)e^{g(x)}$ is non-decreasing.
Step 3: Since $f(x)e^{g(x)}$ is non-decreasing, $f(0)e^{g(0)} \leq f(1)e^{g(1)}$, giving:
$$\frac{f(0)}{f(1)} \leq e^{g(1)-g(0)}$$
Step 4: From Step 1, $g(1) - g(0) \leq \frac{1}{2}$, so:
$$\frac{f(0)}{f(1)} \leq e^{1/2}$$
Thus the maximum value of $\dfrac{f(0)}{f(1)}$ is $e^{1/2}$, confirming option 4.
Correct Answer: 2, 4