ISC Class 12 Mathematics Board Exam Question Paper 2024 with Solutions

CISCE Class 12 Mathematics Board Exam Question Paper 2024 with Solutions

SECTION A - 65 MARKS

Q1. In subparts (i) to (x) choose the correct options and in subparts (xi) to (xv), answer the questions as instructed. [15 Marks]

1.1. Let \( L \) be a set of all straight lines in a plane. The relation \( R \) on \( L \) defined as 'perpendicular to' is ______. [1 Mark]
(A) Symmetric and Transitive
(B) Transitive
(C) Symmetric
(D) Equivalence

Answer: (C) Symmetric

If line \( l_1 \) is perpendicular to line \( l_2 \), then line \( l_2 \) is perpendicular to line \( l_1 \). However, if \( l_1 \perp l_2 \) and \( l_2 \perp l_3 \), then \( l_1 \) and \( l_3 \) are parallel, not perpendicular.

Teacher's Note:
a) A relation is symmetric if \( (a, b) \in R \) implies \( (b, a) \in R \).
b) Students often confuse perpendicularity with parallelism; remember parallelism is an equivalence relation, whereas perpendicularity is only symmetric.

1.2. The order and degree of the differential equation \(\left[1 + \left(\frac{dy}{dx}\right)^2\right] = \frac{d^2y}{dx^2}\) are ______. [1 Mark]
(A) \(2, \frac{3}{2}\)
(B) \(2, 3\)
(C) \(2, 1\)
(D) \(3, 4\)

Answer: (C) \(2, 1\)

The highest-order derivative present in the equation is \( \frac{d^2y}{dx^2} \), so its order is 2. The power of this highest-order derivative is 1, so its degree is 1.

Teacher's Note:
a) Order is determined by the highest derivative present in the differential equation.
b) Degree is the power of the highest-order derivative, provided the differential equation is a polynomial equation in derivatives.

1.3. Let \( A \) be a non-empty set.
Statement 1: Identity relation on \( A \) is Reflexive.
Statement 2: Every Reflexive relation on \( A \) is an Identity relation. [1 Mark]
(A) Both the statements are true.
(B) Both the statements are false.
(C) Statement 1 is true and Statement 2 is false.
(D) Statement 1 is false and Statement 2 is true.

Answer: (C) Statement 1 is true and Statement 2 is false.

An identity relation contains only elements of the form \( (a, a) \) for all \( a \in A \), which makes it reflexive. However, a reflexive relation can contain additional ordered pairs like \( (a, b) \), so it need not be an identity relation.

Teacher's Note:
a) Every identity relation is reflexive, but the converse is not true.
b) Check counterexamples carefully when testing relation properties.

1.4. The graph of the function \( f \) is shown below. Of the following options, at what values of \( x \) is the function \( f \) NOT differentiable? [1 Mark]
(A) At \( x = 0 \) and \( x = 2 \)
(B) At \( x = 1 \) and \( x = 3 \)
(C) At \( x = -1 \) and \( x = 1 \)
(D) At \( x = -1.5 \) and \( x = 1.5 \)

[Figure: Graph of a curve with sharp corners or cusps at \( x = 0 \) and \( x = 2 \) on the coordinate axes with \( x \) ranging from \( -2 \) to \( 4 \).]

Answer: (A) At \( x = 0 \) and \( x = 2 \)

A function is not differentiable at points where the graph has sharp corners or cusps, which occur at \( x = 0 \) and \( x = 2 \).

Teacher's Note:
a) Sharp corners on a graph indicate points where left-hand derivative and right-hand derivative are not equal.
b) Continuity does not imply differentiability at corners.

1.5. The value of \(\operatorname{cosec}\left[\sin^{-1}\left(\frac{-1}{2}\right)\right] - \sec\left[\cos^{-1}\left(\frac{-1}{2}\right)\right]\) is equal to ______. [1 Mark]
(A) \(-4\)
(B) \(0\)
(C) \(-1\)
(D) \(4\)

Answer: (B) \(0\)

\(\operatorname{cosec}\left(-\frac{\pi}{6}\right) - \sec\left(\frac{2\pi}{3}\right) = -2 - (-2) = 0\).

Teacher's Note:
a) Use standard principal value branches for inverse trigonometric functions.
b) Be careful with negative arguments inside inverse secant and cosecant functions.

1.6. \(\int_{1}^{\sqrt{3}} \frac{1}{1 + x^2} \, dx\) is equal to ______. [1 Mark]
(A) \(\frac{\pi}{12}\)
(B) \(\frac{\pi}{6}\)
(C) \(\frac{\pi}{4}\)
(D) \(\frac{\pi}{3}\)

Answer: (A) \(\frac{\pi}{12}\)

\(\int_{1}^{\sqrt{3}} \frac{1}{1 + x^2} \, dx = [\tan^{-1}x]_{1}^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}\).

Teacher's Note:
a) Standard integral formula \(\int \frac{1}{1+x^2} \, dx = \tan^{-1}x + c\).
b) Ensure correct subtraction of lower limit from upper limit value.

1.7. Assertion: Let the matrices \( A = \begin{pmatrix} -3 & 2 \\ -5 & 4 \end{pmatrix} \) and \( B = \begin{pmatrix} 4 & -2 \\ 5 & -3 \end{pmatrix} \) be such that \( A^{100}B = BA^{100} \).
Reason: \( AB = BA \) implies \( A^nB = BA^n \) for all positive integers \( n \). [1 Mark]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Multiplying given matrices shows \( AB = BA \). By mathematical induction, if \( AB = BA \), then \( A^nB = BA^n \) for any positive integer \( n \).

Teacher's Note:
a) Commutative matrices satisfy power exchange properties.
b) Verify matrix multiplication order carefully.

1.8. If \(\int (\cot x - \csc^2 x) e^x \, dx = e^x f(x) + c\) then \(f(x)\) will be ______. [1 Mark]
(A) \(\cot x + \csc x\)
(B) \(\cot^2 x\)
(C) \(\cot x\)
(D) \(\csc x\)

Answer: (C) \(\cot x\)

Let \( g(x) = \cot x \), then \( g'(x) = -\csc^2 x \). Using the integral formula \(\int e^x [g(x) + g'(x)] \, dx = e^x g(x) + c\), we get \( f(x) = \cot x \).

Teacher's Note:
a) Recognize the standard form \(\int e^x [f(x) + f'(x)] \, dx\).
b) Derivative of \(\cot x\) is \(-\csc^2 x\).

1.9. In which one of the following intervals is the function \(f(x) = x^3 - 12x\) increasing? [1 Mark]
(A) \((-2, 2)\)
(B) \((-\infty, -2) \cup (2, \infty)\)
(C) \((-2, \infty)\)
(D) \((-\infty, 2)\)

Answer: (B) \((-\infty, -2) \cup (2, \infty)\)

\(f'(x) = 3x^2 - 12 = 3(x-2)(x+2)\). For increasing function, \( f'(x) \gt 0 \), which yields \( x \lt -2 \) or \( x \gt 2 \).

Teacher's Note:
a) A function is increasing where its first derivative is strictly positive.
b) Set up sign scheme for quadratic inequality correctly.

1.10. If \(A\) and \(B\) are symmetric matrices of the same order, then \(AB - BA\) is ______. [1 Mark]
(A) Skew - symmetric matrix
(B) Symmetric matrix
(C) Diagonal matrix
(D) Identity matrix

Answer: (A) Skew - symmetric matrix

\((AB - BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB - BA)\), which is skew-symmetric.

Teacher's Note:
a) Transpose property \((AB)' = B'A'\) is key here.
b) Difference of two symmetric matrices multiplied in reverse order is always skew-symmetric.

1.11. Find the derivative of \(y = \log x + \frac{1}{x}\) with respect to \(x\). [1 Mark]

Answer:
\(\frac{dy}{dx} = \frac{1}{x} - \frac{1}{x^2}\).

Teacher's Note:
a) Derivative of \(\log x\) is \(\frac{1}{x}\).
b) Derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\).

1.12. Teena is practising for an upcoming Rifle Shooting tournament. The probability of her shooting the target in the \(1^{\text{st}}\), \(2^{\text{nd}}\), \(3^{\text{rd}}\) and \(4^{\text{th}}\) shots are \(0.4\), \(0.3\), \(0.2\) and \(0.1\) respectively. Find the probability of at least one shot of Teena hitting the target. [1 Mark]

Answer:
\(0.6976\)

Teacher's Note:
a) Use the complement rule: \(P(\text{at least one}) = 1 - P(\text{none})\).
b) Multiply individual failure probabilities: \((1 - 0.4)(1 - 0.3)(1 - 0.2)(1 - 0.1)\).

1.13. Which one of the following graphs is a function of \(x\)? [1 Mark]
(A) Graph A
(B) Graph B

[Figure: Graph A representing a wave curve passing the vertical line test; Graph B representing a sideways curve failing the vertical line test.]

Answer: (A) Graph A

Graph A passes the vertical line test, meaning every vertical line intersects the graph at most once, which defines a function of \(x\).

Teacher's Note:
a) Vertical line test is used to determine if a graph represents \(y\) as a function of \(x\).
b) Graph B fails because a single \(x\) corresponds to multiple \(y\) values.

1.14. Evaluate: \(\int_{0}^{6} |x + 3| \, dx\) [1 Mark]

Answer:
\(36\)

Teacher's Note:
a) Since \(x + 3 \gt 0\) for all \(x \in [0, 6]\), the modulus can be removed directly.
b) Integrate \((x + 3)\) from \(0\) to \(6\) to get \(36\).

1.15. Given that \(\frac{1}{y} + \frac{1}{x} = \frac{1}{12}\) and \(y\) decreases at a rate of \(1 \text{ cm s}^{-1}\), find the rate of change of \(x\) when \(x = 5 \text{ cm}\) and \(y = 1 \text{ cm}\). [1 Mark]

Answer:
\(-25 \text{ cm/sec}\)

Teacher's Note:
a) Differentiate implicitly with respect to time \(t\): \(-\frac{1}{y^2}\frac{dy}{dt} - \frac{1}{x^2}\frac{dx}{dt} = 0\).
b) Substitute given values for \(x\), \(y\) and \(\frac{dy}{dt}\) to solve for \(\frac{dx}{dt}\).

Q2. [5 Marks]

2.1. Let \(f : R - \left\{\frac{-1}{3}\right\} \to R - \{0\}\) be defined as \(f(x) = \frac{5}{3x + 1}\) is invertible. Find \(f^{-1}(x)\). [5 Marks]

Answer:
\(f^{-1}(x) = \frac{5 - x}{3x}\)

Teacher's Note:
a) Put \(f(x) = y\) and express \(x\) in terms of \(y\).
b) Replace \(y\) with \(x\) to obtain the inverse function expression.

OR

2.2. If \(f : R \to R\) is defined by \(f(x) = \frac{2x - 7}{4}\), show that \(f(x)\) is one-one and onto. [5 Marks]

Answer:
Proved that \(f(x_1) = f(x_2) \implies x_1 = x_2\) (one-one) and for every \(y \in R\), there exists \(x = \frac{4y+7}{2} \in R\) such that \(f(x) = y\) (onto).

Teacher's Note:
a) To prove one-one, assume \(f(a) = f(b)\) and show \(a = b\).
b) To prove onto, express \(x\) in terms of \(y\) from \(y = f(x)\) and check domain membership.

Q3. Find the value of the determinant given below, without expanding it at any stage. [5 Marks]
\(\begin{vmatrix} \beta\gamma & 1 & \alpha(\beta + \gamma) \\ \gamma\alpha & 1 & \beta(\gamma + \alpha) \\ \alpha\beta & 1 & \gamma(\alpha + \beta) \end{vmatrix}\)

Answer:
\(0\)

Teacher's Note:
a) Expand column 3 as \(\alpha\beta + \beta\gamma + \gamma\alpha\) and add to column 1 or factor out common terms.
b) When two columns are identical after row or column operations, the determinant is zero.

Q4. [5 Marks]

4.1. Determine the value of \(k\) for which the following function is continuous at \(x = 3\):
\(f(x) = \begin{cases} \frac{(x + 3)^2 - 36}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases}\) [5 Marks]

Answer:
\(k = 12\)

Teacher's Note:
a) Evaluate \(\lim_{x \to 3} f(x)\) using algebraic simplification or L'Hopital's rule.
b) Equate the limit value to \(f(3)\) to find \(k\).

OR

4.2. Find a point on the curve \(y = (x - 2)^2\) at which the tangent is parallel to the chord joining the points \((2, 0)\) and \((4, 4)\). [5 Marks]

Answer:
\((3, 1)\)

Teacher's Note:
a) Slope of the chord joining \((2,0)\) and \((4,4)\) is \(\frac{4-0}{4-2} = 2\).
b) Equate the derivative \(\frac{dy}{dx} = 2(x-2)\) to 2 and solve for \(x\).

Q5. Evaluate: \(\int_{0}^{2\pi} \frac{1}{1 + e^{\sin x}} \, dx\) [5 Marks]

Answer:
\(\pi\)

Teacher's Note:
a) Use the definite integral property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\).
b) Add the original integral and transformed integral to simplify evaluation.

Q6. Evaluate \(P(A \cup B)\), if \(2P(A) = P(B) = \frac{5}{13}\) and \(P(A \mid B) = \frac{2}{5}\). [5 Marks]

Answer:
\(\frac{11}{26}\)

Teacher's Note:
a) Use conditional probability formula \(P(A \cap B) = P(A \mid B) \times P(B)\).
b) Apply addition theorem \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

Q7. If \(y = 3\cos(\log x) + 4\sin(\log x)\), show that \(x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0\). [5 Marks]

Answer:
Proved successfully by differentiating twice and substituting into the given differential expression.

Teacher's Note:
a) Find first derivative using chain rule and multiply by \(x\).
b) Differentiate again with respect to \(x\) to form the second derivative equation.

Q8. [5 Marks]

8.1. Solve for \(x\): \(\sin^{-1}\left(\frac{x}{2}\right) + \cos^{-1}x = \frac{\pi}{6}\) [5 Marks]

Answer:
\(x = 1\)

Teacher's Note:
a) Use identity \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) to simplify.
b) Check extraneous roots in inverse trigonometric equations.

OR

8.2. If \(\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi\), show that \(x^2 - y^2 - z^2 + 2yz\sqrt{1 - x^2} = 0\). [5 Marks]

Answer:
Proved successfully by rearranging and squaring terms.

Teacher's Note:
a) Express one inverse sine term on the right side as \(\pi - \sin^{-1}z\).
b) Apply sine and cosine addition formulas after taking sine on both sides.

Q9. [5 Marks]

9.1. Evaluate: \(\int x^2 \cos x \, dx\) [5 Marks]

Answer:
\((x^2 - 2)\sin x + 2x\cos x + c\)

Teacher's Note:
a) Apply integration by parts twice consecutively.
b) Keep track of signs when integrating trigonometric terms.

OR

9.2. Evaluate: \(\int \frac{x + 7}{x^2 + 4x + 7} \, dx\) [5 Marks]

Answer:
\(\frac{1}{2}\log|x^2 + 4x + 7| + \frac{5}{\sqrt{3}}\tan^{-1}\left(\frac{x + 2}{\sqrt{3}}\right) + c\)

Teacher's Note:
a) Express numerator as derivative of denominator plus a constant: \(x + 7 = A(2x + 4) + B\).
b) Split into two integrals: one logarithmic and one standard inverse tangent integral.

Q10. A jewellery seller has precious gems in white and red colour which he has put in three boxes. The distribution of these gems is shown in the table given below: [5 Marks]

He wants to gift two gems to his mother. So, he asks her to select one box at random and pick out any two gems one after the other without replacement from the selected box. The mother selects one white and one red gem. Calculate the probability that the gems drawn are from Box II. [5 Marks]

Answer:
\(\frac{18}{53}\)

Teacher's Note:
a) Use Bayes' Theorem with hypotheses corresponding to selecting Box I, Box II, or Box III.
b) Calculate conditional probabilities of drawing one white and one red gem for each box using combinations.

Q11. A furniture factory uses three types of wood namely, teakwood, rosewood and satinwood for manufacturing three types of furniture, that are, table, chair and cot. The wood requirements (in tonnes) for each type of furniture are given below: [6 Marks]

It is found that 29 tonnes of teakwood, 13 tonnes of rosewood and 16 tonnes of satinwood are available to make all three types of furniture.
Using the above information, answer the following questions:
i. Express the data given in the table above in the form of a set of simultaneous equations.
ii. Solve the set of simultaneous equations formed in subpart (i) by matrix method.
iii. Hence, find the number of table(s), chair(s) and cot(s) produced. [6 Marks]

Answer:
i. \(2x + 3y + 4z = 29\), \(x + y + 2z = 13\), \(3x + 2y + z = 16\)
ii. \(X = A^{-1}B = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}\)
iii. Tables = 2, Chairs = 3, Cots = 4

Teacher's Note:
a) Formulate the matrix equation \(AX = B\) correctly from the word problem.
b) Calculate determinant, cofactor matrix, and adjugate carefully to find \(A^{-1}\).

Q12. [6 Marks]

12.1. Mrs. Roy designs a window in her son's study room so that the room gets maximum sunlight. She designs the window in the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is \(12 \text{ m}\), find the dimensions of the window that will admit maximum sunlight into the room. [6 Marks]

[Figure: Sketch of a rectangular window surmounted by an equilateral triangle with dimensions labeled.]

Answer:
Width \(x = \frac{12}{6 - \sqrt{3}} \text{ m}\) and height \(y = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \text{ m}\)

Teacher's Note:
a) Express total area as a function of single variable \(x\) using the perimeter constraint.
b) Apply first and second derivative tests to confirm maximum area.

OR

12.2. Sumit has bought a closed cylindrical dustbin. The radius of the dustbin is \(r \text{ cm}\) and height is \(h \text{ cm}\). It has a volume of \(20\pi \text{ cm}^3\).
a. Express \(h\) in terms of \(r\), using the given volume.
b. Prove that the total surface area of the dustbin is \(2\pi r^2 + \frac{40\pi}{r}\).
c. Sumit wants to paint the dustbin. The cost of painting the base and top of the dustbin is Rs. \(2 \text{ per cm}^2\) and the cost of painting the curved side is Rs. \(25 \text{ per cm}^2\). Find the total cost in terms of \(r\), for painting the outer surface of the dustbin including the base and top.
d. Calculate the minimum cost for painting the dustbin. [6 Marks]

[Figure: Sketch of a closed cylinder with radius \(r\) and height \(h\).]

Answer:
a. \(h = \frac{20}{r^2}\)
b. Proved total surface area formula.
c. Total cost \(C = \frac{1000\pi}{r} + 4\pi r^2\)
d. Minimum cost = Rs. \(240\pi\)

Teacher's Note:
a) Substitute volume relation into surface area and cost formulas.
b) Use derivative test to find critical radius where cost is minimized.

Q13. [6 Marks]

13.1. Find the particular solution of the differential equation: \(2y e^{x/y} dx + (y - 2x e^{x/y}) dy = 0\) given that \(x = 0\) when \(y = 1\). [6 Marks]

Answer:
\(y = e^2 - 2xe^{x/y}\) (or \(y = e^{2 - 2e^{x/y}}\) depending on rearrangement)

Teacher's Note:
a) Recognize the equation as homogeneous and substitute \(x = vy\).
b) Separate variables and integrate to find general solution, then apply initial conditions to find constant \(C\).

OR

13.2. For the following differential equation, find a particular solution satisfying the given condition: \(x(x^2 - 1)\frac{dy}{dx} = 1\), \(y = 0\) when \(x = 2\). [6 Marks]

Answer:
\(y = \frac{1}{2}\log\left(\frac{x^2 - 1}{x^2}\right) - \frac{1}{2}\log\left(\frac{3}{4}\right)\)

Teacher's Note:
a) Separate variables and use partial fractions to integrate the rational function.
b) Substitute \(x = 2, y = 0\) to determine the constant of integration \(C\).

Q14. A primary school teacher wants to teach the concept of 'larger number' to the students of Class II. To teach this concept, he conducts an activity in his class. He asks the children to select two numbers from a set of numbers given as \(2, 3, 4, 5\) one after the other without replacement. All the outcomes of this activity are tabulated in the form of ordered pairs given below: [6 Marks]

i. Complete the table given above.
ii. Find the total number of ordered pairs having one larger number.
iii. Let the random variable \(X\) denote the larger of two numbers in the ordered pair. Now, complete the probability distribution table for \(X\) given below.
X | 3 | 4 | 5
P(X = x) | | |
i. Find the value of \(P(X \lt 5)\)
ii. Calculate the expected value of the probability distribution. [6 Marks]

Answer:
i. Table completed with ordered pairs without replacement.
ii. 12 ordered pairs.
iii. Probabilities: \(P(X=3) = \frac{2}{12}\), \(P(X=4) = \frac{4}{12}\), \(P(X=5) = \frac{6}{12}\).
iv. \(P(X \lt 5) = \frac{1}{2}\)
v. Expected value \(E(X) = \frac{13}{3}\)

Teacher's Note:
a) Without replacement means pairs like \((2,2)\) might be excluded or counted based on the exact sample space specified; follow standard conditional counting.
b) Calculate expectation using formula \(\sum X_i P(X = X_i)\).

SECTION B - 15 MARKS

Q15. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as instructed. [5 Marks]

15.1. If \(\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}\) and \(\vec{b} = 2\hat{i} - 4\hat{j} - 3\hat{k}\) then the value of \(|\vec{a} - 2\vec{b}|\) will be ______. [1 Mark]
(A) \(\sqrt{85}\)
(B) \(\sqrt{86}\)
(C) \(\sqrt{87}\)
(D) \(\sqrt{88}\)

Answer: (B) \(\sqrt{86}\)

\(\vec{a} - 2\vec{b} = (3 - 4)\hat{i} + (-2 + 8)\hat{j} + (1 + 6)\hat{k} = -\hat{i} + 6\hat{j} + 7\hat{k}\). Magnitude is \(\sqrt{(-1)^2 + 6^2 + 7^2} = \sqrt{1 + 36 + 49} = \sqrt{86}\).

Teacher's Note:
a) Multiply vector \(\vec{b}\) by scalar 2 before subtraction.
b) Calculate vector magnitude using square root of sum of squares of components.

15.2. If a line makes an angle \(\alpha\), \(\beta\) and \(\gamma\) with positive direction of the coordinate axes, then the value of \(\sin^2\alpha + \sin^2\beta + \sin^2\gamma\) will be ______. [1 Mark]
(A) \(1$
(B) \(3$
(C) \(-2$
(D) \(2$

Answer: (D) \(2\)

Since \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\), substituting \(\cos^2\theta = 1 - \sin^2\theta\) gives \(3 - (\sin^2\alpha + \sin^2\beta + \sin^2\gamma) = 1\), hence the sum equals 2.

Teacher's Note:
a) Fundamental direction cosine identity is \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\).
b) Convert cosines to sines using trigonometric identities.

15.3. In the figure given below, if the coordinates of the point P are \((a, b, c)\), then what are the perpendicular distances of P from XY, YZ and ZX planes respectively? [1 Mark]

[Figure: 3D rectangular box with point P\((a,b,c)\) plotted in octant with coordinate axes.]

Answer:
Perpendicular distance from XY-plane = \(c\), from YZ-plane = \(a\), from ZX-plane = \(b\).

Teacher's Note:
a) Distance of point \((a,b,c)\) from XY-plane is the absolute value of its z-coordinate.
b) Similarly, distances from YZ and ZX planes are \(|a|\) and \(|b|\) respectively.

15.4. If \(\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{b} = 5\hat{i} - 3\hat{j} + \hat{k}\), find the projection of \(\vec{b}\) on \(\vec{a}\). [1 Mark]

Answer:
\(3\)

Teacher's Note:
a) Projection formula is \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}\).
b) Dot product is \(10 - 3 + 2 = 9\), and magnitude of \(\vec{a}\) is 3.

15.5. Find a vector of magnitude 20 units parallel to the vector \(2\hat{i} + 5\hat{j} + 4\hat{k}\). [1 Mark]

Answer:
\(\pm \frac{40}{\sqrt{45}}\hat{i} \pm \frac{100}{\sqrt{45}}\hat{j} \pm \frac{80}{\sqrt{45}}\hat{k}\) (or simplified as \(\pm \frac{8\sqrt{5}}{3}\hat{i} \pm \frac{20\sqrt{5}}{3}\hat{j} \pm \frac{16\sqrt{5}}{3}\hat{k}\))

Teacher's Note:
a) Find unit vector in the direction of the given vector.
b) Multiply the unit vector by \(\pm 20\) to get the required parallel vector.

Q16. [4 Marks]

16.1. If \(\vec{a} \times \vec{b} = \vec{a} \times \vec{c}\) where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are non-zero vectors, then prove that either \(\vec{b} = \vec{c}\) or \(\vec{a}\) and \((\vec{b} - \vec{c})\) are parallel. [4 Marks]

Answer:
Proved by rearranging cross product as \(\vec{a} \times (\vec{b} - \vec{c}) = 0\).

Teacher's Note:
a) Cross product of two vectors is zero when they are collinear or parallel.
b) Distributive property of cross product over vector subtraction is utilized.

OR

16.2. If \(\vec{a}\) and \(\vec{b}\) are two non-zero vectors such that \(|\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b}\), find the angle between \(\vec{a}\) and \(\vec{b}\). [4 Marks]

Answer:
\(45^{\circ}\) (or \(\frac{\pi}{4}\))

Teacher's Note:
a) Substitute definitions: \(|\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta\).
b) This leads to \(\tan\theta = 1\), giving \(\theta = 45^{\circ}\).

Q17. A mobile tower is situated at the top of a hill. Consider the surface on which the tower stands as a plane having points \(A(1, 0, 2)\), \(B(3, -1, 1)\) and \(C(1, 2, 1)\) on it. The mobile tower is tied with three cables from the points A, B and C such that it stands vertically on the ground. The top of the tower is at point \(P(2, 3, 1)\) as shown in the figure below. The foot of the perpendicular from the point P on the plane is at the point \(Q\left(\frac{43}{29}, \frac{77}{29}, \frac{9}{29}\right)\).
Answer the following questions:
i. Find the equation of the plane containing the points A, B and C.
ii. Find the equation of the line PQ.
iii. Calculate the height of the tower. [6 Marks]

[Figure: 3D diagram showing plane ABC and tower point P with perpendicular foot Q.]

Answer:
i. \(3x + 2y + 4z = 11$
ii. Vector equation \(\vec{r} = (2\hat{i} + 3\hat{j} + \hat{k}) + \lambda\left(\left(\frac{43}{29}-2\right)\hat{i} + \left(\frac{77}{29}-3\right)\hat{j} + \left(\frac{9}{29}-1\right)\hat{k}\right)\)
iii. Height = \(\frac{5}{\sqrt{29}}\)

Teacher's Note:
a) Equation of plane through three points is found using determinant expansion.
b) Height of the tower is the perpendicular distance from point P to the plane ABC.

Q18. [5 Marks]

18.1. Using integration, find the area bounded by the curve \(y^2 = 4ax\) and the line \(x = a\). [5 Marks]

[Figure: Parabola symmetric about x-axis bounded by line \(x = a\).]

Answer:
\(\frac{8}{3}a^2\) sq. units

Teacher's Note:
a) Use symmetry about the x-axis: Area = \(2 \int_{0}^{a} y \, dx\).
b) Substitute \(y = \sqrt{4ax}\) and integrate with respect to \(x\).

OR

18.2. Using integration, find the area of the region bounded by the curve \(y^2 = 4x\) and \(x^2 = 4y\). [5 Marks]

[Figure: Intersection of two parabolas \(y^2 = 4x\) and \(x^2 = 4y\) in the first quadrant.]

Answer:
\(\frac{16}{3}\) sq. units

Teacher's Note:
a) Solve simultaneous equations to find intersection points \((0,0)\) and \((4,4)\).
b) Integrate the difference of the upper curve and lower curve between limits \(0\) and \(4\).

SECTION C - 15 MARKS

Q19. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as instructed. [5 Marks]

19.1. A company sells hand towels at Rs. \(100\) per unit. The fixed cost for the company to manufacture hand towels is Rs. \(35,000\) and variable cost is estimated to be \(30\%\) of total revenue. What will be the total cost function for manufacturing hand towels? [1 Mark]
(A) \(35000 + 3x$
(B) \(35000 + 30x$
(C) \(35000 + 100x$
(D) \(35000 + 10x\)

Answer: (B) \(35000 + 30x\)

Total revenue for \(x\) units is \(100x\). Variable cost is \(30\%\) of \(100x = 30x\). Total cost = Fixed cost + Variable cost = \(35000 + 30x\).

Teacher's Note:
a) Total cost function is sum of fixed cost and variable cost.
b) Variable cost depends directly on the number of units produced.

19.2. If the correlation coefficient of two sets of variables \((X, Y)\) is \(\frac{-3}{4}\), which one of the following statements is true for the same set of variables? [1 Mark]
(A) Only one of the two regression lines has a negative coefficient.
(B) Both regression coefficients are positive.
(C) Both regression coefficients are negative.
(D) One of the lines of regression is parallel to the x-axis.

Answer: (C) Both regression coefficients are negative.

The correlation coefficient \(r\) and both regression coefficients \(b_{xy}\) and \(b_{yx}\) always share the same algebraic sign.

Teacher's Note:
a) Correlation coefficient is the geometric mean of the two regression coefficients.
b) Negative correlation implies both regression lines slope downwards, making both regression coefficients negative.

19.3. If the total cost function is given by \(C = x + 2x^3 - \frac{7}{2}x^2\), find the Marginal Average Cost function (MAC). [1 Mark]

Answer:
\(4x - \frac{7}{2}\)

Teacher's Note:
a) Average Cost \(AC = \frac{C(x)}{x}\).
b) Marginal Average Cost is the derivative of Average Cost with respect to \(x\).

19.4. The equations of two lines of regression are \(4x + 3y + 7 = 0\) and \(3x + 4y + 8 = 0\). Find the mean value of \(x\) and \(y\). [1 Mark]

Answer:
\(\bar{x} = -\frac{4}{7}\), \(\bar{y} = -\frac{11}{7}\)

Teacher's Note:
a) The point of intersection of two regression lines gives the mean values \(\bar{x}\) and \(\bar{y}\).
b) Solve the simultaneous linear equations in \(x\) and \(y\).

19.5. The manufacturer of a pen fixes its selling price at Rs. \(45\) and the cost function is \(C(x) = 30x + 240\). The manufacturer will begin to earn profit if he sells more than \(16\) pens. Why? Give one reason. [1 Mark]

Answer:
Because the breakeven point occurs at \(x = 16\) pens where revenue equals total cost, and selling more than 16 pens results in total revenue exceeding total cost.

Teacher's Note:
a) Breakeven quantity is found by setting Revenue equal to Cost (\(45x = 30x + 240\)).
b) Solving gives \(x = 16\), beyond which profit is positive.

Q20. [4 Marks]

20.1. The Average Cost function associated with producing and marketing \(x\) units of an item is given by \(AC = x + 5 + \frac{36}{x}\).
a. Find the Total cost function.
b. Find the range of values of \(x\) for which Average Cost is increasing. [4 Marks]

Answer:
a. \(T.C.(x) = x^2 + 5x + 36$
b. \(x \gt 6\)

Teacher's Note:
a) Total cost is product of Average Cost and quantity \(x\).
b) Average cost is increasing where its derivative with respect to \(x\) is positive.

OR

20.2. A monopolist's demand function is \(x = 60 - \frac{p}{5}\). At what level of output will marginal revenue be zero? [4 Marks]

Answer:
\(x = 30\)

Teacher's Note:
a) Express price \(p\) in terms of output \(x\) and form Total Revenue function \(R(x) = px\).
b) Differentiate to find Marginal Revenue (MR) and set it to zero to solve for \(x\).

Q21. [4 Marks]

21.1. A monopolist's demand function is \(x = 60 - \frac{p}{5}\). At what level of output will marginal revenue be zero? [4 Marks]

Answer:
\(x = 30\)

Teacher's Note:
a) Same as Q20.2; derive revenue function from inverse demand equation.
b) Setting derivative of revenue to zero yields the output level.

OR

21.2. For 50 students of a class, the regression equation of marks in statistics \((X)\) on the marks in accountancy \((Y)\) is \(3y - 5x + 180 = 0\). The mean marks in accountancy is \(44\) and the variance of marks in statistics is \(\left(\frac{9}{16}\right)^{\text{th}}\) of the variance of marks in accountancy. Find the mean marks in statistics and the correlation coefficient between marks in the two subjects. [4 Marks]

Answer:
Mean marks in statistics \(\bar{x} = 62.4\), Correlation coefficient \(r = 0.8\)

Teacher's Note:
a) Substitute mean accountancy marks \(\bar{y} = 44\) into regression equation of \(x\) on \(y\) to find \(\bar{x}\).
b) Use regression coefficient \(b_{xy} = r \frac{\sigma_x}{\sigma_y}\) to find correlation coefficient \(r\).

Q22. Aman has Rs. \(1500\) to purchase rice and wheat for his grocery shop. Each sack of rice and wheat costs Rs. \(180\) and Rs. \(120\) respectively. He can store a maximum number of \(10\) bags in his shop. He will earn a profit of Rs. \(11\) per bag of rice and Rs. \(9\) per bag of wheat.
i. Formulate a Linear Programming Problem to maximise Aman's profit.
ii. Calculate the maximum profit. [6 Marks]

Answer:
i. Maximize \(P = 11x + 9y\) subject to \(180x + 120y \le 1500\) (\(3x + 2y \le 25\)), \(x + y \le 10\), \(x \ge 0, y \ge 0\).
ii. Maximum profit = Rs. \(100\) at \((5, 5)\).

Teacher's Note:
a) Define decision variables \(x\) and \(y\) for rice and wheat sacks clearly.
b) Evaluate objective function at all corner points of the feasible region to determine maximum profit.