Find the value of the following: At what points in the interval $[0, 2\pi]$, does the function $\sin 2x$ attain its maximum value?

> [!TLDR] > The function $\sin 2x$ attains its maximum value of $1$ at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ within $[0, 2\pi]$, found by analyzing critical points and checking endpoints. We need to find where $\sin 2x$ reaches its highest value in the closed interval $[0, 2\pi]$. The sine function itself peaks at $1$, so we’re really asking: for which $x$ in $[0, 2\pi]$ does $\sin 2x = 1$? But we must also consider that the maximum could occur at the boundaries of the interval, so a full critical-point analysis is the reliable method. The key idea: for a continuous function on a closed interval, the maximum occurs either at critical points (where derivative is zero or undefined) or at the endpoints. Since $\sin 2x$ is differentiable everywhere, we only need to find where its derivative vanishes and then compare function values. 1. **Find the derivative and critical points.** Let $f(x) = \sin 2x$. Then $f'(x) = 2\cos 2x$. Set $f'(x) = 0$: $$2\cos 2x = 0 \quad \Rightarrow \quad \cos 2x = 0.$$ In the interval $[0, 2\pi]$, $2x$ ranges from $0$ to $4\pi$. The cosine function is zero at odd multiples of $\frac{\pi}{2}$: $$2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}.$$ Solving for $x$: $$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.$$ These four points are the critical points inside $(0, 2\pi)$. 2. **Evaluate $f(x)$ at critical points and endpoints.** Endpoints: $x = 0$ and $x = 2\pi$. $$f(0) = \sin 0 = 0, \quad f(2\pi) = \sin 4\pi = 0.$$ Critical points: $$f\left(\frac{\pi}{4}\right) = \sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\frac{\pi}{2} = 1,$$ $$f\left(\frac{3\pi}{4}\right) = \sin\left(2 \cdot \frac{3\pi}{4}\right) = \sin\frac{3\pi}{2} = -1,$$ $$f\left(\frac{5\pi}{4}\right) = \sin\left(2 \cdot \frac{5\pi}{4}\right) = \sin\frac{5\pi}{2} = \sin\left(2\pi + \frac{\pi}{2}\right) = 1,$$ $$f\left(\frac{7\pi}{4}\right) = \sin\left(2 \cdot \fra

Find the value of the following: At what points in the interval [0, 2π], does the function sin 2x attain its m

Exercise 6.3 · Q8

[0, 2\pi]

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Concept understanding — Critical Points Analysis

Critical Points Analysis: Where Functions Change Direction

Hiking a mountain range, you reach peaks (highest spot around), valleys (bottoms), and flat stretches where the ground doesn't slope. These special locations — peaks, valleys, and flat spots — are critical points.

The Intuition

A function's graph is like that trail. At most points it is rising (positive slope) or falling (negative slope). At a critical point something changes: the slope becomes zero, or the slope doesn't exist (a sharp corner).

Throw a ball straight up: at the very top of its arc it stops for an instant before falling. Its velocity — the rate of change of height — is zero at that moment. That's a critical point.

The Precise Definition

x = c

Why Two Conditions?

Derivative equals zero catches the "flat" spots — peaks, valleys, horizontal plateaus — where the tangent line is horizontal.

Derivative does not exist catches sharp corners (like the tip of $|x|$ at $x=0$ ), vertical tangents, and cusps. Even without a zero slope, these can be peaks or valleys.

Watch out

A common mistake: thinking every critical point is a maximum or minimum. Not true. A critical point could be a "saddle point" — flat but neither. For example, $f(x) = x^3$ at $x=0$ has $f'(0)=0$ , yet the function just passes through with no extremum.

How to Find Critical Points

Find the derivative $f'(x)$ .
Solve $f'(x) = 0$ — these are candidates.
Check where $f'(x)$ does not exist — but only if $f(x)$ exists there (the point must be in the domain).
Collect all such $x$ -values.

Example 1: A Simple Polynomial

Let $f(x) = x^3 - 3x^2 + 1$ .

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ .

$f'(x) = 0 \implies x = 0$ or $x = 2$ . Since $f'$ exists everywhere, the critical points are $x = 0$ and $x = 2$ .

Example 2: A Function with a Corner

Let $f(x) = |x|$ . Here $f'(x)$ does not exist at $x = 0$ (left derivative $-1$ , right derivative $+1$ ), and $f'(x) = 0$ has no solutions. So the only critical point is $x = 0$ .

Note

$x=0$ is actually a minimum of $|x|$ — the sharp corner is a valley.

What Critical Points Tell Us …

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The function $\sin 2x$ attains its maximum value of $1$ at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ within $[0, 2\pi]$ , found by analyzing critical points and checking endpoints.

We need to find where $\sin 2x$ reaches its highest value in the closed interval $[0, 2\pi]$ . The sine function itself peaks at $1$ , so we’re really asking: for which $x$ in $[0, 2\pi]$ does $\sin 2x = 1$ ? But we must also consider that the maximum could occur at the boundaries of the interval, so a full critical-point analysis is the reliable method.

The key idea: for a continuous function on a closed interval, the maximum occurs either at critical points (where derivative is zero or undefined) or at the endpoints. Since $\sin 2x$ is differentiable everywhere, we only need to find where its derivative vanishes and then compare function values.

Find the derivative and critical points. Let $f(x) = \sin 2x$ . Then $f'(x) = 2\cos 2x$ . Set $f'(x) = 0$ :

2\cos 2x = 0 \quad \Rightarrow \quad \cos 2x = 0.

In the interval $[0, 2\pi]$ , $2x$ ranges from $0$ to $4\pi$ . The cosine function is zero at odd multiples of $\frac{\pi}{2}$ :

2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}.

Solving for $x$ :

x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.

These four points are the critical points inside $(0, 2\pi)$ .

Evaluate $f(x)$ at critical points and endpoints. Endpoints: $x = 0$ and $x = 2\pi$ .

f(0) = \sin 0 = 0, \quad f(2\pi) = \sin 4\pi = 0.

Critical points:

f\left(\frac{\pi}{4}\right) = \sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\frac{\pi}{2} = 1,

f\left(\frac{3\pi}{4}\right) = \sin\left(2 \cdot \frac{3\pi}{4}\right) = \sin\frac{3\pi}{2} = -1,

f\left(\frac{5\pi}{4}\right) = \sin\left(2 \cdot \frac{5\pi}{4}\right) = \sin\frac{5\pi}{2} = \sin\left(2\pi + \frac{\pi}{2}\right) = 1,

$f\left(\frac{7\pi}{4}\right) = \sin\left(2 \cdot \frac{7\pi}{4}\right) = \sin\frac{7\pi}{2} = \sin\left(3\pi + \frac{\pi}{2}\right) = -1.$ …

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

AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$f(x) = 2\cos x - \sin 2x$
›Reveal solutionSolution
$f'(x)=0$ factorises to give sign changes only at the two points where $\sin x=-\tfrac12$ , so there are 2 turning points.

Concept and Intuition

A turning point (local max/min) needs $f'(x)=0$ with a genuine sign change of $f'$ . A stationary point where $f'$ merely touches zero without changing sign is not a turning point.

Step-by-Step Solution

$f'(x) = -2\sin x - 2\cos 2x = -2\sin x - 2(1-2\sin^2 x) = 4\sin^2 x - 2\sin x - 2$ .

Set $f'=0$ : $2\sin^2 x - \sin x - 1 = 0 \Rightarrow (2\sin x + 1)(\sin x - 1)=0$ .

Roots: $\sin x = 1$ (i.e. $x=\tfrac{\pi}{2}$ ) and $\sin x = -\tfrac12$ (i.e. $x=-\tfrac{\pi}{6},\,-\tfrac{5\pi}{6}$ in $[-\pi,\pi]$ ).

At $x=\tfrac{\pi}{2}$ the factor $(\sin x - 1)$ only touches zero; since $\sin x - 1 \le 0$ on both sides, $f'$ does not change sign — not a turning point.

At $x=-\tfrac{\pi}{6}$ and $x=-\tfrac{5\pi}{6}$ the factor $(2\sin x + 1)$ changes sign, so $f'$ changes sign — two genuine turning points.

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$2\alpha + \beta = 8$
›Reveal solutionSolution
This tests locating the maxima/minima of a cubic via the second-derivative test and then solving a linear condition on the parameter $a$ . Answer: $a=2$ .

Concept and Intuition

For a cubic $f(x)=2x^3-9ax^2+12a^2x+1$ , the critical points come from $f'(x)=0$ , and whether each is a max or min is determined by the sign of $f''$ there.

Step-by-Step Solution

$f'(x) = 6x^2-18ax+12a^2 = 6(x^2-3ax+2a^2) = 6(x-a)(x-2a)$ .

Critical points: $x=a$ and $x=2a$ .

$f''(x) = 12x-18a$ . At $x=a$ : $f''(a)=12a-18a=-6a$ , which is negative for $a>0$ — so $x=a$ is a local maximum, i.e. $\alpha=a$ .

At $x=2a$ : $f''(2a)=24a-18a=6a$ , positive for $a>0$ — so $x=2a$ is a local minimum, i.e. $\beta=2a$ . …
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AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
$f(x)=x^3-6x^2-12x-3$
›Reveal solutionSolution
$f''(2)=0$ with a change of concavity and $f'''(2)\neq 0$ , so $x=2$ is a point of inflection.

Concept and Intuition

A point of inflection is where the curve changes concavity, i.e. $f''$ changes sign (with $f'''\neq 0$ ). It need not be a stationary point — $f'$ can be non-zero there. A maximum/minimum instead requires $f'=0$ with $f''$ non-zero of the appropriate sign.

Step-by-Step Solution

$f'(x)=3x^2-12x-12$ , so $f'(2)=12-24-12=-24\neq 0$ — not a critical point in the stationary sense.

$f''(x)=6x-12$ , so $f''(2)=12-12=0$ .

For $x<2$ , $f''<0$ (concave down); for $x>2$ , $f''>0$ (concave up) — concavity changes.

$f'''(x)=6\neq 0$ , confirming a genuine change of concavity.

Therefore $x=2$ is a point of inflection. …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$f(x)=\sqrt{x+\sin x}$
›Reveal solutionSolution
The points where $f'(x)=0$ occur exactly where $\cos x=-1$ ; at those points $\sin x=0$ , so $f(x)=\sqrt{x}$ , which traces a parabola $y^2=x$ .

Concept and Intuition

We need to find the locus of points $(x, f(x))$ where the tangent is horizontal ( $f'(x)=0$ ), then recognize what curve those points lie on. The key simplification is that at these special $x$ -values, $\sin x$ vanishes, collapsing $f(x)$ to a simple power of $x$ .

Step-by-Step Solution

Differentiate $f(x)=(x+\sin x)^{1/2}$ using the chain rule:

$f'(x) = \frac{1}{2}(x+\sin x)^{-1/2}\cdot(1+\cos x) = \frac{1+\cos x}{2\sqrt{x+\sin x}}$

Set $f'(x)=0$ . Since the denominator is never zero (whenever defined and nonzero), we need the numerator to vanish:

$1+\cos x = 0 \implies \cos x = -1 \implies x = (2k+1)\pi,\ k\in\mathbb{Z}$

At these values, $\sin x = \sin((2k+1)\pi) = 0$ . So: $f(x) = \sqrt{x + 0} = \sqrt{x}$ …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$f(x) = x^5 - 5x^4 + 5x^3 - 10$
›Reveal solutionSolution
Factor $f'(x)$ fully, identify each critical point's nature via the second derivative (or a sign chart), then discard the non-extremum double root.

Concept and Intuition

A repeated root of $f'(x)$ (even multiplicity) does not change the sign of $f'$ across it, so it is NOT a local extremum — only an inflection point with a horizontal tangent. Only the simple roots correspond to genuine local max/min.

Step-by-Step Solution

$f(x)=x^5-5x^4+5x^3-10\Rightarrow f'(x)=5x^4-20x^3+15x^2=5x^2(x^2-4x+3)=5x^2(x-1)(x-3)$ .

Critical points: $x=0$ (double root), $x=1$ , $x=3$ .

$f''(x)=20x^3-60x^2+30x$ . At $x=1$ : $f''(1)=20-60+30=-10<0\Rightarrow$ local maximum at $x=1$ , so $a=1$ .

At $x=3$ : $f''(3)=540-540+90=90>0\Rightarrow$ local minimum at $x=3$ , so $b=3$ . …
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AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$y = 8x^2 - x^4 - 4$
›Reveal solutionSolution
Stationary points occur where $y'=0$ . Solving the cubic-in- $x$ derivative gives three critical $x$ -values, and substituting back into $y$ gives the three stationary points.

Concept and Intuition

A stationary point of a curve is where the tangent is horizontal, i.e., $\dfrac{dy}{dx}=0$ . For a quartic like $y=8x^2-x^4-4$ , the derivative is cubic in $x$ , so we expect up to three real roots — three turning points (a local minimum and two local maxima, by symmetry since the curve is even in $x$ ).

Step-by-Step Solution

$y = 8x^2 - x^4 - 4$ .

$y' = 16x - 4x^3 = 4x(4 - x^2) = 4x(2-x)(2+x)$ .

Setting $y'=0$ : $x = 0, 2, -2$ .

At $x=0$ : $y = 0 - 0 - 4 = -4$ , giving $(0,-4)$ .

At $x=2$ : $y = 8(4) - 16 - 4 = 32-16-4=12$ , giving $(2,12)$ . …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$y=2x^3-15x^2+36x+c$
›Reveal solutionSolution
The curve's stationary points occur where $y'=0$ , at $x=2$ and $x=3$ ; matching the given $y$ -value at $x=3$ pins down $c$ , then $a$ follows from $y(2)$ , giving $a+b+c=15$ .

Concept and Intuition

Stationary (critical) points of a curve are where its derivative vanishes. Since the cubic here has exactly two stationary points, and we're told they are $(2,a)$ and $(b,19)$ , matching the $x$ -coordinates to the roots of $y'=0$ identifies $b$ , and substituting back into $y(x)$ pins down the unknown constant $c$ (and hence $a$ ).

Step-by-Step Solution

$y=2x^3-15x^2+36x+c \Rightarrow y'=6x^2-30x+36=6(x^2-5x+6)=6(x-2)(x-3)$ .

Stationary points at $x=2$ and $x=3$ . Given points are $(2,a)$ and $(b,19)$ , so $b=3$ . …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$f(x) = ax^3 + bx^2 + cx + d$
›Reveal solutionSolution
A cubic has no extreme (turning) points exactly when its derivative — a quadratic — has no real roots, which translates to $b^2<3ac$ .

Concept and Intuition

Extreme values of $f$ occur where $f'(x)=0$ and the derivative changes sign there. For a cubic, $f'$ is a quadratic; if that quadratic never touches zero (negative discriminant), $f'$ keeps a constant sign everywhere, so $f$ is strictly monotonic and has no local max/min at all.

Step-by-Step Solution

$f(x)=ax^3+bx^2+cx+d \Rightarrow f'(x) = 3ax^2+2bx+c$ .

For no real critical points, the quadratic $f'(x)=0$ must have no real roots: discriminant $<0$ .

Discriminant of $3ax^2+2bx+c$ : $(2b)^2-4(3a)(c) = 4b^2-12ac$ . …
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Q.Find the value of the following: At what points in the interval $[0, 2\pi]$ , does the function $\sin 2x$ attain its maximum value?

Concept understanding — Critical Points Analysis

Critical Points Analysis: Where Functions Change Direction

The Intuition

The Precise Definition

Why Two Conditions?

How to Find Critical Points

Example 1: A Simple Polynomial

Example 2: A Function with a Corner

What Critical Points Tell Us …

Method: Locating Where a Periodic Function Attains Its Maximum on a Closed Interval

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

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

Concept and Intuition

Step-by-Step Solution

Q.Find the value of the following: At what points in the interval [0,2π][0, 2\pi][0,2π], does the function sin⁡2x\sin 2xsin2x attain its maximum value?

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Q.Find the value of the following: At what points in the interval $[0, 2\pi]$ , does the function $\sin 2x$ attain its maximum value?

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