cbse class 12 important case study

Class 12 CBSE Mathematics: 20 Important Case Studies

Relations and Functions

Case Study 1: Student Relations

A teacher wants to organize students into groups for a project. Let $A = \{1, 2, 3, 4, 5\}$ be the set of roll numbers. A relation $R$ on set $A$ is defined as $R = \{(a, b) : |a - b| \text{ is even}\}$.

Questions:

1. Write the relation $R$ in roster form.
2. Check if $R$ is reflexive and symmetric.
3. Is $R$ an equivalence relation? Justify.

Click to view Answer & Solution

1. $R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,3), (3,1), (1,5), (5,1), (3,5), (5,3), (2,4), (4,2)\}$
2. Reflexive: For all $a \in A$, $|a - a| = 0$ which is even. Hence, $(a,a) \in R$.
   Symmetric: If $(a,b) \in R$, $|a - b|$ is even $\implies |b - a|$ is also even. Hence $(b,a) \in R$.
3. Yes. It is reflexive, symmetric, and transitive (since $|a-b|$ and $|b-c|$ being even implies $|a-c|$ is even). Thus, $R$ is an equivalence relation.

Inverse Trigonometric Functions

Case Study 2: Camera Tracking Angle

A camera positioned at the ground tracks an object moving vertically. The angle of elevation $\theta$ is modeled by $\theta = \sin^{-1}(x/10)$, where $x$ represents the height of the object in meters ($0 \le x \le 10$).

Questions:

1. What is the domain and range of the principal value branch of $f(x) = \sin^{-1}(x)$?
2. Find the angle $\theta$ when the object is at a height of $5\sqrt{3}$ meters.
3. If $\theta = \pi/6$, find the height $x$.

Click to view Answer & Solution

1. Domain: $[-1, 1]$, Range: $[-\pi/2, \pi/2]$.
2. $\theta = \sin^{-1}\left(\frac{5\sqrt{3}}{10}\right) = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \text{ radians}$ (or $60^\circ$).
3. $\frac{\pi}{6} = \sin^{-1}\left(\frac{x}{10}\right) \implies \frac{x}{10} = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \implies x = 5 \text{ meters}$.

Matrices

Case Study 3: Bakery Sales Analysis

A bakery sells two types of cakes: Chocolate (C) and Vanilla (V). On Day 1, they sold 30 C and 20 V. On Day 2, they sold 40 C and 10 V. The selling price of C is ₹100 and V is ₹80.

Questions:

1. Represent the sales quantity in a $2 \times 2$ matrix $A$ and prices in a matrix $B$.
2. Find the total revenue collected on each day using matrix multiplication.
3. Find the total sales revenue over both days.

Click to view Answer & Solution

1. Sales Matrix $A = \begin{bmatrix} 30 & 20 \\ 40 & 10 \end{bmatrix}$, Price Matrix $B = \begin{bmatrix} 100 \\ 80 \end{bmatrix}$.
2. $AB = \begin{bmatrix} 30 & 20 \\ 40 & 10 \end{bmatrix} \begin{bmatrix} 100 \\ 80 \end{bmatrix} = \begin{bmatrix} 30(100) + 20(80) \\ 40(100) + 10(80) \end{bmatrix} = \begin{bmatrix} 4600 \\ 4800 \end{bmatrix}$.
   Day 1 Revenue = ₹4,600, Day 2 Revenue = ₹4,800.
3. Total Revenue = ₹4,600 + ₹4,800 = ₹9,400.

Determinants

Case Study 4: Linear System of Equations

Three friends buy notebooks, pens, and erasers. Friend A buys 2 notebooks, 1 pen, 1 eraser for ₹35. Friend B buys 1 notebook, 3 pens, 2 erasers for ₹46. Friend C buys 3 notebooks, 2 pens, 1 eraser for ₹53.

Questions:

1. Express the system in matrix form $AX = B$.
2. Find $|A|$ and verify if the system has a unique solution.
3. Find the cost of 1 notebook using Cramer's rule or $X = A^{-1}B$.

Click to view Answer & Solution

1. $\begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 35 \\ 46 \\ 53 \end{bmatrix}$
2. $|A| = 2(3-4) - 1(1-6) + 1(2-9) = 2(-1) - 1(-5) + 1(-7) = -2 + 5 - 7 = -4 \neq 0$. Since $|A| \neq 0$, it has a unique solution.
3. Solving $AX = B$ yields $x = 12, y = 8, z = 3$. Cost of 1 notebook ($x$) = ₹12.

Continuity and Differentiability

Case Study 5: Roller Coaster Track Design

An engineer designs a smooth path for a roller coaster defined by: $$f(x) = \begin{cases} ax + 1, & x \le 2 \\ x^2 + b, & x > 2 \end{cases}$$

Questions:

1. Find a relation between $a$ and $b$ if $f(x)$ is continuous at $x = 2$.
2. If $f(x)$ is differentiable at $x = 2$, find the values of $a$ and $b$.

Click to view Answer & Solution

1. For continuity at $x=2$: $\lim_{x\to 2^-} f(x) = \lim_{x\to 2^+} f(x) \implies 2a + 1 = 4 + b \implies 2a - b = 3$.
2. For differentiability at $x=2$: Left derivative = Right derivative.
   $L f'(2) = a$, $R f'(2) = 2(2) = 4 \implies a = 4$.
   Substitute $a = 4$ into $2a - b = 3$: $2(4) - b = 3 \implies b = 5$.

Application of Derivatives

Case Study 6: Maximizing Box Volume

An open box is made from a square sheet of tin of side $24\text{ cm}$ by cutting equal squares of side $x\text{ cm}$ from each corner and turning up the sides.

Questions:

1. Express the volume $V$ of the box as a function of $x$.
2. Find critical values of $x$.
3. Find the value of $x$ for maximum volume, and calculate the maximum volume.

Click to view Answer & Solution

1. $V(x) = x(24 - 2x)^2 = 4x(12 - x)^2$.
2. $\frac{dV}{dx} = (24-2x)^2 + 2x(24-2x)(-2) = (24-2x)(24-6x)$. Setting $\frac{dV}{dx} = 0 \implies x = 12$ or $x = 4$. Since $x < 12$, $x = 4$.
3. Using second derivative test: $\frac{d^2V}{dx^2}\Big|_{x=4} < 0$, so $x = 4\text{ cm}$ gives max volume.
   Max Volume $V(4) = 4(24 - 8)^2 = 4(16)^2 = 1024\text{ cm}^3$.

Application of Derivatives

Case Study 7: Profit Maximization

The total cost $C(x)$ and revenue $R(x)$ in Rupees associated with the production and sale of $x$ units of an item are given by $C(x) = 2x^2 + 40x + 100$ and $R(x) = 120x - 3x^2$.

Questions:

1. Write the profit function $P(x)$.
2. How many items should be sold to maximize profit?
3. What is the maximum profit achievable?

Click to view Answer & Solution

1. $P(x) = R(x) - C(x) = (120x - 3x^2) - (2x^2 + 40x + 100) = -5x^2 + 80x - 100$.
2. $P'(x) = -10x + 80 = 0 \implies x = 8$. $P''(x) = -10 < 0$ (Maximum). Sold units = 8.
3. Maximum profit $P(8) = -5(8)^2 + 80(8) - 100 = -320 + 640 - 100 = \text{₹}220$.

Integrals

Case Study 8: Water Flow Rate

Water flows into a storage tank at a rate modeled by $r(t) = 3t^2 + 2t$ liters per minute, where $t$ is the time in minutes ($0 \le t \le 5$).

Questions:

1. Find the total volume of water that flows into the tank between $t = 0$ and $t = 5$ minutes.
2. Find the average rate of water flow during the first 5 minutes.

Click to view Answer & Solution

1. Total Volume $V = \int_{0}^{5} (3t^2 + 2t) dt = \left[ t^3 + t^2 \right]_0^5 = (125 + 25) - 0 = 150 \text{ Liters}$.
2. Average Rate = $\frac{\text{Total Volume}}{\text{Total Time}} = \frac{150}{5} = 30 \text{ Liters/min}$.

Application of Integrals

Case Study 9: Arch Bridge Design

An architect designs a parabolic bridge support modeled by the curve $y = 4 - x^2$ bounded above the ground $y = 0$.

Questions:

1. Find the intersection points of the curve with the x-axis ($y = 0$).
2. Calculate the area under the parabolic arch supported above the x-axis.

Click to view Answer & Solution

1. Set $y = 0 \implies 4 - x^2 = 0 \implies x = -2, 2$. Intersection points: $(-2,0)$ and $(2,0)$.
2. $\text{Area} = \int_{-2}^{2} (4 - x^2) dx = 2 \int_{0}^{2} (4 - x^2) dx = 2 \left[ 4x - \frac{x^3}{3} \right]_0^2 = 2 \left( 8 - \frac{8}{3} \right) = \frac{32}{3} \text{ sq. units}$.

Application of Integrals

Case Study 10: Circular Park and Pathway

A circular park is represented by $x^2 + y^2 = 16$. A straight walking line cuts across it along $y = x$ in the first quadrant.

Questions:

1. Find the point of intersection of the circle and the line in the first quadrant.
2. Find the area of the smaller region bounded by $x^2 + y^2 = 16$, the line $y = x$, and the x-axis in the first quadrant.

Click to view Answer & Solution

1. $x^2 + x^2 = 16 \implies 2x^2 = 16 \implies x = 2\sqrt{2}$. Point is $(2\sqrt{2}, 2\sqrt{2})$.
2. $\text{Area} = \int_0^{2\sqrt{2}} x dx + \int_{2\sqrt{2}}^4 \sqrt{16-x^2} dx = \left[\frac{x^2}{2}\right]_0^{2\sqrt{2}} + \left[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\frac{x}{4} \right]_{2\sqrt{2}}^4 = 4 + [8(\pi/2) - (4 + 8(\pi/4))] = 2\pi \text{ sq. units}$.

Differential Equations

Case Study 11: Bacteria Population Growth

The rate of growth of a bacteria population $P$ is proportional to the population present at any time $t$ (in hours), given by $\frac{dP}{dt} = kP$.

Questions:

1. Solve the differential equation to find $P(t)$ in terms of initial population $P_0$.
2. If the population doubles in 2 hours, find $k$.

Click to view Answer & Solution

1. $\frac{dP}{P} = k dt \implies \ln P = kt + C \implies P(t) = P_0 e^{kt}$.
2. At $t = 2$, $P(2) = 2P_0 \implies 2P_0 = P_0 e^{2k} \implies e^{2k} = 2 \implies k = \frac{\ln 2}{2}$.

Differential Equations

Case Study 12: Cooling Rate

Newton's Law of Cooling states that the rate of change of temperature $T$ of a liquid is proportional to $(T - T_s)$, where $T_s$ is the surrounding temperature ($20^\circ\text{C}$). $\frac{dT}{dt} = -k(T - 20)$.

Questions:

1. Find the general solution for $T(t)$.
2. Find $T(t)$ if $T(0) = 80^\circ\text{C}$.

Click to view Answer & Solution

1. $\frac{dT}{T - 20} = -k dt \implies \ln(T - 20) = -kt + C \implies T - 20 = C e^{-kt} \implies T(t) = 20 + C e^{-kt}$.
2. Using $T(0) = 80$: $80 = 20 + C e^0 \implies C = 60$. Thus, $T(t) = 20 + 60e^{-kt}$.

Vector Algebra

Case Study 13: Aircraft Velocity Vectors

Two drones A and B are moving in space. Drone A's position vector velocity is $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and Drone B's velocity is $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$.

Questions:

1. Calculate the dot product $\vec{a} \cdot \vec{b}$.
2. Find the angle $\theta$ between the flight directions of both drones.
3. Find a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$.

Click to view Answer & Solution

1. $\vec{a} \cdot \vec{b} = 3(1) + 2(-2) + (-1)(2) = 3 - 4 - 2 = -3$.
2. $|\vec{a}| = \sqrt{9+4+1} = \sqrt{14}$, $|\vec{b}| = \sqrt{1+4+4} = 3$.
   $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-3}{3\sqrt{14}} = -\frac{1}{\sqrt{14}} \implies \theta = \cos^{-1}\left(-\frac{1}{\sqrt{14}}\right)$.
3. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & -2 & 2 \end{vmatrix} = 2\hat{i} - 7\hat{j} - 8\hat{k}$.
   Unit Vector = $\frac{2\hat{i} - 7\hat{j} - 8\hat{k}}{\sqrt{4 + 49 + 64}} = \frac{2\hat{i} - 7\hat{j} - 8\hat{k}}{\sqrt{117}}$.

Three Dimensional Geometry

Case Study 14: Flight Paths of Jets

Two jet paths are defined by the lines: $$L_1: \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$$ $$L_2: \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$$

Questions:

1. Are the directional vectors of $L_1$ and $L_2$ parallel?
2. Find the shortest distance between the two lines $L_1$ and $L_2$.

Click to view Answer & Solution

1. Direction vectors are $\vec{b}_1 = \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}_2 = 2\hat{i}+\hat{j}+2\hat{k}$. Since ratios $1/2 \neq -1/1 \neq 1/2$, they are NOT parallel.
2. $\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}$.
   $\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 0\hat{j} + 3\hat{k}$. Magnitude = $\sqrt{9+0+9} = 3\sqrt{2}$.
   Shortest Distance = $\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{| -3 + 0 - 6 |}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} \text{ units}$.

Linear Programming

Case Study 15: Manufacturing Optimization

A factory produces items A and B. Item A requires 2 hours of Machine 1 and 1 hour of Machine 2. Item B requires 1 hour of Machine 1 and 3 hours of Machine 2. Machine 1 is available for at most 8 hours, and Machine 2 for 9 hours. Profit on A is ₹40, and on B is ₹50.

Questions:

1. Formulate the LPP objective function and constraints.
2. Find corner points of the feasible region and maximize profit.

Click to view Answer & Solution

1. Maximize $Z = 40x + 50y$ subject to:
   $2x + y \le 8$
   $x + 3y \le 9$
   $x, y \ge 0$.
2. Corner points of feasible region: $(0,0), (4,0), (3,2), (0,3)$.
   $Z(0,0) = 0$
   $Z(4,0) = 160$
   $Z(3,2) = 40(3) + 50(2) = 220$
   $Z(0,3) = 150$
   Maximum profit is ₹220 at $x=3, y=2$.

Probability

Case Study 16: Medical Test Diagnostic Accuracy

A doctor knows that 1% of the population has a rare condition. A diagnostic test is 95% accurate in detecting it when present, but yields a false positive result for 2% of healthy people.

Questions:

1. Define events and express given probabilities.
2. If a randomly chosen person tests positive, what is the probability that they actually have the disease? (Use Bayes' Theorem)

Click to view Answer & Solution

1. Let $E_1$: Person has disease ($P(E_1) = 0.01$), $E_2$: Person healthy ($P(E_2) = 0.99$).
   $A$: Test result positive. $P(A|E_1) = 0.95$, $P(A|E_2) = 0.02$.
2. $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac{0.01 \times 0.95}{(0.01 \times 0.95) + (0.99 \times 0.02)} = \frac{0.0095}{0.0095 + 0.0198} = \frac{95}{293} \approx 0.324$ (32.4%).

Probability

Case Study 17: Quality Control Inspection

A machine produces items, 10% of which are defective. A sample of 3 items is drawn at random with replacement. Let $X$ be the number of defective items.

Questions:

1. Find the probability distribution of $X$.
2. Calculate the mean/expected value of $X$.

Click to view Answer & Solution

1. $p = 0.1, q = 0.9, n = 3$.
   $P(X=0) = \binom{3}{0}(0.1)^0(0.9)^3 = 0.729$
   $P(X=1) = \binom{3}{1}(0.1)^1(0.9)^2 = 0.243$
   $P(X=2) = \binom{3}{2}(0.1)^2(0.9)^1 = 0.027$
   $P(X=3) = \binom{3}{3}(0.1)^3(0.9)^0 = 0.001$
2. $\text{Mean } E(X) = n \cdot p = 3 \times 0.1 = 0.3$.

Relations and Functions

Case Study 18: Bijective Mapping in Business

A software platform assigns employee IDs to email addresses. Let $A$ be the set of all employees and $B$ be the set of company emails. Function $f: A \to B$ maps each employee to their unique email address.

Questions:

1. Define an injective (one-one) function in this context. Is $f$ injective?
2. If every email in $B$ belongs to at least one employee in $A$, is $f$ surjective (onto)?
3. Is $f$ invertible? Explain.

Click to view Answer & Solution

1. Injective means distinct elements in $A$ have distinct images in $B$. Since no two employees share an email, $f$ is injective.
2. Yes, because Range of $f$ = Co-domain $B$.
3. Yes, because $f$ is both injective and surjective (bijective), making it invertible.

Integrals

Case Study 19: Marginal Cost and Total Cost

The marginal cost of producing $x$ units of a commodity is given by $MC = \frac{dC}{dx} = 3x^2 - 2x + 5$. The fixed cost of production is ₹100.

Questions:

1. Find the total cost function $C(x)$.
2. Calculate total cost when 5 units are produced.

Click to view Answer & Solution

1. $C(x) = \int (3x^2 - 2x + 5) dx = x^3 - x^2 + 5x + K$.
   Given fixed cost $C(0) = 100 \implies K = 100$.
   Thus, $C(x) = x^3 - x^2 + 5x + 100$.
2. $C(5) = (5)^3 - (5)^2 + 5(5) + 100 = 125 - 25 + 25 + 100 = \text{₹}225$.

Vector Algebra

Case Study 20: Triangle Area in Land Survey

A land surveyor marks three triangular plot corners at position vectors $\vec{A} = \hat{i} + \hat{j} + 2\hat{k}$, $\vec{B} = 2\hat{i} + 3\hat{j} + 5\hat{k}$, and $\vec{C} = 1\hat{i} + 5\hat{j} + 5\hat{k}$.

Questions:

1. Find vectors $\vec{AB}$ and $\vec{AC}$.
2. Find the cross product $\vec{AB} \times \vec{AC}$.
3. Calculate the area of the triangular plot.

Click to view Answer & Solution

1. $\vec{AB} = \vec{B} - \vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}$.
   $\vec{AC} = \vec{C} - \vec{A} = 0\hat{i} + 4\hat{j} + 3\hat{k}$.
2. $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = -6\hat{i} - 3\hat{j} + 4\hat{k}$.
3. $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{(-6)^2 + (-3)^2 + 4^2} = \frac{1}{2}\sqrt{36 + 9 + 16} = \frac{\sqrt{61}}{2} \text{ sq. units}$.