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

SECTION A

1. 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 diagonal matrix with entries \(\pm 1\) is its own inverse.
b) Verify by checking that \(A \times A = I\).

2. If vector \(\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}\) and vector \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), then which of the following is correct ? [1 Mark]
(A) \(\vec{a} \| \vec{b}\)
(B) \(\vec{a} \perp \vec{b}\)
(C) \(|\vec{b}| \gt |\vec{a}|\)
(D) \(|\vec{a}| = |\vec{b}|\)

Answer: (B) \(\vec{a} \perp \vec{b}\)

Teacher's Note:
a) Compute \(\vec{a}\cdot\vec{b} = 3-2-1=0\), which shows the vectors are perpendicular.
b) Remember dot product zero means perpendicular vectors.

3. \(\displaystyle\int_{-1}^{1} \frac{|x|}{x}\,dx, x \neq 0\) is equal to [1 Mark]
(A) -1
(B) 0
(C) 1
(D) 2

Answer: (B) 0

Teacher's Note:
a) Split the integral at 0: the function equals -1 for negative x and 1 for positive x.
b) The two equal and opposite areas cancel to give 0.

4. 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 if \(f(\lambda x, \lambda y) = \lambda^n f(x,y)\) for some n.
b) Trigonometric expressions like \(\tan x - \sec y\) do not satisfy this scaling property.

5. 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 changes sign.

6. If A is a square matrix of order 2 such that det (A) = 4, then det (4 adj A) is equal to : [1 Mark]
(A) 16
(B) 64
(C) 256
(D) 512

Answer: (B) 64

Teacher's Note:
a) For a \(2 \times 2\) matrix, \(\det(\text{adj } A) = \det(A)^{n-1} = \det(A)\).
b) Use \(\det(kA) = k^n \det(A)\) for an \(n \times n\) matrix: \(\det(4\,\text{adj }A) = 4^2 \times 4 = 64\).

7. 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) For independent events, \(P(E/\overline{F}) = P(E)\).
b) Do not confuse independence with mutual exclusivity.

8. 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 by setting \(f'(x)=3x^2-3=0 \Rightarrow x=1\) (within [0,2]).
b) Compare f(0)=2, f(1)=0, f(2)=4; the largest value is the absolute maximum.

9. 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 defined only when the number of columns of the first matrix equals the number of rows of the second.
b) A is \(3\times3\) and B is \(3\times1\), so AB (\(3\times1\)) is defined; AC and BA fail this rule.

10. 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 \(u=\dfrac{1}{x}\) so that \(du=-\dfrac{1}{x^2}dx\).
b) The integral becomes \(-\displaystyle\int 2^u\,du = -\dfrac{2^u}{\log 2}+C\).

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

12. The integrating factor of differential equation \((x+2y^3)\dfrac{dy}{dx}=2y\) is [1 Mark]
(A) \(e^{\frac{y^2}{2}}\)
(B) \(\dfrac{1}{\sqrt{y}}\)
(C) \(\dfrac{1}{y^2}\)
(D) \(e^{-\frac{1}{y^2}}\)

Answer: (B) \(\dfrac{1}{\sqrt{y}}\)

Teacher's Note:
a) Rewrite as \(\dfrac{dx}{dy}-\dfrac{x}{2y}=y^2\), a linear equation in x.
b) IF \(=e^{-\int \frac{1}{2y}dy}=e^{-\frac{1}{2}\ln y}=\dfrac{1}{\sqrt{y}}\).

13. If A = \(\begin{bmatrix} 7 & 0 & x \\ 0 & 7 & 0 \\ 0 & 0 & y \end{bmatrix}\) is a scalar matrix, then y^x is equal to [1 Mark]
(A) 0
(B) 1
(C) 7
(D) \(\pm 7\)

Answer: (B) 1

Teacher's Note:
a) A scalar matrix has all diagonal entries equal and off-diagonal entries zero.
b) Here \(x=0\) and \(y=7\), so \(y^x = 7^0 = 1\).

14. 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) The corner point method gives the maximum and minimum directly from these values.

15. 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.
b) The other three identities are standard matrix inverse properties.

16. 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 attains both maximum and minimum of a linear objective function.
b) These extreme values occur at corner points of the feasible region.

17. 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 region between the curve, the line \(x=4\) and the x-axis (for \(y \geq 0\)) shaded, x-axis marked from 0 to 6, y-axis marked from -1 to 2.]

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

Teacher's Note:
a) Since the shaded region is only above the x-axis, express \(y=\sqrt{x}\) and integrate with respect to x.
b) The limits of integration are from \(x=0\) to \(x=4\).

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

[Figure: The main graph shows a cosine-type curve marked with \(-\frac{\pi}{2}\), \(\frac{\pi}{2}\) and \(\pi\) on the x-axis. Option (A) shows an increasing curve from \((-\pi/2,-1)\) through origin to \((\pi/2,1)\). Option (B) shows a similar increasing curve through the origin between \(-\pi/2\) and \(\pi/2\) with values -1 and 1. Option (C) shows a decreasing curve from \((-1,\pi)\) through \((0,\pi/2)\) to \((1,0)\). Option (D) shows a decreasing curve from a value near \(\pi\) down towards 0, marked with \(\pi/2\).]

Answer: (C) the decreasing curve from \((-1,\pi)\) through \((0,\pi/2)\) to \((1,0)\)

Teacher's Note:
a) The inverse graph is obtained by reflecting the given function's graph about the line \(y=x\).
b) The domain and range of the original function interchange for its inverse.

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 one-one on Z but not onto Z since not every integer y gives an integer x.
b) The definition of bijective given in the Reason is correct in general.

20. 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 \(x = a\), \(\displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)\). [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) At \(x=5\), left hand limit is \(3(5)-8=7\), so for continuity \(2k=7\) giving \(k=3.5\), not \(\dfrac{5}{2}\).
b) The Reason correctly states the general condition for continuity at a point.

SECTION - B

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

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

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

Teacher's Note:
a) Use the chain rule formula \(\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}\) for differentiating one function with respect to another.
b) Keep the exponential term unsimplified while differentiating to avoid sign errors.

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 both sides w.r.t. x: \(2x+2y\dfrac{dy}{dx}=0\).
3. Therefore \(\dfrac{dy}{dx}=-\dfrac{x}{y}\).

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

22. Evaluate : \(\tan^{-1}\left[2\sin\left(2\cos^{-1}\dfrac{\sqrt{3}}{2}\right)\right]\) [2 Marks]

Answer:
1. \(\cos^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{6}\), so \(2\cos^{-1}\dfrac{\sqrt3}{2}=\dfrac{\pi}{3}\).
2. \(\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}\), so \(2\sin\dfrac{\pi}{3}=\sqrt3\).
3. \(\tan^{-1}\sqrt3=\dfrac{\pi}{3}\).

Teacher's Note:
a) Simplify the innermost inverse trigonometric value first.
b) Use standard angle values of sine and tangent to complete the evaluation.

23. 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 of parallelogram \(=\dfrac{1}{2}|\vec{a}\times\vec{b}| = \dfrac{1}{2}\sqrt{4+9+49}=\dfrac{\sqrt{62}}{2}\).

Teacher's Note:
a) When diagonals of a parallelogram are given as vectors, area = half the magnitude of their cross product.
b) Compute the cross product carefully using the determinant method.

24. Find the intervals in which function f(x) = \(5x^{\frac{3}{2}} - 3x^{\frac{5}{2}}\) is (i) increasing (ii) decreasing. [2 Marks]

Answer:
1. \(f'(x)=\dfrac{15}{2}\sqrt{x}(1-x)\). Setting \(f'(x)=0\) gives \(x=0,1\).
2. For \(x\in[0,1]\), \(f'(x)\geq 0\), so f is increasing on \([0,1]\).
3. For \(x\in[1,\infty)\), \(f'(x)\leq 0\), so f is decreasing on \([1,\infty)\).

Teacher's Note:
a) Since the domain of f is \(x\geq 0\), test intervals only within this domain.
b) The sign of \(f'(x)\) determines whether the function is increasing or decreasing.

25. (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 formula \(\cos\theta=\dfrac{\vec a \cdot \vec b}{|\vec a||\vec b|}\) to find the angle between two vectors.
b) Simplify the surd carefully before taking the inverse cosine.

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 \(|\overrightarrow{BA}|=\sqrt{36+4+9}=7\).
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 direction of \(\overrightarrow{BA}\).
b) Multiply the unit vector along \(\overrightarrow{BA}\) by the required magnitude.

SECTION - C

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

26. The side of an equilateral triangle is increasing at the rate of 3 cm/s. At what rate its area increasing when the side of the triangle is 15 cm ? [3 Marks]

Answer:
1. Let side = a, so \(\dfrac{da}{dt}=3\) cm/s. Area of equilateral triangle \(A=\dfrac{\sqrt3}{4}a^2\).
2. \(\dfrac{dA}{dt}=\dfrac{\sqrt3}{2}a\dfrac{da}{dt}\).
3. At \(a=15\): \(\dfrac{dA}{dt}=\dfrac{\sqrt3}{2}\times15\times3=\dfrac{45\sqrt3}{2}\) cm²/s.

Teacher's Note:
a) This is a related rates problem: differentiate the area formula with respect to time.
b) Substitute the given side length only after differentiating.

27. Solve the following linear programming problem graphically :
Maximise Z = x + 2y
Subject to the constraints :
\(x - y \geq 0\)
\(x - 2y \geq -2\)
\(x \geq 0, y \geq 0\) [3 Marks]

[Figure: A graph showing lines \(x=y\), \(x-2y=-2\), with the feasible (unbounded) region shaded above these lines in the first quadrant, corner points at O(0,0) and A(2,2).]

Answer:
1. Corner points of the feasible region are O(0,0) and A(2,2).
2. Z(0,0) = 0, Z(2,2) = 2+4 = 6.
3. Since the feasible region is unbounded, plot \(x+2y \gt 6\); it has points common with the feasible region, so Z has no maximum value.

Teacher's Note:
a) For unbounded feasible regions, always check whether the region beyond the highest corner value also lies in the feasible region.
b) If it does, the objective function has no maximum (or minimum) value.

28. (a) Find : \(\displaystyle\int \dfrac{x+\sin x}{1+\cos x}\,dx\) [3 Marks]

Answer:
1. Using \(1+\cos x = 2\cos^2\dfrac{x}{2}\) and \(\sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\), the integral becomes \(\displaystyle\int x\cdot\dfrac{1}{2}\sec^2\dfrac{x}{2}\,dx + \int \tan\dfrac{x}{2}\,dx\).
2. Integrating the first term by parts: \(x\tan\dfrac{x}{2} - \displaystyle\int \tan\dfrac{x}{2}\,dx\).
3. Adding the second term, the two \(\int \tan\frac{x}{2}dx\) terms cancel, giving the result \(x\tan\dfrac{x}{2}+C\).

Teacher's Note:
a) Use the half-angle identities to simplify the denominator before integrating.
b) Integration by parts is required for the term containing x.

OR

(b) Evaluate : \(\displaystyle\int_0^{\frac{\pi}{4}} \dfrac{dx}{\cos^3 x \sqrt{2\sin 2x}}\) [3 Marks]

Answer:
1. Rewrite \(2\sin2x=4\sin x\cos x\), so the integral becomes \(\dfrac{1}{2}\displaystyle\int_0^{\pi/4}\dfrac{\sec^2 x(1+\tan^2 x)}{\sqrt{\tan x}}\,dx\).
2. Substitute \(\tan x=t\), \(\sec^2x\,dx=dt\): integral \(=\dfrac{1}{2}\displaystyle\int_0^1 \left(t^{-1/2}+t^{3/2}\right)dt\).
3. Evaluating gives \(\dfrac12\left[2\sqrt{t}+\dfrac25t^{5/2}\right]_0^1=\dfrac{6}{5}\).

Teacher's Note:
a) Express everything in terms of \(\tan x\) so a simple substitution works.
b) Keep track of the changed limits after substitution.

29. (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: \(\vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda(-\hat{i}+\hat{j}-2\hat{k})\) and \(\vec{r}=(\hat{i}-\hat{j}-\hat{k})+\mu(\hat{i}+2\hat{j}-2\hat{k})\); the direction ratios are not proportional, so the lines are not parallel.
2. \(\vec{a_2}-\vec{a_1}=\hat{j}-4\hat{k}\), \(\vec{b_1}\times\vec{b_2}=2\hat{i}-4\hat{j}-3\hat{k}\); \((\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) = -4+12=8 \neq 0\), 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) Two lines are skew if their direction vectors are not parallel and \((\vec a_2-\vec a_1)\cdot(\vec b_1\times \vec b_2)\neq 0\).
b) Use the shortest distance formula for skew lines directly after verifying.

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 wicket keeper divide the segment BF in ratio \(k:1\), so \(\vec{W}=\dfrac{k\vec{F}+\vec{B}}{k+1}\).
2. \(6\hat{i}+12\hat{j}=\left(\dfrac{12k+2}{k+1}\right)\hat{i}+\left(\dfrac{18k+8}{k+1}\right)\hat{j}\).
3. Solving gives \(k=\dfrac{2}{3}\), so the required ratio is 2 : 3.

Teacher's Note:
a) Use the section formula for a point dividing a line segment internally.
b) Equating either the x or y components is sufficient to find the ratio.

30. (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 \(=\sum X\cdot P(X)=0(p)+2(2p)+4(3p)+5(p)=21p\).
3. Mean \(=21\times\dfrac17=3\).

Teacher's Note:
a) The sum of all probabilities in a probability distribution must equal 1.
b) Mean of a discrete random variable is \(\sum X\cdot P(X)\).

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. \(P(E_1)=\dfrac13, P(E_2)=\dfrac23\), \(P(A|E_1)=0.4, P(A|E_2)=0.35\).
2. By the law of total probability: \(P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)\).
3. \(P(A)=\dfrac13\times0.4+\dfrac23\times0.35=\dfrac{11}{30}\).

Teacher's Note:
a) Identify the partition of the sample space (male/female) before applying the law of total probability.
b) Keep fractions consistent to avoid rounding errors.

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

Answer:
1. Required area \(=\displaystyle\int_{-6}^{0} y\,dx = 2\int_{-3}^{0}(x+3)\,dx\) (using symmetry about \(x=-3\)).
2. \(=2\left[\dfrac{(x+3)^2}{2}\right]_{-3}^{0}\).
3. \(=9\) square units.

Teacher's Note:
a) Split the modulus function at its vertex \(x=-3\) to write it without modulus.
b) The graph is symmetric about the vertical line through the vertex, which simplifies the integration.

SECTION - D

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

32. (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 \(x=\sin A, y=\sin B\), so \(A=\sin^{-1}x, B=\sin^{-1}y\). The equation becomes \(\cos A+\cos B=a(\sin A-\sin B)\).
2. Using sum-to-product formulas: \(2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)=2a\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)\).
3. This simplifies to \(\cot\left(\dfrac{A-B}{2}\right)=a \Rightarrow A-B=2\cot^{-1}a\).
4. So \(\sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a\), a constant.
5. Differentiating both sides w.r.t. x: \(\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{\sqrt{1-y^2}}\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}\).

Teacher's Note:
a) Substituting \(x=\sin A, y=\sin B\) converts the equation to a trigonometric identity.
b) Recognising that \(A-B\) is a constant is the key step before differentiating.

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 \(\dfrac{dy}{dx}=\dfrac{\cos\theta}{a\cot\theta\cos\theta}=\dfrac{\tan\theta}{a}\).
3. Differentiating w.r.t. \(\theta\): \(\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)=\dfrac{\sec^2\theta}{a}\).
4. \(\dfrac{d^2y}{dx^2}=\dfrac{\sec^2\theta}{a}\times\dfrac{d\theta}{dx}=\dfrac{\sec^3\theta\tan\theta}{a^2}\).
5. At \(\theta=\dfrac{\pi}{4}\): \(\dfrac{d^2y}{dx^2}=\dfrac{2\sqrt2}{a^2}\).

Teacher's Note:
a) Compute \(\dfrac{dx}{d\theta}\) and \(\dfrac{dy}{d\theta}\) separately for parametric differentiation.
b) Use \(\dfrac{d^2y}{dx^2}=\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)\Big/\dfrac{dx}{d\theta}\) and substitute the given value of \(\theta\) at the end.

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 \(f'(x)=0\) gives \(x=2,3\), both lying in [1,5].
3. Compute \(f(1)=24, f(2)=29, f(3)=28, f(5)=56\).
4. The absolute maximum value is 56 (at \(x=5\)) and the absolute minimum value is 24 (at \(x=1\)).

Teacher's Note:
a) Always evaluate the function at critical points as well as the endpoints of the interval.
b) The largest and smallest of these values give the absolute maximum and minimum.

34. (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 \(M(\lambda, 2\lambda+1, 3\lambda+2)\), so 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 to the line.
4. Since M is the midpoint of AA', \(A'=(1,0,7)\).
5. Equation of line AA': \(\dfrac{x-1}{0}=\dfrac{y-6}{-3}=\dfrac{z-3}{2}\).

Teacher's Note:
a) Find the foot of the perpendicular first, using the fact that direction ratios of AM must be perpendicular to the line's direction ratios.
b) 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 the line is \(P(\lambda-5, 4\lambda-3, -9\lambda+6)\).
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. So the 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) Parametrise a general point on the given line and apply the distance formula.
b) Solving the resulting quadratic in \(\lambda\) gives the required point.

35. 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 number of students in Sports, Music and Drama clubs. Given: \(x=y+z\), \(y=\dfrac{x}{2}+20\), \(x+y+z=180\), which gives \(x-y-z=0\), \(x-2y=-40\), \(x+y+z=180\).
2. This system is 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. \(|A|=-4\neq0\), so \(A^{-1}\) exists; \(\text{adj }A=\begin{bmatrix}-2 & 0 & -2\\ -1 & 2 & -1\\ 3 & -2 & -1\end{bmatrix}\), \(A^{-1}=\dfrac{1}{4}\begin{bmatrix}2 & 0 & 2\\ 1 & -2 & 1\\ -3 & 2 & 1\end{bmatrix}\).
4. \(X=A^{-1}B=\begin{bmatrix}90\\65\\25\end{bmatrix}\).
5. So the number of students allocated in Sports, Music and Drama clubs are 90, 65 and 25 respectively.

Teacher's Note:
a) Translate each condition into a linear equation carefully before forming the matrix equation.
b) Verify that \(|A|\neq 0\) to confirm that \(A^{-1}\) exists before proceeding.

SECTION - E

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

36. 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.

[Figure: A rectangular solar panel installation on a roof, divided into two sections by a partition running parallel to one side.]

Based on this information, answer the following questions :
(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) Count the two sides of length x, two sides of length y, and one extra partition of length y.
b) This gives a total of 2 sides of x and 3 sides of y.

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

Answer: From \(2x+3y=300\), \(y=\dfrac{300-2x}{3}\), so \(A(x)=xy=\dfrac{x}{3}(300-2x)=\dfrac{1}{3}(300x-2x^2)\).

Teacher's Note:
a) Express y in terms of x using the boundary equation, then substitute into A=xy.
b) This converts the two-variable problem into a single-variable optimisation problem.

(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. \(A(x)=\dfrac13(300x-2x^2) \Rightarrow \dfrac{dA}{dx}=\dfrac13(300-4x)\).
2. Setting \(\dfrac{dA}{dx}=0\) gives \(x=75\).
3. \(\dfrac{d^2A}{dx^2}=-\dfrac43 \lt 0\), so A is maximum at \(x=75\).
4. Maximum area \(=\dfrac{75}{3}(300-150)=3750\) m².

Teacher's Note:
a) A negative second derivative at the critical point confirms a maximum.
b) Substitute \(x=75\) back into the area function to find the actual 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. \(\dfrac{dA}{dx}=\dfrac13(300-4x)=0 \Rightarrow x=75\).
2. As x passes through 75 from left to right, \(\dfrac{dA}{dx}\) changes sign from positive to negative, so \(x=75\) is a point of maximum.
3. Maximum area \(=\dfrac{75}{3}(300-150)=3750\) m².

Teacher's Note:
a) The first derivative test checks the sign change of the derivative around the critical point.
b) A change from positive to negative confirms a local maximum.

37. 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: \(R_4\) is reflexive and transitive but not symmetric.

Teacher's Note:
a) Check that all pairs (a,a) are present for reflexivity.
b) Verify that no pair (a,b) with \(a \neq b\) has its reverse (b,a) present, to rule out symmetry.

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

Answer: \(R_5\) is reflexive and symmetric but not transitive.

Teacher's Note:
a) Check all diagonal pairs are present for reflexivity and all pairs have their reverse for symmetry.
b) Find a pair (a,b) and (b,c) present but (a,c) missing to show it is not transitive.

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

Answer: \(R_1\) and \(R_3\) are symmetric but neither reflexive nor transitive.

Teacher's Note:
a) A relation missing at least one diagonal pair is not reflexive.
b) Check for symmetry pair-by-pair and test transitivity by chaining pairs.

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 make the relation reflexive.
b) Then add the missing reverse pairs for every existing pair to make it symmetric, ensuring it remains transitive.

38. 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.

[Figure: An illustration showing stacks of coins decreasing in height from left to right, with a downward arrow labelled "% Interest Rates".]

Based on the above information, answer the following :
(i) What is the probability that a customer after availing the loan will default on the loan repayment ? [2 Marks]

Answer:
1. Let \(E_1, E_2, E_3\) be the events that the customer avails loan at fixed, floating and variable rate, with \(P(E_1)=\dfrac{1}{10}, P(E_2)=\dfrac{2}{10}, P(E_3)=\dfrac{7}{10}\).
2. Let A be the event that the customer defaults, with \(P(A|E_1)=\dfrac{5}{100}, P(A|E_2)=\dfrac{3}{100}, P(A|E_3)=\dfrac{1}{100}\).
3. By the law of total probability, \(P(A)=\dfrac{1}{10}\times\dfrac{5}{100}+\dfrac{2}{10}\times\dfrac{3}{100}+\dfrac{7}{10}\times\dfrac{1}{100}=\dfrac{18}{1000}=\dfrac{9}{500}\).

Teacher's Note:
a) Identify the three mutually exclusive events for the type of interest rate before applying the law of total probability.
b) Multiply each prior probability by its corresponding conditional probability of default and sum them.

(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. By Bayes' theorem, \(P(E_3|A)=\dfrac{P(E_3)P(A|E_3)}{P(A)}\).
2. \(P(E_3|A)=\dfrac{\frac{7}{10}\times\frac{1}{100}}{\frac{18}{1000}}\).
3. \(P(E_3|A)=\dfrac{7}{18}\).

Teacher's Note:
a) Use Bayes' theorem to reverse the conditional probability from cause to effect.
b) The denominator is the total probability of default calculated in part (i).

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