(a) Find the intervals in which the function $f(x) = \dfrac{\log x}{x}$ is strictly increasing or strictly decreasing. OR (b) Find the absolute maximum and absolute minimum values of the function $f$ given by $f(x) = \dfrac{x}{2} + \dfrac{2}{x}$, on the interval $[1, 2]$.

(a) Find the intervals in which the function f(x) = (log x)/(x) is strictly increasing or strictly decreasing.

Question

f(x) = \dfrac{\log x}{x}

(OR)

(b) Find the absolute maximum and absolute minimum values of the function

f

given by

f(x) = \dfrac{x}{2} + \dfrac{2}{x}

, on the interval

[1, 2]

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Derivative Sign Analysis

Derivative Sign Analysis: What the Slope Tells You

Imagine walking along a hilly road — sometimes uphill, sometimes downhill, occasionally flat. The derivative at any point is simply the slope of the road under your feet at that instant.

Derivative sign analysis figures out where a function is increasing, where it is decreasing, and where it has flat spots (critical points) — all from the sign of its derivative.

The Intuition First

If $f'(x)$ is positive, the function is increasing — the graph rises as you move right. If $f'(x)$ is negative, it is decreasing. If $f'(x) = 0$ , there is a horizontal tangent — a potential peak, valley, or flat inflection.

The key: a single point tells you little; you look at intervals. If $f'(x) > 0$ for all $x$ in $(a, b)$ , the function is strictly increasing on that whole interval. Same logic for negative.

Note

The analysis is local — it describes behaviour on intervals, not isolated points. A zero derivative at a single point doesn't guarantee a max or min; check the sign change across that point.

The Precise Statement

Let $f$ be differentiable on an open interval $I$ . Then:

If $f'(x) > 0$ for all $x$ in $I$ , then $f$ is strictly increasing on $I$ .
If $f'(x) < 0$ for all $x$ in $I$ , then $f$ is strictly decreasing on $I$ .
If $f'(x) = 0$ for all $x$ in $I$ , then $f$ is constant on $I$ .

Points where $f'(x) = 0$ (or where $f'$ does not exist) are critical points — the candidates for local maxima and minima.

\text{If } f'(x) > 0 \text{ on } (a,b) \implies f \text{ increasing on } (a,b)

How to Perform It (Step-by-Step)

Find the derivative $f'(x)$ .
Find critical points: solve $f'(x) = 0$ and check where $f'(x)$ is undefined (but $f$ is defined).
Plot these on a number line — they split the domain into intervals.
Pick a test point inside each interval and evaluate $f'$ ; only the sign matters.
Record the sign in each interval and interpret: + means increasing, – means decreasing.

A Concrete Example

Take $f(x) = x^3 - 3x$ .

Step 1: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ .

Step 2: Critical points: $x = -1$ and $x = 1$ .

Step 3: Intervals: $(-\infty, -1)$ , $(-1, 1)$ , $(1, \infty)$ .

Step 4: Test points:

$x = -2$ : $f'(-2) = 3(4-1) = 9 > 0$ .
$x = 0$ : $f'(0) = -3 < 0$ .
$x = 2$ : $f'(2) = 9 > 0$ .

Step 5: So $f$ increases on $(-\infty, -1)$ , decreases on $(-1, 1)$ , increases on $(1, \infty)$ . Thus $x = -1$ is a local maximum (sign changes + to –), and $x = 1$ is a local minimum (– to +). …

Part (b)Concept understanding — Critical Points Analysis

Critical Points Analysis: Where Functions Change Direction

Hiking a mountain range, you reach peaks (highest spot around), valleys (bottoms), and flat stretches where the ground doesn't slope. These special locations — peaks, valleys, and flat spots — are critical points.

The Intuition

A function's graph is like that trail. At most points it is rising (positive slope) or falling (negative slope). At a critical point something changes: the slope becomes zero, or the slope doesn't exist (a sharp corner).

Throw a ball straight up: at the very top of its arc it stops for an instant before falling. Its velocity — the rate of change of height — is zero at that moment. That's a critical point.

The Precise Definition

x = c

Why Two Conditions?

Derivative equals zero catches the "flat" spots — peaks, valleys, horizontal plateaus — where the tangent line is horizontal.

Derivative does not exist catches sharp corners (like the tip of $|x|$ at $x=0$ ), vertical tangents, and cusps. Even without a zero slope, these can be peaks or valleys.

Watch out

A common mistake: thinking every critical point is a maximum or minimum. Not true. A critical point could be a "saddle point" — flat but neither. For example, $f(x) = x^3$ at $x=0$ has $f'(0)=0$ , yet the function just passes through with no extremum.

How to Find Critical Points

Find the derivative $f'(x)$ .
Solve $f'(x) = 0$ — these are candidates.
Check where $f'(x)$ does not exist — but only if $f(x)$ exists there (the point must be in the domain).
Collect all such $x$ -values.

Example 1: A Simple Polynomial

Let $f(x) = x^3 - 3x^2 + 1$ .

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ .

$f'(x) = 0 \implies x = 0$ or $x = 2$ . Since $f'$ exists everywhere, the critical points are $x = 0$ and $x = 2$ .

Example 2: A Function with a Corner

Let $f(x) = |x|$ . Here $f'(x)$ does not exist at $x = 0$ (left derivative $-1$ , right derivative $+1$ ), and $f'(x) = 0$ has no solutions. So the only critical point is $x = 0$ .

Note

$x=0$ is actually a minimum of $|x|$ — the sharp corner is a valley.

What Critical Points Tell Us …

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Part (a): $f(x)=\dfrac{\log x}{x}$ is strictly increasing on $(0,e)$ and strictly decreasing on $(e,\infty)$ . Part (b): on $[1,2]$ , $f(x)=\dfrac{x}{2}+\dfrac{2}{x}$ has absolute maximum $\dfrac52$ at $x=1$ and absolute minimum $2$ at $x=2$ .

Part (a)

1. Domain. $\log x$ needs $x>0$ , so the domain is $(0,\infty)$ .

2. Derivative (quotient rule, $u=\log x$ , $v=x$ ):

f'(x)=\frac{\frac1x\cdot x-\log x\cdot1}{x^2}=\frac{1-\log x}{x^2}.

3. Critical point. $f'(x)=0\Rightarrow1-\log x=0\Rightarrow\log x=1\Rightarrow x=e$ .

4. Sign analysis. Since $x^2>0$ throughout the domain, $\operatorname{sign}f'(x)=\operatorname{sign}(1-\log x)$ :

On $(0,e)$ : $\log x<1$ , so $f'(x)>0$ — strictly increasing. …

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

CBSE 2026Set 65/1/11 markMCQ
$f(x) = x^3 - 12x$
›Reveal solutionSolution
To find the least value of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the interval's endpoints. The smallest of these values is the global minimum. For $f(x) = x^3 - 12x$ on $[0, 3]$ , the least value is $\boxed{-16}$ .

When we need to find the maximum or minimum value of a function over a specific interval, especially a closed one, we are looking for its global extrema. For a continuous function on a closed interval, these global extrema are guaranteed to exist. The key insight from calculus is that these extreme values can only occur at two types of points:

Critical points: These are points where the derivative of the function is zero or undefined. At such points, the function might have a local maximum or a local minimum.

Endpoints of the interval: Even if the function is increasing or decreasing throughout the interval, its maximum or minimum value could be at one of the boundaries.

Therefore, our strategy is to find all potential candidates for the minimum (critical points within the interval and the endpoints) and then compare the function's value at each of these candidates. The smallest value will be the least value of the function on the given interval.

Let's apply this method to $f(x) = x^3 - 12x$ on the interval $x \in [0, 3]$ .

Find the derivative of the function.

To locate critical points, we first need the first derivative of $f(x)$ .

$f(x) = x^3 - 12x$

$f'(x) = \frac{d}{dx}(x^3 - 12x) = 3x^2 - 12$

Find the critical points.

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x) = 3x^2 - 12$ is a polynomial, it is defined for all real $x$ . So, we only need to set $f'(x) = 0$ :

$3x^2 - 12 = 0$

$3x^2 = 12$

$x^2 = 4$

Solving for $x$ , we get $x = \pm 2$ .

Identify critical points within the given interval.

The given interval is $[0, 3]$ . We must check which of our critical points fall within this range.

$x = 2$ is in $[0, 3]$ . This is a candidate for the minimum.

$x = -2$ is not in $[0, 3]$ . We discard this critical point for this problem, as it's outside our domain of interest. …
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CBSE 2026Set V11 markMCQ
$f(x) = x^2$
(a) Both the Statements I and II are true
(b) Both the Statements I and II are false
(c) Statement I is true and Statement II is false
(d) Statement I is false and Statement II is true
›Reveal solutionSolution

Statement I is false and Statement II is true, so the answer is (d).

Statement I: For $f(x)=x^2$ , $f'(x)=2x$ . On $(0,\infty)$ we have $f'(x)=2x>0$ , so $f$ is increasing there, not decreasing. False. …
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CBSE 2026Set ANNUAL1 markMCQ
$f(x)=\int e^x (x-1)(x-2)\,dx$
(a) $(-\infty,-2)$
(b) $(-2,-1)$
(c) $(1,2)$
(d) $(2,+\infty)$
›Reveal solutionSolution
Since $f(x)=\int e^x(x-1)(x-2)\,dx$ , we get $f'(x)=e^x(x-1)(x-2)$ ; $f$ decreases where $f'(x)<0$ , i.e. on $(1,2)$ .

By the Fundamental Theorem of Calculus, if $f(x)=\int e^x(x-1)(x-2)\,dx$ , then

$f'(x)=e^x(x-1)(x-2)$

A function decreases on an interval where its derivative is negative: $f'(x)<0$ .

Since $e^x>0$ for every real $x$ , the sign of $f'(x)$ is entirely determined by the sign of $(x-1)(x-2)$ :

For $x<1$ : both factors negative $\Rightarrow$ product positive $\Rightarrow f'(x)>0$ . …
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CBSE 2025Set 65/1/11 markMCQ
$f(x) = x^3 - 3x + 2$
›Reveal solutionSolution
To find the absolute maximum of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the interval's endpoints, then pick the largest value. For $f(x) = x^3 - 3x + 2$ on $[0, 2]$ , the absolute maximum value is $\boxed{4}$ .

When we need to find the absolute maximum (or minimum) value of a continuous function over a closed interval, we rely on a fundamental concept from calculus called the Extreme Value Theorem. This theorem guarantees that such a maximum and minimum must exist.

The intuition behind finding these extreme values is that they can occur in one of two places:

At a "peak" or "valley" within the interval: These are points where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum). At such points, if the function is differentiable, its derivative will be zero. These are called critical points.

At the boundaries of the interval: Even if the function is steadily increasing or decreasing throughout the interval, its highest or lowest value might simply be at one of the endpoints.

Therefore, our strategy is to check all these potential locations: the critical points that fall within our interval, and the two endpoints of the interval. We then compare the function values at all these points to find the absolute maximum.

Here's how we apply this to $f(x) = x^3 - 3x + 2$ on the interval $[0, 2]$ :

Find the derivative of the function. The derivative $f'(x)$ tells us about the slope of the tangent line to the function at any point $x$ . Critical points occur where the tangent line is horizontal, meaning $f'(x) = 0$ .

$f(x) = x^3 - 3x + 2$ $f'(x) = \frac{d}{dx}(x^3 - 3x + 2)$ $f'(x) = 3x^2 - 3$

Find the critical points by setting the derivative to zero. We solve $f'(x) = 0$ to find the $x$ -values where the function might have a local maximum or minimum.

$3x^2 - 3 = 0$ $3(x^2 - 1) = 0$ $x^2 - 1 = 0$
This is a difference of squares, which factors as $(x-1)(x+1)=0$. So, the critical points are $x = 1$ and $x = -1$.

3. Identify which critical points lie within the given interval.

The given interval is $[0, 2]$ . We must only consider critical points that are inside or on the boundary of this interval.

* $x = 1$ is in $[0, 2]$ . This is a relevant critical point. …
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CBSE 2025Set 65/4/11 markMCQ
$\lambda$
›Reveal solutionSolution
A function decreases everywhere when its derivative is non-positive for all $x$ . Here $f'(x) = \cos x + \sin x - \lambda$ must satisfy $\cos x + \sin x \le \lambda$ for all $x$ , which requires $\lambda \ge \sqrt{2}$ (the maximum of $\cos x + \sin x$ ).

A function decreases for all real $x$ when its rate of change is never positive. This translates to the condition $f'(x) \le 0$ for all $x \in \mathbb{R}$ . The question asks us to find which values of the parameter $\lambda$ enforce this condition.

The key insight is that we need to understand the range of the trigonometric expression in the derivative, then choose $\lambda$ large enough to dominate it everywhere.

Finding the derivative

Differentiate $f(x) = \sin x - \cos x - \lambda x + C$ :

$f'(x) = \cos x + \sin x - \lambda$

For $f$ to be decreasing everywhere, we need:

$f'(x) \le 0 \quad \text{for all } x \in \mathbb{R}$
This means:
$\cos x + \sin x - \lambda \le 0$ $\cos x + \sin x \le \lambda \quad \text{for all } x$
Finding the maximum of $\cos x + \sin x$

The condition $\cos x + \sin x \le \lambda$ for all $x$ is equivalent to requiring:

$\lambda \ge \max_{x \in \mathbb{R}} (\cos x + \sin x)$

To find this maximum, we can express the sum as a single sinusoid. Using the identity:

$\cos x + \sin x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$
›Proof

Derivation of the identity:

We write $\cos x + \sin x = R \sin(x + \phi)$ for some amplitude $R$ and phase $\phi$ .

Expanding: $R \sin(x + \phi) = R(\sin x \cos \phi + \cos x \sin \phi) = R \cos \phi \cdot \sin x + R \sin \phi \cdot \cos x$

Comparing coefficients:

Coefficient of $\sin x$ : $R \cos \phi = 1$

Coefficient of $\cos x$ : $R \sin \phi = 1$

Squaring and adding: $R^2(\cos^2 \phi + \sin^2 \phi) = 1 + 1 = 2$ , so $R = \sqrt{2}$ .

…
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CBSE 2024Set 65/2/11 markMCQ
$f(x) = x^3 - 3x^2 + 12x - 18$
›Reveal solutionSolution
The derivative $f'(x) = 3x^2 - 6x + 12$ is always positive (its discriminant is negative and leading coefficient positive), so $f(x)$ is strictly increasing on $\mathbb{R}$ . The correct option is (B).

The core question here is about monotonicity — whether a function is always increasing, always decreasing, or neither. For a polynomial, the sign of its derivative tells us everything. If the derivative is positive everywhere, the function is strictly increasing; if negative everywhere, strictly decreasing; if it changes sign, the function is neither.

Let’s see what $f'(x)$ looks like.

Find the derivative. $f(x) = x^3 - 3x^2 + 12x - 18$ Differentiating term by term:

$f'(x) = 3x^2 - 6x + 12$

Analyze the sign of $f'(x)$ . This is a quadratic: $3x^2 - 6x + 12$ . To check if it ever becomes negative or zero, compute its discriminant:

$D = (-6)^2 - 4 \cdot 3 \cdot 12 = 36 - 144 = -108$
Since $D < 0$ , the quadratic has no real roots — it never touches or crosses the x-axis.

What does a negative discriminant mean for sign? The leading coefficient $3 > 0$ , so the parabola opens upward. A quadratic that opens upward and has no real roots is always positive. Therefore, $f'(x) > 0$ for every real $x$ . …
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CBSE 2024Set A1 mark
$C$
›Reveal solutionSolution

A point where $f'(c)=0$ or $f$ is not differentiable is called a critical point of $f$ .

In the study of maxima/minima (Application of Derivatives), a point $c$ in the domain of $f$ at which either $f'(c)=0$ (a stationary point) or $f$ fails to be differentiable i …

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CBSE 2019Set ANNUAL1 mark
$\phi(x) = f(x) + f(1-x)$
›Reveal solutionSolution

$f''(x)=0$ forces $f$ to be linear, which makes $\phi(x)=f(x)+f(1-x)$ constant on $[0,1]$ — so $x=\tfrac12$ is neither a genuine maximum nor minimum.

$\phi(x) = f(x)+f(1-x)$ .

$\phi'(x) = f'(x) - f'(1-x)$

At $x=\tfrac12$ : $\phi'\left(\tfrac12\right) = f'\left(\tfrac12\right) - f'\left(\tfrac12\right) = 0$ , so $x=\tfrac12$ is a critical point.

$\phi''(x) = f''(x) + f''(1-x)$

We are given $f''(x)=0$ for all $x\in[0,1]$ , so $f''(1-x)=0$ too, giving $\phi''(x) = 0+0 = 0$ for every $x\in[0,1]$ — the second-derivative test is inconclusive.

In fact, since $f''(x)=0$ throughout $[0,1]$ , $f$ must be a linear function, say $f(x)=mx+c$ . Then

$\phi(x) = (mx+c) + \big(m(1-x)+c\big) = mx+c+m-mx+c = m+2c$ ,

…

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CBSE 2019Set ANNUAL1 markMCQ
$f(x) = \frac{\log x}{x}$
(a) $1$
(b) $\frac{2}{e}$
(c) $e$
(d) $\frac{1}{e}$
›Reveal solutionSolution

Maximum value $=\frac{1}{e}$ at $x=e$ .

…

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Q.(a) Find the intervals in which the function f(x)=log⁡xxf(x) = \dfrac{\log x}{x}f(x)=xlogx​ is strictly increasing or strictly decreasing.

Part (a)

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Q.(a) Find the intervals in which the function $f(x) = \dfrac{\log x}{x}$ is strictly increasing or strictly decreasing.

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