At x = (5π)/(6), f(x) = 2sin 3x + 3cos 3x is: (A) maximum (B) minimum (C) zero (D) neither maximum nor minimum

NCERT Exemplar · Q39

x = \dfrac{5\pi}{6}

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Concept understanding — Maximum Value Sine Cosine

Maximum Value of Sine and Cosine – The Core Idea

Imagine a point moving around a unit circle centred at the origin. Its coordinates are $(\cos\theta, \sin\theta)$ , where $\theta$ is measured from the positive x-axis.

The farthest right the point reaches is $(1,0)$ — $\cos\theta = 1$ ; the farthest left is $(-1,0)$ — $\cos\theta = -1$ . The highest is $(0,1)$ — $\sin\theta = 1$ ; the lowest is $(0,-1)$ — $\sin\theta = -1$ . So sine and cosine never exceed $1$ or fall below $-1$ : they are bounded by the unit circle.

Important

For any real angle $\theta$ ,

$-1 \le \sin \theta \le 1 \quad \text{and} \quad -1 \le \cos \theta \le 1$

The Precise Statement

Maximum value: $1$ ; minimum value: $-1$ . Both are achieved at specific angles.

For sine:

$\sin \theta = 1$ when $\theta = 90^\circ + 360^\circ n$ (i.e. $\frac{\pi}{2} + 2\pi n$ )
$\sin \theta = -1$ when $\theta = 270^\circ + 360^\circ n$ (i.e. $\frac{3\pi}{2} + 2\pi n$ )

For cosine:

$\cos \theta = 1$ when $\theta = 0^\circ + 360^\circ n$ (i.e. $2\pi n$ )
$\cos \theta = -1$ when $\theta = 180^\circ + 360^\circ n$ (i.e. $\pi + 2\pi n$ )

Here $n$ is any integer — the pattern repeats every full rotation.

Why This Matters in Exams

Many problems ask for the maximum or minimum of expressions like $3\sin x + 4\cos x$ or $2 - 5\sin x$ . Since sine and cosine are individually trapped between $-1$ and $1$ , you can bound any linear combination.

Tip

For $a\sin\theta + b\cos\theta$ , the maximum is $\sqrt{a^2 + b^2}$ and the minimum is $-\sqrt{a^2 + b^2}$ . Derive it by rewriting as $R\sin(\theta + \phi)$ .

Common Mistake to Avoid …

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At $x=\frac{5\pi}{6}$ , $f'(x)=-9\neq 0$ , so the point is not a critical point and the function is neither a maximum nor a minimum — option (D).

The idea

A smooth function can only have a local maximum or minimum where its derivative is zero. So the first thing to test is whether $f'$ vanishes at the given point. If $f'\neq 0$ there, the graph is still climbing or falling through the point and it cannot be a turning point.

Set up

f(x)=2\sin 3x+3\cos 3x.

Work the steps

Differentiate (chain rule, since the angle is $3x$ ):

f'(x)=2\cdot 3\cos 3x+3\cdot(-\sin 3x)\cdot 3=6\cos 3x-9\sin 3x.

Plug in $x=\frac{5\pi}{6}$ , so $3x=\frac{5\pi}{2}$ . Since $\frac{5\pi}{2}=2\pi+\frac{\pi}{2}$ ,

\cos\tfrac{5\pi}{2}=\cos\tfrac{\pi}{2}=0,\qquad \sin\tfrac{5\pi}{2}=\sin\tfrac{\pi}{2}=1.

Therefore

$f'\left(\tfrac{5\pi}{6}\right)=6(0)-9(1)=-9.$ …

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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 14 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$f(x) = \cos(5x)\cos(3x) - \sin(5x)\sin(3x),\ 0 \le x \le \dfrac{\pi}{4}$
›Reveal solutionSolution

The expression collapses to $\cos 8x$ ; on $[0,\tfrac{\pi}{4}]$ it hits its minimum $-1$ at $x=\tfrac{\pi}{8}$ .

Using $\cos A\cos B-\sin A\sin B=\cos(A+B)$ with $A=5x$ , $B=3x$ :

$f(x)=\cos(5x+3x)=\cos 8x.$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$f(x) = \sin x \sin\left(x + \dfrac{\pi}{3}\right)$
›Reveal solutionSolution

Use product-to-sum; the minimum is set by the maximum of the cosine term.

$f(x)=\sin x\sin\!\left(x+\frac{\pi}{3}\right)=\frac12\Big[\cos\frac{\pi}{3}-\cos\!\Big(2x+\frac{\pi}{3}\Big)\Big]=\frac12\Big[\frac12-\cos\!\Big(2x+\frac{\pi}{3}\Big)\Big].$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$f(x)=2-7\sin\left(\dfrac{2x}{7}\right)$
›Reveal solutionSolution

$f(x)=2-7\sin(2x/7)$ is largest when $\sin=-1$ . …

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KEAM 2026Set eng-2026-04174 marksMCQ
$f(x)=\dfrac{10}{7+4\sin x+3\cos x}, x \in \mathbb{R}$
›Reveal solutionSolution

The denominator ranges over $[2,12]$ , so $f$ ranges over $\left[\dfrac{5}{6},5\right]$ .

The key fact is that for $a\sin x+b\cos x$ , the amplitude is $\sqrt{a^2+b^2}$ . Here $a=4,\;b=3$ , so
$4\sin x+3\cos x\in[-\sqrt{16+9},\sqrt{16+9}]=[-5,5].$
Hence the denominator

$D=7+4\sin x+3\cos x\in[7-5,\,7+5]=[2,12],$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$f(x) = \sin x + \cos x + 3$
›Reveal solutionSolution

$\sin x+\cos x=\sqrt2\sin\!\left(x+\frac{\pi}{4}\right)$ ranges over $[-\sqrt2,\sqrt2]$ , so $f=\sin x+\cos x+3$ ranges over $[3-\sqrt2,\,3+\sqrt2]$ .

Write

$\sin x+\cos x=\sqrt2\left(\tfrac{1}{\sqrt2}\sin x+\tfrac{1}{\sqrt2}\cos x\right)=\sqrt2\sin\!\left(x+\tfrac{\pi}{4}\right).$ …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$x$
›Reveal solutionSolution
The maximum value of $5\sin x\sin y$ is $5$ .

Concept and Intuition

The constraint $x+y=\pi$ makes $\sin y=\sin(\pi-x)=\sin x$ , reducing the two-variable expression to a single-variable one.

Step-by-Step Solution

$\sin y=\sin(\pi-x)=\sin x$ .

$5\sin x\sin y=5\sin^2 x$ .

$\sin^2 x\le 1$ , maximum $1$ at $x=\frac{\pi}{2}$ (then $y=\frac{\pi}{2}$ ), giving maximum $5$ . …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$4\cos x - 3\sin x + 5$
›Reveal solutionSolution
The expression 4cos x - 3sin x oscillates in [-5, 5], so 4cos x - 3sin x + 5 has least value 0.

Concept and Intuition

Any expression a cos x + b sin x has range [-sqrt(a^2+b^2), sqrt(a^2+b^2)]. Adding a constant just shifts that range.

Step-by-Step Solution

Amplitude = sqrt(4^2 + (-3)^2) = sqrt(16+9) = 5.

So 4cos x - 3sin x ranges over [-5, 5]. …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$f(x) = 2\sin(3x) + 1$
›Reveal solutionSolution
The range of $f(x)=2\sin(3x)+1$ is $[-1,3]$ .

Concept and Intuition

The frequency (the 3 inside) does not affect the range of a sine function, only its period. The amplitude 2 and vertical shift 1 set the range.

Step-by-Step Solution

$\sin(3x)$ ranges over $[-1,1]$ .

Multiply by 2: $2\sin(3x)\in[-2,2]$ .

Add 1: $f(x)\in[-2+1,\,2+1]=[-1,3]$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$f(x) = \dfrac{1}{7 - \cos x}, \; x \in \mathbb{R}$
›Reveal solutionSolution

The denominator ranges over $[6,8]$ , so $f$ ranges over $[1/8,\,1/6]$ .

Since $\cos x\in[-1,1]$ for all real $x$ , the denominator satisfies
$7-\cos x\in[7-1,\ 7+1]=[6,8].$
As $f(x)=\dfrac{1}{7-\cos x}$ is a decreasing function of the denominator, its extreme values are …
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KEAM 2026Set eng-2026-04224 marksMCQ
$f(x) = (8\sin x + 15\cos x + 3)^{2} - 15$
›Reveal solutionSolution

Maximize the inner linear-trig expression, then evaluate the squared form.

$8\sin x+15\cos x$ has maximum $\sqrt{8^2+15^2}=\sqrt{289}=17$ .

So $8\sin x+15\cos x+3$ has maximum $17+3=20$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$f(x) = \left(\frac{1}{3}\right)^{3 + \sin x}$
›Reveal solutionSolution

Track the exponent's range through the decreasing base $\frac13$ .

Since $\sin x\in[-1,1]$ , the exponent $3+\sin x\in[2,4]$ .

$f(x)=(1/3)^{3+\sin x}$ is a decreasing function of the exponent, so the maximum occurs at exponent $2$ and minimum at exponent $4$ : …
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KEAM 2024Set eng-2024-06074 marksMCQ
$f(x)=7\cos(10x+4\pi)$
›Reveal solutionSolution

Amplitude times the range of cosine gives the range.

For any argument, $\cos(10x+4\pi)\in[-1,1]$ . Multiplying by the amplitude $7$ :

$f(x)=7\cos(10x+4\pi)\in[-7,7].$ …
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Q.At $x = \dfrac{5\pi}{6}$ , $f(x) = 2\sin 3x + 3\cos 3x$ is:
(A) maximum
(B) minimum
(C) zero
(D) neither maximum nor minimum

Concept understanding — Maximum Value Sine Cosine

Maximum Value of Sine and Cosine – The Core Idea

The Precise Statement

Why This Matters in Exams

Common Mistake to Avoid …

The idea

Set up

Work the steps

Method: Testing Whether a Given Point Is a Local Extremum

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.At x=5π6x = \dfrac{5\pi}{6}x=65π​, f(x)=2sin⁡3x+3cos⁡3xf(x) = 2\sin 3x + 3\cos 3xf(x)=2sin3x+3cos3x is: (A) maximum (B) minimum (C) zero (D) neither maximum nor minimum

The idea

Set up

Work the steps

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Q.At $x = \dfrac{5\pi}{6}$ , $f(x) = 2\sin 3x + 3\cos 3x$ is:
(A) maximum
(B) minimum
(C) zero
(D) neither maximum nor minimum

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