The maximum value of $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1 \end{vmatrix}$ is ($\theta$ is real number) (A) $\dfrac{1}{2}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\sqrt{2}$ (D) $\dfrac{2\sqrt{3}}{4}$

The maximum value of Δ = 1 1 1 \1 1 + sinθ 1 \1 + cosθ 1 1 is (θ is real number) (A) (1)/(2) (B) dfrac√(3)2 (C

NCERT Exemplar · Q31

\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1 \end{vmatrix}

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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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$\Delta = -\sin\theta\cos\theta = -\tfrac12\sin 2\theta$ , whose maximum value is $\tfrac12$ . Correct option: (A).

Apply the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ (these do not change the value of the determinant):

\Delta = \begin{vmatrix} 1 & 1 & 1 \\0 & \sin\theta & 0 \\\cos\theta & 0 & 0 \end{vmatrix}.

Expand along the third row, which has two zeros:

$\Delta = \cos\theta\,(+1)\begin{vmatrix} 1 & 1 \\\sin\theta & 0 \end{vmatrix} = \cos\theta\,(0 - \sin\theta) = -\sin\theta\cos\theta.$ …

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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-2026-04174 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 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 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 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 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 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 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 2023Set eng-2023-P2-B24 marksMCQ
$|r| > \sqrt{p^2 + q^2}$
›Reveal solutionSolution
Since p cos(theta) + q sin(theta) can be at most sqrt(p^2+q^2) in magnitude, and |r| exceeds that, the equation has no real solution.

Concept and Intuition

p cos(theta) + q sin(theta) has maximum magnitude sqrt(p^2+q^2). If r lies strictly outside that bound, the line is unreachable.

Step-by-Step Solution

Max of |p cos(theta) + q sin(theta)| = sqrt(p^2 + q^2).

Given |r| > sqrt(p^2 + q^2), so |r| exceeds the maximum possible LHS value. …
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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
$f(x) = \left(\frac{1}{3}\right)^{3 + \sin x}$
›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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Q.The maximum value of $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 + \cos\theta & 1 & 1 \end{vmatrix}$ is ( $\theta$ is real number)
(A) $\dfrac{1}{2}$
(B) $\dfrac{\sqrt{3}}{2}$
(C) $\sqrt{2}$
(D) $\dfrac{2\sqrt{3}}{4}$

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 …

Method: Simplifying a Determinant to Bound Its Maximum or Minimum Value

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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