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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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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$f(x) = 1 + x\log(x + \sqrt{x^2 + 1}) - \sqrt{x^2 + 1}, x \geq 0$
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Differentiate; the $x/\sqrt{x^2+1}$ terms cancel, leaving $f'=\log(x+\sqrt{x^2+1})\ge0$ .

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Using $\dfrac{d}{dx}\log(x+\sqrt{x^2+1})=\dfrac{1}{\sqrt{x^2+1}}$ and $\dfrac{d}{dx}\sqrt{x^2+1}=\dfrac{x}{\sqrt{x^2+1}}$ : …
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KEAM 2026Set eng-2026-04204 marksMCQ
$f(x) = \frac{\log(e + x)}{\log(\pi + x)}$
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The derivative's sign reduces to comparing $(\pi+x)\log(\pi+x)$ with $(e+x)\log(e+x)$ ; $t\log t$ is increasing here, so $f'>0$ everywhere.

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Multiplying by $(e+x)(\pi+x)>0$ , the sign of $N$ equals the sign of $(\pi+x)\log(\pi+x)-(e+x)\log(e+x)$ . …
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KEAM 2026Set eng-2026-04214 marksMCQ
$f(x) = x^4 - 2x^2$
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$f'(x)=4x(x-1)(x+1)>0$ on $(-1,0)$ and $(1,\infty)$ ; hence strictly increasing on $[-1,0]$ and $[1,\infty)$ .

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Sign chart with roots $-1,0,1$ :

$x<-1$ : negative (decreasing)

$-1<x<0$ : positive (increasing)

$0<x<1$ : negative (decreasing) …
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$f'(x)=e^x-1>0$ exactly when $x>0$ .

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$f(x)=6x^4-3x^2-5$
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Factor $f'$ and take a sign chart across its roots $0,\pm\tfrac12$ .

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$-\tfrac12<x<0$ : positive,

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$x>\tfrac12$ : positive. …
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$f$
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$f$ is increasing on $(5,\infty)$ .

Concept and Intuition

A function increases where its derivative is positive. Here the positive denominator means the sign of $f'$ is controlled entirely by the numerator $x-5$ .

Step-by-Step Solution

$\sqrt{x^2+4}>0$ for all $x$ .

So $f'(x)>0\iff x-5>0\iff x>5$ .

Therefore $f$ is increasing on $(5,\infty)$ .

Common Mistakes …
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Concept and Intuition

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Step-by-Step Solution

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$f(x) = x^5 e^{-x}$
›Reveal solutionSolution
$f$ is increasing exactly where $5-x>0$ , i.e. on $(-\infty,5)$ .

Concept and Intuition

A function increases where its derivative is positive. Differentiate the product $x^5e^{-x}$ and factor to read off the sign.

Step-by-Step Solution

$f(x)=x^5e^{-x}$ .

$f'(x)=5x^4e^{-x}-x^5e^{-x}=x^4e^{-x}(5-x)$ .

$x^4\geq0$ and $e^{-x}>0$ for all $x$ , so the sign of $f'$ equals the sign of $(5-x)$ .

$f'(x)>0\iff 5-x>0\iff x<5$ .

Hence $f$ is increasing on $(-\infty,5)$ .

Common Mistakes …
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$f(x)=x^{3/5}(5x-12)$
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The factor $x^{-2/5}$ is always positive, so the sign of $f'$ follows $8x-\frac{36}{5}$ ; increasing on $\left(\frac{9}{10},\infty\right)$ .

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$f'(x)=8x^{3/5}-\frac{36}{5}x^{-2/5}=x^{-2/5}\left(8x-\frac{36}{5}\right)$ .

Since $x^{-2/5}=\left(x^2\right)^{-1/5}>0$ for all $x\neq0$ , the sign of $f'$ equals the sign of $8x-\frac{36}{5}$ . …
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KEAM 2026Set eng-2026-04214 marksMCQ
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Require $f'(x)=2x+a\ge 0$ throughout $[1,2]$ ; the minimum of $2x$ there is at $x=1$ , giving $a\ge -2$ .

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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

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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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