Board 2024 Class 12 Maths Q4 Solution: When is f strictly increasing in (a, b)?

The question

Let $f(x)$ be a continuous function on $[a, b]$ and differentiable on $(a, b)$ ⁠. Then, this function $f(x)$ is strictly increasing in $(a, b)$ if

(A) $f'(x) < 0,\ \forall\ x \in (a, b)$

(B) $f'(x) > 0,\ \forall\ x \in (a, b)$

(C) $f'(x) = 0,\ \forall\ x \in (a, b)$

(D) $f(x) > 0,\ \forall\ x \in (a, b)$

Exam answer

For $f$ continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠: if $f'(x) > 0$ for each $x \in (a, b)$ ⁠, then $f$ is strictly increasing in $(a, b)$ ⁠.

Why?

Strictly increasing means that a larger input always gives a larger output:

x_1 < x_2 \Rightarrow f(x_1) < f(x_2)

$f'(x)$ is the slope of the tangent to the graph at $x$ ⁠. If the slope is positive at every point of $(a, b)$ ⁠, the graph rises all the way from left to right, so $f$ is strictly increasing there. This is the standard test.

Checking the other options:

(A) $f'(x) < 0$ makes $f$ strictly decreasing.

(C) $f'(x) = 0$ makes $f$ constant.

(D) $f(x) = 2 - x$ is positive on $(0, 1)$ ⁠, but it is decreasing.

Why?

Watch out: option (D) is about the value $f(x)$ ⁠, not the slope $f'(x)$ ⁠. A function can stay positive while it falls.

For $f(x) = 2 - x$ on $(0, 1)$ ⁠: $f(x) > 1 > 0$ ⁠, yet $f'(x) = -1 < 0$ ⁠, so $f$ is decreasing. One example like this is enough to rule (D) out.

Answer: (B) $f'(x) > 0,\ \forall\ x \in (a, b)$

Where this comes from

Chapter 6, Application of Derivatives: Increasing and decreasing functions.

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