Show that $2\sin x + \tan x \ge 3x$ for all $x \in \left(0, \dfrac{\pi}{2}\right)$.

Show that 2sin x + tan x ≥ 3x for all x ∈ (0, (π)/(2)).

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2\sin x + \tan x \ge 3x

Odisha ChseOdisha CHSE +2 Science Board Exam 2024Subjective· 4mImportance★★★★★

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Concept understanding — Monotonicity of Trigonometric Functions

Monotonicity of Trigonometric Functions

The trigonometric functions rise and fall in a repeating pattern, so unlike a polynomial they are not monotonic over the whole real line — but on each piece of a period they are strictly increasing or strictly decreasing. Derivatives pin down exactly which piece is which.

The Idea

Picture the unit circle. As the angle $x$ grows, $\sin x$ (the height) climbs from $-1$ up to $1$ and back down, while $\cos x$ (the horizontal coordinate) does the same shifted by a quarter turn. Because the motion reverses at the top and bottom, each function alternates between increasing and decreasing stretches.

Sine

$\dfrac{d}{dx}\sin x = \cos x$ , so the sign of $\cos x$ decides the monotonicity of $\sin x$ :

$\cos x > 0$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ , so $\sin x$ is strictly increasing there.
$\cos x < 0$ on $\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right)$ , so $\sin x$ is strictly decreasing there.

This pattern repeats every $2\pi$ .

Cosine

$\dfrac{d}{dx}\cos x = -\sin x$ , so the sign of $-\sin x$ governs $\cos x$ :

On $(0, \pi)$ , $\sin x > 0$ , hence $-\sin x < 0$ : $\cos x$ is strictly decreasing.
On $(\pi, 2\pi)$ , $\sin x < 0$ , hence $-\sin x > 0$ : $\cos x$ is strictly increasing.

Tangent

$\dfrac{d}{dx}\tan x = \sec^2 x > 0$ wherever it is defined. So $\tan x$ is strictly increasing on every interval $\left(-\tfrac{\pi}{2} + n\pi,\ \tfrac{\pi}{2} + n\pi\right)$ between its vertical asymptotes — but it does not carry that increase across an asymptote, so it is not monotonic on the whole line. …

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Define $f(x)=2\sin x+\tan x-3x$ . AM-GM shows $f'(x)\ge0$ on $(0,\pi/2)$ , so $f$ is increasing from $f(0)=0$ , proving the inequality.

Let $f(x) = 2\sin x+\tan x-3x$ . Note $f(0)=0$ .

$f'(x) = 2\cos x+\sec^2x-3$

For $x\in\left(0,\dfrac\pi2\right)$ , $\cos x>0$ and $\sec^2x>0$ . By AM-GM on the three positive terms $\cos x,\cos x,\sec^2x$ :

$\dfrac{\cos x+\cos x+\sec^2x}{3} \ge \left(\cos x\cdot\cos x\cdot\sec^2x\right)^{1/3} = \left(\cos^2x\cdot\sec^2x\right)^{1/3} = 1$

So $2\cos x+\sec^2x \ge 3$ , i.e. $f'(x)\ge0$ , with equality only when $\cos x=\sec^2x$ , i.e. $\cos^3x=1$ , i.e. $x=0$ (excluded from the open interval). So $f'(x)>0$ strictly for $x\in\left(0,\dfrac\pi2\right)$ .

…

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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 ANNUAL1 markMCQ
$Q. The function 'f' given by f(x) = log(cos x) is strictly decreasing on$
(a) (0, π/2)
(b) (π/2, π)
(c) (3π/2, 2π)
(d) None of the above
›Reveal solutionSolution

$f$ is decreasing exactly where $f'(x)<0$ AND the function is actually defined there (needs $\cos x>0$ ).

$f(x) = \log(\cos x)$ , so $f'(x) = \dfrac{-\sin x}{\cos x} = -\tan x$ .

For $f$ to be strictly decreasing, we need $f'(x) < 0 \Rightarrow \tan x > 0$ .

But we also need $\cos x > 0$ for $\log(\cos x)$ to even be defined (real-valued).

…

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CBSE 2025Set ANNUAL1 markMCQ
$\left(0,\dfrac{\pi}{2}\right)$
(a) $\sin x$
(b) $\cos x$
(c) $\tan x$
(d) $\sin 2x$
›Reveal solutionSolution

A function is strictly decreasing on an interval where its derivative is negative there.

On $\left(0,\dfrac{\pi}{2}\right)$ : $\dfrac{d}{dx}(\sin x)=\cos x>0$ (increasing); $\dfrac{d}{dx}(\cos x)=-\sin x<0$ (decreasing); $\dfrac{d}{dx}(\tan x)=\sec^2 x>0$ (increasing); $\dfrac{d}{dx}(\sin 2x)=2\cos 2x$ , which changes sign in this interval (not strictly monotonic throughout).

…

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CBSE 2025Set ANNUAL1 markMCQ
$f(x)=\sin x$
(a) $\left(0,\dfrac{\pi}{2}\right)$
(b) $\left(\dfrac{\pi}{2},\pi\right)$
(c) $(0,\pi)$
(d) $(0,2\pi)$
›Reveal solutionSolution

$\sin x$ increases exactly where $\cos x>0$ ; among the options that is $(0,\pi/2)$ .

Let $f(x)=\sin x$ . Then $f'(x)=\cos x$ .

$f$ is increasing on an interval where $f'(x)=\cos x>0$ .

On $(0,\pi/2)$ : $\cos x>0$ throughout, so $f$ is increasing here.

On $(\pi/2,\pi)$ : $\cos x<0$ , so $f$ is decreasing.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The function f(x) = sin x is increasing in -$
(i) (π/2, π)
(ii) (π, 3π/2)
(iii) (0, π)
(iv) (−π/2, π/2)
›Reveal solutionSolution
$\sin x$ increases exactly where $\cos x>0$ ; check each given interval against that sign.

$f(x)=\sin x \Rightarrow f'(x)=\cos x$ . $f$ is increasing on an interval where $\cos x>0$ .

$(\pi/2,\pi)$ : $\cos x<0$ here (second quadrant) — decreasing, not this one.

$(\pi,3\pi/2)$ : $\cos x<0$ here (third quadrant) — decreasing, not this one. …
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CBSE 2024Set ANNUAL1 markMCQ
$f(x)=\sin x$
(a) f(x) is a constant function
(b) f(x) is an increasing function
(c) f(x) is a decreasing function
(d) f(x) is neither increasing nor decreasing function
›Reveal solutionSolution

From the graph, $\sin x$ rises on $(0,\pi/2)$ and falls on $(\pi/2,\pi)$ , so it is not monotonic on the whole interval $(0,\pi)$ .

From the given graph of $f(x) = \sin x$ : on $(0, \pi/2)$ the curve rises from $0$ to its maximum value $1$ at $x=\pi/2$ , so $f'(x) = \cos x > 0$ there and $f$ is increasing on $(0,\pi/2)$ .

On $(\pi/2, \pi)$ the curve falls from $1$ back down to $0$ at $x=\pi$ , so $f'(x) = \cos x < 0$ there and $f$ is decreasing on $(\pi/2,\pi)$ .

…

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CBSE 2023Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Since $f'(x)=\cos x - 1 \le 0$ for every real $x$ , the function never increases; it is monotonically decreasing throughout $\mathbb{R}$ .

Differentiate: $f'(x) = \cos x - 1$ .

Since $\cos x \le 1$ for all real $x$ , we get $f'(x) = \cos x - 1 \le 0$ for every $x \in \mathbb{R}$ , with equality only at the isolated points $x = 2n\pi$ , $n \in \mathbb{Z}$ .

…

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CBSE 2020Set ANNUAL1 markMCQ
$f(x) = \log(\sin x)$
(a) $\left(0, \frac{\pi}{2}\right)$
(b) $\left(\frac{\pi}{2}, \pi\right)$
(c) $(0, \pi)$
(d) None of these
›Reveal solutionSolution

$f'(x)=\cot x$ , which is negative on $\left(\frac{\pi}{2},\pi\right)$ , making $f$ decreasing there.

$f'(x) = \dfrac{\cos x}{\sin x} = \cot x$

On $\left(0,\dfrac{\pi}{2}\right)$ , $\cot x>0$ so $f$ is increasing. …

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Q.Show that 2sin⁡x+tan⁡x≥3x2\sin x + \tan x \ge 3x2sinx+tanx≥3x for all x∈(0,π2)x \in \left(0, \dfrac{\pi}{2}\right)x∈(0,2π​).

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Q.Show that $2\sin x + \tan x \ge 3x$ for all $x \in \left(0, \dfrac{\pi}{2}\right)$ .

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