CBSE Class 12 Maths Question Paper 2025 Solved Code 65-1-2

SECTION - A

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

1. If E and F are two independent events such that \( P(E) = \dfrac{2}{3} \), \( P(F) = \dfrac{3}{7} \), then \( P(E/\overline{F}) \) is equal to : [1 Mark]
(A) \( \dfrac{1}{6} \)
(B) \( \dfrac{1}{2} \)
(C) \( \dfrac{2}{3} \)
(D) \( \dfrac{7}{9} \)

Answer: (C) \( \dfrac{2}{3} \)

Teacher's Note:
a) Since E and F are independent, E and \(\overline{F}\) are also independent.
b) For independent events, \( P(E/\overline{F}) = P(E) \), so no calculation with F is needed.

2. If \( \vec{\alpha} = \hat{i} - 4\hat{j} + 9\hat{k} \) and \( \vec{\beta} = 2\hat{i} - 8\hat{j} + \lambda\hat{k} \) are two mutually parallel vectors, then \( \lambda \) is equal to : [1 Mark]
(A) -18
(B) 18
(C) \( \dfrac{-34}{9} \)
(D) \( \dfrac{34}{9} \)

Answer: (B) 18

Teacher's Note:
a) For parallel vectors, ratios of corresponding components must be equal.
b) Use \( \dfrac{1}{2} = \dfrac{-4}{-8} = \dfrac{9}{\lambda} \) to get \( \lambda = 18 \).

3. \( \displaystyle\int \dfrac{1-2\sin x}{\cos^2 x} \, dx \) is equal to : [1 Mark]
(A) \( \tan x - 2\sec x + C \)
(B) \( -\tan x + 2\sec x + C \)
(C) \( -\tan x - 2\sec x + C \)
(D) \( \tan x + 2\sec x + C \)

Answer: (A) \( \tan x - 2\sec x + C \)

Teacher's Note:
a) Split the integrand as \( \sec^2 x - 2\sec x \tan x \) before integrating.
b) Remember \( \int \sec x \tan x\, dx = \sec x + C \).

4. 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) Use \( |\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b}\cdot\vec{c} \) obtained by squaring \( \vec{a} = -(\vec{b}+\vec{c}) \).
b) Substitute values to get \( \cos\theta = \dfrac{1}{2} \), giving \( \theta = \dfrac{\pi}{3} \).

5. The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse ? [1 Mark]
(A) Graph of curve
(B) Graph of curve
(C) Graph of curve
(D) Graph of curve

[Figure: The given graph shows a hump-shaped trigonometric curve (cosine-type) on the Y-axis with x-values marked at \(-\dfrac{\pi}{2}\), \(\dfrac{\pi}{2}\) and \(\pi\), rising from the left, peaking near the y-axis and falling towards \(\pi\). Option (A) shows an increasing S-shaped curve through the origin between \(-\pi/2\) and \(\pi/2\), y from -1 to 1. Option (B) shows a similar increasing curve with different concavity. Option (C) shows a decreasing curve starting at point \((-1,\pi)\), passing through \((0,\pi/2)\) and ending at \((1,0)\). Option (D) shows a decreasing curve starting near \(\pi\) on the y-axis and falling to the right, marked at \(\pi/2\).]

Answer: (C) - decreasing curve starting at \((-1,\pi)\), passing through \((0,\pi/2)\) and ending at \((1,0)\)

Teacher's Note:
a) The inverse of a function is obtained by reflecting its graph about the line \(y=x\).
b) Since the given function has range \([0,\pi]\), its inverse must have domain \([-1,1]\) and range \([0,\pi]\), matching option (C).

6. If A is a square matrix of order 3 such that \( \det(A) = 9 \), then \( \det(9A^{-1}) \) is equal to [1 Mark]
(A) 9
(B) \( 9^2 \)
(C) \( 9^3 \)
(D) \( 9^4 \)

Answer: (B) \( 9^2 \)

Teacher's Note:
a) Use \( \det(kA) = k^n \det(A) \) for an \(n \times n\) matrix.
b) Also use \( \det(A^{-1}) = \dfrac{1}{\det(A)} \) to compute the final value.

7. 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.
b) They fail to be differentiable at points where the expression inside the modulus becomes zero (here \(x=0\) and \(x=1\)).

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 function is homogeneous of degree n if \( f(\lambda x, \lambda y) = \lambda^n f(x,y) \).
b) Options (A), (B) and (C) satisfy this condition but (D) does not, as it does not reduce to a pure function of \(y/x\).

9. Let A be a matrix of order \( m \times n \) and B is a matrix such that \( A^T B \) and \( BA^T \) are defined. Then, the order of B is : [1 Mark]
(A) \( m \times m \)
(B) \( n \times n \)
(C) \( m \times n \)
(D) \( n \times m \)

Answer: (C) \( m \times n \)

Teacher's Note:
a) \( A^T \) has order \( n \times m \), so for \( A^T B \) to be defined, B must have m rows.
b) For \( BA^T \) to be defined, B must have n columns, giving order \( m \times n \).

10. If the feasible region of a linear programming problem with objective function \( Z = ax + by \), 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 has both a maximum and minimum value of the objective function.
b) These extreme values occur at the corner points of the feasible region.

11. 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) For a diagonal matrix, the inverse is obtained by taking the reciprocal of each diagonal entry.
b) Since A is its own inverse here (as \(A^2 = I\)), verify by multiplication if in doubt.

12. The integrating factor of the differential equation \( \dfrac{dy}{dx} + y = \dfrac{1+y}{x} \) is [1 Mark]
(A) \( xe^x \)
(B) \( \dfrac{e^x}{x} \)
(C) \( \dfrac{x}{e^x} \)
(D) \( xe^{\frac{1}{x}} \)

Answer: (B) \( \dfrac{e^x}{x} \)

Teacher's Note:
a) First rewrite the equation in standard linear form \( \dfrac{dy}{dx} + \left(1-\dfrac{1}{x}\right)y = \dfrac{1}{x} \).
b) Integrating factor \( = e^{\int (1-1/x)\,dx} = e^{x-\ln x} = \dfrac{e^x}{x} \).

13. Let \( A = [a_{ij}] \) be a square matrix of order 3 such that \( a_{ij} = \hat{j} - 2\hat{i} \). Then which of the following is true ? [1 Mark]
(A) \( a_{12} \gt 0 \)
(B) all \( a_{ij} \lt 0 \)
(C) \( a_{13} + a_{31} = -6 \)
(D) \( a_{23} \gt a_{32} \)

Answer: (D) \( a_{23} \gt a_{32} \)

Teacher's Note:
a) Here \( a_{ij} = j - 2i \), so compute each entry using row and column numbers.
b) \( a_{23} = 3-4 = -1 \) and \( a_{32} = 2-6 = -4 \), so \( a_{23} \gt a_{32} \).

14. 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 critical points using \( f'(x) = 3x^2 - 3 = 0 \Rightarrow x = 1 \) (in the interval).
b) Compare \( f(0)=2, f(1)=0, f(2)=4 \) to find the absolute maximum value.

15. 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 \( -\int 2^t\, dt = -\dfrac{2^t}{\log 2}+C \), giving \( k = \dfrac{-1}{\log 2} \).

16. 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) \( \text{adj}(A) = |A|A^{-1} \)
(D) \( |A|^{-1} = |A^{-1}| \)

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

Teacher's Note:
a) There is no standard formula for the inverse of a sum of matrices.
b) The other three options are standard identities for invertible matrices.

17. The corner points of the feasible region in graphical representation of a L.P.P. are (2, 72), (15, 20) and (40, 15). If \( Z = 18x + 9y \) 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 at each corner point: \( Z(2,72)=684, Z(15,20)=450, Z(40,15)=855 \).
b) Compare all values to identify maximum and minimum.

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: Graph of the parabola \(y^2=x\) opening rightwards, with the region above the x-axis, below the curve, and to the left of the vertical line \(x=4\) shaded. Axes marked with x from 1 to 6 and y from -1 to 2.]

Answer: (D) \( \displaystyle\int_0^4 \sqrt{x}\, dx \)

Teacher's Note:
a) Since only the region above the x-axis is shaded, use \( y = \sqrt{x} \) as the upper boundary.
b) The area is found by integrating \( y \) with respect to \( x \) from 0 to 4.

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) : Let Z be the set of integers. A function \( f : Z \to Z \) defined as \( f(x) = 3x - 5, \forall x \in Z \) 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(x)=3x-5 is injective but not surjective over integers, since not every integer is of the form 3x-5.
b) So the function is not bijective, making Assertion false while the Reason statement itself is a true definition.

20. Assertion (A) : \( f(x) = \begin{cases} 3x-8, & x \le 5 \\ 2k, & x \gt 5 \end{cases} \) is continuous at \( x = 5 \) for \( k = \dfrac{5}{2} \).
Reason (R) : For a function f to be continuous at \( x = a \), \( \displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \). [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) \( f(5) = 3(5)-8 = 7 \) and \( \lim_{x\to5^+} f(x) = 2k \).
b) For continuity, \( 2k = 7 \Rightarrow k = 7/2 \), not \( 5/2 \), so the Assertion is false while the Reason is a correct statement.

SECTION - B

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

21. (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. \( \cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{(3)(2)+(1)(-2)+(2)(4)}{\sqrt{9+1+4}\sqrt{4+4+16}} = \dfrac{12}{\sqrt{336}} = \dfrac{3}{\sqrt{21}} \)
2. \( \theta = \cos^{-1}\left(\dfrac{3}{\sqrt{21}}\right) \)

Teacher's Note:
a) Use the dot product formula for the angle between two vectors.
b) Carefully compute the magnitudes before finding the cosine value.

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. \( \overrightarrow{BA} = -6\hat{i} + 2\hat{j} + 3\hat{k} \), \( |\overrightarrow{BA}| = 7 \)
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 vector opposite to \( \overrightarrow{AB} \) is \( \overrightarrow{BA} \), so compute B - A.
b) Multiply the unit vector by the required magnitude to get the final answer.

22. Find the values of 'a' for which \( f(x) = x^2 - 2ax + b \) is an increasing function for \( x \gt 0 \). [2 Marks]

Answer:
1. \( f'(x) = 2x - 2a \). For increasing function, \( f'(x) \ge 0 \) for all \( x \gt 0 \).
2. This requires \( 2a \le 2x \) for all \( x \gt 0 \), which holds only if \( a \le 0 \), i.e., \( a \in (-\infty, 0] \).

Teacher's Note:
a) A function is increasing on an interval if its derivative is non-negative throughout that interval.
b) Take the infimum condition as \( x \to 0^+ \) to find the boundary value of a.

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

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

Teacher's Note:
a) Treat v as the independent variable and differentiate u directly with respect to v.
b) Use the standard rule \( \dfrac{d}{dv}(a^v) = a^v \log a \).

OR

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

Answer:
1. \( \tan^{-1}(x^2+y^2)=a^2 \Rightarrow x^2+y^2 = \tan a^2 \) (a constant).
2. Differentiating, \( 2x + 2y\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{y} \)

Teacher's Note:
a) Since \( a^2 \) is a constant, \( \tan a^2 \) is also a constant, simplifying the equation.
b) Differentiate implicitly with respect to x, treating y as a function of x.

24. Evaluate : \( \sin^{-1}\left(\sin \dfrac{3\pi}{5}\right) \) [2 Marks]

Answer:
1. \( \sin^{-1}\left(\sin\dfrac{3\pi}{5}\right) = \sin^{-1}\left(\sin\left(\pi - \dfrac{2\pi}{5}\right)\right) = \sin^{-1}\left(\sin\dfrac{2\pi}{5}\right) \)
2. Since \( \dfrac{2\pi}{5} \) lies in \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \), the value is \( \dfrac{2\pi}{5} \).

Teacher's Note:
a) The range of \( \sin^{-1} \) is \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \), so bring the angle into this range first.
b) Use the identity \( \sin(\pi - \theta) = \sin\theta \) to simplify.

25. 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. \( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = -2\hat{i}+3\hat{j}+7\hat{k} \)
2. Area \( = \dfrac{1}{2}|\vec{a}\times\vec{b}| = \dfrac{1}{2}\sqrt{4+9+49} = \dfrac{\sqrt{62}}{2} \)

Teacher's Note:
a) The area of a parallelogram in terms of its diagonals is \( \dfrac{1}{2}|\vec{d_1}\times\vec{d_2}| \).
b) Compute the cross product carefully using the determinant method.

SECTION - C

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

26. (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. Rewrite lines as \( \vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda(-\hat{i}+\hat{j}-2\hat{k}) \) and \( \vec{r}=(\hat{i}-\hat{j}-\hat{k})+\mu(\hat{i}+2\hat{j}-2\hat{k}) \).
2. The direction ratios are not proportional, so the lines are not parallel. Checking \( (\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) = (\hat{j}-4\hat{k})\cdot(2\hat{i}-4\hat{j}-3\hat{k}) = 8 \ne 0 \), so the lines do not intersect and are skew.
3. Shortest distance \( = \dfrac{|8|}{\sqrt{4+16+9}} = \dfrac{8}{\sqrt{29}} \)

Teacher's Note:
a) Lines are skew if they are neither parallel nor intersecting.
b) Use the shortest distance formula \( \dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} \) for skew lines.

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 W divide BF in ratio \( k:1 \), so \( \vec{W} = \dfrac{k\vec{F}+\vec{B}}{k+1} \).
2. \( 6\hat{i}+12\hat{j} = \dfrac{(12k+2)\hat{i}+(18k+8)\hat{j}}{k+1} \)
3. Solving, \( k = \dfrac{2}{3} \), so 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) Equate the x-components (or y-components) to solve for the ratio k.

27. Solve the following linear programming problem graphically :
Maximise \( Z = 20x + 30y \)
Subject to the constraints :
\( x + y \le 80 \)
\( 2x + 3y \ge 100 \)
\( x \ge 14 \)
\( y \ge 14 \) [3 Marks]

[Figure: Feasible region bounded by lines x+y=80, 2x+3y=100, x=14 and y=14, forming a quadrilateral with corner points A(14,66), B(14,24), C(29,14) and D(66,14).]

Answer:
1. Corner points of feasible region: A(14,66), B(14,24), C(29,14), D(66,14).
2. Values of Z: Z(A)=2260, Z(B)=1000, Z(C)=1740, Z(D)=1000.
3. Maximum value of Z is 2260 at (14, 66).

Teacher's Note:
a) Plot all constraint lines accurately and shade the common feasible region.
b) Evaluate Z at every corner point to identify the maximum value, as per the corner point theorem.

28. The area of an expanding rectangle is increasing at the rate of 48 cm²/s. The length of the rectangle is always square of its breadth. At what rate the length of rectangle increasing at an instant, when breadth = 4.5 cm ? [3 Marks]

Answer:
1. Let length x and breadth y with \( x = y^2 \), so Area \( A = xy = x^{3/2} \).
2. \( \dfrac{dA}{dt} = \dfrac{3}{2}\sqrt{x}\dfrac{dx}{dt} \)
3. When \( y=4.5, x = (4.5)^2 = 20.25 \); substituting, \( 48 = \dfrac{3}{2}\sqrt{20.25}\dfrac{dx}{dt} \Rightarrow \dfrac{dx}{dt} = \dfrac{64}{9} \) cm/s

Teacher's Note:
a) Express area purely in terms of x using the given relation between length and breadth.
b) Differentiate with respect to time and substitute the given rate to find the required rate.

29. (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 \( = \sum X \cdot P(X) = 0(p)+2(2p)+4(3p)+5(p) = 21p = 21\left(\dfrac{1}{7}\right) = 3 \)

Teacher's Note:
a) The sum of all probabilities in a probability distribution must equal 1.
b) Mean is calculated as \( \sum X_i P(X_i) \).

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.
2. \( P(E_1)=\dfrac{1}{3}, P(E_2)=\dfrac{2}{3}, P(A|E_1)=0.4, P(A|E_2)=0.35 \)
3. \( P(A) = P(E_1)P(A|E_1)+P(E_2)P(A|E_2) = \dfrac{1}{3}(0.4)+\dfrac{2}{3}(0.35) = \dfrac{11}{30} \)

Teacher's Note:
a) This is a direct application of the Law of Total Probability.
b) Carefully identify the proportion of male and female applicants from the given data.

30. (a) Find : \( \displaystyle\int \dfrac{\cos 2x}{(\sin x + \cos x)^2}\, dx \) [3 Marks]

Answer:
1. \( \cos 2x = \cos^2 x - \sin^2 x \), so the integral becomes \( \displaystyle\int \dfrac{\cos x - \sin x}{\sin x + \cos x}\, dx \)
2. Let \( t = \sin x + \cos x \), so \( dt = (\cos x - \sin x)dx \).
3. Integral \( = \displaystyle\int \dfrac{dt}{t} = \log|\sin x + \cos x| + C \)

Teacher's Note:
a) Factorise \( \cos 2x \) as a difference of squares before simplifying.
b) Recognise that the numerator is the derivative of the denominator to use substitution.

OR

(b) Evaluate : \( \displaystyle\int_0^{\pi/2} \dfrac{5\sin x + 3\cos x}{\sin x + \cos x}\, dx \) [3 Marks]

Answer:
1. Let \( I = \displaystyle\int_0^{\pi/2} \dfrac{5\sin x+3\cos x}{\sin x+\cos x}dx \). Using \( x \to \pi/2 - x \), \( I = \displaystyle\int_0^{\pi/2}\dfrac{5\cos x+3\sin x}{\cos x+\sin x}dx \)
2. Adding both expressions of I: \( 2I = \displaystyle\int_0^{\pi/2} 8\, dx = 8 \times \dfrac{\pi}{2} = 4\pi \)
3. Therefore \( I = 2\pi \)

Teacher's Note:
a) Use the property \( \int_0^a f(x)dx = \int_0^a f(a-x)dx \) for definite integrals.
b) Adding the two forms of the integral eliminates the fraction, simplifying the calculation.

31. 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: V-shaped graph of y=|x+3| with vertex at (-3,0), rising on both sides; the shaded region lies between x=-6 and x=0, above the x-axis and below the two line segments.]

Answer:
1. By symmetry, Required Area \( = 2\displaystyle\int_{-3}^{0}(x+3)\,dx \)
2. \( = 2\left[\dfrac{(x+3)^2}{2}\right]_{-3}^{0} \)
3. \( = 9 \) square units

Teacher's Note:
a) Sketch the graph correctly to identify the vertex at \( x=-3 \) where the expression changes sign.
b) Use symmetry of the V-shaped graph about \( x=-3 \) to simplify the integration.

SECTION - D

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

32. (a) Differentiate \( \tan^{-1}\dfrac{\sqrt{1-x^2}}{x} \) w.r.t. \( \cos^{-1}(2x\sqrt{1-x^2}) \), \( x \in \left(\dfrac{1}{\sqrt{2}},1\right) \) [5 Marks]

Answer:
1. Put \( x = \cos\theta \), so \( \theta = \cos^{-1}x \).
2. Let \( u = \tan^{-1}\dfrac{\sqrt{1-x^2}}{x} = \tan^{-1}(\tan\theta) = \theta = \cos^{-1}x \), so \( \dfrac{du}{dx} = -\dfrac{1}{\sqrt{1-x^2}} \)
3. Let \( v = \cos^{-1}(2x\sqrt{1-x^2}) = \cos^{-1}(\sin 2\theta) = \dfrac{\pi}{2} - 2\cos^{-1}x \), so \( \dfrac{dv}{dx} = \dfrac{2}{\sqrt{1-x^2}} \)
4. Therefore, \( \dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = -\dfrac{1}{2} \)

Teacher's Note:
a) Use the substitution \( x = \cos\theta \) to simplify both inverse trigonometric expressions.
b) Ensure the correct range of \(\theta\) is used based on the given domain of x.

OR

(b) Find \( \dfrac{dy}{dx} \), if \( y = x^{\tan x} + \dfrac{\sqrt{x^2+1}}{2} \). [5 Marks]

Answer:
1. Let \( y = u+v \), where \( u=x^{\tan x} \) and \( v=\dfrac{\sqrt{x^2+1}}{2} \).
2. For u: \( \log u = \tan x \log x \). Differentiating, \( \dfrac{1}{u}\dfrac{du}{dx} = \dfrac{\tan x}{x}+\sec^2 x \log x \), so \( \dfrac{du}{dx} = x^{\tan x}\left(\dfrac{\tan x}{x}+\sec^2 x \log x\right) \)
3. For v: \( \dfrac{dv}{dx} = \dfrac{x}{2\sqrt{x^2+1}} \)
4. Therefore, \( \dfrac{dy}{dx} = x^{\tan x}\left(\dfrac{\tan x}{x}+\sec^2 x\log x\right) + \dfrac{x}{2\sqrt{x^2+1}} \)

Teacher's Note:
a) Use logarithmic differentiation for terms with variable base and exponent.
b) Differentiate each term separately and combine results using the sum rule.

33. Find the absolute maximum and absolute minimum of function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \). [5 Marks]

Answer:
1. \( f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3) \)
2. Setting \( f'(x)=0 \), we get \( x=2, 3 \), both lying in \( [1,5] \).
3. Evaluate: \( f(1)=24, f(2)=29, f(3)=28, f(5)=56 \)
4. The absolute maximum value is 56 (at x=5) and the absolute minimum value is 24 (at x=1).

Teacher's Note:
a) Find critical points by setting the first derivative equal to zero.
b) Compare function values at all critical points and endpoints to determine absolute extrema.

34. 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 students in Sports, Music, Drama respectively. Equations: \( x-y-z=0 \), \( x-2y=-40 \), \( x+y+z=180 \)
2. Writing as \( AX=B \), where \( A = \begin{bmatrix} 1 & -1 & -1 \\ 1 & -2 & 0 \\ 1 & 1 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 0 \\ -40 \\ 180 \end{bmatrix} \)
3. \( |A| = -4 \ne 0 \), so \( A^{-1} \) exists. \( \text{adj}(A) = \begin{bmatrix} -2 & 0 & -2 \\ -1 & 2 & -1 \\ 3 & -2 & -1 \end{bmatrix} \)
4. \( A^{-1} = \dfrac{1}{4}\begin{bmatrix} 2 & 0 & 2 \\ 1 & -2 & 1 \\ -3 & 2 & 1 \end{bmatrix} \)
5. \( X = A^{-1}B = \begin{bmatrix} 90 \\ 65 \\ 25 \end{bmatrix} \). So the number of students allocated to Sports, Music and Drama are 90, 65 and 25 respectively.

Teacher's Note:
a) Translate the given word conditions carefully into three linear equations.
b) Compute the adjoint and inverse of A systematically, then multiply to find X.

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 \( \langle \lambda-1, 2\lambda-5, 3\lambda-1\rangle \)
2. Since AM is perpendicular to the line: \( 1(\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=0 \Rightarrow \lambda=1 \)
3. So \( M(1,3,5) \) is the foot of perpendicular from A.
4. Let image be \( A'(\alpha,\beta,\gamma) \). Since M is midpoint of AA', \( M\left(\dfrac{1+\alpha}{2},\dfrac{6+\beta}{2},\dfrac{3+\gamma}{2}\right) = (1,3,5) \Rightarrow A'(1,0,7) \)
5. Equation of line AA': \( \dfrac{x-1}{0} = \dfrac{y-6}{-3} = \dfrac{z-3}{2} \)

Teacher's Note:
a) The foot of the perpendicular is found using the perpendicularity condition between AM and the line's direction vector.
b) The image point is obtained since the foot of perpendicular is the midpoint of A and its image 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 line: \( P(\lambda-5, 4\lambda-3, -9\lambda+6) \)
2. Since PQ = 7: \( \sqrt{(\lambda-7)^2+(4\lambda-7)^2+(-9\lambda+7)^2}=7 \)
3. Simplifying: \( 98(\lambda^2-2\lambda+1)=0 \Rightarrow \lambda=1 \)
4. Required point is \( P(-4,1,-3) \)
5. Equation of line PQ: \( \dfrac{x+4}{6} = \dfrac{y-1}{3} = \dfrac{z+3}{2} \)

Teacher's Note:
a) Write a general point on the given line using the parameter \(\lambda\).
b) Use the distance formula to set up an equation and solve for \(\lambda\).

SECTION - E

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

36. 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 illustrative image showing stacked coins of decreasing height with an arrow labelled "% INTEREST RATES" pointing downward.]

(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 availing loan at fixed, floating, variable rate. \( P(E_1)=0.1, P(E_2)=0.2, P(E_3)=0.7 \); \( P(A|E_1)=0.05, P(A|E_2)=0.03, P(A|E_3)=0.01 \)
2. \( P(A) = 0.1(0.05)+0.2(0.03)+0.7(0.01) = \dfrac{18}{1000} = \dfrac{9}{500} \)

Teacher's Note:
a) This uses the Law of Total Probability with three mutually exclusive events.
b) Convert percentages to decimals 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, \( P(E_3|A) = \dfrac{P(E_3)P(A|E_3)}{P(A)} \)
2. \( = \dfrac{0.7 \times 0.01}{9/500} = \dfrac{7/1000}{18/1000} = \dfrac{7}{18} \)

Teacher's Note:
a) Bayes' theorem is used to find the reverse conditional probability.
b) Use the value of P(A) computed in part (i) to simplify the calculation.

37. 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 rectangular solar panel installation mounted on a frame, with a central dividing line (partition) running parallel to one side, dividing the panel into two rectangular sections.]

(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 total material used is \( 2x + 3y = 300 \) (two sides of length x, two sides of length y, and one internal partition of length y).

Teacher's Note:
a) Count the boundary sides (2x + 2y) plus one extra partition of length y.
b) This gives the total material equation \( 2x+3y=300 \).

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

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

Teacher's Note:
a) Express y in terms of x using the boundary equation from part (i).
b) Substitute into \( A=xy \) to get area purely as a function of x.

(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. \( A = \dfrac{1}{3}(300x-2x^2) \Rightarrow \dfrac{dA}{dx} = \dfrac{1}{3}(300-4x) \)
2. Setting \( \dfrac{dA}{dx}=0 \Rightarrow x=75 \)
3. \( \dfrac{d^2A}{dx^2} = -\dfrac{4}{3} \lt 0 \), so A is maximum at x=75.
4. Maximum area \( = \dfrac{75}{3}(300-150) = 3750 \text{ m}^2 \)

Teacher's Note:
a) A negative second derivative confirms a point of maximum.
b) Substitute the critical value of x back into the area function to find the maximum area.

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. \( \dfrac{dA}{dx} = \dfrac{1}{3}(300-4x) = 0 \Rightarrow x=75 \)
2. Since \( \dfrac{dA}{dx} \) changes sign from positive to negative as x passes through 75, x=75 is a point of maximum.
3. Maximum area \( = \dfrac{75}{3}(300-150) = 3750 \text{ m}^2 \)

Teacher's Note:
a) The sign change of the first derivative around the critical point confirms it is a maximum.
b) Full credit is also given if students take the boundary equation as \(2x+2y=300\) or similar valid forms, with corresponding adjusted answers.

38. 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} :
\( R_1 = \{(2, 3), (3, 2)\} \)
\( R_2 = \{(1, 2), (1, 3), (3, 2)\} \)
\( R_3 = \{(1, 2), (2, 1), (1, 1)\} \)
\( R_4 = \{(1, 1), (1, 2), (3, 3), (2, 2)\} \)
\( R_5 = \{(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)\} \)
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: \( R_4 \)

Teacher's Note:
a) Check that all elements (1,1),(2,2),(3,3) are present for reflexivity.
b) Verify there is no pair (b,a) present when (a,b) is present, except the reflexive pairs, to confirm it is not symmetric.

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

Answer: \( R_5 \)

Teacher's Note:
a) Reflexive requires all (1,1),(2,2),(3,3); symmetric requires pairs to occur in both orders.
b) Find a case like (1,2) and (2,3) present but (1,3) absent, showing transitivity fails.

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

Answer: \( R_1 \) and \( R_3 \)

Teacher's Note:
a) Both relations lack all reflexive pairs (1,1),(2,2),(3,3).
b) Check each relation is symmetric but fails the transitivity condition for at least one pair.

OR

(iii) (b) What pairs should be added to the relation \( R_2 \) 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 reflexive pairs (1,1),(2,2),(3,3).
b) Then add the missing symmetric and transitive pairs so that the relation becomes an equivalence relation on the full set.

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