Find the sub-interval of $\left(0, \frac{\pi}{2}\right)$ in which the function $f(x) = \tan x - 4x$ is increasing.

Find the sub-interval of (0, (π)/(2)) in which the func — CBSE Class 12 Mathematics 2026 PYQ

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\left(0, \frac{\pi}{2}\right)

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

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Concept understanding — Increasing Function Test

The Intuition: What Does "Increasing" Really Mean?

Imagine walking along the graph of a function from left to right. If the function is increasing, then as you step right (increasing $x$ ), you always move upward — your height $f(x)$ never drops. You might stay flat briefly, but you never go down.

That's the visual idea. But we need a precise way to check it without drawing the entire graph — that's the Increasing Function Test, which uses the derivative to tell you where a function is rising.

Note

A function $f$ is increasing on an interval if, for any two points $x_1 < x_2$ in it, $f(x_1) \le f(x_2)$ . With strict inequality ( $<$ ), it's strictly increasing.

The Core Idea: Derivative as a Slope Detector

The derivative $f'(x)$ gives the slope of the tangent line — the instantaneous rate of change. Positive slope means the function is rising at that instant; negative means falling. So the natural question: if the derivative is positive everywhere on an interval, does that guarantee the function is increasing on that whole interval? The answer is yes — and that's the Increasing Function Test.

The Precise Statement

f

The conditions "continuous on the closed interval" and "differentiable on the open interval" ensure there are no jumps or corners that could break the logic.

Why Does This Work? (A Quick Proof Sketch)

The proof relies on the Mean Value Theorem. For $x_1 < x_2$ in $[a, b]$ , there exists some $c$ between them such that:

f(x_2) - f(x_1) = f'(c)(x_2 - x_1)

Since $x_2 - x_1 > 0$ , if $f'(c) > 0$ the right-hand side is positive, so $f(x_2) > f(x_1)$ . This holds for any pair $x_1 < x_2$ — exactly the definition of strictly increasing.

Watch out

The converse is not true. A function can be strictly increasing even if its derivative is zero at some isolated points. Example: $f(x) = x^3$ is strictly increasing everywhere, but $f'(0) = 0$ . The test gives a sufficient condition, not a necessary one.

How to Use It in Practice

Compute $f'(x)$ .
Solve $f'(x) > 0$ — the solution intervals tell you where $f$ is strictly increasing.
Check endpoints if needed. …

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A function is increasing where its derivative is positive. Since $f'(x) = \sec^2 x - 4$ , we solve $\sec^2 x > 4$ on $(0, \frac{\pi}{2})$ to find $f$ is increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ .

Why the derivative tells us where a function increases

When we want to know where a function climbs upward as we move left to right, we look at its derivative. If $f'(x) > 0$ at every point in an interval, then $f$ is increasing there—each tiny step forward in $x$ produces a positive change in $f(x)$ .

For $f(x) = \tan x - 4x$ , we need to find where $f'(x) > 0$ within the given domain $(0, \frac{\pi}{2})$ .

Step-by-step solution

1. Compute the derivative

Differentiating term by term:

f'(x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(4x) = \sec^2 x - 4

2. Set up the inequality for increasing behavior

We want $f'(x) > 0$ :

\sec^2 x - 4 > 0

\sec^2 x > 4

3. Solve the inequality

Recall that $\sec x = \frac{1}{\cos x}$ , so $\sec^2 x = \frac{1}{\cos^2 x}$ . The inequality becomes:

\frac{1}{\cos^2 x} > 4

Since $\cos x > 0$ throughout $(0, \frac{\pi}{2})$ , we can safely take reciprocals and reverse the inequality:

\cos^2 x < \frac{1}{4}

Taking square roots (and keeping $\cos x > 0$ in our interval):

\cos x < \frac{1}{2}

4. Find where $\cos x < \frac{1}{2}$ on $(0, \frac{\pi}{2})$

We know $\cos x = \frac{1}{2}$ when $x = \frac{\pi}{3}$ . Since cosine is a decreasing function on $(0, \pi)$ , we have:

For $0 < x < \frac{\pi}{3}$ : $\cos x > \frac{1}{2}$ , so $f'(x) < 0$ (function is decreasing) …

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

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set CX1 markMCQ
$f(x)=x^{2}-4x+6$
(a) $(2,10)$
(b) $(2,\infty)$
(c) $(-2,\infty)$
(d) $(0,\infty)$
›Reveal solutionSolution

$f'(x)=2x-4$ is positive for $x>2$ , so $f$ is increasing on $(2,\infty)$ — option (b).

Concept: A differentiable function increases where its derivative is positive.

$f(x)=x^{2}-4x+6\;\Rightarrow\;f'(x)=2x-4.$

Set $f'(x)>0$ :

$2x-4>0\;\Rightarrow\;x>2.$ …

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CBSE 2026Set ANNUAL1 mark
$Q. Show that the function f(x) = x³ - 3x² + 3x + 10 is always increasing.$
›Reveal solutionSolution

A function is increasing on an interval where its derivative is non-negative there; we show $f'(x)\ge 0$ for every real $x$ .

$f(x)=x^3-3x^2+3x+10$

$f'(x) = 3x^2-6x+3 = 3(x^2-2x+1) = 3(x-1)^2$

…

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CBSE 2026Set ANNUAL1 markMCQ
$f(x)=x^{4}$
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true and R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
›Reveal solutionSolution

$f(x)=x^4$ has $f'(x)=4x^3>0$ on $(0,\infty)$ , so it is increasing (A false). The Reason (negative derivative $\Rightarrow$ decreasing) is a true criterion (R true).

Assertion: $f'(x)=4x^{3}$ . For $x\in(0,\infty)$ , $f'(x)>0$ , so $f$ is increasing, not decreasing. Hence A is false.

…

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CBSE 2025Set 65/2/11 markMCQ
$f(x) = x^2 - 4x + 6$
›Reveal solutionSolution
The function $f(x) = x^2 - 4x + 6$ is a parabola opening upward, so it decreases until its vertex and then increases. The vertex is at $x = 2$ , so the function is increasing on $[2, \infty)$ . The correct option is (D).

The key idea here is the Increasing Function Test from calculus: a function $f(x)$ is increasing on an interval if its derivative $f'(x) \geq 0$ for all $x$ in that interval (and strictly increasing if $f'(x) > 0$ ). But before we dive into derivatives, let's think about what this function looks like.

$f(x) = x^2 - 4x + 6$ is a quadratic — a parabola. The coefficient of $x^2$ is positive (it's $1$ ), so the parabola opens upward. That means it has a single minimum point (the vertex), falls to the left of that vertex, and rises to the right. So the function is decreasing on $(-\infty, \text{vertex}]$ and increasing on $[\text{vertex}, \infty)$ . The question is simply: where is the vertex?

Let's work through it step by step.

Find the derivative.

$f'(x) = 2x - 4$ . This is a linear function. The sign of $f'(x)$ tells us where $f$ is increasing or decreasing.

Set the derivative to zero to find the critical point.

$2x - 4 = 0 \implies x = 2$ . This is the vertex of the parabola — the point where the function stops decreasing and starts increasing.

Test the sign of $f'(x)$ on either side of $x = 2$ .

For $x < 2$ , say $x = 0$ : $f'(0) = -4 < 0$ . So $f$ is decreasing on $(-\infty, 2)$ .

For $x > 2$ , say $x = 3$ : $f'(3) = 2 > 0$ . So $f$ is increasing on $(2, \infty)$ .

What about at $x = 2$ itself?

$f'(2) = 0$ . The function is neither increasing nor decreasing at that single point, but by convention, we include the endpoint where the derivative is zero when describing intervals of monotonicity. So the function is increasing on $[2, \infty)$ . …
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CBSE 2025Set 65/4/11 markMCQ
$f(x) = 2x + \cos x$
›Reveal solutionSolution
A function is increasing when its derivative is always positive. Since $f'(x) = 2 - \sin x \geq 1 > 0$ for all $x$ , the function is strictly increasing everywhere.

The question asks about the monotonicity and extrema of $f(x) = 2x + \cos x$ . To understand the behavior of any function, we look at its derivative: the sign of $f'(x)$ tells us whether the function is climbing or falling at each point.

A function has a local maximum or minimum only where $f'(x) = 0$ (critical points), and even then only if the derivative changes sign. If $f'(x)$ never changes sign—if it's always positive or always negative—the function marches steadily in one direction without any peaks or valleys.

Let me find the derivative and analyze its sign.

Differentiate $f(x)$ :

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

Examine the range of $f'(x)$ :

We know that $\sin x$ oscillates between $-1$ and $1$ for all real $x$ . Therefore:

$-1 \leq \sin x \leq 1$
Multiplying by $-1$ (which reverses inequalities):
$-1 \leq -\sin x \leq 1$
Adding $2$ throughout:
$1 \leq 2 - \sin x \leq 3$

Interpret the result:

The derivative $f'(x) = 2 - \sin x$ satisfies $1 \leq f'(x) \leq 3$ for all $x$ . In particular, $f'(x) \geq 1 > 0$ everywhere.

Conclude about monotonicity: …
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CBSE 2025Set A1 mark
$f$
›Reveal solutionSolution

Find $f'(x)$ and determine where it is positive.

Given $f(x) = x^2-2x$ :

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

$f$ is increasing where $f'(x) > 0$ :

$2(x-1) > 0 \implies x > 1$

…

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CBSE 2025Set ANNUAL1 markMCQ
$y=\ln x,\ x\in R^{+}$
(i) $(0,\infty)$
(ii) $(-\infty,\infty)$
(iii) $(-\infty,0)$
(iv) $(-1,\infty)$
›Reveal solutionSolution

$y=\ln x$ has $y' = 1/x > 0$ on its entire domain, so it is increasing on $(0,\infty)$ .

$y = \ln x \implies \frac{dy}{dx} = \frac{1}{x}$

For $x \in R^{+}$ (i.e. $x>0$ ), $\dfrac{1}{x} > 0$ always. A function whose derivative is positive throughout an interval is strictly increasing on that interval.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The function f(x) = log x is :$
(a) Strictly increasing on (0, ∞)
(b) Strictly decreasing on (0, ∞)
(c) Neither increasing nor decreasing on (0, ∞)
(d) None of these
›Reveal solutionSolution

$f'(x)=1/x>0$ for every $x>0$ , so $\log x$ is strictly increasing throughout $(0,\infty)$ .

A function $f$ is strictly increasing on an interval if $f'(x)>0$ for all $x$ in that interval. For $f(x)=\log x$ (domain $x>0$ ),

$f'(x) = \frac{1}{x}.$ …

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CBSE 2025Set ANNUAL1 mark
$f(x) = 3x+17$
›Reveal solutionSolution

Show the derivative is positive throughout $\mathbb{R}$ .

A differentiable function is strictly increasing on an interval where $f'(x)>0$ .

For $f(x)=3x+17$ ,

$f'(x)=\frac{d}{dx}(3x+17)=3.$

This is positive for every real $x$ :

$f'(x)=3>0\quad\forall x\in\mathbb{R}.$ …

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CBSE 2025Set ANNUAL1 markMCQ
$f(x) = x^2e^{-x}$
(a) $(-\infty,\infty)$
(b) $(-2,0)$
(c) $(2,\infty)$
(d) $(0,2)$
›Reveal solutionSolution
Differentiate, factor $f'(x)$ , and find where it is positive.

Given $f(x) = x^2e^{-x}$ .

$f'(x) = 2xe^{-x} + x^2(-e^{-x}) = e^{-x}(2x-x^2) = e^{-x}x(2-x)$

Since $e^{-x}>0$ always, the sign of $f'(x)$ is the sign of $x(2-x)$ .

$x(2-x) > 0$ when both factors have the same sign:

$x>0$ and $2-x>0 \Rightarrow 0<x<2$ …
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CBSE 2025Set ANNUAL1 mark
$Q. Find the interval in which the function f(x) = 2x² + 12x + 1 is increasing.$
›Reveal solutionSolution

A function is increasing where its first derivative is positive. Find $f'(x)$ , set it $>0$ , and solve for $x$ .

Given: $f(x) = 2x^2 + 12x + 1$

Step 1 — differentiate:

$f'(x) = 4x + 12$

Step 2 — set $f'(x) > 0$ for increasing:

$4x + 12 > 0 \Rightarrow 4x > -12 \Rightarrow x > -3$

…

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CBSE 2024Set ANNUAL1 markMCQ
$y = x^2 e^{-x}$
(a) $(1, 0)$
(b) $(2, 0)$
(c) $(2, -\infty)$
(d) $(0, 2)$
›Reveal solutionSolution

Find $y'$ by the product rule and determine where it is positive.

$y = x^2 e^{-x}$

$y' = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2 - x)$

Since $e^{-x} > 0$ for all $x$ , the sign of $y'$ matches the sign of $x(2-x)$ .

…

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Q.Find the sub-interval of (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​) in which the function f(x)=tan⁡x−4xf(x) = \tan x - 4xf(x)=tanx−4x is increasing.

Why the derivative tells us where a function increases

Step-by-step solution

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Q.Find the sub-interval of $\left(0, \frac{\pi}{2}\right)$ in which the function $f(x) = \tan x - 4x$ is increasing.

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