The function $f(x) = a^x$ is increasing on R, if : (A) $a > 0$ (B) $a > 1$ (C) $a < 0$ (D) $0 < a < 1$

> [!TLDR] > $a^x$ is increasing on $\mathbb{R}$ iff $a>1$, since that makes $\ln a>0$ and hence $f'(x)>0$ everywhere. > [!FORMULA] > $\dfrac{d}{dx}(a^x) = a^x \ln a$; a function is increasing where its derivative is positive. 1. Differentiate: $f'(x) = a^x \ln a$. …

The function f(x) = ax is increasing on R, if : (A) a > — CBSE Class 12 Applied Mathematics 2025 PYQ

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f(x) = a^x

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Concept understanding — Monotonic Function Analysis

Monotonic Function Analysis

The Intuition

Imagine you are walking along a path that only ever goes uphill, or only ever goes downhill. You never have to go up and then down, or down and then up. That path is monotonic — it moves in one consistent direction.

A function is monotonic when its output (the $y$ -value) never reverses direction as its input (the $x$ -value) increases. If it always goes up (or stays flat), it is increasing. If it always goes down (or stays flat), it is decreasing. If it does both — rises, then falls — it is not monotonic.

Note

The word "monotonic" comes from Greek monotonos — "one tone." Just as a monotone voice stays on a single pitch, a monotonic function stays on a single trend.

The Precise Definition

Let $f$ be a function defined on an interval $I$ . We say:

$f$ is increasing (or non-decreasing) on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) \leq f(x_2)$ .
$f$ is strictly increasing on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) < f(x_2)$ .
$f$ is decreasing (or non-increasing) on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) \geq f(x_2)$ .
$f$ is strictly decreasing on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) > f(x_2)$ .

If a function is either increasing or decreasing on an interval, it is called monotonic on that interval.

Watch out

| Common Mistake | Correct Understanding |

|----------------|----------------------|

| "Increasing means $f'(x) > 0$ everywhere" | $f'(x) > 0$ implies strictly increasing, but a function can be increasing even where $f'(x) = 0$ at isolated points (e.g., $f(x) = x^3$ at $x=0$ ). |

| "Monotonic means the whole domain" | A function can be monotonic on a sub-interval but not on its entire domain. For example, $f(x) = x^2$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$ , but not monotonic on $\mathbb{R}$ . |

How to Determine Monotonicity (The Derivative Test)

For differentiable functions, the derivative tells you the direction:

If $f'(x) \geq 0$ for all $x$ in an interval, then $f$ is increasing on that interval.
If $f'(x) \leq 0$ for all $x$ in an interval, then $f$ is decreasing on that interval.
If $f'(x) > 0$ for all $x$ in an interval (except possibly at isolated points), then $f$ is strictly increasing.
If $f'(x) < 0$ for all $x$ in an interval (except possibly at isolated points), then $f$ is strictly decreasing.

f'(x) \geq 0 \implies f \text{ increasing} \qquad f'(x) \leq 0 \implies f \text{ decreasing}

Why It Matters

Monotonic functions are predictable. They have at most one root (if strictly monotonic), they are invertible (if strictly monotonic), and their extreme values occur only at the endpoints of the interval. In exam problems, you will often be asked to:

Find the intervals where a given function is increasing or decreasing.
Use monotonicity to prove inequalities (e.g., show $e^x > 1 + x$ for $x > 0$ ).
Determine the number of real roots of an equation.

A Worked Example …

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$a^x$ is increasing on $\mathbb{R}$ iff $a>1$ , since that makes $\ln a>0$ and hence $f'(x)>0$ everywhere.

\dfrac{d}{dx}(a^x) = a^x \ln a

Differentiate: $f'(x) = a^x \ln a$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/S/WXYZ/41 markMCQ
$f(x) = a^x$
›Reveal solutionSolution
$a^x$ is increasing on $\mathbb{R}$ iff $a>1$ , since that makes $\ln a>0$ and hence $f'(x)>0$ everywhere.

$\dfrac{d}{dx}(a^x) = a^x \ln a$

Differentiate: $f'(x) = a^x \ln a$ . …
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CBSE 2024Set 465/RQPS/41 markMCQ
$f(x) = x^2 - x + 1$
›Reveal solutionSolution
The turning point is at $x=\tfrac12$ ; $f$ decreases on $(0,\tfrac12)$ and increases on $(\tfrac12,1)$ .

$f$

$f(x)=x^2-x+1\Rightarrow f'(x)=2x-1$ .

$f'(x)=0\Rightarrow x=\tfrac12$ (the critical point). …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$Q. Questions number 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b),$
(c) and
(d) as given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true. Assertion (A) : The function $f(x) = (x+2)e^{-x}$ is increasing in the interval $(-1, \infty)$ . Reason (R) : A function $f(x)$ is increasing, if $f'(x) > 0$ .
›Reveal solutionSolution
$f'(x)=-(x+1)e^{-x}>0$ only for $x<-1$ , so $f$ is decreasing on $(-1,\infty)$ — A is false; R is a correct general statement — so answer (d).

$f'(x)>0$

$f(x)=(x+2)e^{-x}$ . Differentiate: $f'(x)=(1)e^{-x}+(x+2)(-e^{-x})=e^{-x}[1-(x+2)]=-(x+1)e^{-x}$ .

Since $e^{-x}>0$ always, $\text{sign}(f')=\text{sign}(-(x+1))$ . …
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Q.The function f(x)=axf(x) = a^xf(x)=ax is increasing on R, if : (A) a>0a > 0a>0 (B) a>1a > 1a>1 (C) a<0a < 0a<0 (D) 0<a<10 < a < 10<a<1

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Q.The function $f(x) = a^x$ is increasing on R, if : (A) $a > 0$ (B) $a > 1$ (C) $a < 0$ (D) $0 < a < 1$

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