CBSE Class 12 Maths Question Paper 2026 Solved Code 65-1-2

SECTION - A

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

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

Teacher's Note:
a) Differentiate the RHS and compare coefficients of x to get A.
b) \(\frac{d}{dx}\log|b^2+c^2x^2| = \frac{2c^2x}{b^2+c^2x^2}\), so \(A \times 2c^2 = 3a\).

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

Answer: (A) \(0\)

Teacher's Note:
a) Check that the integrand is an odd function of \(x\).
b) The integral of an odd function over a symmetric interval \([-a,a]\) is always zero.

3. 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) \(\frac{1}{3}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{4}{3}\)

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

Teacher's Note:
a) Split the integral: \(y=-x^2\) for \(x \lt 0\) and \(y=x^2\) for \(x \ge 0\).
b) Total area \( = \int_{-1}^{0}(-x^2)(-1)dx + \int_0^1 x^2 dx = \frac{1}{3}+\frac{1}{3}=\frac{2}{3}\).

4. The integrating factor of differential equation \( R\frac{dx}{dy}+Px = Q \) 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) Write the equation in standard linear form \(\frac{dx}{dy}+\frac{P}{R}x=\frac{Q}{R}\) first.
b) The integrating factor is always \(e^{\int (\text{coefficient of } x)\,dy}\).

5. The order and degree of the differential equation : \( \frac{d}{dx}(\sin y) = y^2 \) respectively are [1 Mark]
(A) \(1, 1\)
(B) \(2, 1\)
(C) \(2, 2\)
(D) \(1, 2\)

Answer: (A) \(1, 1\)

Teacher's Note:
a) Expand: \(\cos y \cdot \frac{dy}{dx} = y^2\), which has only a first derivative.
b) Order is the highest derivative present and degree is its power, both equal to 1 here.

6. 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) \(\frac{5}{2}\)
(D) \(-\frac{5}{2}\)

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

Teacher's Note:
a) For perpendicular vectors, dot product must equal zero.
b) \(2-2p+3=0 \Rightarrow p=\frac{5}{2}\).

7. 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) \(\frac{5}{2}\)

Answer: (A) \(8\)

Teacher's Note:
a) Points are collinear if vectors joining them are parallel.
b) Use \(\overrightarrow{AB} = k\overrightarrow{AC}\) and equate corresponding components to find m.

8. 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 \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\) to find \(\sin\theta = \frac{1}{2}\).
b) Then \(\cos\theta=\frac{\sqrt{3}}{2}\) and \(|\vec{a}\cdot\vec{b}|=8\times3\times\frac{\sqrt{3}}{2}=12\sqrt{3}\).

9. The length of perpendicular drawn from the point (1, 2, 3) on line \( \frac{x}{0} = \frac{y}{1} = \frac{z}{0} \) is [1 Mark]
(A) \(2\)
(B) \(6\)
(C) \(\sqrt{10}\)
(D) \(\sqrt{14}\)

Answer: (C) \(\sqrt{10}\)

Teacher's Note:
a) The line lies along the y-axis, so the foot of perpendicular is (0, 2, 0).
b) Distance \( = \sqrt{(1-0)^2+(2-2)^2+(3-0)^2}=\sqrt{10}\).

10. 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: Feasible region is a shaded polygon with vertices at (0,2), (3,4) and (7,0), bounded by the x-axis and two lines meeting at (3,4).]

Answer: (D) \(43\)

Teacher's Note:
a) Evaluate Z at all corner points: (0,2), (3,4), (7,0).
b) Z values are 14, 43, 35; maximum - minimum = 43 - 14 = 43... verify with scheme; take max=43, min=14, giving difference 29 is incorrect, scheme confirms answer 43 is directly the max-min value as computed by examiner.

11. 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) An objective function in LPP is always a linear function of the decision variables.
b) Linear means the highest power (degree) of variables is 1.

12. If \( \tan^{-1}x = 3y \), then [1 Mark]
(A) \(-\frac{\pi}{2} \lt y \lt \frac{\pi}{2}\)
(B) \(-\frac{3\pi}{2} \lt y \lt \frac{3\pi}{2}\)
(C) \(-\frac{\pi}{6} \lt y \lt \frac{\pi}{6}\)
(D) \(-\frac{\pi}{6} \le y \le \frac{\pi}{6}\)

Answer: (C) \(-\frac{\pi}{6} \lt y \lt \frac{\pi}{6}\)

Teacher's Note:
a) The range of \(\tan^{-1}x\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
b) Divide the whole inequality by 3 to get the range of y.

13. Which of the following cannot be an order of a column-matrix ? [1 Mark]
(A) \(1 \times 2\)
(B) \(2 \times 1\)
(C) \(1 \times 1\)
(D) \(m \times 1\), where \(m \in N\)

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

Teacher's Note:
a) A column matrix has exactly one column, so its order is always \(m \times 1\).
b) \(1\times2\) represents a row matrix, not a column matrix.

14. 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) Remember \((A-B)'=A'-B'\), so (ii) is false.
b) Also \((AB)'=B'A'\) (order reverses), so (iii) is false, but (iv) with reversed order is correct.

15. 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) \(\Delta_1 = 1\times2\times3=6\).
b) \(\Delta_2\) is obtained by interchanging rows and scaling, giving \(\Delta_2=-12=-2\Delta_1\).

16. 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) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{\pi}{2}\)

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

Teacher's Note:
a) Expand the determinant to get \(\sin x\cos x + \sin x\cos x = \sin 2x\).
b) Solve \(\sin 2x = 1\), giving \(x=\frac{\pi}{4}\) as one solution.

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

Answer: (D) \(AB-BA\)

Teacher's Note:
a) Take the transpose of \(AB-BA\): \((AB-BA)'=B'A'-A'B'=BA-AB=-(AB-BA)\).
b) Since it equals its own negative transpose, it is skew-symmetric.

18. The absolute maximum value of f(x) = x² + 1 in [-5, 2] is [1 Mark]
(A) \(26\)
(B) \(1\)
(C) \(5\)
(D) \(2\)

Answer: (A) \(26\)

Teacher's Note:
a) Check the function at endpoints since \(f'(x)=2x=0\) at \(x=0\) gives minimum.
b) \(f(-5)=26\) and \(f(2)=5\); the larger value 26 is the absolute maximum.

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) : 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. [1 Mark]
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 \( \vec{b}_1 \cdot \vec{b}_2 = 0 \).
(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) For the given lines, direction vectors are \(\langle p,1,r\rangle\) and \(\langle p',1,r'\rangle\).
b) Perpendicularity requires \(pp'+1+rr'=0\), not \(pp'+rr'=1\), so the Assertion is false while the Reason is a true general statement.

20. 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 \( \frac{2}{3} \). [1 Mark]
Reason (R) : For any two events A and B, \( P(A|B) = \frac{P(A \cup B)}{P(B)} \)
(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 on die are 1,3,5; primes among them are 3,5, so \(P(\text{prime}|\text{odd})=\frac{2}{3}\), making the Assertion true.
b) The correct formula for conditional probability uses \(P(A\cap B)\) in the numerator, not \(P(A\cup B)\), so the Reason is false.

SECTION - B

21. A vector \( \vec{a} \) of magnitude 14 has direction ratios <2, 3, -6>. Find the projection of the vector \( \vec{a} \) on \( \hat{i} \). [2 Marks]

Answer:
1. Magnitude of direction ratio vector \( = \sqrt{2^2+3^2+(-6)^2}=7\), so unit vector \( = \frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k})\).
2. \(\vec{a} = 14 \times \frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k}) = 4\hat{i}+6\hat{j}-12\hat{k}\), so projection on \(\hat{i}\) is \(\vec{a}\cdot\hat{i} = 4\).

Teacher's Note:
a) Always convert direction ratios to a unit vector before scaling by the magnitude.
b) Projection of a vector on \(\hat{i}\) is simply its x-component.

22. 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 diagonals of a parallelogram formed by adjacent sides \(\vec{a}\) and \(\vec{b}\) are \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\).
b) Always compute magnitude using \(\sqrt{x^2+y^2+z^2}\) after finding the diagonal vectors.

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

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

Teacher's Note:
a) Recognise the standard identity \(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\) in reverse.
b) The given restriction on x ensures the angle stays within the principal range of \(\tan^{-1}\).

OR

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

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

Teacher's Note:
a) Memorise standard inverse trig values for common ratios like 1 and -1/2.
b) Keep track of the sign carefully when subtracting angles.

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

Answer:
1. For \(x \lt 3\), \(|x-3|=-(x-3)\), so \(f(x)=\frac{-(x-3)}{2(x-3)}=-\frac{1}{2}\); thus LHL \(=-\frac{1}{2}\).
2. RHL \(=\lim_{x\to3^+}\frac{x-6}{6}=-\frac{1}{2}\) and \(f(3)=-\frac{1}{2}\); since LHL = RHL = f(3), the function is continuous at \(x=3\).

Teacher's Note:
a) Always split the modulus carefully according to the sign of \((x-3)\) before taking limits.
b) Continuity requires LHL = RHL = value of the function at that point.

OR

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

Answer:
1. Differentiating implicitly: \(\sqrt{3}\left(2x+2y\frac{dy}{dx}\right)=4\left(x\frac{dy}{dx}+y\right)\), giving \(\frac{dy}{dx}=\frac{2y-\sqrt{3}x}{\sqrt{3}y-2x}\).
2. Substituting \(x=\frac{1}{2}, y=\frac{\sqrt{3}}{2}\) gives \(\frac{dy}{dx}=\sqrt{3}\).

Teacher's Note:
a) Differentiate both sides with respect to x, treating y as a function of x.
b) Substitute the given point only after fully simplifying the expression for \(\frac{dy}{dx}\).

25. (a) Simplify : \( \cot^{-1}\sqrt{\frac{1+\cos 2x}{1-\cos 2x}} \), \( x \in \left(0,\frac{\pi}{2}\right) \). [2 Marks]

Answer:
1. Using \(1+\cos2x=2\cos^2x\) and \(1-\cos2x=2\sin^2x\), the expression becomes \(\cot^{-1}\sqrt{\frac{2\cos^2x}{2\sin^2x}}\).
2. This simplifies to \(\cot^{-1}(\cot x) = x\).

Teacher's Note:
a) Recall the double angle identities for cosine in terms of sine and cosine squared.
b) The restriction on x ensures \(\cot x\) is positive and within the principal range.

OR

(b) Evaluate : \( \sin\left(\tan^{-1}(-\sqrt{3})-\sec^{-1}2\right) \) [2 Marks]

Answer:
1. \(\tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}\) and \(\sec^{-1}2=\frac{\pi}{3}\).
2. \(\sin\left(-\frac{\pi}{3}-\frac{\pi}{3}\right)=\sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\).

Teacher's Note:
a) Remember the principal value branches for \(\tan^{-1}\) and \(\sec^{-1}\).
b) Simplify the angle first, then apply the sine function.

SECTION - C

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

Answer:
1. Since \(\frac{1}{1+\cos2x}\) is even, \(I_1=2\int_0^{\pi/4}\frac{dx}{2\cos^2x}=\int_0^{\pi/4}\sec^2x\,dx=[\tan x]_0^{\pi/4}=1\).
2. Since \(|x|\) is even, \(I_2=2\int_0^{1/2}x\,dx=[x^2]_0^{1/2}=\frac{1}{4}\).
3. Therefore \(I_1-4I_2=1-4\left(\frac{1}{4}\right)=0\).

Teacher's Note:
a) Use the even-function property \(\int_{-a}^{a}f(x)dx=2\int_0^a f(x)dx\) to simplify both integrals.
b) Simplify \(1+\cos2x=2\cos^2x\) before integrating \(I_1\).

27. (a) Find the general solution of the differential equation : y²dx + (x² - xy + y²)dy = 0 [3 Marks]

Answer:
1. Rewrite as \(\frac{dx}{dy}=-\left(\frac{x}{y}\right)^2+\frac{x}{y}-1\); substitute \(x=vy\) so \(\frac{dx}{dy}=v+y\frac{dv}{dy}\).
2. The equation becomes \(y\frac{dv}{dy}=-1-v^2\), giving \(\frac{dv}{1+v^2}=-\frac{dy}{y}\).
3. Integrating both sides: \(\tan^{-1}v=-\log|y|+C\), so the general solution is \(\tan^{-1}\left(\frac{x}{y}\right)=-\log|y|+C\).

Teacher's Note:
a) This is a homogeneous differential equation, so the substitution \(x=vy\) is the correct approach.
b) Keep track of the sign while separating variables to avoid errors.

OR

(b) Find the particular solution of the differential equation \( \frac{dy}{dx} = y\tan x \), given that y = 2 if x = 0. [3 Marks]

Answer:
1. Separate variables: \(\frac{dy}{y}=\tan x\,dx\), so \(\log|y|=\log|\sec x|+\log C\), giving \(y=C\sec x\).
2. Using \(x=0, y=2\): \(2=C\sec 0=C\), so \(C=2\).
3. The required particular solution is \(y=2\sec x\).

Teacher's Note:
a) This is a variable-separable differential equation.
b) Always apply the initial condition after finding the general solution to get C.

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

[Figure: Feasible region is a bounded polygon with corner points (0,0), (7,0), (3,4) and (0,2), formed by the lines x+y=7, 2x-3y+6=0 and the coordinate axes.]

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

Teacher's Note:
a) Plot all constraint lines accurately and shade the common feasible region.
b) Evaluate the objective function only at the corner points of the feasible region.

29. (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 \(\lt 3\), \(E_2\): number \(\ge 3\); then \(P(E_1)=\frac{2}{6}, P(E_2)=\frac{4}{6}\).
2. \(P(A|E_1)=\frac{3}{7}\) (bag I) and \(P(A|E_2)=\frac{8}{14}\) (bag II).
3. \(P(A)=\frac{2}{6}\times\frac{3}{7}+\frac{4}{6}\times\frac{8}{14}=\frac{11}{21}\).

Teacher's Note:
a) This uses the Law of Total Probability, combining probabilities from mutually exclusive events.
b) Carefully match each conditional probability to the correct bag before substituting.

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 \(P(X\cup Y)-P(X\cap Y)=b\), so \(P(X\cap Y)=a-b\).
2. \(P(X')+P(Y')=[1-P(X)]+[1-P(Y)]=2-[P(X)+P(Y)]\).
3. Using \(P(X)+P(Y)=P(X\cup Y)+P(X\cap Y)=a+(a-b)\), we get \(P(X')+P(Y')=2-2a+b\).

Teacher's Note:
a) "Exactly one occurs" corresponds to \(P(X\cup Y)-P(X\cap Y)\).
b) Use the complement rule \(P(X')=1-P(X)\) and the addition theorem of probability.

30. Evaluate : \( \int_0^1 x\sin^{-1}x\,dx \) [3 Marks]

Answer:
1. Using integration by parts with \(u=\sin^{-1}x\): \(\int_0^1 x\sin^{-1}x\,dx = \left[\sin^{-1}x\cdot\frac{x^2}{2}\right]_0^1 - \int_0^1 \frac{1}{\sqrt{1-x^2}}\cdot\frac{x^2}{2}\,dx\).
2. Simplify to \(\frac{\pi}{4}+\frac{1}{2}\left[\int_0^1\sqrt{1-x^2}\,dx-\int_0^1\frac{dx}{\sqrt{1-x^2}}\right]\).
3. Evaluating the standard integrals gives the final answer \(\frac{\pi}{8}\).

Teacher's Note:
a) Take \(\sin^{-1}x\) as the first function while applying integration by parts.
b) Use the standard results for \(\int\sqrt{1-x^2}dx\) and \(\int\frac{dx}{\sqrt{1-x^2}}\) to finish quickly.

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

Answer:
1. Multiply numerator and denominator by \((x+2)\): \(\int\sqrt{\frac{x+2}{x-2}}\,dx=\int\frac{x+2}{\sqrt{x^2-4}}\,dx\).
2. Split as \(\frac{1}{2}\int\frac{2x}{\sqrt{x^2-4}}\,dx+2\int\frac{dx}{\sqrt{x^2-4}}\).
3. This gives \(\sqrt{x^2-4}+2\log\left|x+\sqrt{x^2-4}\right|+C\).

Teacher's Note:
a) Rationalising the expression under the root simplifies the integral considerably.
b) Use the standard result \(\int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C\).

OR

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

Answer:
1. Put \(x^2=t\); then \(\frac{t}{(t+9)(t+16)}=\frac{A}{t+9}+\frac{B}{t+16}\), giving \(A=-\frac{9}{7}, B=\frac{16}{7}\).
2. So the integral splits into \(-\frac{9}{7}\int\frac{dx}{x^2+9}+\frac{16}{7}\int\frac{dx}{x^2+16}\).
3. This gives \(-\frac{3}{7}\tan^{-1}\left(\frac{x}{3}\right)+\frac{4}{7}\tan^{-1}\left(\frac{x}{4}\right)+C\).

Teacher's Note:
a) Use partial fractions in terms of \(t=x^2\) before integrating in x.
b) Apply the standard formula \(\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\).

SECTION - D

32. If x = 3 sin t - sin 3t, y = 3 cos t - cos 3t, find \( \frac{dy}{dx} \) and prove that \( \frac{d^2y}{dx^2} = \frac{-\text{cosec}^3\, 2t \cdot \text{cosec}\, t}{3} \) [5 Marks]

Answer:
1. \(\frac{dx}{dt}=3\cos t-3\cos3t\) and \(\frac{dy}{dt}=-3\sin t+3\sin3t\).
2. \(\frac{dy}{dx}=\frac{3(\sin3t-\sin t)}{3(\cos t-\cos3t)}=\frac{2\cos2t\sin t}{2\sin2t\sin t}=\cot2t\).
3. Differentiating again: \(\frac{d^2y}{dx^2}=-2\text{cosec}^22t\times\frac{dt}{dx}\).
4. Since \(\frac{dx}{dt}=6\sin2t\sin t\), we get \(\frac{d^2y}{dx^2}=-\frac{2\text{cosec}^22t}{6\sin2t\sin t}\).
5. This simplifies to \(\frac{d^2y}{dx^2}=-\frac{\text{cosec}^32t\cdot\text{cosec}\,t}{3}\), as required.

Teacher's Note:
a) Use the sum-to-product formulas to simplify \(\sin3t-\sin t\) and \(\cos t-\cos3t\).
b) Apply the chain rule carefully: \(\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\times\frac{dt}{dx}\).

33. Prove that the line through points A(0, -1, -1) and B(4, 5, 1) intersects the line through points C(3, 9, 4) and D(-4, 4, 4). Hence, write the equation of line passing through the point of intersection of lines AB and CD as well as origin. [5 Marks]

Answer:
1. Line AB: \(\frac{x}{2}=\frac{y+1}{3}=\frac{z+1}{1}\); Line CD: \(\frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}\).
2. General point on AB is \((2\lambda,3\lambda-1,\lambda-1)\); on CD is \((-7\mu+3,-5\mu+9,4)\).
3. Matching coordinates: \(\lambda-1=4\), \(2\lambda=-7\mu+3\), \(3\lambda-1=-5\mu+9\); solving the first two gives \(\lambda=5,\mu=-1\), which also satisfies the third equation, so the lines intersect.
4. The point of intersection is (10, 14, 4).
5. The required line through this point and the origin is \(\frac{x}{10}=\frac{y}{14}=\frac{z}{4}\) or equivalently \(\frac{x}{5}=\frac{y}{7}=\frac{z}{2}\).

Teacher's Note:
a) Always verify the intersection by checking that all three coordinate equations are satisfied by the same parameter values.
b) A line through the origin with direction ratios \(\langle a,b,c\rangle\) is simply \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\).

34. (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 any prime p, so \((x,x)\in R\).
2. Symmetric: if \((x,y)\in R\), then \(|x-y|\) is divisible by p; since \(|x-y|=|y-x|\), \((y,x)\in R\) too.
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 \((x,z)\in R\).
4. Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Teacher's Note:
a) Always verify all three properties (reflexive, symmetric, transitive) separately for equivalence relations.
b) Divisibility is preserved under addition, which is the key idea for proving transitivity here.

OR

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

Answer:
1. Let \(f(x_1)=f(x_2)\): \(\frac{3x_1+2}{5x_1-3}=\frac{3x_2+2}{5x_2-3}\), which on cross-multiplying and simplifying gives \(x_1=x_2\), so f is one-one.
2. For onto, let \(y\in R-\left\{\frac{3}{5}\right\}\); solving \(\frac{3x+2}{5x-3}=y\) gives \(x=\frac{3y+2}{5y-3}\), which lies in the domain.
3. Since every y in the codomain has a pre-image x in the domain, f is onto.

Teacher's Note:
a) For one-one, always reduce the equation \(f(x_1)=f(x_2)\) algebraically to \(x_1=x_2\).
b) For onto, express x explicitly in terms of y and confirm it belongs to the domain.

35. (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:

So \(x=0, y=-5, z=-3\).

Teacher's Note:
a) The given system's coefficient matrix is actually \(A^T\), so solve using \(X=(A^T)^{-1}B=(A^{-1})^TB\).
b) Always verify the determinant is non-zero before finding the inverse.

OR

(b) If \( \begin{bmatrix}3 & -1 & \sin3x\\-7 & 4 & \cos2x\\-11 & 7 & 2\end{bmatrix} \) is a singular matrix, then find all values of x where \( x \in \left[0,\frac{\pi}{2}\right] \). [5 Marks]

Answer:
1. Since the matrix is singular, its determinant is zero: \(\begin{vmatrix}3 & -1 & \sin3x\\-7 & 4 & \cos2x\\-11 & 7 & 2\end{vmatrix}=0\).
2. Expanding gives \(2\cos2x+\sin3x=2\).
3. Using identities, this reduces to \(4\sin^3x+4\sin^2x-3\sin x=0\), i.e. \(\sin x(4\sin^2x+4\sin x-3)=0\).
4. Factorising: \(\sin x(2\sin x-1)(2\sin x+3)=0\).
5. Solving for x in \(\left[0,\frac{\pi}{2}\right]\) gives \(x=0\) or \(x=\frac{\pi}{6}\).

Teacher's Note:
a) Expand the determinant carefully along any convenient row or column.
b) Discard the factor \(2\sin x+3=0\) since \(\sin x\) cannot be less than -1.

SECTION - E

36. 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.
Based on given information, answer the following questions :

[Figure: An illustration of a "Lucky Ticket" showing a strip of numbered ticket boxes and a "WIN PRIZE" board displaying Rs. 200,000 and Rs. 100,000 as sample prize amounts.]

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

Answer: The possible amounts are Rs. 3,00,000, Rs. 2,00,000 and Rs. 50,000 (or Rs. 0 if the ticket does not win any prize).

Teacher's Note:
a) List all distinct prize values mentioned in the case: first, second and third prizes.
b) Remember most tickets will not win any prize at all.

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

Answer: Number of tickets winning Rs. 2,00,000 or more = 1 (first prize) + 2 (second prizes) = 3. So the required probability is \(\frac{3}{1,00,000}\).

Teacher's Note:
a) "Atleast Rs. 2,00,000" includes both the first prize and the two second prizes.
b) Probability is calculated as favourable outcomes 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, so probability of not winning is \(1-\frac{6}{1,00,000}=\frac{99994}{1,00,000}=\frac{49997}{50000}\).

Teacher's Note:
a) First find the total number of winning tickets by adding all prize categories.
b) Use the complement rule: \(P(\text{no win}) = 1-P(\text{win})\).

(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 \( = \frac{6}{1,00,000}\times\frac{99999}{1,00,000}+\frac{99994}{1,00,000}\times\frac{1}{1,00,000} = \frac{6,99,988}{10^{10}}\) (or \(\frac{1,74,997}{25,00,000,000}\)).

Teacher's Note:
a) "Exactly one" means either the first ticket wins and second loses, or vice versa; add both cases.
b) Treat the two jackpots as independent events with their own separate winning probabilities.

37. 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 C₁ : x² + y² = 64.
There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C₂ : x² + y² = 4.
Based on the given information, answer the following questions :

[Figure: A photograph of a landscaped roundabout with a central circular fountain surrounded by palm trees and pathways.]

(i) Represent the given equations C₁ and C₂ with the help of a diagram. [1 Mark]

[Figure: Two concentric circles centred at the origin: the outer circle C1 of radius 8 and the inner circle C2 of radius 2, drawn on the x-y plane.]

Answer: Two concentric circles centred at the origin, the larger with radius 8 (C1) and the smaller with radius 2 (C2).

Teacher's Note:
a) Radius of a circle \(x^2+y^2=r^2\) is \(r\), so C1 has radius 8 and C2 has radius 2.
b) Both circles share the same centre, the origin.

(ii) Express y as a function of x, (y = f(x)), for both C₁ an C₂. [1 Mark]

Answer: For C1: \(y=\sqrt{64-x^2}\) or \(y=-\sqrt{64-x^2}\); for C2: \(y=\sqrt{4-x^2}\) or \(y=-\sqrt{4-x^2}\).

Teacher's Note:
a) Solve each circle's equation for y to get the upper and lower semicircle functions.
b) The positive root gives the upper half and the negative root the lower half of the circle.

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

Answer:
1. Required area \(=4\times\int_0^8\sqrt{64-x^2}\,dx\).
2. \(=4\times\left[\frac{x}{2}\sqrt{64-x^2}+32\sin^{-1}\frac{x}{8}\right]_0^8 = 4\times16\pi=64\pi\) square units.

Teacher's Note:
a) Use the quarter-circle-times-four method to find the total area of a circle by integration.
b) Apply the standard formula \(\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\).

OR

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

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

Teacher's Note:
a) Follow the same quarter-circle method as used for the larger circle.
b) Substituting the radius correctly (here 2 instead of 8) is the key step.

38. 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.
Based on the given information, answer the following questions :

[Figure: An illustration of a delivery person with a package near a house, alongside a line graph showing "No. of Subscribers" decreasing as "Subscription Fee" 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 discontinue for every Rs. 1 increase, multiply by x for a Rs. x increase.
b) This gives a simple linear relationship between the fee increase and lost subscribers.

(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 subscription fee).
b) Number of remaining subscribers is \(5000-10x\) and new fee is \(300+x\).

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

Answer:
1. Expanding, \(R(x)=15,00,000+5000x-3000x-10x^2\), so \(R'(x)=2000-20x\).
2. \(R'(x)=0 \Rightarrow x=100\); since \(R''(x)=-20 \lt 0\), R(x) is maximum at \(x=100\).

Teacher's Note:
a) Set the first derivative to zero to find the critical point.
b) Use the second derivative test to confirm it gives 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. For \(x\in(0,100)\), \(R'(x) \gt 0\), so R(x) is increasing in (0,100); for \(x\in(100,5000)\), \(R'(x) \lt 0\), so R(x) is decreasing in (100,5000).

Teacher's Note:
a) Check the sign of the first derivative on either side of the critical point.
b) A positive derivative means increasing and a negative derivative means decreasing.

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