Q.Let $f(x)$ be a continuous function in the interval $[a, b]$ and differentiable in the interval $(a, b)$. Then $f(x)$ is strictly increasing in the interval $(a, b)$, if:
(A) $f'(x) < 0$, for all $x \in (a, b)$
(B) $f'(x) > 0$, for all $x \in (a, b)$
(C) $f'(x) = 0$, for all $x \in (a, b)$
(D) $f(x) > 0$, for all $x \in (a, b)$

Concept: Monotonicity Condition — A differentiable function is strictly increasing on an interval when its derivative is positive at every interior point.

Step 1: For a function $f$ that is continuous on $[a,b]$ and differentiable on $(a,b)$, the sign of $f'(x)$ determines monotonicity.

Step 2: If $f'(x) > 0$ for all $x \in (a,b)$, then by the Mean Value Theorem, for any $x_1 < x_2$ in $(a,b)$, there exists $c \in (x_1, x_2)$ such that

$f(x_2) - f(x_1) = f'(c)(x_2 - x_1) > 0$, so $f(x_2) > f(x_1)$ — strictly increasing. …

The key idea here is the Monotonicity Condition from differential calculus. When a function is continuous on $[a, b]$ and differentiable on $(a, b)$, the sign of its derivative tells us exactly how the function behaves — whether it rises, falls, or stays flat.

Think of the derivative $f'(x)$ as the instantaneous slope. If at every point the slope is positive, the function must be climbing as you move right. That’s the intuitive meaning of “strictly increasing”: for any two points $x_1 < x_2$, we have $f(x_1) < f(x_2)$.

Now let’s examine each option carefully.

1. Option (A): $f'(x) < 0$ for all $x \in (a, b)$

   A negative derivative everywhere means the function is strictly decreasing, not increasing. So this is the opposite of what we want.

2. Option (B): $f'(x) > 0$ for all $x \in (a, b)$

   This is the standard sufficient condition for strict increase. By the Mean Value Theorem, for any $x_1 < x_2$ in $(a, b)$, there exists some $c \in (x_1, x_2)$ such that

Since $f'(c) > 0$ and $x_2 - x_1 > 0$, the difference $f(x_2) - f(x_1)$ is positive. Hence $f(x_2) > f(x_1)$, proving strict increase.

3. Option (C): $f'(x) = 0$ for all $x \in (a, b)$

   A zero derivative everywhere forces the function to be constant on the interval. A constant function is neither strictly increasing nor strictly decreasing — it’s flat.

4. Option (D): $f(x) > 0$ for all $x \in (a, b)$ …

Method: Recognizing the Sufficient Condition for Strict Monotonicity

This method is for MCQs that test the theoretical statement connecting the sign of a derivative to whether a function is increasing or decreasing — no computation is needed, only recalling and matching the correct theorem.

Steps

Step 1: Recall the precise statement of the test

For $f$ continuous on $[a,b]$ and differentiable on $(a,b)$:

Step 2: Eliminate options that describe the wrong direction or no direction at all …

Common Mistakes

Mistake 1: Confusing the sign of $f(x)$ with the sign of $f'(x)$

Option (D), $f(x)>0$, tests whether a student mixes up the function's value being positive with the function's slope being positive. A function can be positive everywhere and still be strictly decreasing (e.g. $f(x)=1-x$ on $(0,1)$) — positivity of $f(x)$ says nothing about monotonicity.

Mistake 2: Treating $f'(x)=0$ (option C) as consistent with "strictly increasing" …