(a) Find the absolute maximum value of f(x) = cos x + sin² x,x ∈ [0, π]. OR (b) If the volume of a solid hemis

Question

f(x) = \cos x + \sin^2 x,\ x \in [0, \pi]

(OR)

(b) If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

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Part (a)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 …

Part (b)Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

Many real situations involve two or more quantities that all vary with time, linked by a fixed relationship. Inflate a balloon and its radius and volume both grow; slide a ladder down a wall and the top's height and the foot's distance both change. A related-rates problem gives you the rate at which one quantity is changing and asks for the rate of another, at some instant.

The key insight: if the quantities are tied together by an equation, then their rates are tied together too. We uncover that link by differentiating the equation with respect to time $t$ .

The core mechanism: differentiate with respect to time

Every variable is a function of $t$ , so differentiating brings in the chain rule — each variable's derivative picks up a factor of its own rate. For example, if the volume of a sphere is $V=\tfrac{4}{3}\pi r^{3}$ , then differentiating both sides with respect to $t$ gives

\frac{dV}{dt} = 4\pi r^{2}\,\frac{dr}{dt}.

This single equation connects the rate the volume grows, $\dfrac{dV}{dt}$ , to the rate the radius grows, $\dfrac{dr}{dt}$ . Knowing one (and the current $r$ ) gives the other.

The standard procedure

t

Worked example

Air is pumped into a spherical balloon at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$ . How fast is the radius increasing when $r=5\ \text{cm}$ ?

From $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$ , substitute $\dfrac{dV}{dt}=100$ and $r=5$ :

$100 = 4\pi(5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}.$ …

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(a) The absolute maximum of $f(x)=\cos x+\sin^2x$ on $[0,\pi]$ is $\tfrac54$ (at $x=\tfrac\pi3$ ). (b) If $V$ of a solid hemisphere grows at a constant rate, then $\tfrac{dS}{dt}=\tfrac{3k}{r}\propto\tfrac1r$ .

Part (a)

Convert to a single trigonometric quantity using $\sin^2x=1-\cos^2x$ :

f(x)=\cos x+1-\cos^2x=-\cos^2x+\cos x+1.

Let $t=\cos x$ . On $[0,\pi]$ , $\cos x$ decreases from $1$ to $-1$ , so $t\in[-1,1]$ , and

g(t)=-t^2+t+1.

This is a downward parabola with vertex at $t=-\dfrac{b}{2a}=-\dfrac{1}{2(-1)}=\dfrac12\in[-1,1]$ :

$g\!\left(\tfrac12\right)=-\tfrac14+\tfrac12+1=\tfrac54.$ …

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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 25 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. The edge of a variable cube is increasing at the rate of 3 cm/s. The volume of the cube is increasing at the rate of __________ while the edge is 10 cm long.$
›Reveal solutionSolution

Use $V=e^3$ and the chain rule $dV/dt = 3e^2\,de/dt$ .

Let $e$ be the edge; $V=e^3$ , so $\dfrac{dV}{dt}=3e^2\dfrac{de}{dt}$ .

Given $\dfrac{de}{dt}=3$ cm/s and $e=10$ cm:

…

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CBSE 2026Set ANNUAL1 mark
$1/2$
›Reveal solutionSolution

Use $V=\dfrac43\pi r^3$ and $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt}=\dfrac12$ cm/s, at $r=1$ cm:

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Radius of a circle is increasing at the rate of 1/π m/s. Rate of change of its circumference is:$
(a) 4π m/s
(b) 2 m/s
(c) 2π m/s
(d) 4 m/s
›Reveal solutionSolution

Since $C = 2\pi r$ , differentiating both sides w.r.t. time gives $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$ directly.

The circumference of a circle of radius $r$ is $C = 2\pi r$ .

Differentiating with respect to time $t$ :

$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

Given $\frac{dr}{dt} = \frac{1}{\pi}$ m/s, substitute: …

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CBSE 2025Set 65/2/11 markMCQ
$100\pi$
›Reveal solutionSolution
The volume of a cylinder is $V = \pi r^2 h$ . Since the radius is constant, the rate of change of volume with respect to time is $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ . Given $\frac{dV}{dt} = 100\pi$ cm³/s and $r = 10$ cm, solving gives $\frac{dh}{dt} = 1$ cm/s. The correct option is (C).

This is a classic Related Rates problem. The core idea is that when two quantities are linked by a geometric formula (here, volume and height of a cylinder), their rates of change with respect to time are also linked. You differentiate the relationship with respect to time, plug in what you know, and solve for the unknown rate.

The key insight: the tank’s radius is fixed at 10 cm. So as sugar pours in, the height increases, but the cross-sectional area stays the same. That means the volume increases at a constant rate per unit height — specifically, each 1 cm rise in height adds $\pi (10)^2 = 100\pi$ cm³ of volume. Since sugar is being added at exactly $100\pi$ cm³/s, the height must be rising at 1 cm/s.

Let’s work it out formally.

Write the relationship between volume and height. For a cylinder, $V = \pi r^2 h$ . Here $r = 10$ cm, so

$V = \pi (10)^2 h = 100\pi h.$

Differentiate both sides with respect to time $t$ .

Since $r$ is constant, $\frac{dV}{dt} = 100\pi \frac{dh}{dt}$ .

This is the related rates equation — it tells us how fast the volume changes in terms of how fast the height changes.

Substitute the given rate.

We know $\frac{dV}{dt} = 100\pi$ cm³/s. So:

$100\pi = 100\pi \frac{dh}{dt}.$

Solve for $\frac{dh}{dt}$ . Divide both sides by $100\pi$ : …
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CBSE 2025Set ANNUAL1 markMCQ
$\begin{vmatrix} \sin x & 2\cos x \\ -\cos x & 1+2\sin x \end{vmatrix}$
(i) 0
(ii) 1
(iii) 3
(iv) 2
›Reveal solutionSolution

Expand the determinant to get $\sin x + 2$ ; since $\sin x \le 1$ , the maximum value is $3$ .

We are given $\begin{vmatrix} \sin x & 2\cos x \\ -\cos x & 1+2\sin x \end{vmatrix}$ .

Expanding along the first row:

$\Delta = \sin x(1+2\sin x) - 2\cos x(-\cos x) = \sin x + 2\sin^2 x + 2\cos^2 x$

Since $\sin^2 x + \cos^2 x = 1$ :

$\Delta = \sin x + 2(\sin^2 x + \cos^2 x) = \sin x + 2$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Radius of a circle is increasing at the rate of 2 m/s. Rate of change of its circumference is:$
(a) 4π m/s
(b) 2 m/s
(c) 2π m/s
(d) 4 m/s
›Reveal solutionSolution

Differentiate the circumference formula $C=2\pi r$ with respect to time and plug in $\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt} = 2$ m/s. Circumference $C = 2\pi r$ .

Differentiating both sides with respect to time $t$ : …

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CBSE 2025Set sz1 mark
$Q. cos x = 0.6 for some value of x in its domain. (True/False)$
›Reveal solutionSolution

The statement is True: the range of $\cos x$ is $[-1, 1]$ , and $0.6$ lies inside this range.

The domain of $\cos x$ is all real numbers, and its range is exactly $[-1, 1]$ — cosine never exceeds 1 or goes below $-1$ , but it does take every value in between (it is a continuous function).

…

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CBSE 2025Set ANNUAL1 markMCQ
$y = \sin x$
(a) $-1 < y \le 1$
(b) $-1 < y < 1$
(c) $-1 \le y \le 1$
(d) $-1 \le y < 1$
›Reveal solutionSolution

$\sin x$ attains every value between $-1$ and $1$ , including both endpoints, so its range is the closed interval $[-1, 1]$ .

…

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CBSE 2024Set ANNUAL1 mark
$\begin{vmatrix} \sin x & \cos x \\ -\cos x & 1+\sin x \end{vmatrix}$
›Reveal solutionSolution

Expanding the determinant gives $1+\sin x$ , whose maximum value over $x$ is $2$ .

$\begin{vmatrix}\sin x & \cos x\\ -\cos x & 1+\sin x\end{vmatrix} = \sin x(1+\sin x) - \cos x(-\cos x) = \sin x + \sin^2x + \cos^2x = \sin x + 1$

…

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CBSE 2024Set ANNUAL1 mark
$Q. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.$
›Reveal solutionSolution

Use related rates: differentiate $A=\pi r^2$ w.r.t. time and substitute the given $\dfrac{dr}{dt}$ and $r$ .

Given $\dfrac{dr}{dt} = 3$ cm/s, find $\dfrac{dA}{dt}$ at $r = 10$ cm.

$A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$

…

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CBSE 2024Set ANNUAL1 mark
$0.7$
›Reveal solutionSolution

Differentiate the circumference formula $C=2\pi r$ with respect to time.

The circumference of a circle of radius $r$ is

$C = 2\pi r.$

Given $\dfrac{dr}{dt} = 0.7$ cm/s. Differentiating with respect to $t$ : …

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CBSE 2024Set hz1 markMCQ
$\cos \theta$
(a) $-1$
(b) $0$
(c) $1$
(d) None of these
›Reveal solutionSolution

$\cos\theta \in [-1, 1]$ for all real $\theta$ ; its maximum value is $1$ .

For every real number $\theta$ , the cosine function satisfies $-1 \le \cos\theta \le 1$ . This bound comes directly from the definition of $\cos\theta$ as the $x$ -coordinate of a point on the unit circle, which can never leave the interval $[-1, 1]$ . The value $\cos\theta = 1$ …

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Q.(a) Find the absolute maximum value of f(x)=cos⁡x+sin⁡2x, x∈[0,π]f(x) = \cos x + \sin^2 x,\ x \in [0, \pi]f(x)=cosx+sin2x, x∈[0,π].

Part (a)

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Q.(a) Find the absolute maximum value of $f(x) = \cos x + \sin^2 x,\ x \in [0, \pi]$ .

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