Show that $f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)$ is increasing in $\mathbb{R}$.

> [!TLDR] > Differentiating gives $f'(x)=2-\dfrac{1}{1+x^2}-\dfrac{1}{\sqrt{1+x^2}}=\dfrac{(2t+1)(t-1)}{t^2}$ with $t=\sqrt{1+x^2}\ge1$, which is $\ge0$ everywhere (zero only at $x=0$), so $f$ is increasing on $\mathbb{R}$. ## The idea A differentiable function is increasing on an interval when its derivative never goes negative there. So the whole task is: compute $f'(x)$ and show $f'(x)\ge0$ for every real $x$. ## Step 1 — differentiate the easy terms $$f(x)=2x+\cot^{-1}x+\log\!\left(\sqrt{1+x^2}-x\right).$$ - $\dfrac{d}{dx}(2x)=2$ - $\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}$ ## Step 2 — the logarithmic term (the one to be careful with) Let $u=\sqrt{1+x^2}-x$. Then $$u'=\frac{x}{\sqrt{1+x^2}}-1=\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}=-\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}}.$$ So $$\frac{d}{dx}\log u=\frac{u'}{u}=\frac{-\left(\sqrt{1+x^2}-x\right)}{\sqrt{1+x^2}\,\left(\sqrt{1+x^2}-x\right)}=-\frac{1}{\sqrt{1+x^2}}.$$ Note the $\sqrt{1+x^2}$ stays in the denominator — the derivative of this term is $-\dfrac{1}{\sqrt{1+x^2}}$, **not** $-1$. ## Step 3 — assemble $f'(x)$ $$f'(x)=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}.$$ ## Step 4 — show it is non-negative …

Show that f(x) = 2x + cot⁻¹x + log(√(1+x²) - x) is – Class 12 Mathematics

NCERT Exemplar · Q10

f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)

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Appeared in past exams:AP EAPCET 2026· Set eng-2026-05-14-AN· 1mexactCOMEDK 2023· Set 2023-E· 1mreworded

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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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Differentiating gives $f'(x)=2-\dfrac{1}{1+x^2}-\dfrac{1}{\sqrt{1+x^2}}=\dfrac{(2t+1)(t-1)}{t^2}$ with $t=\sqrt{1+x^2}\ge1$ , which is $\ge0$ everywhere (zero only at $x=0$ ), so $f$ is increasing on $\mathbb{R}$ .

The idea

A differentiable function is increasing on an interval when its derivative never goes negative there. So the whole task is: compute $f'(x)$ and show $f'(x)\ge0$ for every real $x$ .

Step 1 — differentiate the easy terms

f(x)=2x+\cot^{-1}x+\log\!\left(\sqrt{1+x^2}-x\right).

$\dfrac{d}{dx}(2x)=2$
$\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}$

Step 2 — the logarithmic term (the one to be careful with)

Let $u=\sqrt{1+x^2}-x$ . Then

u'=\frac{x}{\sqrt{1+x^2}}-1=\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}=-\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}}.

\frac{d}{dx}\log u=\frac{u'}{u}=\frac{-\left(\sqrt{1+x^2}-x\right)}{\sqrt{1+x^2}\,\left(\sqrt{1+x^2}-x\right)}=-\frac{1}{\sqrt{1+x^2}}.

Note the $\sqrt{1+x^2}$ stays in the denominator — the derivative of this term is $-\dfrac{1}{\sqrt{1+x^2}}$ , not $-1$ .

Step 3 — assemble $f'(x)$

f'(x)=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}.

Step 4 — show it is non-negative …

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

AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL6 marks
$x$
(i) and (ii):
(i) Show that the function given by $f(x) = \sin x$ is strictly decreasing in $(\pi/2, \pi)$ . [3 marks]
(ii) A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. [3 marks]
›Reveal solutionSolution
Differentiate, factor, and sign-test to find the increasing intervals; OR uses the sign of $\cos x$ and related rates via the sphere volume formula.

Main question: $y=x^2(x-2)^2$ . By the product rule,

$y'=2x(x-2)^2+x^2\cdot2(x-2)=2x(x-2)\big[(x-2)+x\big]=2x(x-2)(2x-2)=4x(x-1)(x-2).$

We need $y'>0$ , i.e. $x(x-1)(x-2)>0$ . The critical points are $x=0,1,2$ . Testing each interval:

$x<0$ : $(-)(-)(-)=-$ (negative)

$0<x<1$ : $(+)(-)(-)=+$ (positive)

$1<x<2$ : $(+)(+)(-)=-$ (negative)

$x>2$ : $(+)(+)(+)=+$ (positive)

So $y'>0$ on $(0,1)$ and $(2,\infty)$ — these are the intervals where $y$ is increasing.

OR (i) $f(x)=\sin x\Rightarrow f'(x)=\cos x$ . For $x\in(\pi/2,\pi)$ , $\cos x<0$ , so $f'(x)<0$ throughout the interval, meaning $f$ is strictly decreasing on $(\pi/2,\pi)$ .

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Q.Show that $f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)$ is increasing in $\mathbb{R}$ .

Concept understanding — Increasing Function Test

The Intuition: What Does "Increasing" Really Mean?

The Core Idea: Derivative as a Slope Detector

The Precise Statement

Why Does This Work? (A Quick Proof Sketch)

How to Use It in Practice

The idea

Step 1 — differentiate the easy terms

Step 2 — the logarithmic term (the one to be careful with)

Step 3 — assemble $f'(x)$

Step 4 — show it is non-negative …

Method: Proving a Function Is Increasing (or Decreasing) Everywhere Using the Sign of $f'(x)$

Steps

Common Mistakes

Q.Show that f(x)=2x+cot⁡−1x+log⁡(1+x2−x)f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)f(x)=2x+cot−1x+log(1+x2​−x) is increasing in R\mathbb{R}R.

The idea

Step 1 — differentiate the easy terms

Step 2 — the logarithmic term (the one to be careful with)

Step 3 — assemble f′(x)f'(x)f′(x)

Step 4 — show it is non-negative …

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Q.Show that $f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)$ is increasing in $\mathbb{R}$ .

Step 3 — assemble $f'(x)$

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