CBSE Class 12 Maths Question Paper 2026 Solved Code 65-1-1

SECTION - A

Question 1 to 20 are multiple choice questions of 1 mark each.

1. If \( 2\cos^{-1}x = y \), then [1 Mark]
(A) \( 0 \leq y \leq \pi \)
(B) \( -\pi \leq y \leq \pi \)
(C) \( 0 \leq y \leq 2\pi \)
(D) \( -\pi \leq y \leq 0 \)

Answer: (C) \( 0 \leq y \leq 2\pi \)

Teacher's Note:
a) The range of \( \cos^{-1}x \) is \( [0, \pi] \), so multiplying by 2 gives \( [0, 2\pi] \).
b) Do not confuse the range of \( \cos^{-1}x \) with the range of \( y \) here.

2. Which of the following cannot be the order of a row-matrix ? [1 Mark]
(A) \( 2 \times 1 \)
(B) \( 1 \times 2 \)
(C) \( 1 \times 1 \)
(D) \( 1 \times n \)

Answer: (A) \( 2 \times 1 \)

Teacher's Note:
a) A row matrix always has exactly one row, so its order must be \( 1 \times n \).
b) \( 2 \times 1 \) represents a column matrix, not a row matrix.

3. Which of the following properties is/are true for two matrices of suitable orders ?
(i) \( (A+B)' = A' + B' \)
(ii) \( (A-B)' = B' - A' \)
(iii) \( (AB)' = A'B' \)
(iv) \( (kAB)' = kB'A' \) (k is a scalar) [1 Mark]
(A) (i) only
(B) (i), (ii) and (iii)
(C) (i) and (ii)
(D) (i) and (iv)

Answer: (D) (i) and (iv)

Teacher's Note:
a) \( (A-B)' = A' - B' \), not \( B' - A' \), so (ii) is false.
b) \( (AB)' = B'A' \), not \( A'B' \), so (iii) is false.
c) Remember the reversal law for transpose of a product of matrices.

4. If \( \Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} \) and \( \Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix} \), then [1 Mark]
(A) \( \Delta_1 = 2\Delta_2 \)
(B) \( \Delta_2 = -2\Delta_1 \)
(C) \( \Delta_1 = \Delta_2 \)
(D) \( \Delta_2 = -\Delta_1 \)

Answer: (B) \( \Delta_2 = -2\Delta_1 \)

Teacher's Note:
a) \( \Delta_1 = 1 \times 2 \times 3 = 6 \) (diagonal matrix).
b) \( \Delta_2 \) is obtained by swapping rows 1 and 2 of a similar diagonal matrix, which changes the sign and gives \( \Delta_2 = -12 \).
c) Swapping two rows of a determinant changes only its sign.

5. One of the values of x for which \( \begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1 \) is [1 Mark]
(A) 0
(B) \( \dfrac{\pi}{4} \)
(C) \( \dfrac{\pi}{3} \)
(D) \( \dfrac{\pi}{2} \)

Answer: (B) \( \dfrac{\pi}{4} \)

Teacher's Note:
a) Expanding gives \( \sin x \cos x + \sin x \cos x = \sin 2x = 1 \).
b) \( \sin 2x = 1 \) gives \( 2x = \dfrac{\pi}{2} \), so \( x = \dfrac{\pi}{4} \).

6. If A and B are skew symmetric matrices of same order, then which of the following matrices is also skew symmetric ? [1 Mark]
(A) AB
(B) AB + BA
(C) \( (A+B)^2 \)
(D) A - B

Answer: (D) A - B

Teacher's Note:
a) A linear combination of skew-symmetric matrices is always skew-symmetric.
b) Products like AB or AB+BA are generally symmetric, not skew-symmetric, when A and B are skew-symmetric.

7. The least value of \( f(x) = x^3 - 12x, x \in [0, 3] \) is [1 Mark]
(A) -16
(B) -9
(C) 0
(D) 16

Answer: (A) -16

Teacher's Note:
a) \( f'(x) = 3x^2 - 12 = 0 \) gives \( x = 2 \) within \( [0,3] \).
b) Compare \( f(0) = 0, f(2) = -16, f(3) = -9 \); the least value is \( -16 \).

8. If \( \displaystyle\int \dfrac{3ax}{b^2 + c^2x^2}\,dx = A \log |b^2 + c^2x^2| + K \), then the value of A is [1 Mark]
(A) 3a
(B) \( \dfrac{3a}{2b^2} \)
(C) \( \dfrac{3a}{b^2c^2} \)
(D) \( \dfrac{3a}{2c^2} \)

Answer: (D) \( \dfrac{3a}{2c^2} \)

Teacher's Note:
a) Put \( t = b^2 + c^2x^2 \), so \( dt = 2c^2x\,dx \).
b) This gives \( A = \dfrac{3a}{2c^2} \) after matching coefficients.

9. The value of \( \displaystyle\int_{-1}^{1} \dfrac{x^3}{x^2 + 2|x| + 1}\,dx \) is [1 Mark]
(A) 0
(B) \( \log 2 \)
(C) \( 2\log 2 \)
(D) \( \dfrac{1}{2}\log 2 \)

Answer: (A) 0

Teacher's Note:
a) The numerator \( x^3 \) is odd and the denominator is an even function of x.
b) An odd function integrated over a symmetric interval \( [-1,1] \) gives 0.

10. The area bounded by the curve \( y = x|x| \), x-axis and the ordinates \( x = -1 \) and \( x = 1 \) is given by [1 Mark]
(A) 0
(B) \( \dfrac{1}{3} \)
(C) \( \dfrac{2}{3} \)
(D) \( \dfrac{4}{3} \)

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

Teacher's Note:
a) For \( x \geq 0 \), \( y = x^2 \); for \( x \lt 0 \), \( y = -x^2 \).
b) Area \( = \int_{-1}^{0} x^2\,dx + \int_{0}^{1} x^2\,dx = \dfrac{1}{3}+\dfrac{1}{3} = \dfrac{2}{3} \).

11. The integrating factor of differential equation \( R\dfrac{dx}{dy} + Px = Q \) where P, Q, R are functions of y is [1 Mark]
(A) \( e^{\int \frac{P}{Q}\,dy} \)
(B) \( e^{\int P\,dy} \)
(C) \( e^{\int \frac{P}{R}\,dy} \)
(D) \( e^{\int \frac{P}{R}\,dx} \)

Answer: (C) \( e^{\int \frac{P}{R}\,dy} \)

Teacher's Note:
a) Divide throughout by R to get standard form \( \dfrac{dx}{dy} + \dfrac{P}{R}x = \dfrac{Q}{R} \).
b) Integrating factor is \( e^{\int (\text{coefficient of } x)\,dy} \).

12. The order and degree of the differential equation \( \dfrac{d}{dx}(e^y) = 0 \) respectively are [1 Mark]
(A) 0, 1
(B) 1, 1
(C) 2, 1
(D) 1, not defined

Answer: (B) 1, 1

Teacher's Note:
a) \( \dfrac{d}{dx}(e^y) = e^y \dfrac{dy}{dx} = 0 \), so highest derivative present is \( \dfrac{dy}{dx} \), giving order 1.
b) Since \( \dfrac{dy}{dx} \) occurs to the power 1, the degree is 1.

13. The value of p for which vectors \( \hat{i}+2\hat{j}+3\hat{k} \) and \( 2\hat{i}-p\hat{j}+\hat{k} \) are perpendicular to each other is [1 Mark]
(A) 0
(B) 1
(C) \( \dfrac{5}{2} \)
(D) \( -\dfrac{5}{2} \)

Answer: (C) \( \dfrac{5}{2} \)

Teacher's Note:
a) For perpendicular vectors, dot product = 0.
b) \( 2 - 2p + 3 = 0 \) gives \( p = \dfrac{5}{2} \).

14. The value of m for which the points with position vectors \( -\hat{i}-\hat{j}+2\hat{k} \), \( 2\hat{i}+m\hat{j}+5\hat{k} \) and \( 3\hat{i}+11\hat{j}+6\hat{k} \) are collinear, is [1 Mark]
(A) 8
(B) -8
(C) 2
(D) \( \dfrac{5}{2} \)

Answer: (A) 8

Teacher's Note:
a) Find vectors joining consecutive points and check that they are proportional.
b) Equating ratios of corresponding components gives \( m = 8 \).

15. If \( |\vec{a}| = 8, |\vec{b}| = 3 \) and \( |\vec{a} \times \vec{b}| = 12 \), then the value of \( |\vec{a} \cdot \vec{b}| \) [1 Mark]
(A) \( 6\sqrt{3} \)
(B) \( 8\sqrt{3} \)
(C) \( 12\sqrt{3} \)
(D) \( 3\sqrt{12} \)

Answer: (C) \( 12\sqrt{3} \)

Teacher's Note:
a) Use the identity \( |\vec{a}\times\vec{b}|^2 + |\vec{a}\cdot\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 \).
b) Substituting gives \( |\vec{a}\cdot\vec{b}|^2 = 576-144=432 \), so \( |\vec{a}\cdot\vec{b}| = 12\sqrt{3} \).

16. The length of perpendicular drawn from point (2, 5, 7) on line \( \dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{0} \) is [1 Mark]
(A) 2
(B) 5
(C) \( \sqrt{74} \)
(D) \( \sqrt{78} \)

Answer: (C) \( \sqrt{74} \)

Teacher's Note:
a) The line is the x-axis, so the foot of perpendicular from (2,5,7) is (2,0,0).
b) Distance \( = \sqrt{0+25+49} = \sqrt{74} \).

17. The feasible region of a linear programming problem with objective function \( Z = 5x+7y \) is shown below :
The maximum value of Z - minimum value of Z is [1 Mark]
(A) 8
(B) 29
(C) 35
(D) 43

[Figure: Graph of a feasible region shaded with vertical lines, with vertices at (0,2), (3,4) marked on the boundary and the boundary meeting the x-axis at (7,0); origin (0,0) also lies in the shaded region.]

Answer: (D) 43

Teacher's Note:
a) Evaluate Z at all corner points: (0,0)=0, (7,0)=35, (3,4)=43, (0,2)=14.
b) Maximum Z = 43 at (3,4), minimum Z = 0 at (0,0), so the difference is 43.

18. The degree of an objective function of a linear programming problem is [1 Mark]
(A) 0
(B) 1
(C) 2
(D) Any natural number

Answer: (B) 1

Teacher's Note:
a) The objective function of an LPP is always a linear function of the decision variables.
b) A linear function has degree 1.

Assertion - Reason Based Questions

Direction : Questions number 19 and 20 are Assertion and Reason 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 codes (A), (B), (C) and (D) as given below.

19. Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \dfrac{2}{3} \).
Reason (R) : For any two events A and B, \( P(A|B) = \dfrac{P(A \cup B)}{P(B)} \) [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 and Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is true.

Answer: (C) Assertion (A) is true and Reason (R) is false.

Teacher's Note:
a) Odd numbers are {1,3,5}, prime numbers are {2,3,5}; primes among odd numbers are {3,5}, so \( P(\text{prime}|\text{odd}) = \dfrac{2/6}{3/6} = \dfrac{2}{3} \), so (A) is true.
b) The correct formula is \( P(A|B) = \dfrac{P(A\cap B)}{P(B)} \), not \( P(A\cup B) \), so (R) is false.

20. Assertion (A) : Lines given by \( x = py+q, z = ry+s \) and \( x = p'y+q', z = r'y+s' \) are perpendicular to each other when \( pp' + rr' = 1 \).
Reason (R) : Two lines \( \vec{r} = \vec{a_1} + \lambda \vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu \vec{b_2} \) are perpendicular to each other if \( \vec{b_1} \cdot \vec{b_2} = 0 \). [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 and Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is true.

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

Teacher's Note:
a) Direction ratios of the lines are (p,1,r) and (p',1,r'); perpendicularity requires \( pp'+1+rr'=0 \), i.e. \( pp'+rr'=-1 \), not 1, so (A) is false.
b) The stated condition in Reason (R) for perpendicularity of two lines using dot product of direction vectors is the standard correct result.

SECTION - B

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

21. (a) Check whether function f(x) defined as
f(x) = \( \dfrac{|x-3|}{2(x-3)} \), \( x \lt 3 \)
f(x) = \( \dfrac{x-6}{6} \), \( x \geq 3 \)
is continuous at \( x = 3 \) or not ? [2 Marks]

Answer:
1. For \( x \lt 3 \), \( f(x) = \dfrac{-(x-3)}{2(x-3)} = -\dfrac{1}{2} \), so LHL \( = -\dfrac{1}{2} \).
2. RHL \( = \displaystyle\lim_{x\to 3^+} \dfrac{x-6}{6} = -\dfrac{1}{2} \), and \( f(3) = \dfrac{3-6}{6} = -\dfrac{1}{2} \).
3. Since LHL = RHL = f(3), the function is continuous at \( x = 3 \).

Teacher's Note:
a) Always simplify the modulus expression first based on the sign of \( x-3 \) for \( x \lt 3 \).
b) Continuity requires LHL = RHL = value of the function at that point.

OR

(b) If \( \sqrt{3}(x^2+y^2) = 4xy \), then find \( \dfrac{dy}{dx} \) at \( \left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right) \). [2 Marks]

Answer:
1. Differentiating implicitly: \( \sqrt{3}(2x+2y y') = 4(xy'+y) \).
2. Solving, \( \dfrac{dy}{dx} = \dfrac{2y-\sqrt{3}x}{\sqrt{3}y-2x} \).
3. Substituting the point: \( \dfrac{dy}{dx} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3} \).

Teacher's Note:
a) Differentiate both sides with respect to x, treating y as a function of x.
b) Collect all \( \dfrac{dy}{dx} \) terms on one side before substituting the given point.

22. A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm³/min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is \( \dfrac{\pi}{6} \). [2 Marks]

[Figure: An inverted cone-shaped bottle spraying perfume mist, with a second cone diagram showing semi-vertical angle 30 degrees and the perfume level marked as 10 mm from the vertex.]

Answer:
1. Given \( \dfrac{dv}{dt} = -1 \) mm³/min, and \( \dfrac{r}{h} = \tan\dfrac{\pi}{6} \), so \( r = \dfrac{h}{\sqrt{3}} \).
2. Volume \( v = \dfrac{1}{3}\pi r^2 h = \dfrac{\pi h^3}{9} \), so \( \dfrac{dv}{dt} = \dfrac{\pi h^2}{3}\dfrac{dh}{dt} \).
3. At \( h = 10 \): \( \dfrac{dh}{dt} = -\dfrac{3}{\pi(10)^2} = -\dfrac{3}{100\pi} \) mm/min.

Teacher's Note:
a) Express radius in terms of height using the semi-vertical angle before substituting into the volume formula.
b) The negative sign shows the level is dropping; state the final answer with correct units.

23. Find the vector of magnitude 14 in the direction of \( \overrightarrow{QP} \), where P and Q are the points (1, 3, 2) and (-1, 0, 8) respectively. [2 Marks]

Answer:
1. \( \overrightarrow{QP} = 2\hat{i}+3\hat{j}-6\hat{k} \), so \( |\overrightarrow{QP}| = \sqrt{4+9+36} = 7 \).
2. Unit vector \( = \dfrac{2}{7}\hat{i}+\dfrac{3}{7}\hat{j}-\dfrac{6}{7}\hat{k} \).
3. Required vector \( = 14 \times \) unit vector \( = 4\hat{i}+6\hat{j}-12\hat{k} \).

Teacher's Note:
a) Always find \( \overrightarrow{QP} = \) Position vector of P minus Position vector of Q, not the other way round.
b) Multiply the unit vector by the required magnitude to get the final answer.

24. Vectors \( \vec{a} = 3\hat{i}-2\hat{j}+2\hat{k} \) and \( \vec{b} = \hat{i}+2\hat{k} \) represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths. [2 Marks]

Answer:
1. One diagonal \( = \vec{a}+\vec{b} = 4\hat{i}-2\hat{j}+4\hat{k} \), with length \( \sqrt{16+4+16} = 6 \).
2. Other diagonal \( = \vec{a}-\vec{b} = 2\hat{i}-2\hat{j} \), with length \( \sqrt{4+4} = 2\sqrt{2} \).

Teacher's Note:
a) The two diagonals of a parallelogram formed by adjacent sides \( \vec{a}, \vec{b} \) are \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \).
b) Find the magnitude of each diagonal vector separately using the distance formula.

25. (a) Simplify : \( \tan^{-1}\left(\dfrac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right), 0 \lt x \lt \dfrac{\pi}{4} \). [2 Marks]

Answer:
1. Dividing numerator and denominator by \( \cos 2x \): expression \( = \tan^{-1}\left(\dfrac{1-\tan 2x}{1+\tan 2x}\right) \).
2. This equals \( \tan^{-1}\left[\tan\left(\dfrac{\pi}{4}-2x\right)\right] = \dfrac{\pi}{4}-2x \).

Teacher's Note:
a) Use the identity \( \tan(A-B) = \dfrac{\tan A - \tan B}{1+\tan A \tan B} \) with \( A=\pi/4 \).
b) The given range of x ensures the angle lies within the principal value range of \( \tan^{-1} \).

OR

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

Answer:
1. \( \sin^{-1}1 = \dfrac{\pi}{2} \) and \( \cos^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3} \).
2. Expression \( = \tan\left(\dfrac{\pi}{2}-\dfrac{2\pi}{3}\right) = \tan\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{\sqrt{3}} \).

Teacher's Note:
a) Recall standard values: \( \sin^{-1}1 = \pi/2 \) and \( \cos^{-1}(-1/2) = 2\pi/3 \).
b) Simplify the angle before applying the tangent function.

SECTION - C

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

26. Evaluate : \( \displaystyle\int_{0}^{1} x\tan^{-1}x\,dx \). [3 Marks]

Answer:
1. Using integration by parts: \( = \left[\tan^{-1}x \cdot \dfrac{x^2}{2}\right]_0^1 - \displaystyle\int_0^1 \dfrac{1}{1+x^2}\cdot\dfrac{x^2}{2}\,dx \).
2. This gives \( \dfrac{\pi}{8} - \dfrac{1}{2}\displaystyle\int_0^1\left(1-\dfrac{1}{1+x^2}\right)dx = \dfrac{\pi}{8} - \dfrac{1}{2}\left[x-\tan^{-1}x\right]_0^1 \).
3. Simplifying, the value \( = \dfrac{\pi}{8} - \dfrac{1}{2} + \dfrac{\pi}{8} = \dfrac{\pi}{4} - \dfrac{1}{2} \).

Teacher's Note:
a) Take \( \tan^{-1}x \) as the first function (ILATE rule) for integration by parts.
b) Simplify \( \dfrac{x^2}{1+x^2} \) as \( 1 - \dfrac{1}{1+x^2} \) before integrating.

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

Answer:
1. Rationalising: \( \displaystyle\int \dfrac{x+2}{\sqrt{x^2-4}}\,dx = \dfrac{1}{2}\displaystyle\int\dfrac{2x}{\sqrt{x^2-4}}\,dx + 2\displaystyle\int\dfrac{1}{\sqrt{x^2-4}}\,dx \).
2. First integral \( = \sqrt{x^2-4} \).
3. Second integral \( = 2\log\left|x+\sqrt{x^2-4}\right| \); so total \( = \sqrt{x^2-4} + 2\log\left|x+\sqrt{x^2-4}\right| + C \).

Teacher's Note:
a) Multiply numerator and denominator by \( \sqrt{x+2} \) to simplify the surd.
b) Split into a standard form \( \int \dfrac{2x}{\sqrt{x^2-a^2}}dx \) and \( \int \dfrac{1}{\sqrt{x^2-a^2}}dx \).

OR

(b) Find : \( \displaystyle\int \dfrac{x^2}{(x^2+9)(x^2+16)}\,dx \) [3 Marks]

Answer:
1. Put \( t = x^2 \): \( \dfrac{t}{(t+9)(t+16)} = \dfrac{A}{t+9}+\dfrac{B}{t+16} \), giving \( A = -\dfrac{9}{7}, B = \dfrac{16}{7} \).
2. Integral \( = -\dfrac{9}{7}\displaystyle\int\dfrac{1}{x^2+9}dx + \dfrac{16}{7}\displaystyle\int\dfrac{1}{x^2+16}dx \).
3. This gives \( -\dfrac{3}{7}\tan^{-1}\left(\dfrac{x}{3}\right) + \dfrac{4}{7}\tan^{-1}\left(\dfrac{x}{4}\right) + C \).

Teacher's Note:
a) Substitute \( t = x^2 \) to convert into a partial fractions problem.
b) Use the standard result \( \int \dfrac{1}{x^2+a^2}dx = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} \) for each term.

28. If \( I_1 = \displaystyle\int_{-\pi/4}^{\pi/4} \dfrac{dx}{1+\cos 2x} \) and \( I_2 = \displaystyle\int_{-1/2}^{1/2} |x|\,dx \), then show that \( I_1 - 4I_2 = 0 \). [3 Marks]

Answer:
1. Since \( 1+\cos 2x = 2\cos^2x \), \( I_1 = \displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{1}{2}\sec^2x\,dx = \left[\tan x\right]_0^{\pi/4} \times 2 \times \dfrac{1}{2} = 1 \) (using the even function property).
2. \( I_2 = 2\displaystyle\int_0^{1/2} x\,dx = 2\left[\dfrac{x^2}{2}\right]_0^{1/2} = \dfrac{1}{4} \) (since \( |x| \) is even).
3. Hence \( I_1 - 4I_2 = 1 - 4\left(\dfrac{1}{4}\right) = 0 \), as required.

Teacher's Note:
a) Use the property that for even functions, \( \int_{-a}^{a}f(x)dx = 2\int_0^a f(x)dx \) to simplify both integrals.
b) Convert \( 1+\cos2x \) to \( 2\cos^2x \) before integrating.

29. (a) Find the general solution of the following differential equation :
\( x^2\dfrac{dy}{dx} = x^2+xy+y^2 \) [3 Marks]

Answer:
1. Rewriting: \( \dfrac{dy}{dx} = 1+\dfrac{y}{x}+\left(\dfrac{y}{x}\right)^2 \); this is a homogeneous equation, so put \( y = vx \), giving \( v+x\dfrac{dv}{dx} = 1+v+v^2 \).
2. This gives \( x\dfrac{dv}{dx} = 1+v^2 \), so \( \dfrac{dv}{1+v^2} = \dfrac{dx}{x} \).
3. Integrating both sides: \( \tan^{-1}v = \log|x|+C \), so \( \tan^{-1}\left(\dfrac{y}{x}\right) = \log|x|+C \).

Teacher's Note:
a) Recognise this as a homogeneous equation and substitute \( y = vx \).
b) Do not forget to replace v back with \( y/x \) in the final answer.

OR

(b) Find the particular solution of the differential equation \( xy\dfrac{dy}{dx} = (x+2)(y+2) \), given that \( y(1) = -1 \). [3 Marks]

Answer:
1. Separating variables: \( \dfrac{y}{y+2}dy = \dfrac{x+2}{x}dx \), i.e., \( \left(1-\dfrac{2}{y+2}\right)dy = \left(1+\dfrac{2}{x}\right)dx \).
2. Integrating: \( y - 2\log|y+2| = x + 2\log|x| + C \).
3. Using \( y(1) = -1 \): \( C = -2 \), so the particular solution is \( y - 2\log|y+2| = x + 2\log|x| - 2 \).

Teacher's Note:
a) Split each fraction so that variables are separated before integrating.
b) Substitute the initial condition after integrating to find the constant C.

30. Solve the following linear programming problem graphically :
Minimize Z = 13x - 15y
Subject to constraints
\( x + y \leq 7 \),
\( 2x - 3y + 6 \geq 0 \),
\( x \geq 0, y \geq 0 \) [3 Marks]

Answer:
1. The corner points of the feasible region are (0,0), (7,0), (3,4) and (0,2).
2. Evaluating Z at these points: Z(0,0) = 0, Z(7,0) = 91, Z(3,4) = -21, Z(0,2) = -30.
3. The minimum value of Z is -30, attained at \( x=0, y=2 \).

Teacher's Note:
a) Plot the boundary lines \( x+y=7 \) and \( 2x-3y+6=0 \) and identify the common shaded (feasible) region.
b) Evaluate the objective function only at the corner points of the feasible region.

31. (a) Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball. [3 Marks]

Answer:
1. Let \( E_1 \): number less than 3 (probability \( 2/6 \)), \( E_2 \): number \( \geq 3 \) (probability \( 4/6 \)).
2. \( P(\text{red}|E_1) = 3/7 \) (bag I has 7 balls) and \( P(\text{red}|E_2) = 8/14 \) (bag II has 14 balls).
3. By total probability, \( P(\text{red}) = \dfrac{2}{6}\times\dfrac{3}{7} + \dfrac{4}{6}\times\dfrac{8}{14} = \dfrac{11}{21} \).

Teacher's Note:
a) Identify the two mutually exclusive events based on the die outcome before applying the law of total probability.
b) Keep the conditional probabilities of drawing red from each bag separate and clear.

OR

(b) The probability of simultaneous occurrence of atleast one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that \( P(X') + P(Y') = 2 - 2a + b \). [3 Marks]

Answer:
1. Given \( P(X \cup Y) = a \) and \( P(X\cup Y) - P(X\cap Y) = b \) (exactly one occurs), so \( P(X\cap Y) = a - b \).
2. \( P(X') + P(Y') = [1-P(X)]+[1-P(Y)] = 2 - [P(X)+P(Y)] \).
3. Since \( P(X)+P(Y) = P(X\cup Y)+P(X\cap Y) = a+(a-b) \), we get \( P(X')+P(Y') = 2-2a+b \).

Teacher's Note:
a) "Exactly one occurs" corresponds to \( P(X\cup Y) - P(X\cap Y) \), a key relation to remember.
b) Use the complement formula \( P(X)+P(Y) = P(X\cup Y)+P(X\cap Y) \) to finish the proof.

SECTION - D

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

32. (a) A relation R is defined on Z, the set of integers, as
R = {(x, y) : |x - y| is divisible by a prime number 'p', x, y \( \in \) Z}
check whether R is an equivalence relation or not. [5 Marks]

Answer:
1. Reflexive: For any \( x \in Z \), \( |x-x| = 0 \), which is divisible by p, so \( (x,x) \in R \).
2. Symmetric: If \( (x,y) \in R \), then \( |x-y| \) is divisible by p; since \( |x-y| = |y-x| \), \( (y,x) \in R \).
3. Transitive: If \( (x,y), (y,z) \in R \), then \( x-y \) and \( y-z \) are both divisible by p, so their sum \( x-z \) is divisible by p, giving \( |x-z| \) divisible by p, so \( (x,z) \in R \).
4. Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Teacher's Note:
a) Check all three properties separately with clear justification for full marks.
b) Use the fact that a sum of two numbers divisible by p is also divisible by p, to prove transitivity.

OR

(b) A function \( f : R - \left\{\dfrac{3}{5}\right\} \longrightarrow R - \left\{\dfrac{3}{5}\right\} \) is defined as \( f(x) = \dfrac{3x+2}{5x-3} \). Show that f is one-one and onto. [5 Marks]

Answer:
1. Let \( f(x_1) = f(x_2) \): \( \dfrac{3x_1+2}{5x_1-3} = \dfrac{3x_2+2}{5x_2-3} \).
2. Cross-multiplying and simplifying gives \( x_1 = x_2 \), so f is one-one.
3. For any \( y \in R-\{3/5\} \) (codomain), solving \( y = \dfrac{3x+2}{5x-3} \) gives \( x = \dfrac{3y+2}{5y-3} \), which lies in the domain.
4. Since every y in the codomain has a corresponding x in the domain, f is onto.

Teacher's Note:
a) For one-one, simplify the cross-multiplied equation carefully to show \( x_1=x_2 \).
b) For onto, express x explicitly in terms of y and confirm it lies in the domain.

33. (a) If \( A = \begin{bmatrix} 0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0 \end{bmatrix} \), find \( A^{-1} \) and use it to solve the following system of equations :
\( -2y+z=7, 2x-y-z=8, x-2y=10 \) [5 Marks]

Answer:
1. \( |A| = -1 \) (non-zero, so \( A^{-1} \) exists).
2. \( \text{adj}\,A = \begin{bmatrix} -2 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} \), so \( A^{-1} = \dfrac{1}{|A|}\text{adj}\,A = \begin{bmatrix} 2 & 1 & 3 \\ 2 & 1 & 2 \\ -3 & -2 & -4 \end{bmatrix} \).
3. The given system corresponds to \( A^T X = B \), where \( B = \begin{bmatrix}7\\8\\10\end{bmatrix} \); so \( X = (A^{-1})^T B \).
4. Computing, \( x = 0, y = -5, z = -3 \).

Teacher's Note:
a) Notice that the coefficient matrix of the given system is \( A^T \), not A, so use \( (A^{-1})^T \) to solve.
b) Verify the final solution by substituting back into the original equations.

OR

(b) If \( \begin{bmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{bmatrix} \) is a singular matrix, then find all values of x where \( x \in \left[0, \dfrac{\pi}{2}\right] \). [5 Marks]

Answer:
1. Since the matrix is singular, its determinant equals 0, which simplifies to \( 2\cos 2x + \sin 3x = 2 \).
2. Using double and triple angle formulas, this reduces to \( 4\sin^3x + 4\sin^2x - 3\sin x = 0 \).
3. Factorising: \( \sin x(2\sin x-1)(2\sin x+3) = 0 \).
4. Since \( \sin x \neq -\dfrac{3}{2} \), the solutions in \( [0,\pi/2] \) are \( x = 0 \) and \( x = \dfrac{\pi}{6} \).

Teacher's Note:
a) Expand the determinant carefully and set it equal to zero for a singular matrix.
b) Reject any root of the factorised equation that gives \( |\sin x| \gt 1 \).

34. If \( x = \cos t, y = \cos mt \), prove that \( (1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} + m^2y = 0 \). [5 Marks]

Answer:
1. \( \dfrac{dx}{dt} = -\sin t \), \( \dfrac{dy}{dt} = -m\sin(mt) \), so \( \sin t\dfrac{dy}{dx} = m\sin(mt) \).
2. Differentiating both sides with respect to x: \( \sin t\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx}\cos t\dfrac{dt}{dx} = m^2\cos(mt)\dfrac{dt}{dx} \).
3. Using \( \dfrac{dt}{dx} = -\dfrac{1}{\sin t} \) and \( \cos t = x \), this simplifies to \( (1-\cos^2t)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = -m^2\cos(mt) \).
4. Since \( \cos^2t = x^2 \) and \( \cos(mt) = y \), we get \( (1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} + m^2y = 0 \), as required.

Teacher's Note:
a) Differentiate the relation \( \sin t \, dy/dx = m\sin(mt) \) again with respect to x, applying the chain rule to \( dt/dx \).
b) Substitute \( \cos t = x \) and \( \cos(mt) = y \) at the end to match the required form.

35. Check whether the lines given by \( \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4} \) and \( \dfrac{x-4}{5} = \dfrac{y-1}{2} = z \) are parallel or not. If parallel, find the distance between them, otherwise find their point of intersection, if the lines are intersecting. [5 Marks]

Answer:
1. Direction ratios of the lines are \( (2,3,4) \) and \( (5,2,1) \); since these are not proportional, the lines are not parallel.
2. A general point on the first line is \( (2\lambda+1, 3\lambda+2, 4\lambda+3) \) and on the second is \( (5\mu+4, 2\mu+1, \mu) \).
3. Equating coordinates gives \( 2\lambda-5\mu=3 \) and \( 3\lambda-2\mu=-1 \); solving these two equations gives \( \lambda=-1, \mu=-1 \).
4. This satisfies the third equation \( 4\lambda+3=\mu \), so the lines intersect at the point \( (-1,-1,-1) \).

Teacher's Note:
a) First check proportionality of direction ratios to decide between parallel and intersecting cases.
b) Always verify the values of the parameters in the third equation to confirm the lines actually intersect.

SECTION - E

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

36. An online delivery company in a city has 5000 subscribers and collects annual subscription fees of Rs. 300 per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of Rs. 1, ten subscribers will discontinue. Assume that the company increased the annual fee by Rs. x.

[Figure: An illustration of a delivery person handing a package at a doorstep, alongside a graph showing the number of subscribers on the y-axis decreasing linearly as the subscription fee on the x-axis increases.]

(i) How many subscribers will discontinue after an increase of Rs. x in annual fee ? [1 Mark]

Answer: \( 10x \) subscribers will discontinue.

Teacher's Note:
a) Since 10 subscribers leave for every Rs. 1 increase, for Rs. x increase, \( 10x \) subscribers leave.
b) This forms the basis for setting up the revenue function.

(ii) If R(x) denotes the total revenue collected after the increase of Rs. x in subscription fee, express R(x) as a function of x. [1 Mark]

Answer: \( R(x) = (5000-10x)(300+x) \).

Teacher's Note:
a) Revenue = (number of remaining subscribers) \( \times \) (new fee per subscriber).
b) Remaining subscribers = \( 5000-10x \) and new fee = \( 300+x \).

(iii) (a) Find the value of x for which R(x) is maximum. [2 Marks]

Answer:
1. \( R(x) = 1500000 + 2000x - 10x^2 \), so \( R'(x) = 2000-20x \).
2. Setting \( R'(x)=0 \) gives \( x = 100 \); since \( R''(x) = -20 \lt 0 \), R(x) is maximum at \( x = 100 \).

Teacher's Note:
a) Simplify R(x) first before differentiating to avoid errors.
b) Use the second derivative test to confirm it is a maximum, not a minimum.

OR

(iii) (b) Find the sub-intervals of (0, 5000) in which R(x) is increasing and decreasing. [2 Marks]

Answer:
1. \( R'(x) = 2000-20x \), which is zero at \( x=100 \).
2. \( R'(x) \gt 0 \) for \( x \in (0,100) \), so R(x) is increasing there; \( R'(x) \lt 0 \) for \( x \in (100,5000) \), so R(x) is decreasing there.

Teacher's Note:
a) The sign of \( R'(x) \) on either side of the critical point \( x=100 \) determines increasing/decreasing behaviour.
b) State both sub-intervals clearly with correct inequality signs.

37. In an online jackpot, there is one first prize of Rs. 3,00,000, two second prizes of Rs. 2,00,000 each and three third prizes of Rs. 50,000 each.
A total of 1,00,000 jackpot tickets each costing Rs. 100 were sold there by raising a fund of Rs. 1,00,00,000.
Rohan bought one ticket.

[Figure: An illustration of a "Lucky Ticket" booth with a ticket showing "WIN" and prize amounts of Rs. 200,000 and Rs. 100,000.]

(i) What are the possible amounts, the person can win ? [1 Mark]

Answer: The person can win Rs. 3,00,000, Rs. 2,00,000, or Rs. 50,000.

Teacher's Note:
a) List the amounts as given directly by the three prize categories.
b) A ticket that is not a winning ticket earns no prize amount.

(ii) (a) What is the probability that the person wins atleast Rs. 2,00,000 ? [2 Marks]

Answer: The number of tickets winning at least Rs. 2,00,000 is \( 1+2=3 \) out of 1,00,000, so the required probability \( = \dfrac{3}{1,00,000} \).

Teacher's Note:
a) "At least Rs. 2,00,000" includes both the first prize and second prizes.
b) Probability = favourable winning tickets divided by total tickets sold.

OR

(ii) (b) What is the probability that the person does not win any amount ? [2 Marks]

Answer: Total winning tickets = \( 1+2+3=6 \); probability of not winning \( = 1-\dfrac{6}{1,00,000} = \dfrac{99994}{1,00,000} = \dfrac{49997}{50000} \).

Teacher's Note:
a) Add all winning tickets from all three prize categories first.
b) Use the complement rule: P(no win) = 1 - P(win any prize).

(iii) In another jackpot, Rohan also bought a ticket having a prize money of Rs. 5,00,000. The chances of winning the jackpot are 1 in 1,00,000. Find the probability that on exactly one of tickets he wins the jackpot. [1 Mark]

Answer: Required probability \( = \dfrac{6}{1,00,000}\times\dfrac{99999}{1,00,000} + \dfrac{99994}{1,00,000}\times\dfrac{1}{1,00,000} = \dfrac{699988}{10^{10}} \).

Teacher's Note:
a) "Exactly one" means either he wins the first jackpot and loses the second, or loses the first and wins the second.
b) Add the probabilities of these two mutually exclusive cases.

38. Roundabouts are often made on busy roads to ease the traffic and avoid red lights.
One such round-about is made such that equation representing its boundary is given by \( C_1 : x^2+y^2 = 64 \).
There is a circular pond with a fountain in the middle of the roundabout whose equation is given by \( C_2 : x^2+y^2 = 4 \).

[Figure: An aerial illustration of a road roundabout with a central circular garden containing palm trees and a fountain, surrounded by a circular road.]

(i) Represent the given equations \( C_1 \) and \( C_2 \) with the help of a diagram. [1 Mark]

Answer: Two concentric circles centred at the origin: \( C_1 \) with radius 8 units (the outer boundary, the roundabout) and \( C_2 \) with radius 2 units (the inner circle, the pond).

Teacher's Note:
a) Both circles are centred at the origin since there is no shift in their equations.
b) The radius is found by writing the equation in the form \( x^2+y^2=r^2 \).

(ii) Express y as a function of x, (y = f(x)), for both \( C_1 \) and \( C_2 \). [1 Mark]

Answer: For \( C_1 \): \( y = \pm\sqrt{64-x^2} \); for \( C_2 \): \( y = \pm\sqrt{4-x^2} \).

Teacher's Note:
a) Each circle equation gives two branches for y, an upper semicircle and a lower semicircle.
b) This form is essential for setting up the integral for area.

(iii) (a) Using integration find the area of region covered by the roundabout. [2 Marks]

Answer:
1. Required area \( = 4\displaystyle\int_0^8 \sqrt{64-x^2}\,dx \).
2. \( = 4\left[\dfrac{x}{2}\sqrt{64-x^2} + 32\sin^{-1}\dfrac{x}{8}\right]_0^8 = 4\left[0+32\times\dfrac{\pi}{2}\right] = 64\pi \) square units.

Teacher's Note:
a) Use the standard formula \( \int\sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} \).
b) Multiply by 4 to account for all four quadrants of the circle.

OR

(iii) (b) Using integration, find the area of region covered by circular pond. [2 Marks]

Answer:
1. Required area \( = 4\displaystyle\int_0^2 \sqrt{4-x^2}\,dx \).
2. \( = 4\left[\dfrac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\dfrac{x}{2}\right]_0^2 = 4\left[0+2\times\dfrac{\pi}{2}\right] = 4\pi \) square units.

Teacher's Note:
a) This follows the same integration method as part (a) but with radius 2 instead of 8.
b) The result matches the standard area formula \( \pi r^2 = \pi(2)^2 = 4\pi \), which can be used to verify the answer.

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