Limits, Continuity & Differentiability Question - None

Question

Given that
(a) $x - g'(x) \geq 0 \; \forall x$
(b) $f'(x) + f(x)g'(x) \geq 0 \; \forall x$
Which of the following are correct?
(a) $g(1) - g(0) \leq \dfrac{1}{2}$
(b) $f(x) \cdot e^{g(x)}$ is an increasing function $\forall x$
(c) $f(0) \cdot e^{g(0)} \leq f(1) \cdot e^{g(1)}$
(d) $\dfrac{f(0)}{f(1)} \leq e^{g(1)-g(0)} \leq e^{1/2}$

Accepted Answer

Step 1: Analyze the given conditions

Given: (i) x - g'(x) ≥ 0 for all x, which means g'(x) ≤ x

(ii) f'(x) + f(x)g'(x) ≥ 0 for all x

Step 2: Check option (a): g(1) - g(0) ≤ 1/2

From condition (i): g'(x) ≤ x

Integrating from 0 to 1:

∫₀¹ g'(x)dx ≤ ∫₀¹ x dx

g(1) - g(0) ≤ [x²/2]₀¹ = 1/2

✓ Option (a) is CORRECT

Step 3: Check option (b): f(x)·e^{g(x)} is increasing

Let h(x) = f(x)·e^{g(x)}

h'(x) = f'(x)·e^{g(x)} + f(x)·g'(x)·e^{g(x)}

h'(x) = e^{g(x)}[f'(x) + f(x)g'(x)]

From condition (ii): f'(x) + f(x)g'(x) ≥ 0

Since e^{g(x)} > 0, we have h'(x) ≥ 0

However, this means h(x) is non-decreasing (monotone increasing or constant), not strictly increasing ∀x.

✗ Option (b) is AMBIGUOUS/INCORRECT (the function may not be strictly increasing)

Step 4: Check option (c): f(0)·e^{g(0)} ≤ f(1)·e^{g(1)}

From Step 3, h'(x) = e^{g(x)}[f'(x) + f(x)g'(x)] ≥ 0

This means h(x) = f(x)·e^{g(x)} is non-decreasing.

Therefore: h(0) ≤ h(1), i.e., f(0)·e^{g(0)} ≤ f(1)·e^{g(1)}

✓ Option (c) is CORRECT

Step 5: Check option (d): f(0)/f(1) ≤ e^{g(1)-g(0)} ≤ e^{1/2}

From option (c): f(0)·e^{g(0)} ≤ f(1)·e^{g(1)}

Dividing both sides by f(1)·e^{g(0)} (assuming f(1) > 0):

f(0)/f(1) ≤ e^{g(1)-g(0)}

From option (a): g(1) - g(0) ≤ 1/2

Therefore: e^{g(1)-g(0)} ≤ e^{1/2}

Combined: f(0)/f(1) ≤ e^{g(1)-g(0)} ≤ e^{1/2}

✓ Option (d) is CORRECT

∴ Answer: A,D

Question:

Step-by-Step Solution

Master Limits, Continuity & Differentiability with Mathbee