Board Exam-Pattern Questions

About This Question Bank

This section is a Board-style practice bank: 16 fully solved written questions in the CBSE/State Board pattern, organised by marks — 2-mark short answers (Q1-5), 3-mark standard problems (Q6-10) and 5-mark long answers with proofs (Q11-16) — followed by a 15-question MCQ quiz in the 1-mark pattern.

These questions are modelled on the pattern and difficulty of Board examinations; they are practice questions in the exam style, not reproductions of specific past papers.

How Boards typically distribute this chapter: a 1-2 mark question on rate of change or marginal cost, a 3-mark question on intervals of increase/decrease or local extrema, and a near-certain 5-mark optimisation question (open box, inscribed figures, wire problems, minimum distance). The 5-mark answers below show the full presentation — variable definition, constraint, derivative test, and a stated conclusion with units — that earns complete marks.

[Board Important] In 5-mark optimisation answers, examiners award specific marks for: defining variables with a figure (1), forming the one-variable function (1), the derivative computation (1), the max/min test (1), and the final answer with units (1). Skipping the second-derivative check typically costs a full mark.

2-Mark Questions

Q1. The radius of a spherical soap bubble is increasing at 0.2 cm/s. Find the rate of increase of its volume when the radius is 5 cm.

Solution:

1. $V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.
2. At $r = 5$, $\frac{dr}{dt} = 0.2$: $\frac{dV}{dt} = 4\pi(25)(0.2) = 20\pi$ cm³/s.

Final Answer: $20\pi$ cm³/s.

Q2. The total revenue received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.

Solution:

1. MR $= \frac{dR}{dx} = 26x + 26$.
2. At $x = 7$: MR $= 26(7) + 26 = 208$.

Final Answer: ₹208.

Q3. Show that the function $f(x) = x^3 - 3x^2 + 3x - 100$ is increasing on $\mathbb{R}$.

Solution:

1. $f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$.
2. $(x-1)^2 \geq 0$ for all $x$, so $f'(x) \geq 0$ on $\mathbb{R}$ (zero only at the single point $x = 1$).
3. Hence $f$ is increasing on $\mathbb{R}$. ∎

Q4. Find the maximum and minimum values of $f(x) = \sin 2x + 5$.

Solution:

1. $-1 \leq \sin 2x \leq 1$ for all $x$.
2. Therefore $4 \leq f(x) \leq 6$; both bounds are attained.

Final Answer: Maximum value 6; minimum value 4.

Q5. Show that $f(x) = x - \sin x$ is increasing on $\mathbb{R}$.

Solution:

1. $f'(x) = 1 - \cos x$.
2. Since $\cos x \leq 1$ always, $f'(x) \geq 0$ on $\mathbb{R}$, vanishing only at the isolated points $x = 2k\pi$.
3. Hence $f$ is increasing on $\mathbb{R}$. ∎

Takeaway (2-markers): One derivative, one sign argument or one substitution — never write more than four lines.

3-Mark Questions

Q6. The length $x$ of a rectangle is decreasing at 5 cm/min and the width $y$ is increasing at 4 cm/min. When $x = 8$ cm and $y = 6$ cm, find the rates of change of (a) the perimeter (b) the area.

Solution:

1. Given (with signs): $\frac{dx}{dt} = -5$, $\frac{dy}{dt} = 4$ (cm/min).
2. (a) $P = 2(x + y)$: $\frac{dP}{dt} = 2(-5 + 4) = -2$ cm/min — the perimeter is decreasing at 2 cm/min.
3. (b) $A = xy$: $\frac{dA}{dt} = \frac{dx}{dt}y + x\frac{dy}{dt} = (-5)(6) + (8)(4) = 2$ cm²/min — the area is increasing at 2 cm²/min.

Final Answer: (a) decreasing at 2 cm/min; (b) increasing at 2 cm²/min.

Q7. Find the intervals in which the function $f(x) = 2x^3 - 9x^2 + 12x + 15$ is (a) increasing (b) decreasing.

Solution:

1. $f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)$; critical points $x = 1, 2$.
2. Sign table: $(-\infty, 1)$: $(+)$; $(1, 2)$: $(-)$; $(2, \infty)$: $(+)$.

Final Answer: Increasing on $(-\infty, 1)$ and $(2, \infty)$; decreasing on $(1, 2)$.

Q8. Find the local maximum and local minimum values of $f(x) = x^3 - 12x^2 + 36x + 17$.

Solution:

1. $f'(x) = 3x^2 - 24x + 36 = 3(x - 2)(x - 6)$; critical points $x = 2, 6$.
2. $f''(x) = 6x - 24$: $f''(2) = -12 < 0$ (local max); $f''(6) = 12 > 0$ (local min).
3. Values: $f(2) = 8 - 48 + 72 + 17 = 49$; $f(6) = 216 - 432 + 216 + 17 = 17$.

Final Answer: Local maximum value 49 at $x = 2$; local minimum value 17 at $x = 6$.

Q9. Find the absolute maximum and absolute minimum values of $f(x) = \sin x + \cos x$ on $[0, \pi]$.

Solution:

1. $f'(x) = \cos x - \sin x = 0 \Rightarrow x = \frac{\pi}{4}$ (in the interval).
2. Candidates: $f(0) = 1$; $f\left(\frac{\pi}{4}\right) = \sqrt{2}$; $f(\pi) = -1$.

Final Answer: Absolute maximum $\sqrt{2}$ at $x = \frac{\pi}{4}$; absolute minimum $-1$ at $x = \pi$.

Q10. Find the values of $a$ for which $f(x) = x^3 - ax$ is increasing on $\mathbb{R}$.

Solution:

1. $f'(x) = 3x^2 - a$. We need $f'(x) \geq 0$ for all $x$.
2. The minimum of $3x^2$ is 0 (at $x = 0$), so we need $-a \geq 0$, i.e. $a \leq 0$.
3. For $a \leq 0$: $f'(x) = 3x^2 - a \geq 0$ everywhere (zero only at $x = 0$ when $a = 0$) — increasing.

Final Answer: $a \leq 0$.

Takeaway (3-markers): Show the factorised derivative and either a sign table or second-derivative values — that's where the middle mark lives.

5-Mark Questions

Q11. Show that of all rectangles with a given perimeter, the square has the maximum area.

Solution:

1. Variables: sides $x$ and $y$ with fixed perimeter $2(x + y) = P$, so $y = \frac{P}{2} - x$, $0 < x < \frac{P}{2}$.
2. Target: $A(x) = x\left(\frac{P}{2} - x\right) = \frac{P}{2}x - x^2$.
3. Differentiate: $A'(x) = \frac{P}{2} - 2x = 0 \Rightarrow x = \frac{P}{4}$.
4. Test: $A''(x) = -2 < 0$: maximum.
5. Conclude: $x = \frac{P}{4}$ gives $y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4} = x$ — the rectangle is a square. ∎

Q12. Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

Solution:

1. Constraint: total surface $S = 2\pi r^2 + 2\pi r h$ (fixed), so $h = \frac{S - 2\pi r^2}{2\pi r}$.
2. Target: $V = \pi r^2 h = \pi r^2 \cdot \frac{S - 2\pi r^2}{2\pi r} = \frac{Sr - 2\pi r^3}{2}$.
3. Differentiate: $V'(r) = \frac{S - 6\pi r^2}{2} = 0 \Rightarrow S = 6\pi r^2$.
4. Test: $V''(r) = -6\pi r < 0$: maximum.
5. Conclude: $6\pi r^2 = 2\pi r^2 + 2\pi rh \Rightarrow 4\pi r^2 = 2\pi r h \Rightarrow h = 2r$ — height equals diameter. ∎

Q13. Show that the semi-vertical angle of the cone of maximum volume and given slant height is $\tan^{-1}\sqrt{2}$.

Solution:

1. Variables: slant height $l$ (fixed), height $h$, base radius $r$ with $r^2 = l^2 - h^2$.
2. Target: $V(h) = \frac{1}{3}\pi r^2 h = \frac{\pi}{3}(l^2 h - h^3)$, $0 < h < l$.
3. Differentiate: $V'(h) = \frac{\pi}{3}(l^2 - 3h^2) = 0 \Rightarrow h = \frac{l}{\sqrt{3}}$.
4. Test: $V''(h) = -2\pi h < 0$: maximum.
5. Conclude: $r^2 = l^2 - \frac{l^2}{3} = \frac{2l^2}{3}$, so $\tan\alpha = \frac{r}{h} = \frac{l\sqrt{2/3}}{l/\sqrt{3}} = \sqrt{2}$, i.e. $\alpha = \tan^{-1}\sqrt{2}$. ∎

Q14. Show that the right circular cone of least curved surface area and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

Solution:

1. Constraint: $V = \frac{1}{3}\pi r^2 h$ (fixed) $\Rightarrow h = \frac{3V}{\pi r^2}$.
2. Target: curved surface $S = \pi r l = \pi r\sqrt{r^2 + h^2}$. Minimise $S^2$ instead: $S^2 = \pi^2 r^2(r^2 + h^2) = \pi^2 r^4 + \frac{9V^2}{r^2}$.
3. Differentiate: $\frac{d(S^2)}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3} = 0 \Rightarrow r^6 = \frac{9V^2}{2\pi^2}$, i.e. $V^2 = \frac{2\pi^2 r^6}{9}$.
4. Test: $\frac{d^2(S^2)}{dr^2} = 12\pi^2 r^2 + \frac{54V^2}{r^4} > 0$: minimum.
5. Conclude: since $V = \frac{1}{3}\pi r^2 h$, $V^2 = \frac{\pi^2 r^4 h^2}{9}$. Equating: $\frac{\pi^2 r^4 h^2}{9} = \frac{2\pi^2 r^6}{9} \Rightarrow h^2 = 2r^2 \Rightarrow h = \sqrt{2}\,r$. ∎

Takeaway: Minimising $S^2$ (not $S$) removes the square root — the single most useful trick in surface-minimisation proofs.

Q15. Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is $\sin^{-1}\left(\frac{1}{3}\right)$.

Solution:

1. Constraint: total surface $S = \pi r^2 + \pi r l$ (fixed) $\Rightarrow l = \frac{S - \pi r^2}{\pi r}$.
2. Target: $V = \frac{1}{3}\pi r^2 h$ with $h^2 = l^2 - r^2$, so work with $V^2 = \frac{\pi^2 r^4}{9}(l^2 - r^2)$.
3. Substitute $l$: $l^2 - r^2 = \frac{(S - \pi r^2)^2 - \pi^2 r^4}{\pi^2 r^2} = \frac{S^2 - 2S\pi r^2}{\pi^2 r^2}$. Hence $V^2 = \frac{r^2(S^2 - 2S\pi r^2)}{9} = \frac{S^2 r^2 - 2S\pi r^4}{9}$.
4. Differentiate (in $u = r^2$): $V^2 = \frac{S^2 u - 2S\pi u^2}{9}$, so $\frac{d(V^2)}{du} = \frac{S^2 - 4S\pi u}{9} = 0 \Rightarrow u = r^2 = \frac{S}{4\pi}$, i.e. $S = 4\pi r^2$. Second derivative $= -\frac{4S\pi}{9} < 0$: maximum.
5. Conclude: $\pi r^2 + \pi r l = 4\pi r^2 \Rightarrow l = 3r$, so $\sin\alpha = \frac{r}{l} = \frac{1}{3}$, i.e. $\alpha = \sin^{-1}\left(\frac{1}{3}\right)$. ∎

Takeaway: Substituting $u = r^2$ turned a quartic optimisation into a parabola. Change of variable is a legitimate, mark-earning step — just state it.

Q16. Find the point on the curve $y^2 = 2x$ which is at the minimum distance from the point $(1, 4)$.

Solution:

1. Parametrise by $y = t$: the point is $\left(\frac{t^2}{2}, t\right)$.
2. Squared distance: $D^2(t) = \left(\frac{t^2}{2} - 1\right)^2 + (t - 4)^2$.
3. Differentiate: $\frac{d(D^2)}{dt} = 2\left(\frac{t^2}{2} - 1\right)t + 2(t - 4) = t^3 - 2t + 2t - 8 = t^3 - 8$.
4. Solve: $t^3 = 8 \Rightarrow t = 2$ (only real root).
5. Test: $\frac{d^2(D^2)}{dt^2} = 3t^2 = 12 > 0$: minimum. The point is $\left(\frac{4}{2}, 2\right) = (2, 2)$, at distance $\sqrt{1 + 4} = \sqrt{5}$.

Final Answer: The nearest point is $(2, 2)$ (minimum distance $\sqrt{5}$).

Takeaway: Parametrising by $y$ made the derivative collapse to $t^3 - 8$ — always choose the parameter that makes the curve's equation linear in the other coordinate.