CBSE Class 12 Maths Question Paper 2025 Solved Code 65-1-3

SECTION - A

This section comprises of 20 Multiple Choice Questions (MCQs) of 1 mark each.

1. If the feasible region of a linear programming problem with objective function Z = a\(x\) + by, is bounded, then which of the following is correct ? [1 Mark]
(A) It will only have a maximum value.
(B) It will only have a minimum value.
(C) It will have both maximum and minimum values.
(D) It will have neither maximum nor minimum value.

Answer: (C) It will have both maximum and minimum values.

Teacher's Note:
a) A bounded feasible region always guarantees existence of both maximum and minimum values of the objective function.
b) Remember this differs from an unbounded region, where a maximum or minimum may not exist.

2. The unit vector perpendicular to the vectors \( \hat{i} - \hat{j} \) and \( \hat{i} + \hat{j} \) is [1 Mark]
(A) \( \hat{k} \)
(B) \( -\hat{k} + \hat{j} \)
(C) \( \dfrac{\hat{i} - \hat{j}}{\sqrt{2}} \)
(D) \( \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \)

Answer: (A) \( \hat{k} \)

Teacher's Note:
a) Use the cross product of the two given vectors to get a perpendicular vector, then normalise it.
b) \( (\hat{i}-\hat{j}) \times (\hat{i}+\hat{j}) = 2\hat{k} \), so the unit vector is \( \hat{k} \).

3. If \( \displaystyle\int_{0}^{1} \dfrac{e^{x}}{1+x}\, dx = \alpha \), then \( \displaystyle\int_{0}^{1} \dfrac{e^{x}}{(1+x)^{2}}\, dx \) is equal to [1 Mark]
(A) \( \alpha - 1 + \dfrac{e}{2} \)
(B) \( \alpha + 1 - \dfrac{e}{2} \)
(C) \( \alpha - 1 - \dfrac{e}{2} \)
(D) \( \alpha + 1 + \dfrac{e}{2} \)

Answer: (B) \( \alpha + 1 - \dfrac{e}{2} \)

Teacher's Note:
a) Split \( \dfrac{1}{(1+x)^2} = \dfrac{1}{1+x} - \dfrac{x}{(1+x)^2} \) and integrate by parts.
b) Carefully apply the limits 0 to 1 to relate the new integral to \( \alpha \).

4. If \( \displaystyle\int \dfrac{2^{\frac{1}{x}}}{x^{2}}\, dx = k \cdot 2^{\frac{1}{x}} + C \), then k is equal to [1 Mark]
(A) \( \dfrac{-1}{\log 2} \)
(B) \( -\log 2 \)
(C) \( -1 \)
(D) \( \dfrac{1}{2} \)

Answer: (A) \( \dfrac{-1}{\log 2} \)

Teacher's Note:
a) Substitute \( t = \dfrac{1}{x} \) so that \( dt = -\dfrac{1}{x^2}dx \).
b) The integral becomes \( -\int 2^t\, dt = -\dfrac{2^t}{\log 2} \), giving k.

5. If \( A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \), then \( A^{-1} \) is [1 Mark]
(A) \( \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \)
(B) \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \)
(C) \( \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
(D) \( \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Answer: (D) \( \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

Teacher's Note:
a) A is a diagonal matrix, so its inverse is a diagonal matrix with reciprocals of the diagonal entries.
b) Since A is its own inverse here, verify by checking \( AA = I \).

6. If \( \begin{bmatrix} x+y & 3y \\ 3x & x+3 \end{bmatrix} = \begin{bmatrix} 9 & 4x+y \\ x+6 & y \end{bmatrix} \), then \( (x-y) \) = ? [1 Mark]
(A) \( -7 \)
(B) \( -3 \)
(C) \( 3 \)
(D) \( 7 \)

Answer: (B) \( -3 \)

Teacher's Note:
a) Equate corresponding elements to form equations in \(x\) and y.
b) Solve simultaneously; use \(x+3=y\) and \(3x=x+6\) type relations to get \(x-y\).

7. Let M and N be two events such that P(M) = 0.6, P(N) = 0.2 and \( P(M \cap N) = 0.5 \), then \( P(M'/N') \) is [1 Mark]
(A) \( \dfrac{7}{8} \)
(B) \( \dfrac{2}{5} \)
(C) \( \dfrac{1}{2} \)
(D) \( \dfrac{2}{3} \)

Answer: (A) \( \dfrac{7}{8} \)

Teacher's Note:
a) Use \( P(M'/N') = \dfrac{P(M' \cap N')}{P(N')} = \dfrac{1-P(M \cup N)}{1-P(N)} \).
b) First find \( P(M \cup N) = P(M)+P(N)-P(M \cap N) \).

8. Which of the following is not a homogeneous function of \(x\) and y ? [1 Mark]
(A) \( y^2 - xy \)
(B) \( x - 3y \)
(C) \( \sin^2\dfrac{y}{x} + \dfrac{y}{x} \)
(D) \( \tan x - \sec y \)

Answer: (D) \( \tan x - \sec y \)

Teacher's Note:
a) A homogeneous function must satisfy \( f(\lambda x, \lambda y) = \lambda^n f(x,y) \) for some n.
b) Option (D) cannot be written in this form since tan and sec do not scale homogeneously.

9. If \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), \( |\vec{a}| = \sqrt{37} \), \( |\vec{b}| = 3 \) and \( |\vec{c}| = 4 \), then angle between \( \vec{b} \) and \( \vec{c} \) is : [1 Mark]
(A) \( \dfrac{\pi}{6} \)
(B) \( \dfrac{\pi}{4} \)
(C) \( \dfrac{\pi}{3} \)
(D) \( \dfrac{\pi}{2} \)

Answer: (C) \( \dfrac{\pi}{3} \)

Teacher's Note:
a) Since \( \vec{a} = -(\vec{b}+\vec{c}) \), square both sides: \( |\vec{a}|^2 = |\vec{b}|^2+|\vec{c}|^2+2\vec{b}\cdot\vec{c} \).
b) Substitute values to find \( \cos\theta \) between b and c.

10. If f(\(x\)) = \( |x| + |x-1| \), then which of the following is correct ? [1 Mark]
(A) f(\(x\)) is both continuous and differentiable, at \(x\) = 0 and \(x\) = 1.
(B) f(\(x\)) is differentiable but not continuous, at \(x\) = 0 and \(x\) = 1.
(C) f(\(x\)) is continuous but not differentiable, at \(x\) = 0 and \(x\) = 1.
(D) f(\(x\)) is neither continuous nor differentiable, at \(x\) = 0 and \(x\) = 1.

Answer: (C) f(\(x\)) is continuous but not differentiable, at \(x\) = 0 and \(x\) = 1.

Teacher's Note:
a) Modulus functions are always continuous everywhere, being a sum of continuous functions.
b) They fail to be differentiable exactly at the points where the expression inside modulus is zero, here \(x\) = 0 and \(x\) = 1.

11. A system of linear equations is represented as AX = B, where A is coefficient matrix, X is variable matrix and B is the constant matrix. Then the system of equations is [1 Mark]
(A) Consistent, if \( |A| \neq 0 \), solution is given by \( X = BA^{-1} \).
(B) Inconsistent if \( |A| = 0 \) and (adj A) B = 0
(C) Inconsistent if \( |A| \neq 0 \)
(D) May or may not be consistent if \( |A| = 0 \) and (adj A) B = 0

Answer: (D) May or may not be consistent if \( |A| = 0 \) and (adj A) B = 0

Teacher's Note:
a) When \( |A|=0 \) and (adj A)B = 0, the system may have infinitely many solutions or no solution, so it is not certain.
b) Remember the correct solution formula when \( |A| \neq 0 \) is \( X = A^{-1}B \), not \( BA^{-1} \).

12. The absolute maximum value of function f(\(x\)) = \( x^3 - 3x + 2 \) in [0, 2] is : [1 Mark]
(A) 0
(B) 2
(C) 4
(D) 5

Answer: (C) 4

Teacher's Note:
a) Find f'(\(x\)) = 3\(x^2\)-3, critical point at \(x\)=1 inside [0,2].
b) Compare f(0), f(1), f(2) to find the absolute maximum: f(0)=2, f(1)=0, f(2)=4.

13. The order and degree of differential function \( \left[1+\left(\dfrac{dy}{dx}\right)^2\right]^5 = \dfrac{d^2y}{dx^2} \) are : [1 Mark]
(A) order 1, degree 1
(B) order 1, degree 2
(C) order 2, degree 1
(D) order 2, degree 2

Answer: (C) order 2, degree 1

Teacher's Note:
a) Order is the highest derivative present, here \( d^2y/dx^2 \), so order = 2.
b) Degree is the power of the highest order derivative after removing radicals/fractions; here it appears to power 1.

14. The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse ? [1 Mark]

[Figure: The given graph shows a hump-shaped curve rising to a peak between x = -\(\pi/2\) and x = \(\pi/2\), then falling below the x-axis as it approaches x = \(\pi\), with marked points at x = -\(\pi/2\), \(\pi/2\) and \(\pi\) on the x-axis. Option (A): an increasing curve through the origin, rising from about -1 at x = -\(\pi/2\) to 1 at x = \(\pi/2\). Option (B): a similar increasing S-shaped curve, flatter near the origin, from -1 to 1 between x = -\(\pi/2\) and \(\pi/2\). Option (C): a decreasing curve starting near (-1, \(\pi\)) marked with a dashed line at height \(\pi\), passing through (0, \(\pi/2\)), and ending at (1, 0). Option (D): a decreasing curve starting near (0, \(\pi\)) with a dashed line at height \(\pi\), passing through (\(\pi/2\), 0) region, flattening towards y = 0.]

Answer: (C) the decreasing curve from (-1, \(\pi\)) through (0, \(\pi/2\)) to (1, 0), representing the inverse cosine type graph.

Teacher's Note:
a) The inverse of a function is obtained by reflecting its graph about the line y = \(x\).
b) Since the original graph resembles the cosine curve restricted suitably, its inverse resembles the arccosine graph, which is decreasing.

15. The corner points of the feasible region in graphical representation of a L.P.P. are (2, 72), (15, 20) and (40, 15). If Z = 18\(x\) + 9y be the objective function, then [1 Mark]
(A) Z is maximum at (2, 72), minimum at (15, 20)
(B) Z is maximum at (15, 20) minimum at (40, 15)
(C) Z is maximum at (40, 15), minimum at (15, 20)
(D) Z is maximum at (40, 15), minimum at (2, 72)

Answer: (C) Z is maximum at (40, 15), minimum at (15, 20)

Teacher's Note:
a) Evaluate Z = 18\(x\)+9y at each corner point and compare.
b) Z(2,72)=684, Z(15,20)=450, Z(40,15)=855; the largest and smallest give the answer.

16. Let \( A = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} -2 \\ -5 \\ -7 \end{bmatrix} \), C = [9 8 7], which of the following is defined ? [1 Mark]
(A) Only AB
(B) Only AC
(C) Only BA
(D) All AB, AC and BA

Answer: (A) Only AB

Teacher's Note:
a) Matrix multiplication is possible only when the number of columns of the first matrix equals the number of rows of the second.
b) Check the orders: A is 3x3, B is 3x1, C is 1x3; only AB (3x3)(3x1) is valid.

17. If A and B are invertible matrices, then which of the following is not correct ? [1 Mark]
(A) \( (A+B)^{-1} = B^{-1}+A^{-1} \)
(B) \( (AB)^{-1} = B^{-1}A^{-1} \)
(C) adj (A) = \( |A| A^{-1} \)
(D) \( |A|^{-1} = |A^{-1}| \)

Answer: (A) \( (A+B)^{-1} = B^{-1}+A^{-1} \)

Teacher's Note:
a) There is no general formula for the inverse of a sum of matrices; this rule does not hold in general.
b) The other three are standard matrix identities and are always true for invertible matrices.

18. The area of the shaded region bounded by the curves \( y^2 = x \), \(x\) = 4 and the \(x\)-axis is given by [1 Mark]
(A) \( \displaystyle\int_{0}^{4} x\, dx \)
(B) \( \displaystyle\int_{0}^{2} y^2\, dy \)
(C) \( 2\displaystyle\int_{0}^{4} \sqrt{x}\, dx \)
(D) \( \displaystyle\int_{0}^{4} \sqrt{x}\, dx \)

[Figure: A parabola \(y^2 = x\) opening rightwards, with the shaded region between the curve, the line \(x\) = 4, and the \(x\)-axis, from \(x\) = 0 to \(x\) = 4, above the x-axis.]

Answer: (D) \( \displaystyle\int_{0}^{4} \sqrt{x}\, dx \)

Teacher's Note:
a) Since \( y^2 = x \) gives \( y = \sqrt{x} \) for the upper half, the area under the curve from 0 to 4 is \( \int \sqrt{x}\,dx \).
b) Do not confuse this with integrating with respect to y, which would need different limits.

Assertion - Reason Based Questions

Direction : Question numbers 19 and 20 are Assertion (A) and Reason (R) based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and other labelled Reason (R). Select the correct answer from the options (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

19. Assertion (A) : f(\(x\)) = \( \begin{cases} 3x-8, & x \leq 5 \\ 2k, & x \gt 5 \end{cases} \) is continuous at \(x\) = 5 for k = \( \dfrac{5}{2} \).
Reason (R) : For a function f to be continuous at \(x\) = a, \( \displaystyle\lim_{x \to a^-} f(x) = \displaystyle\lim_{x \to a^+} f(x) = f(a) \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) Check the left hand limit: at \(x\)=5, f(5) = 3(5)-8=7, and right hand limit = 2k.
b) For continuity 2k must equal 7, giving k = 7/2, not 5/2, so the Assertion is false while the Reason (the continuity condition) is a true statement.

20. Assertion (A) : Let Z be the set of integers. A function f : Z \(\to\) Z defined as f(\(x\)) = 3\(x\)-5, \( \forall x \in Z \) is a bijective.
Reason (R) : A function is a bijective if it is both surjective and injective. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) f is injective, but it is not surjective on Z since values like 1 or 2 have no pre-image in Z.
b) The Reason correctly defines bijective functions in general, but the Assertion's specific claim is false.

SECTION - B

This section comprises 5 Very Short Answer (VSA) type questions of 2 marks each.

21. The diagonals of a parallelogram are given by \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 3\hat{j} - \hat{k} \). Find the area of the parallelogram. [2 Marks]

Answer:
1. Compute \( \vec{a} \times \vec{b} = -2\hat{i} + 3\hat{j} + 7\hat{k} \).
2. Area of parallelogram \( = \dfrac{1}{2}|\vec{a}\times\vec{b}| = \dfrac{1}{2}\sqrt{4+9+49} = \dfrac{\sqrt{62}}{2} \) square units.

Teacher's Note:
a) Remember the area formula uses half the cross product of the diagonals, not the full magnitude.
b) Show the determinant expansion clearly for full marks.

22. Find the values of 'a' for which f(\(x\)) = \( \sqrt{3}\sin x - \cos x - 2ax + b \) is decreasing on \( \mathbb{R} \). [2 Marks]

Answer:
1. f is decreasing iff \( f'(x) \leq 0 \): \( \sqrt{3}\cos x + \sin x - 2a \leq 0 \) for all \(x\).
2. This gives \( \sin\left(x+\dfrac{\pi}{3}\right) \leq a \); since the maximum value of the sine expression is 1, we need \( a \geq 1 \), i.e. \( a \in [1,\infty) \).

Teacher's Note:
a) Convert the expression into a single sine function using \( R\sin(x+\phi) \) form.
b) The inequality must hold for all \(x\), so use the maximum possible value of sine, which is 1.

23. (a) Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors \( \vec{a} = 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k} \). Determine the angle formed between the kite strings. Assume there is no slack in the strings. [2 Marks]

Answer:
1. \( \cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{6-2+8}{\sqrt{14}\sqrt{24}} = \dfrac{12}{\sqrt{336}} = \dfrac{3}{\sqrt{21}} \).
2. Therefore, \( \theta = \cos^{-1}\left(\dfrac{3}{\sqrt{21}}\right) \).

Teacher's Note:
a) Use the dot product formula for angle between two vectors.
b) Simplify the surd carefully to match the marking scheme's form.

OR

(b) Find a vector of magnitude 21 units in the direction opposite to that of \( \overrightarrow{AB} \) where A and B are the points A(2, 1, 3) and B(8, -1, 0) respectively. [2 Marks]

Answer:
1. \( \overrightarrow{BA} = -6\hat{i}+2\hat{j}+3\hat{k} \), with \( |\overrightarrow{BA}| = 7 \).
2. Required vector \( = 21\times\dfrac{-6\hat{i}+2\hat{j}+3\hat{k}}{7} = -18\hat{i}+6\hat{j}+9\hat{k} \).

Teacher's Note:
a) The direction opposite to \( \overrightarrow{AB} \) is the same as the direction of \( \overrightarrow{BA} \).
b) Multiply the unit vector by the required magnitude, here 21.

24. Solve for \(x\), \( 2\tan^{-1}x + \sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = 4\sqrt{3} \) [2 Marks]

Answer:
1. Using \( \sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = 2\tan^{-1}x \), the equation becomes \( 4\tan^{-1}x = 4\sqrt{3} \), so \( \tan^{-1}x = \sqrt{3} \).
2. Since \( \sqrt{3} \notin \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right) \), this equation has no solution.

Teacher's Note:
a) Always check the range restriction of \( \tan^{-1}x \) before concluding a value of \(x\).
b) This is a common trap question where the algebra looks solvable but the range makes it invalid.

25. (a) Differentiate \( 2^{\cos^2 x} \) w.r.t \( \cos^2 x \). [2 Marks]

Answer:
1. Let \( u = 2^{\cos^2x} \) and \( v = \cos^2x \).
2. \( \dfrac{du}{dv} = 2^{\cos^2x}\log 2 \).

Teacher's Note:
a) Use the parametric differentiation approach: find \( du/dx \) and \( dv/dx \) separately, then divide.
b) Recall \( \dfrac{d}{dt}(a^t) = a^t \log a \).

OR

(b) If \( \tan^{-1}(x^2+y^2) = a^2 \), then find \( \dfrac{dy}{dx} \). [2 Marks]

Answer:
1. \( x^2+y^2 = \tan(a^2) \), a constant, so differentiate both sides with respect to \(x\): \( 2x+2y\dfrac{dy}{dx}=0 \).
2. Hence \( \dfrac{dy}{dx} = -\dfrac{x}{y} \).

Teacher's Note:
a) Since \( a^2 \) is a constant, \( \tan(a^2) \) is also a constant.
b) Implicit differentiation of a constant on the right hand side gives zero.

SECTION - C

This section comprises 6 Short Answer (SA) type questions of 3 marks each.

26. Solve the following linear programming problem graphically :
Maximize Z = 8\(x\) + 9y
Subject to the constraints :
\( 2x+3y \leq 6 \)
\( 3x-2y \leq 6 \)
\( y \leq 1 \)
\( x \geq 0, y \geq 0 \) [3 Marks]

[Figure: A graph showing the feasible region bounded by the lines \( 2x+3y=6 \), \( 3x-2y=6 \), y = 1, and the axes, forming a shaded polygon with corner points O(0,0), A(0,1), B(1.5,1), C(30/13, 6/13), D(2,0).]

Answer:
1. The corner points of the feasible region are O(0,0), A(0,1), B\(\left(\dfrac{3}{2},1\right)\), C\(\left(\dfrac{30}{13},\dfrac{6}{13}\right)\), D(2,0).
2. Value of Z = 8\(x\)+9y at each point: O = 0, A = 9, B = 21, C = \( \dfrac{294}{13} \), D = 16.
3. The maximum value is \( Z = \dfrac{294}{13} \), attained at \( \left(\dfrac{30}{13},\dfrac{6}{13}\right) \).

Teacher's Note:
a) Always find the intersection points of the boundary lines accurately using simultaneous equations.
b) Evaluate Z at all corner points; the largest value gives the maximum.
c) A neat, correctly shaded graph is essential for full marks.

27. (a) Find : \( \displaystyle\int \dfrac{2x-1}{(x-1)(x+2)(x-3)}\, dx \) [3 Marks]

Answer:
1. Using partial fractions, \( \dfrac{2x-1}{(x-1)(x+2)(x-3)} = \dfrac{-1/6}{x-1} + \dfrac{-1/3}{x+2} + \dfrac{1/2}{x-3} \).
2. Integrating gives \( -\dfrac{1}{6}\log|x-1| - \dfrac{1}{3}\log|x+2| + \dfrac{1}{2}\log|x-3| + C \).

Teacher's Note:
a) Set up partial fractions by equating coefficients or substituting convenient values of \(x\).
b) Double check the signs of each fraction's coefficient before integrating.

OR

(b) Evaluate : \( \displaystyle\int_{0}^{5} (|x-1|+|x-2|+|x-5|)\, dx \) [3 Marks]

Answer:
1. Split the integral at the critical points \(x\)=1, 2, 5 and remove the modulus signs accordingly on each subinterval.
2. Evaluating each part gives \( \dfrac{17}{2}+\dfrac{13}{2}+\dfrac{25}{2} = \dfrac{55}{2} \).

Teacher's Note:
a) Break the integral at each point where the expression inside modulus changes sign.
b) Be careful with signs when the expression is negative on a subinterval.

28. A spherical medicine ball when dropped in water dissolves in such a way that the rate of decrease of volume at any instant is proportional to its surface area. Calculate the rate of decrease of its radius. [3 Marks]

Answer:
1. Let V and S be the volume and surface area with radius r; given \( \dfrac{dV}{dt} = -kS \), k \(\gt\) 0.
2. Since \( V = \dfrac{4}{3}\pi r^3 \), \( \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} \), so \( -k(4\pi r^2) = 4\pi r^2 \dfrac{dr}{dt} \).
3. This gives \( \dfrac{dr}{dt} = -k \), meaning the radius decreases at a constant rate.

Teacher's Note:
a) Relate volume and surface area formulas of a sphere carefully before differentiating.
b) The \( r^2 \) terms cancel out, leading to a constant rate, which is the key insight examiners look for.

29. Sketch the graph of y = \( |x+3| \) and find the area of the region enclosed by the curve, \(x\)-axis, between \(x\) = -6 and \(x\) = 0, using integration. [3 Marks]

[Figure: A V-shaped graph of y = |x+3|, with the vertex at (-3, 0), rising on both sides, shaded region between x = -6 and x = 0 above the x-axis.]

Answer:
1. Since the graph is symmetric about \(x\) = -3, required area \( = 2\displaystyle\int_{-3}^{0}(x+3)\,dx \).
2. This equals \( 2\left[\dfrac{(x+3)^2}{2}\right]_{-3}^{0} = 9 \) square units.

Teacher's Note:
a) Sketching the V-shaped graph correctly helps identify the symmetry that simplifies the integral.
b) Use the symmetry about the vertex \(x\) = -3 to halve the calculation work.

30. (a) Verify that lines given by \( \vec{r} = (1-\lambda)\hat{i} + (\lambda-2)\hat{j} + (3-2\lambda)\hat{k} \) and \( \vec{r} = (\mu+1)\hat{i} + (2\mu-1)\hat{j} - (2\mu+1)\hat{k} \) are skew lines. Hence, find shortest distance between the lines. [3 Marks]

Answer:
1. Rewriting, \( \vec{a_1}=\hat{i}-2\hat{j}+3\hat{k} \), \( \vec{b_1}=-\hat{i}+\hat{j}-2\hat{k} \); \( \vec{a_2}=\hat{i}-\hat{j}-\hat{k} \), \( \vec{b_2}=\hat{i}+2\hat{j}-2\hat{k} \). Direction ratios are not proportional, so lines are not parallel.
2. \( \vec{a_2}-\vec{a_1} = \hat{j}-4\hat{k} \) and \( \vec{b_1}\times\vec{b_2} = 2\hat{i}-4\hat{j}-3\hat{k} \); their dot product = 8 \(\neq\) 0, so the lines do not intersect and are skew.
3. Shortest distance \( = \dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} = \dfrac{8}{\sqrt{29}} \).

Teacher's Note:
a) First check that direction ratios are not proportional to rule out parallel lines.
b) A non-zero scalar triple product confirms the lines are skew and gives the numerator for shortest distance.

OR

(b) During a cricket match, the position of the bowler, the wicket keeper and the leg slip fielder are in a line given by \( \vec{B} = 2\hat{i}+8\hat{j} \), \( \vec{W} = 6\hat{i}+12\hat{j} \) and \( \vec{F} = 12\hat{i}+18\hat{j} \) respectively. Calculate the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder. [3 Marks]

Answer:
1. Let the wicketkeeper divide BF in ratio \(k\):1, so \( \vec{W} = \dfrac{k\vec{F}+\vec{B}}{k+1} \).
2. Equating components: \( 6 = \dfrac{12k+2}{k+1} \), which gives \( k = \dfrac{2}{3} \).
3. Hence the required ratio is 2 : 3.

Teacher's Note:
a) Use the section formula for points dividing a line segment in a given ratio.
b) Solve using either the x or y coordinate; both should give the same value of k.

31. (a) The probability distribution for the number of students being absent in a class on a Saturday is as follows :
X: 0 | 2 | 4 | 5
P(X): p | 2p | 3p | p
Where X is the number of students absent.
(i) Calculate p. [1 Mark]
(ii) Calculate the mean of the number of absent students on Saturday. [2 Marks]

Answer:
1. Since \( \sum P(X)=1 \), \( p+2p+3p+p=1 \Rightarrow p=\dfrac{1}{7} \).
2. Mean \( = \sum X\cdot P(X) = 0(p)+2(2p)+4(3p)+5(p) = 21p = 21\times\dfrac{1}{7} = 3 \).

Teacher's Note:
a) The sum of all probabilities in a distribution must always equal 1.
b) Mean of a discrete random variable is \( \sum X_iP(X_i) \).

OR

(b) For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test. [3 Marks]

Answer:
1. Let \( E_1 \): applicant is male, \( E_2 \): applicant is female, A: candidate has distinction. \( P(E_1)=\dfrac{1}{3} \), \( P(E_2)=\dfrac{2}{3} \), \( P(A|E_1)=0.4 \), \( P(A|E_2)=0.35 \).
2. By total probability, \( P(A) = P(E_1)P(A|E_1)+P(E_2)P(A|E_2) = \dfrac{1}{3}\times0.4+\dfrac{2}{3}\times0.35 \).
3. This gives \( P(A) = \dfrac{11}{30} \).

Teacher's Note:
a) Use the Theorem of Total Probability when an event can occur through multiple mutually exclusive causes.
b) Carefully assign the correct fractions to male and female applicants.

SECTION - D

This section comprises 4 Long Answer (LA) type questions of 5 marks each.

32. A school wants to allocate students into three clubs : Sports, Music and Drama, under following conditions :
The number of students in Sports club should be equal to the sum of the number of students in Music and Drama club.
The number of students in Music club should be 20 more than half the number of students in Sports club.
The total number of students to be allocated in all three clubs are 180.
Find the number of students allocated to different clubs, using matrix method. [5 Marks]

Answer:
1. Let \(x\), y, z be the numbers in Sports, Music, Drama clubs. The conditions give \( x-y-z=0 \), \( x-2y=-40 \), \( x+y+z=180 \).
2. Writing as AX = B, we find \( |A| = -4 \neq 0 \), so \( A^{-1} \) exists.
3. Computing adj A and then \( A^{-1} = \dfrac{1}{|A|}\text{adj }A \).
4. \( X = A^{-1}B \) gives \( x=90 \), \( y=65 \), \( z=25 \).
5. Hence, 90, 65 and 25 students are allocated to Sports, Music and Drama clubs respectively.

Teacher's Note:
a) Translate the word conditions into three linear equations very carefully before forming the matrix.
b) Check \( |A| \neq 0 \) before proceeding to find the inverse.
c) Verify the final answer satisfies the original conditions of the problem.

33. Find : \( \displaystyle\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\, dx \) [5 Marks]

Answer:
1. Substitute \( x = a\tan^2\theta \), so \( dx = 2a\tan\theta\sec^2\theta\, d\theta \).
2. The integral becomes \( 2a\displaystyle\int \theta\tan\theta\sec^2\theta\, d\theta \).
3. Using integration by parts, this simplifies to \( a\left[\theta\tan^2\theta - (\tan\theta-\theta)\right]+C \).
4. Substituting back \( \theta = \tan^{-1}\sqrt{x/a} \), the answer is \( (a+x)\tan^{-1}\sqrt{\dfrac{x}{a}} - \sqrt{ax} + C \).

Teacher's Note:
a) The substitution \( x=a\tan^2\theta \) is standard for expressions of the form \( \sqrt{x/(a+x)} \).
b) Apply integration by parts carefully treating \(\theta\) as the first function.

34. (a) If \( \sqrt{1-x^2}+\sqrt{1-y^2} = a(x-y) \), then prove that \( \dfrac{dy}{dx} = \sqrt{\dfrac{1-y^2}{1-x^2}} \). [5 Marks]

Answer:
1. Let \( x = \sin A \), \( y = \sin B \); the equation becomes \( \cos A + \cos B = a(\sin A - \sin B) \).
2. Using sum-to-product formulas, this simplifies to \( \cot\left(\dfrac{A-B}{2}\right) = a \), so \( A - B = 2\cot^{-1}a \), a constant.
3. So \( \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a \). Differentiating with respect to \(x\): \( \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx} = 0 \).
4. This gives \( \dfrac{dy}{dx} = \sqrt{\dfrac{1-y^2}{1-x^2}} \), as required.

Teacher's Note:
a) The substitution \( x=\sin A, y=\sin B \) converts the surd expression into a manageable trigonometric identity.
b) The sum-to-product formulas for cosine and sine are key steps that examiners check for marks.

OR

(b) If \( x = a\left(\cos\theta+\log\tan\dfrac{\theta}{2}\right) \) and y = \( \sin\theta \), then find \( \dfrac{d^2y}{dx^2} \) at \( \theta = \dfrac{\pi}{4} \). [5 Marks]

Answer:
1. \( \dfrac{dx}{d\theta} = a\left(-\sin\theta+\dfrac{1}{\sin\theta}\right) = a\cdot\dfrac{1-\sin^2\theta}{\sin\theta} = a\cot\theta\cos\theta \).
2. \( \dfrac{dy}{d\theta} = \cos\theta \), so \( \dfrac{dy}{dx} = \dfrac{\tan\theta}{a} \).
3. Differentiating again with respect to \(x\): \( \dfrac{d^2y}{dx^2} = \dfrac{\sec^2\theta}{a}\times\dfrac{d\theta}{dx} = \dfrac{\sec^3\theta\tan\theta}{a^2} \).
4. At \( \theta=\pi/4 \), this equals \( \dfrac{2\sqrt{2}}{a^2} \).

Teacher's Note:
a) For parametric differentiation, always find \( dy/dx \) first as a function of \(\theta\), then differentiate again with respect to \(x\) using the chain rule.
b) Simplify \( dx/d\theta \) fully before proceeding, as this step is error-prone.

35. (a) Find the image A' of the point A(1, 6, 3) in the line \( \dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3} \). Also, find the equation of the line joining A and A'. [5 Marks]

Answer:
1. Any point on the line is \( M(\lambda, 2\lambda+1, 3\lambda+2) \); direction ratios of AM are \( \lt \lambda-1, 2\lambda-5, 3\lambda-1 \gt \).
2. Since AM is perpendicular to the line, \( 1(\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=0 \Rightarrow \lambda=1 \), giving foot of perpendicular M(1,3,5).
3. Since M is the midpoint of AA', \( A'(\alpha,\beta,\gamma) \) satisfies \( M\left(\dfrac{1+\alpha}{2},\dfrac{6+\beta}{2},\dfrac{3+\gamma}{2}\right)=M(1,3,5) \), giving A'(1,0,7).
4. Equation of line AA' is \( \dfrac{x-1}{0}=\dfrac{y-6}{-3}=\dfrac{z-3}{2} \).

Teacher's Note:
a) The foot of perpendicular is found by using the perpendicularity condition between AM and the given line's direction ratios.
b) The image point A' is obtained since M is the midpoint of A and A'.

OR

(b) Find a point P on the line \( \dfrac{x+5}{1}=\dfrac{y+3}{4}=\dfrac{z-6}{-9} \) such that its distance from point Q(2, 4, -1) is 7 units. Also, find the equation of line joining P and Q. [5 Marks]

Answer:
1. Any point on the line is \( P(\lambda-5, 4\lambda-3, -9\lambda+6) \).
2. Setting PQ = 7: \( \sqrt{(\lambda-7)^2+(4\lambda-7)^2+(-9\lambda+7)^2}=7 \), which simplifies to \( 98(\lambda^2-2\lambda+1)=0 \Rightarrow \lambda=1 \).
3. Hence, the required point is P(-4, 1, -3).
4. Equation of line PQ is \( \dfrac{x+4}{6}=\dfrac{y-1}{3}=\dfrac{z+3}{2} \).

Teacher's Note:
a) Express the general point on the line in terms of the parameter, then use the distance formula with Q.
b) Solving the resulting quadratic in \(\lambda\) should give exactly one valid solution here.

SECTION - E

This section comprises 3 case study-based questions of 4 marks each.

36. A class-room teacher is keen to assess the learning of her students the concept of "relations" taught to them. She writes the following five relations each defined on the set A = {1, 2, 3} :
R₁ = {(2, 3), (3, 2)}
R₂ = {(1, 2), (1, 3), (3, 2)}
R₃ = {(1, 2), (2, 1), (1, 1)}
R₄ = {(1, 1), (1, 2), (3, 3), (2, 2)}
R₅ = {(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)}
The students are asked to answer the following questions about the above relations :

(i) Identify the relation which is reflexive, transitive but not symmetric. [1 Mark]

Answer: R₄ is reflexive and transitive but not symmetric.

Teacher's Note:
a) Check that all pairs (1,1), (2,2), (3,3) are present for reflexivity.
b) Verify (1,2) is present but (2,1) is not, confirming it is not symmetric.

(ii) Identify the relation which is reflexive and symmetric but not transitive. [1 Mark]

Answer: R₅ is reflexive and symmetric but not transitive.

Teacher's Note:
a) All the diagonal pairs (1,1), (2,2), (3,3) are present, giving reflexivity.
b) Check a triple like (1,2) and (2,3): if (1,3) is missing, transitivity fails.

(iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive. [2 Marks]

Answer: R₁ and R₃ are symmetric but neither reflexive nor transitive.

Teacher's Note:
a) R₁ has only the pair (2,3) and its reverse (3,2), so it is symmetric but missing diagonal elements.
b) Check each relation systematically against all three properties: reflexive, symmetric, transitive.

OR

(iii) (b) What pairs should be added to the relation R₂ to make it an equivalence relation ? [2 Marks]

Answer: The pairs to be added are (1,1), (2,2), (3,3), (2,1), (3,1) and (2,3).

Teacher's Note:
a) First add all missing diagonal pairs to ensure reflexivity.
b) Then add the reverse of every existing pair to ensure symmetry, and check transitivity is also satisfied.

37. A bank offers loan to its customers on different types of interest namely, fixed rate, floating rate and variable rate. From the past data with the bank, it is known that a customer avails loan on fixed rate, floating rate or variable rate with probabilities 10%, 20% and 70% respectively. A customer after availing loan can pay the loan or default on loan repayment. The bank data suggests that the probability that a person defaults on loan after availing it at fixed rate, floating rate and variable rate is 5%, 3% and 1% respectively.
Based on the above information, answer the following :

[Figure: An image of stacked coins of decreasing height with an arrow labelled "% INTEREST RATES" pointing downward, representing declining interest rates.]

(i) What is the probability that a customer after availing the loan will default on the loan repayment ? [2 Marks]

Answer:
1. Let \( E_1, E_2, E_3 \) denote fixed, floating and variable rate loans, with \( P(E_1)=0.1 \), \( P(E_2)=0.2 \), \( P(E_3)=0.7 \), and \( P(A|E_1)=0.05 \), \( P(A|E_2)=0.03 \), \( P(A|E_3)=0.01 \).
2. By total probability, \( P(A) = 0.1(0.05)+0.2(0.03)+0.7(0.01) = 0.018 \), i.e. \( \dfrac{9}{500} \).

Teacher's Note:
a) This is a direct application of the Theorem of Total Probability with three mutually exclusive causes.
b) Convert percentages to decimals carefully before multiplying.

(ii) A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest ? [2 Marks]

Answer:
1. Using Bayes' theorem, \( P(E_3|A) = \dfrac{P(E_3)P(A|E_3)}{P(A)} = \dfrac{0.7\times0.01}{0.018} \).
2. This simplifies to \( P(E_3|A) = \dfrac{7}{18} \).

Teacher's Note:
a) Use Bayes' theorem to reverse the conditional probability.
b) The denominator is the total probability found in part (i); reuse it to save time.

38. A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.
Let the length of the side perpendicular to the partition be \(x\) metres and with parallel to the partition be y metres.
Based on this information, answer the following questions :

[Figure: A photograph of a large rectangular solar panel installation mounted on a rooftop, divided into sections by grid lines and a central partition.]

(i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of \(x\) and y. [1 Mark]

Answer: The equation is \( 2x+3y=300 \).

Teacher's Note:
a) The boundary uses 2 lengths of \(x\) (perpendicular sides) and the y-direction material is used 3 times: twice for the outer boundary and once for the partition.
b) Total material used must equal exactly 300 metres.

(ii) Write the area of the solar panel as a function of \(x\). [1 Mark]

Answer: From \( y=\dfrac{300-2x}{3} \), area \( A(x) = xy = \dfrac{x(300-2x)}{3} = \dfrac{300x-2x^2}{3} \).

Teacher's Note:
a) Express y in terms of \(x\) using the constraint from part (i), then substitute into A = \(xy\).
b) Keep the area as a single-variable function of \(x\) for the optimisation that follows.

(iii) (a) Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area. [2 Marks]

Answer:
1. \( \dfrac{dA}{dx} = \dfrac{1}{3}(300-4x) \); setting this to zero gives \( x=75 \).
2. \( \dfrac{d^2A}{dx^2} = -\dfrac{4}{3} \lt 0 \), so A is maximum at \(x\)=75.
3. Maximum area \( = \dfrac{75}{3}(300-150) = 3750 \) square metres.

Teacher's Note:
a) A negative second derivative confirms a maximum at the critical point.
b) Substitute \(x\)=75 back into the area function, not just the derivative, to find the maximum value.

OR

(iii) (b) Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition. [2 Marks]

Answer:
1. \( \dfrac{dA}{dx} = \dfrac{1}{3}(300-4x) \); setting this to zero gives \( x=75 \).
2. As \( \dfrac{dA}{dx} \) changes sign from positive to negative as \(x\) passes through 75, \(x\)=75 is a point of maximum.
3. Maximum area \( = \dfrac{75}{3}(300-150) = 3750 \) square metres.

Teacher's Note:
a) The first derivative test requires checking the sign change of \( dA/dx \) around the critical point.
b) Both the first and second derivative tests should give the same maximum area value.

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