Question

Suppose that \( f, g : \mathbb{R} \to \mathbb{R} \) are differentiable functions such that \( f \) is strictly increasing and \( g \) is strictly decreasing. Define \( p(x) = f(g(x)) \) and \( q(x) = g(f(x)) \), \( \forall x \in \mathbb{R} \). Then, for \( t > 0 \), the sign of \( \int_0^t p'(x) (q'(x)-3) \, dx \) is 
 

• A. positive
• B. negative
• C. dependent on \( t \)
• D. dependent on \( f \) and \( g \)

Hint:

When dealing with integrals involving products of functions, the signs of the derivatives and the behavior of the functions are crucial in determining the sign of the integral.

Answer 1

The Correct Option is A

Step 1: Find $p'(x)$

Using the chain rule: $$p'(x) = \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

Since $f'(g(x)) > 0$ (as $f$ is strictly increasing) and $g'(x) < 0$ (as $g$ is strictly decreasing): $$p'(x) = f'(g(x)) \cdot g'(x) < 0$$

Step 2: Find $q'(x)$

Using the chain rule: $$q'(x) = \frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x)$$

Since $g'(f(x)) < 0$ (as $g$ is strictly decreasing) and $f'(x) > 0$ (as $f$ is strictly increasing): $$q'(x) = g'(f(x)) \cdot f'(x) < 0$$

Step 3: Analyze $p'(x) \cdot q'(x)$

$$p'(x) \cdot q'(x) = [f'(g(x)) \cdot g'(x)] \cdot [g'(f(x)) \cdot f'(x)]$$

$$= f'(g(x)) \cdot f'(x) \cdot g'(x) \cdot g'(f(x))$$

Since:

• $f'(g(x)) > 0$ and $f'(x) > 0$, their product is positive
• $g'(x) < 0$ and $g'(f(x)) < 0$, their product is positive

Therefore: $$p'(x) \cdot q'(x) = (\text{positive}) \cdot (\text{positive}) > 0$$

Step 4: Evaluate the integral

$$\int_0^t p'(x)(q'(x) - 3) dx = \int_0^t [p'(x) \cdot q'(x) - 3p'(x)] dx$$

$$= \int_0^t p'(x) \cdot q'(x) dx - 3\int_0^t p'(x) dx$$

From Step 3: $p'(x) \cdot q'(x) > 0$, so $\int_0^t p'(x) \cdot q'(x) dx > 0$

From Step 1: $p'(x) < 0$, so $\int_0^t p'(x) dx < 0$, which means $-3\int_0^t p'(x) dx > 0$

Step 5: Determine the sign

We have: $$\int_0^t p'(x)(q'(x) - 3) dx = \underbrace{\int_0^t p'(x) \cdot q'(x) dx}{>0} + \underbrace{(-3)\int_0^t p'(x) dx}{>0}$$

Both terms are positive, so their sum is positive.

Therefore, for $t > 0$: $$\int_0^t p'(x)(q'(x) - 3) dx > 0$$

Answer: Option (A)