Questions and Answers (Board Exam)

Introduction to Board Exam PYQs

This section contains 30 highly important board-style questions based on the Applications of Derivatives chapter. The main focus areas are:

• Rate of Change of Quantities
• Increasing and Decreasing Functions
• Local/Absolute Maxima and Minima
• Optimization Problems

In board exams, marks are awarded not only for the final answer but also for the method, so always:

1. Define the variables clearly.
2. Write the governing formula or relation.
3. Differentiate carefully with respect to the required variable (usually time or $x$).
4. Substitute the given values only after differentiation.
5. State whether the quantity is increasing or decreasing wherever required.

Question 1 [CBSE 2026]

A spherical balloon is being inflated so that its volume is increasing at the rate of $8 \text{ cm}^3/\text{s}$. Find the rate at which its surface area is increasing when the radius of the balloon is $12 \text{ cm}$.

Solution: Step 1: Let $V$ be the volume, $S$ the surface area, and $r$ the radius of the balloon at time $t$. Given: $$\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$$ and we need to find $\frac{dS}{dt}$ when $r = 12 \text{ cm}$.

Step 2: Use the volume formula of a sphere: $$V = \frac{4}{3}\pi r^3.$$ Differentiating with respect to $t$, $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.$$ Substituting $\frac{dV}{dt} = 8$, $$8 = 4\pi r^2 \frac{dr}{dt}$$ $$\frac{dr}{dt} = \frac{8}{4\pi r^2} = \frac{2}{\pi r^2}.$$

Step 3: Use the surface area formula of a sphere: $$S = 4\pi r^2.$$ Differentiating with respect to $t$, $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}.$$ Now substitute $\frac{dr}{dt} = \frac{2}{\pi r^2}$: $$\frac{dS}{dt} = 8\pi r \left( \frac{2}{\pi r^2} \right) = \frac{16}{r}.$$

Step 4: Put $r = 12$ cm. $$\frac{dS}{dt} = \frac{16}{12} = \frac{4}{3} \text{ cm}^2/\text{s}.$$

Answer: The surface area is increasing at the rate of $\frac{4}{3} \text{ cm}^2/\text{s}$.

Question 2 [CBSE 2025]

A ladder $5 \text{ m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2 \text{ m/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $4 \text{ m}$ away from the wall?

Solution: Step 1: Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall at time $t$. Given: $$\frac{dx}{dt} = 2 \text{ m/s}$$ and the ladder length is constant, $5$ m.

Step 2: By the Pythagorean theorem, $$x^2 + y^2 = 25.$$ When $x = 4$ m, $$4^2 + y^2 = 25$$ $$16 + y^2 = 25$$ $$y^2 = 9 \Rightarrow y = 3 \text{ m}.$$

Step 3: Differentiate the relation $x^2 + y^2 = 25$ with respect to $t$. $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0.$$

Step 4: Substitute $x = 4$, $y = 3$, and $\frac{dx}{dt} = 2$. $$4(2) + 3\frac{dy}{dt} = 0$$ $$8 + 3\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = -\frac{8}{3} \text{ m/s}.$$ The negative sign shows that the height is decreasing.

Answer: The height is decreasing at the rate of $\frac{8}{3} \text{ m/s}$.

Question 3 [CBSE 2026]

A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the y-coordinate is changing $8$ times as fast as the x-coordinate.

Solution: Step 1: It is given that the y-coordinate is changing 8 times as fast as the x-coordinate. So, $$\frac{dy}{dt} = 8\frac{dx}{dt}.$$

Step 2: The equation of the curve is $$6y = x^3 + 2.$$ Differentiate both sides with respect to $t$: $$6\frac{dy}{dt} = 3x^2\frac{dx}{dt}.$$

Step 3: Substitute $\frac{dy}{dt} = 8\frac{dx}{dt}$. $$6\left(8\frac{dx}{dt}\right) = 3x^2\frac{dx}{dt}.$$ Assuming the particle is actually moving, $\frac{dx}{dt} \neq 0$, so we divide both sides by $\frac{dx}{dt}$: $$48 = 3x^2$$ $$x^2 = 16$$ $$x = \pm 4.$$

Step 4: Find the corresponding $y$-coordinates. If $x = 4$: $$6y = 4^3 + 2 = 64 + 2 = 66$$ $$y = 11.$$ So one point is $(4,11)$.

If $x = -4$: $$6y = (-4)^3 + 2 = -64 + 2 = -62$$ $$y = -\frac{31}{3}.$$ So the other point is $\left(-4,-\frac{31}{3}\right)$.

Answer: The required points are $(4,11)$ and $\left(-4,-\frac{31}{3}\right)$.

Question 4 [CBSE 2024]

The length $x$ of a rectangle is decreasing at the rate of $5 \text{ cm/minute}$ and the width $y$ is increasing at the rate of $4 \text{ cm/minute}$. When $x = 8 \text{ cm}$ and $y = 6 \text{ cm}$, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Solution: Step 1: Since the length is decreasing, $$\frac{dx}{dt} = -5 \text{ cm/min}.$$ Since the width is increasing, $$\frac{dy}{dt} = 4 \text{ cm/min}.$$

Step 2: For perimeter, $$P = 2(x+y).$$ Differentiate with respect to $t$: $$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right).$$ Substitute the given values: $$\frac{dP}{dt} = 2(-5 + 4) = 2(-1) = -2 \text{ cm/min}.$$ Thus, the perimeter is decreasing at the rate of $2 \text{ cm/min}$.

Step 3: For area, $$A = xy.$$ Differentiate using the product rule: $$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}.$$ Substitute $x = 8$, $y = 6$: $$\frac{dA}{dt} = 8(4) + 6(-5) = 32 - 30 = 2 \text{ cm}^2/\text{min}.$$ Thus, the area is increasing at the rate of $2 \text{ cm}^2/\text{min}$.

Answer: (a) Perimeter is decreasing at $2 \text{ cm/min}$. (b) Area is increasing at $2 \text{ cm}^2/\text{min}$.

Question 5 [CBSE 2026]

Water is dripping out from a conical funnel of semi-vertical angle $\pi/4$ at the uniform rate of $2 \text{ cm}^2/\text{s}$ in its curved surface area through a tiny hole at the vertex. When the slant height of the water cone is $4 \text{ cm}$, find the rate of decrease of the slant height of the water.

Solution: Step 1: Let $r$ be the radius, $l$ the slant height, and $\alpha$ the semi-vertical angle of the cone. Given: $$\alpha = \frac{\pi}{4}.$$ From the geometry of the cone, $$r = l\sin\alpha = l\sin\frac{\pi}{4} = \frac{l}{\sqrt{2}}.$$

Step 2: The curved surface area of the water cone is $$S = \pi r l.$$ Substitute $r = \frac{l}{\sqrt{2}}$: $$S = \pi\left(\frac{l}{\sqrt{2}}\right)l = \frac{\pi l^2}{\sqrt{2}}.$$

Step 3: Differentiate with respect to time. $$\frac{dS}{dt} = \frac{\pi}{\sqrt{2}} \cdot 2l \frac{dl}{dt} = \sqrt{2}\pi l \frac{dl}{dt}.$$

Step 4: Since the curved surface area is decreasing at the rate of $2 \text{ cm}^2/\text{s}$, $$\frac{dS}{dt} = -2.$$ When $l = 4$ cm, $$-2 = \sqrt{2}\pi(4)\frac{dl}{dt} = 4\sqrt{2}\pi\frac{dl}{dt}.$$ So, $$\frac{dl}{dt} = \frac{-2}{4\sqrt{2}\pi} = -\frac{1}{2\sqrt{2}\pi} \text{ cm/s}.$$ The negative sign shows the slant height is decreasing.

Answer: The slant height is decreasing at the rate of $\frac{1}{2\sqrt{2}\pi} \text{ cm/s}$.

Question 6 [CBSE 2023]

Find the intervals in which the function $f$ given by $f(x) = 2x^3 - 3x^2 - 36x + 7$ is (a) strictly increasing, (b) strictly decreasing.

Solution: Step 1: Differentiate the function: $$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6).$$ Factorize: $$f'(x) = 6(x-3)(x+2).$$

Step 2: Find critical points by setting $f'(x)=0$. $$6(x-3)(x+2) = 0$$ $$x = 3, -2.$$ These divide the number line into three intervals: $$(-\infty,-2), \quad (-2,3), \quad (3,\infty).$$

Step 3: Check the sign of $f'(x)$.

• For $x \in (-\infty,-2)$, take $x=-3$: $$f'(-3) = 6(-6)(-1) > 0.$$ So $f(x)$ is strictly increasing there.

• For $x \in (-2,3)$, take $x=0$: $$f'(0) = 6(-3)(2) < 0.$$ So $f(x)$ is strictly decreasing there.

• For $x \in (3,\infty)$, take $x=4$: $$f'(4) = 6(1)(6) > 0.$$ So $f(x)$ is strictly increasing there.

Answer: (a) Strictly increasing on $(-\infty,-2) \cup (3,\infty)$. (b) Strictly decreasing on $(-2,3)$.

Question 7 [CBSE 2025]

Find the intervals in which the function $f(x) = \sin x + \cos x$, $0 \le x \le 2\pi$ is strictly increasing or strictly decreasing.

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

Step 2: Set $f'(x)=0$. $$\cos x - \sin x = 0$$ $$\tan x = 1.$$ In $[0,2\pi]$, this happens at $$x = \frac{\pi}{4}, \quad \frac{5\pi}{4}.$$

Step 3: These points divide the interval into: $$(0,\pi/4), \quad (\pi/4,5\pi/4), \quad (5\pi/4,2\pi).$$

Step 4: Test the sign of $f'(x)$.

• In $(0,\pi/4)$, take $x=0$: $$f'(0)=1>0.$$ So $f$ is strictly increasing.

• In $(\pi/4,5\pi/4)$, take $x=\pi$: $$f'(\pi)=\cos\pi-\sin\pi=-1<0.$$ So $f$ is strictly decreasing.

• In $(5\pi/4,2\pi)$, take $x=3\pi/2$: $$f'\left(\frac{3\pi}{2}\right)=0-(-1)=1>0.$$ So $f$ is strictly increasing.

Answer: Strictly increasing on $(0,\pi/4) \cup (5\pi/4,2\pi)$ and strictly decreasing on $(\pi/4,5\pi/4)$.

Question 8 [CBSE 2026]

Show that the function $f(\theta) = \frac{4\sin\theta}{2+\cos\theta} - \theta$ is an increasing function of $\theta$ in $[0,\pi/2]$.

Solution: Step 1: Differentiate using the quotient rule. $$f'(\theta)=\frac{(2+\cos\theta)(4\cos\theta)-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2}-1.$$

Step 2: Simplify the numerator of the fraction: $$(2+\cos\theta)(4\cos\theta)+4\sin^2\theta = 8\cos\theta + 4\cos^2\theta + 4\sin^2\theta.$$ Using $\sin^2\theta + \cos^2\theta = 1$, $$8\cos\theta + 4\cos^2\theta + 4\sin^2\theta = 8\cos\theta + 4.$$ So, $$f'(\theta)=\frac{8\cos\theta+4}{(2+\cos\theta)^2}-1.$$

Step 3: Write as a single fraction: $$f'(\theta)=\frac{8\cos\theta+4-(2+\cos\theta)^2}{(2+\cos\theta)^2}.$$ Now, $$(2+\cos\theta)^2 = 4+4\cos\theta+\cos^2\theta.$$ Hence, $$f'(\theta)=\frac{8\cos\theta+4-(4+4\cos\theta+\cos^2\theta)}{(2+\cos\theta)^2}$$ $$= \frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}$$ $$= \frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}.$$

Step 4: Analyze the sign in $[0,\pi/2]$. In this interval,

• $\cos\theta \ge 0$,
• $4-\cos\theta > 0$,
• $(2+\cos\theta)^2 > 0$. Therefore, $$f'(\theta) \ge 0$$ for all $\theta \in [0,\pi/2]$. So the function is increasing on $[0,\pi/2]$.

Answer: Proved analytically.

Question 9 [CBSE 2024]

Find the values of $x$ for which $y = [x(x-2)]^2$ is an increasing function.

Solution: Step 1: Here the square brackets are used only as grouping symbols, so $$y = (x(x-2))^2 = (x^2-2x)^2.$$

Step 2: Differentiate using the chain rule: $$\frac{dy}{dx} = 2(x^2-2x)\cdot(2x-2).$$ Factorize: $$\frac{dy}{dx} = 2x(x-2)\cdot 2(x-1) = 4x(x-1)(x-2).$$

Step 3: Critical points are obtained from $$4x(x-1)(x-2)=0 \Rightarrow x=0,1,2.$$ These divide the number line into: $$(-\infty,0), \quad (0,1), \quad (1,2), \quad (2,\infty).$$

Step 4: Check the sign of $\frac{dy}{dx}$.

• For $x<0$, take $x=-1$: $$\frac{dy}{dx}<0.$$
• For $0<x<1$, take $x=\frac12$: $$\frac{dy}{dx}>0.$$
• For $1<x<2$, take $x=\frac32$: $$\frac{dy}{dx}<0.$$
• For $x>2$, take $x=3$: $$\frac{dy}{dx}>0.$$ Thus, the function is increasing where the derivative is positive.

Answer: The function is increasing on $(0,1) \cup (2,\infty)$.

Question 10 [CBSE 2026]

Find the intervals in which $f(x) = -2x^3 - 9x^2 - 12x + 1$ is strictly increasing or strictly decreasing.

Solution: Step 1: Differentiate: $$f'(x) = -6x^2 - 18x - 12.$$ Factorize: $$f'(x) = -6(x^2+3x+2) = -6(x+1)(x+2).$$

Step 2: Critical points occur when $$-6(x+1)(x+2)=0 \Rightarrow x=-1,-2.$$ So the intervals are: $$(-\infty,-2), \quad (-2,-1), \quad (-1,\infty).$$

Step 3: Test the sign of $f'(x)$.

• In $(-\infty,-2)$, take $x=-3$: $$f'(-3) = -6(-2)(-1) < 0.$$ So $f$ is strictly decreasing.

• In $(-2,-1)$, take $x=-\frac32$: $$f'\left(-\frac32\right) = -6\left(-\frac12\right)\left(\frac12\right) > 0.$$ So $f$ is strictly increasing.

• In $(-1,\infty)$, take $x=0$: $$f'(0)=-12<0.$$ So $f$ is strictly decreasing.

Answer: Strictly increasing on $(-2,-1)$; strictly decreasing on $(-\infty,-2) \cup (-1,\infty)$.

Question 11 [CBSE 2025]

Find the intervals in which the function $f(x) = \log(1+x) - x$ is increasing or decreasing in its domain.

Solution: Step 1: First find the domain. For $\log(1+x)$ to be defined, $$1+x > 0 \Rightarrow x>-1.$$ So the domain is $$(-1,\infty).$$

Step 2: Differentiate: $$f'(x)=\frac{1}{1+x}-1 = \frac{1-(1+x)}{1+x} = \frac{-x}{1+x}.$$

Step 3: Analyze the sign of $f'(x)$. In the domain, $1+x>0$ always. So the sign of $f'(x)$ depends only on $-x$.

• If $-1 < x < 0$, then $x<0$, so $-x>0$. Hence, $$f'(x)>0.$$ Therefore, $f(x)$ is increasing on $(-1,0)$.

• If $x>0$, then $-x<0$. Hence, $$f'(x)<0.$$ Therefore, $f(x)$ is decreasing on $(0,\infty)$.

At $x=0$, $$f'(0)=0.$$ So $x=0$ is the turning point.

Step 4: Find the maximum value at $x=0$. $$f(0)=\log(1+0)-0=0.$$

Answer: The function is increasing on $(-1,0)$ and decreasing on $(0,\infty)$. It has a maximum value $0$ at $x=0$.

Question 12 [CBSE 2023]

Find the local maxima and local minima, if any, of the function $f(x) = \frac{x}{2} + \frac{2}{x}, x > 0$. Also find the local maximum and local minimum values.

Solution: Step 1: Differentiate: $$f'(x)=\frac12 - \frac{2}{x^2}.$$

Step 2: Find critical points by setting $f'(x)=0$. $$\frac12 - \frac{2}{x^2}=0$$ $$\frac12 = \frac{2}{x^2}$$ $$x^2=4.$$ Since $x>0$, we get $$x=2.$$

Step 3: Use the second derivative test. $$f''(x)=\frac{d}{dx}\left(\frac12 - 2x^{-2}\right)=4x^{-3}=\frac{4}{x^3}.$$ Now, $$f''(2)=\frac{4}{8}=\frac12 >0.$$ So $x=2$ is a point of local minimum.

Step 4: Find the corresponding function value. $$f(2)=\frac{2}{2}+\frac{2}{2}=1+1=2.$$ Since there is only one critical point in the domain and it is a minimum, there is no local maximum.

Answer: Local minimum at $x=2$ with minimum value $2$. There is no local maximum.

Question 13 [CBSE 2026]

Find the absolute maximum and absolute minimum values of the function $f(x) = 12x^{4/3} - 6x^{1/3}$ on the interval $[-1,1]$.

Solution: Step 1: Differentiate: $$f'(x)=12\cdot\frac43 x^{1/3} - 6\cdot\frac13 x^{-2/3} = 16x^{1/3} - 2x^{-2/3}.$$ Write it as a single fraction: $$f'(x)=\frac{16x-2}{x^{2/3}}.$$

Step 2: Critical points occur where $f'(x)=0$ or $f'(x)$ is undefined.

• $f'(x)=0$ when $$16x-2=0 \Rightarrow x=\frac18.$$
• $f'(x)$ is undefined at $$x=0.$$ Both lie in $[-1,1]$.

Step 3: Evaluate $f(x)$ at the critical points and endpoints. $$f(-1)=12(-1)^{4/3}-6(-1)^{1/3}=12(1)-6(-1)=18,$$ $$f(0)=0,$$ $$f\left(\frac18\right)=12\left(\frac18\right)^{4/3}-6\left(\frac18\right)^{1/3}=12\left(\frac1{16}\right)-6\left(\frac12\right)=\frac34-3=-\frac94,$$ $$f(1)=12-6=6.$$

Step 4: Compare these values. The largest value is $18$ and the smallest value is $-\frac94$.

Answer: Absolute maximum is $18$ at $x=-1$; absolute minimum is $-\frac94$ at $x=\frac18$.

Question 14 [CBSE 2024]

Find the maximum value of the function $p(x) = 41 - 72x - 18x^2$.

Solution: Step 1: Differentiate: $$p'(x) = -72 - 36x.$$

Step 2: Find the critical point by setting $p'(x)=0$. $$-72 - 36x = 0$$ $$36x = -72$$ $$x = -2.$$

Step 3: Apply the second derivative test. $$p''(x) = -36.$$ Since $$p''(-2) = -36 < 0,$$ $x=-2$ gives a maximum value.

Step 4: Find the maximum value. $$p(-2)=41-72(-2)-18(-2)^2 = 41+144-72 = 113.$$

Answer: The maximum value is $113$.

Note: If $x$ represents a number of units produced, then the physical domain should be specified separately. As a pure function of a real variable, the maximum value is $113$ at $x=-2$.

Question 15 [CBSE 2025]

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution: Step 1: Let the radius of the circle be $a$. Suppose the inscribed rectangle has length $x$ and breadth $y$. Its diagonal equals the diameter of the circle, so $$\sqrt{x^2+y^2} = 2a.$$ Hence, $$x^2+y^2=4a^2.$$

Step 2: The area of the rectangle is $$A = xy.$$ To maximize $A$, it is convenient to maximize $A^2$: $$A^2 = x^2y^2.$$ From the relation above, $$y^2 = 4a^2 - x^2.$$ So, $$A^2 = x^2(4a^2-x^2)=4a^2x^2-x^4.$$

Step 3: Differentiate with respect to $x$. $$\frac{d}{dx}(A^2)=8a^2x - 4x^3 = 4x(2a^2-x^2).$$ Set equal to zero: $$4x(2a^2-x^2)=0.$$ Ignoring $x=0$ (degenerate rectangle), $$x^2 = 2a^2.$$

Step 4: Then $$y^2 = 4a^2 - 2a^2 = 2a^2.$$ So, $$x^2 = y^2 \Rightarrow x=y.$$ Thus, the rectangle is a square.

Step 5: Verify maximum. $$\frac{d^2}{dx^2}(A^2)=8a^2-12x^2.$$ At $x^2=2a^2$, $$8a^2-24a^2=-16a^2<0,$$ so the area is maximum.

Answer: Proved analytically: among all rectangles inscribed in a fixed circle, the square has the maximum area.

Question 16 [CBSE 2026]

Show that a closed right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of its base.

Solution: Step 1: Let $r$ be the radius and $h$ the height of the cylinder. Let the given total surface area be $S$. For a closed cylinder, $$S = 2\pi r^2 + 2\pi rh.$$ So, $$h = \frac{S - 2\pi r^2}{2\pi r}.$$

Step 2: Volume of cylinder is $$V = \pi r^2 h.$$ Substitute $h$: $$V = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) = \frac{r}{2}(S - 2\pi r^2) = \frac{Sr}{2} - \pi r^3.$$

Step 3: Differentiate with respect to $r$. $$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2.$$ Set this equal to zero: $$\frac{S}{2} - 3\pi r^2 = 0$$ $$S = 6\pi r^2.$$

Step 4: Apply the second derivative test. $$\frac{d^2V}{dr^2} = -6\pi r < 0$$ for $r>0$, hence the volume is maximum.

Step 5: Substitute $S = 6\pi r^2$ back into the expression for $h$. $$h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r.$$ So the height is equal to the diameter of the base.

Answer: Proved analytically.

Question 17 [CBSE 2023]

A wire of length $28 \text{ m}$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum?

Solution: Step 1: Let the length of wire used for the square be $x$ m. Then the length used for the circle is $(28-x)$ m.

Step 2: For the square, perimeter = $x$, so side = $x/4$. Hence area of square: $$A_1 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}.$$

Step 3: For the circle, circumference = $28-x$. So, $$2\pi r = 28-x \Rightarrow r = \frac{28-x}{2\pi}.$$ Thus area of circle: $$A_2 = \pi r^2 = \pi\left(\frac{28-x}{2\pi}\right)^2 = \frac{(28-x)^2}{4\pi}.$$

Step 4: Total area: $$A = \frac{x^2}{16} + \frac{(28-x)^2}{4\pi}.$$ Differentiate with respect to $x$: $$\frac{dA}{dx} = \frac{2x}{16} + \frac{2(28-x)(-1)}{4\pi} = \frac{x}{8} - \frac{28-x}{2\pi}.$$

Step 5: Set derivative equal to zero: $$\frac{x}{8} = \frac{28-x}{2\pi}.$$ Cross-multiplying, $$2\pi x = 8(28-x)$$ $$2\pi x = 224 - 8x$$ $$x(2\pi+8)=224$$ $$x = \frac{112}{\pi+4}.$$ So the other piece has length $$28 - \frac{112}{\pi+4} = \frac{28\pi}{\pi+4}.$$

Step 6: Check that this gives a minimum. $$\frac{d^2A}{dx^2} = \frac18 + \frac{1}{2\pi} > 0.$$ Hence the combined area is minimum.

Answer: Square piece $= \frac{112}{\pi+4}$ m; circle piece $= \frac{28\pi}{\pi+4}$ m.

Question 18 [CBSE 2026]

Show that the semi-vertical angle of the cone of maximum volume and of given slant height is $\tan^{-1}\sqrt{2}$.

Solution: Step 1: Let $l$ be the fixed slant height, $r$ the radius, $h$ the height, and $\theta$ the semi-vertical angle. From the right triangle in the cone, $$r = l\sin\theta, \qquad h = l\cos\theta.$$

Step 2: Volume of the cone is $$V = \frac13 \pi r^2 h.$$ Substitute for $r$ and $h$: $$V = \frac13 \pi (l\sin\theta)^2 (l\cos\theta) = \frac{\pi l^3}{3}\sin^2\theta\cos\theta.$$

Step 3: Differentiate with respect to $\theta$. $$\frac{dV}{d\theta} = \frac{\pi l^3}{3}\left[2\sin\theta\cos\theta\cdot\cos\theta + \sin^2\theta(-\sin\theta)\right].$$ So, $$\frac{dV}{d\theta} = \frac{\pi l^3}{3}\left(2\sin\theta\cos^2\theta - \sin^3\theta\right)$$ $$= \frac{\pi l^3}{3}\sin\theta(2\cos^2\theta - \sin^2\theta).$$

Step 4: Set $\frac{dV}{d\theta}=0$. Since $0<\theta<\pi/2$, $\sin\theta \neq 0$. Hence, $$2\cos^2\theta - \sin^2\theta = 0$$ $$\sin^2\theta = 2\cos^2\theta$$ $$\tan^2\theta = 2.$$ Since $\theta$ is acute, $$\tan\theta = \sqrt2$$ $$\theta = \tan^{-1}\sqrt2.$$

Answer: Proved analytically.

Question 19 [CBSE 2025]

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Solution: Step 1: Let the cone have height $h$ and base radius $r$. Let the sphere have radius $R$. If the vertex of the cone is at the top of the sphere and the base plane is at distance $(h-R)$ below the center, then from geometry, $$r^2 = R^2 - (h-R)^2.$$ Expanding: $$r^2 = R^2 - (h^2 - 2Rh + R^2) = 2Rh - h^2.$$

Step 2: Volume of the cone is $$V = \frac13 \pi r^2 h = \frac13 \pi (2Rh - h^2)h = \frac{\pi}{3}(2Rh^2 - h^3).$$

Step 3: Differentiate with respect to $h$. $$\frac{dV}{dh} = \frac{\pi}{3}(4Rh - 3h^2).$$ Set equal to zero: $$4Rh - 3h^2 = 0$$ $$h(4R - 3h)=0.$$ Ignoring $h=0$, we get $$h = \frac{4R}{3}.$$

Step 4: Verify maximum. $$\frac{d^2V}{dh^2} = \frac{\pi}{3}(4R - 6h).$$ At $h = \frac{4R}{3}$, $$\frac{d^2V}{dh^2} = \frac{\pi}{3}(4R - 8R) < 0,$$ so the volume is maximum.

Step 5: Find the maximum volume. Substitute $h = \frac{4R}{3}$ into $$V = \frac{\pi}{3}(2Rh^2 - h^3).$$ We get $$V_{\max} = \frac{32\pi R^3}{81}.$$ Now the volume of the sphere is $$V_s = \frac{4}{3}\pi R^3.$$ Therefore, $$\frac{V_{\max}}{V_s} = \frac{\frac{32\pi R^3}{81}}{\frac{4}{3}\pi R^3} = \frac{32}{81}\cdot\frac{3}{4} = \frac{8}{27}.$$

Answer: Proved analytically: the largest cone has volume $\frac{8}{27}$ of the volume of the sphere.

Question 20 [CBSE 2024]

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$.

Solution: Step 1: Let the cylinder have radius $r$ and height $h$. In the cross-section through the axis, $$r^2 + \left(\frac{h}{2}\right)^2 = R^2.$$ Thus, $$r^2 = R^2 - \frac{h^2}{4}.$$

Step 2: Volume of the cylinder: $$V = \pi r^2 h = \pi\left(R^2 - \frac{h^2}{4}\right)h = \pi R^2 h - \frac{\pi}{4}h^3.$$

Step 3: Differentiate with respect to $h$. $$\frac{dV}{dh} = \pi R^2 - \frac{3\pi}{4}h^2.$$ Set equal to zero: $$\pi R^2 = \frac{3\pi}{4}h^2$$ $$h^2 = \frac{4R^2}{3}$$ $$h = \frac{2R}{\sqrt3}.$$

Step 4: Verify maximum. $$\frac{d^2V}{dh^2} = -\frac{3\pi}{2}h < 0$$ for $h>0$, so the volume is maximum.

Answer: Proved analytically.

Question 21 [CBSE 2026]

A box with a square base and an open top is to be made from a given quantity of cardboard of area $c^2$. Show that the maximum volume of the box is $\frac{c^3}{6\sqrt{3}}$.

Solution: Step 1: Let the side of the square base be $x$ and the height be $y$. Since the box has no top, its surface area is $$x^2 + 4xy = c^2.$$ So, $$y = \frac{c^2 - x^2}{4x}.$$

Step 2: Volume of the box is $$V = x^2y = x^2\left(\frac{c^2 - x^2}{4x}\right)=\frac14(c^2x - x^3).$$

Step 3: Differentiate with respect to $x$. $$\frac{dV}{dx}=\frac14(c^2 - 3x^2).$$ Set equal to zero: $$c^2 - 3x^2 = 0$$ $$x^2 = \frac{c^2}{3}$$ $$x = \frac{c}{\sqrt3}.$$

Step 4: Verify maximum. $$\frac{d^2V}{dx^2} = -\frac{6x}{4} = -\frac{3x}{2} < 0$$ for positive $x$. So volume is maximum.

Step 5: Substitute $x = \frac{c}{\sqrt3}$ into $V$: $$V_{\max} = \frac14\left(c^2\cdot\frac{c}{\sqrt3} - \left(\frac{c}{\sqrt3}\right)^3\right)$$ $$= \frac14\left(\frac{c^3}{\sqrt3} - \frac{c^3}{3\sqrt3}\right)$$ $$= \frac14\cdot\frac{2c^3}{3\sqrt3} = \frac{c^3}{6\sqrt3}.$$

Answer: Proved analytically.

Question 22 [CBSE 2025]

Show that the right circular cone of least curved surface area and given volume has altitude equal to $\sqrt{2}$ times the radius of the base.

Solution: Step 1: Let $r$ be the base radius, $h$ the height, and $l$ the slant height. The volume is fixed: $$V = \frac13 \pi r^2 h.$$ Hence, $$h = \frac{3V}{\pi r^2}.$$

Step 2: Curved surface area is $$S = \pi r l = \pi r\sqrt{r^2+h^2}.$$ To minimize $S$, it is enough to minimize $S^2$: $$S^2 = \pi^2 r^2(r^2+h^2)=\pi^2 r^4 + \pi^2 r^2 h^2.$$ Substitute $h = \frac{3V}{\pi r^2}$: $$S^2 = \pi^2 r^4 + \pi^2 r^2\left(\frac{9V^2}{\pi^2 r^4}\right)$$ $$= \pi^2 r^4 + \frac{9V^2}{r^2}.$$ Let $$Z = \pi^2 r^4 + \frac{9V^2}{r^2}.$$

Step 3: Differentiate with respect to $r$. $$\frac{dZ}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3}.$$ Set equal to zero: $$4\pi^2 r^3 = \frac{18V^2}{r^3}$$ $$4\pi^2 r^6 = 18V^2.$$ Now use $$V = \frac13 \pi r^2 h \Rightarrow V^2 = \frac{\pi^2 r^4 h^2}{9}.$$ Substitute this into the equation: $$4\pi^2 r^6 = 18\cdot \frac{\pi^2 r^4 h^2}{9} = 2\pi^2 r^4 h^2.$$ Cancel $2\pi^2 r^4$: $$2r^2 = h^2.$$ Hence, $$h = \sqrt2\,r.$$

Step 4: Verify minimum. $$\frac{d^2Z}{dr^2} = 12\pi^2 r^2 + \frac{54V^2}{r^4} > 0,$$ so the curved surface area is minimum.

Answer: Proved analytically.