MathsCBSE Class 12 · Board exam 2026 (session 2025-26)

CBSE Class 12 Mathematics: 2026 board paper, set 65/1/1

The question paper set in the CBSE Class 12 board exam of 2026, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2026/XII/Mathematics.zip (a zip; this paper is the file 2413-1_65-1-1_Mathematics.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-20	MCQ 1-18, assertion-reason 19-20
B	21-25	VSA
C	26-31	SA
D	32-35	LA
E	36-38	Case study based

Instructions in brief: All questions compulsory; no overall choice, internal choice provided in a few questions in Sections B to E; calculator not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQInverse Trigonometric Functions
If $2 cos^{- 1} x = y$ , then
1. (A) $0 \leq y \leq π$
2. (B) $- π \leq y \leq π$
3. (C) $0 \leq y \leq 2 π$
4. (D) $- π \leq y \leq 0$
Try it first, then show our solutionSolution
1. Principal range of $cos^{- 1} x$ is $[0, π]$ , so $0 \leq cos^{- 1} x \leq π \Rightarrow 0 \leq 2 cos^{- 1} x \leq 2 π$ , i.e. $0 \leq y \leq 2 π$ . Option (C).
  1 mark
Answer: $0 \leq y \leq 2 π$ , option (C)
Where the marks are
- Uses range $[0, π]$ of $cos^{- 1} x$ and doubles it (1 mark)
Q2 · 1 mark · MCQMatrices
Which of the following cannot be the order of a row-matrix ?
1. (A) $2 \times 1$
2. (B) $1 \times 2$
3. (C) $1 \times 1$
4. (D) $1 \times n$
Try it first, then show our solutionSolution
1. A row matrix has exactly one row, so its order is $1 \times n$ . $2 \times 1$ has two rows (a column matrix), so it cannot be a row matrix. Option (A).
  1 mark
Answer: $2 \times 1$ , option (A)
Where the marks are
- Identifies that a row matrix must have order $1 \times n$ (1 mark)
Q3 · 1 mark · MCQMatrices
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) $(A B)^{'} = A^{'} B^{'}$ (iv) $(k A B)^{'} = k B^{'} A^{'}$ (k is a scalar)
1. (A)(i) only
2. (B)(i), (ii) and (iii)
3. (C)(i) and (ii)
4. (D)(i) and (iv)
Try it first, then show our solutionSolution
1. $(A + B)^{'} = A^{'} + B^{'}$ is true; $(A - B)^{'} = A^{'} - B^{'}$ , not $B^{'} - A^{'}$ ; $(A B)^{'} = B^{'} A^{'}$ , not $A^{'} B^{'}$ .
2. $(k A B)^{'} = k (A B)^{'} = k B^{'} A^{'}$ is true. So (i) and (iv): option (D).
  1 mark
Answer: (i) and (iv), option (D)
Where the marks are
- Applies transpose rules, reversal law for products (1 mark)
Q4 · 1 mark · MCQDeterminants
If $Δ_{1} = 100020003$ and $Δ_{2} = 010200006$ , then
1. (A) $Δ_{1} = 2 Δ_{2}$
2. (B) $Δ_{2} = - 2 Δ_{1}$
3. (C) $Δ_{1} = Δ_{2}$
4. (D) $Δ_{2} = - Δ_{1}$
Try it first, then show our solutionSolution
1. $Δ_{1} = 1 \cdot 2 \cdot 3 = 6$ (diagonal). $Δ_{2} = 0 - 2 (6 - 0) + 0 = - 12$ .
2. So $Δ_{2} = - 12 = - 2 Δ_{1}$ . Option (B).
  1 mark
Answer: $Δ_{2} = - 2 Δ_{1}$ , option (B)
Where the marks are
- Evaluates $Δ_{1} = 6$ , $Δ_{2} = - 12$ and relates them (1 mark)
Q5 · 1 mark · MCQDeterminants
One of the values of $x$ for which $cos x - cos x sin x sin x = 1$ is
1. (A) $0$
2. (B) $\frac{π}{4}$
3. (C) $\frac{π}{3}$
4. (D) $\frac{π}{2}$
Try it first, then show our solutionSolution
1. Determinant $= cos x sin x + sin x cos x = sin 2 x = 1 \Rightarrow 2 x = \frac{π}{2} \Rightarrow x = \frac{π}{4}$ . Option (B).
  1 mark
Answer: $x = \frac{π}{4}$ , option (B)
Where the marks are
- Reduces determinant to $sin 2 x = 1$ and solves (1 mark)
Q6 · 1 mark · MCQMatrices
If A and B are skew symmetric matrices of same order, then which of the following matrices is also skew symmetric ?
1. (A)AB
2. (B)AB + BA
3. (C) $(A + B)^{2}$
4. (D)A – B
Try it first, then show our solutionSolution
1. $A^{'} = - A, B^{'} = - B$ . Then $(A - B)^{'} = A^{'} - B^{'} = - A + B = - (A - B)$ , so $A - B$ is skew symmetric.
  1 mark
2. $(A B)^{'} = B^{'} A^{'} = B A$ , $(A B + B A)^{'} = A B + B A$ (symmetric), $((A + B)^{2})^{'} = (A + B)^{2}$ (symmetric); these fail. Option (D).
Answer: $A - B$ , option (D)
Where the marks are
- Shows $(A - B)^{'} = - (A - B)$ (1 mark)
Open Q6 on its own page
Q7 · 1 mark · MCQApplication of Derivatives
The least value of $f (x) = x^{3} - 12 x$ , $x \in [0, 3]$ is
1. (A)–16
2. (B)–9
3. (C)0
4. (D)16
Try it first, then show our solutionSolution
1. $f^{'} (x) = 3 x^{2} - 12 = 0 \Rightarrow x = 2$ in $[0, 3]$ .
2. $f (0) = 0$ , $f (2) = 8 - 24 = - 16$ , $f (3) = 27 - 36 = - 9$ . Least value $- 16$ . Option (A).
  1 mark
Answer: $- 16$ , option (A)
Where the marks are
- Finds critical point $x = 2$ and compares values at endpoints (1 mark)
Q8 · 1 mark · MCQIntegrals
If $\int \frac{3 a x}{b ^{2} + c ^{2} x ^{2}} d x = A lo g ∣ b^{2} + c^{2} x^{2} ∣ + K$ , then the value of A is
1. (A) $3 a$
2. (B) $\frac{3 a}{2 b ^{2}}$
3. (C) $\frac{3 a}{b ^{2} c ^{2}}$
4. (D) $\frac{3 a}{2 c ^{2}}$
Try it first, then show our solutionSolution
1. Put $t = b^{2} + c^{2} x^{2}$ , $d t = 2 c^{2} x d x$ . Integral $= \frac{3 a}{2 c ^{2}} \int \frac{d t}{t} = \frac{3 a}{2 c ^{2}} lo g ∣ b^{2} + c^{2} x^{2} ∣ + K$ . So $A = \frac{3 a}{2 c ^{2}}$ . Option (D).
  1 mark
Answer: $A = \frac{3 a}{2 c ^{2}}$ , option (D)
Where the marks are
- Substitution $t = b^{2} + c^{2} x^{2}$ giving coefficient $\frac{3 a}{2 c ^{2}}$ (1 mark)
Q9 · 1 mark · MCQIntegrals
The value of $\int_{- 1}^{1} \frac{x ^{3}}{x ^{2} + 2∣ x ∣ + 1} d x$ is
1. (A)0
2. (B) $lo g 2$
3. (C) $2 lo g 2$
4. (D) $\frac{1}{2} lo g 2$
Try it first, then show our solutionSolution
1. $f (- x) = \frac{- x ^{3}}{x ^{2} + 2∣ x ∣ + 1} = - f (x)$ , so the integrand is odd; over $[- 1, 1]$ the integral is $0$ . Option (A).
  1 mark
Answer: 0, option (A)
Where the marks are
- Recognises odd integrand over symmetric limits (1 mark)
Q10 · 1 mark · MCQApplication of Integrals
The area bounded by the curve $y = x ∣ x ∣$ , $x$ -axis and the ordinates $x = - 1$ and $x = 1$ is given by
1. (A)0
2. (B) $\frac{1}{3}$
3. (C) $\frac{2}{3}$
4. (D) $\frac{4}{3}$
Try it first, then show our solutionSolution
1. $y = x ∣ x ∣$ is $- x^{2}$ for $x < 0$ , $x^{2}$ for $x \geq 0$ ; the region is symmetric.
2. Area $= 2 \int_{0}^{1} x^{2} d x = 2 \cdot \frac{1}{3} = \frac{2}{3}$ . Option (C).
  1 mark
Answer: $\frac{2}{3}$ , option (C)
Where the marks are
- Takes absolute area on both sides, $2 \int_{0}^{1} x^{2} d x = \frac{2}{3}$ (1 mark)
Q11 · 1 mark · MCQDifferential Equations
The integrating factor of differential equation $R \frac{d x}{d y} + P x = Q$ where P, Q, R are functions of y is
1. (A) $e^{\int \frac{P}{Q} d y}$
2. (B) $e^{\int P d y}$
3. (C) $e^{\int \frac{P}{R} d y}$
4. (D) $e^{\int \frac{P}{R} d x}$
Try it first, then show our solutionSolution
1. Divide by R: $\frac{d x}{d y} + \frac{P}{R} x = \frac{Q}{R}$ , linear in $x$ . IF $= e^{\int \frac{P}{R} d y}$ . Option (C).
  1 mark
Answer: $e^{\int \frac{P}{R} d y}$ , option (C)
Where the marks are
- Writes standard form dividing by R and IF w.r.t. y (1 mark)
Q12 · 1 mark · MCQDifferential Equations
The order and degree of the differential equation $\frac{d}{d x} (e^{y}) = 0$ respectively are
1. (A)0, 1
2. (B)1, 1
3. (C)2, 1
4. (D)1, not defined
Try it first, then show our solutionSolution
1. $\frac{d}{d x} (e^{y}) = e^{y} \frac{d y}{d x} = 0$ ; highest derivative is $\frac{d y}{d x}$ with power 1. Order 1, degree 1. Option (B).
  1 mark
Answer: 1, 1, option (B)
Where the marks are
- Expands to $e^{y} y^{'} = 0$ , order 1 degree 1 (1 mark)
Q13 · 1 mark · MCQVector Algebra
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. (A)0
2. (B)1
3. (C) $\frac{5}{2}$
4. (D) $- \frac{5}{2}$
Try it first, then show our solutionSolution
1. Perpendicular ⇒ dot product $= 2 - 2 p + 3 = 0 \Rightarrow p = \frac{5}{2}$ . Option (C).
  1 mark
Answer: $p = \frac{5}{2}$ , option (C)
Where the marks are
- Sets dot product zero and solves (1 mark)
Q14 · 1 mark · MCQVector Algebra
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. (A)8
2. (B)–8
3. (C)2
4. (D) $\frac{5}{2}$
Try it first, then show our solutionSolution
1. $A B = 3 \hat{i} + (m + 1) \hat{j} + 3 \hat{k}$ , $A C = 4 \hat{i} + 12 \hat{j} + 4 \hat{k}$ .
2. Collinear ⇒ proportional: $\frac{3}{4} = \frac{m + 1}{12} \Rightarrow m = 8$ . Option (A).
  1 mark
Answer: $m = 8$ , option (A)
Where the marks are
- Equates ratios of $A B$ and $A C$ to get $m = 8$ (1 mark)
Q15 · 1 mark · MCQVector Algebra
If $∣ a ∣ = 8$ , $∣ b ∣ = 3$ and $∣ a \times b ∣ = 12$ , then the value of $∣ a \cdot b ∣$ is
1. (A) $63$
2. (B) $83$
3. (C) $123$
4. (D) $312$
Try it first, then show our solutionSolution
1. $∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ : $144 + (a \cdot b)^{2} = 576 \Rightarrow ∣ a \cdot b ∣ = 432 = 123$ . Option (C).
  1 mark
Answer: $123$ , option (C)
Where the marks are
- Uses Lagrange identity to get $123$ (1 mark)
Q16 · 1 mark · MCQThree Dimensional Geometry
The length of perpendicular drawn from point (2, 5, 7) on line $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ is
1. (A)2
2. (B)5
3. (C) $74$
4. (D) $78$
Try it first, then show our solutionSolution
1. The line is the x-axis; foot of perpendicular from $(2, 5, 7)$ is $(2, 0, 0)$ . Distance $= 5^{2} + 7^{2} = 74$ . Option (C).
  1 mark
Answer: $74$ , option (C)
Where the marks are
- Recognises x-axis and computes $y^{2} + z^{2}$ (1 mark)
Q17 · 1 mark · MCQLinear Programming
The feasible region of a linear programming problem with objective function $Z = 5 x + 7 y$ is shown below :
The maximum value of Z – minimum value of Z is
1. (A)8
2. (B)29
3. (C)35
4. (D)43
Figure in the paper: Shaded feasible region in the first quadrant with corner points (0, 0), (0, 2), (3, 4) and (7, 0). The axes are marked at 1 to 7 on x-axis and 1, 2 on y-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. Z at corners: $(0, 0) \to 0$ , $(0, 2) \to 14$ , $(3, 4) \to 43$ , $(7, 0) \to 35$ .
2. Max $43$ , min $0$ ; difference $43$ . Option (D).
  1 mark
Answer: 43, option (D)
Where the marks are
- Evaluates Z at all corner points and subtracts (1 mark)
Q18 · 1 mark · MCQLinear Programming
The degree of an objective function of a linear programming problem is
1. (A)0
2. (B)1
3. (C)2
4. (D)Any natural number
Try it first, then show our solutionSolution
1. The objective function $Z = a x + b y$ is linear, so its degree is 1. Option (B).
  1 mark
Answer: 1, option (B)
Where the marks are
- Objective function is linear, degree 1 (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
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 $\frac{2}{3}$ .
Reason (R) : For any two events A and B, $P (A ∣ B) = \frac{P ( A \cup B )}{P ( B )}$
1. (A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true and Reason (R) is false.
4. (D)Assertion (A) is false and Reason (R) is true.
Try it first, then show our solutionSolution
1. Odd: ${1, 3, 5}$ ; prime and odd: ${3, 5}$ . $P = \frac{2}{3}$ , so A is true.
2. Correct formula is $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , not with $\cup$ , so R is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Computes $\frac{2}{3}$ and identifies wrong formula in R (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedThree Dimensional Geometry
Assertion (A) : Lines given by $x = p y + q$ , $z = r y + s$ and $x = p^{'} y + q^{'}$ , $z = r^{'} y + s^{'}$ are perpendicular to each other when $p p^{'} + r r^{'} = 1$ .
Reason (R) : Two lines $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ are perpendicular to each other if $b_{1} \cdot b_{2} = 0$ .
1. (A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true and Reason (R) is false.
4. (D)Assertion (A) is false and Reason (R) is true.
Try it first, then show our solutionSolution
1. First line: $\frac{x - q}{p} = \frac{y}{1} = \frac{z - s}{r}$ , DRs $p, 1, r$ ; second: $p^{'}, 1, r^{'}$ .
2. Perpendicular ⇔ $p p^{'} + 1 + r r^{'} = 0$ , i.e. $p p^{'} + r r^{'} = - 1$ , so A is false. R is the standard true condition. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Derives DRs and condition $p p^{'} + r r^{'} = - 1$ ; R true (1 mark)
Open Q20 on its own page

Section B

VSA

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerContinuity and Differentiability
Check whether function $f (x)$ defined as
$f (x) = {\frac{∣ x - 3∣}{2 ( x - 3 )}, \frac{x - 6}{6}, x < 3 x \geq 3$
is continuous at $x = 3$ or not ?
Try it first, then show our solutionSolution
1. For $x < 3$ , $x - 3 < 0$ , so $∣ x - 3∣ = - (x - 3)$ . Then $f (x) = \frac{- ( x - 3 )}{2 ( x - 3 )} = - \frac{1}{2}$ for $x < 3$ , and $f (x) = \frac{x - 6}{6}$ for $x \geq 3$ .
  1 mark
2. $f (3) = \frac{3 - 6}{6} = - \frac{1}{2}$ .
  LHL $= lim_{x \to 3^{-}} (- \frac{1}{2}) = - \frac{1}{2}$ .
  RHL $= lim_{x \to 3^{+}} \frac{x - 6}{6} = \frac{- 3}{6} = - \frac{1}{2}$ .
  ½ mark
3. Since LHL = RHL = $f (3) = - \frac{1}{2}$ , $f$ is continuous at $x = 3$ .
  ½ mark
Answer: f is continuous at x = 3 (LHL = RHL = f(3) = −1/2)
Where the marks are
- Removes the modulus for $x < 3$ to get $f (x) = - \frac{1}{2}$ (1 mark)
- Finds $f (3)$ and both one-sided limits equal to $- \frac{1}{2}$ (½ mark)
- Concludes $f$ is continuous at $x = 3$ (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerContinuity and Differentiability
If $3 (x^{2} + y^{2}) = 4 x y$ , then find $\frac{d y}{d x}$ at $(\frac{1}{2}, \frac{3}{2})$ .
Try it first, then show our solutionSolution
1. Differentiate both sides with respect to $x$ :
  $3 (2 x + 2 y \frac{d y}{d x}) = 4 (x \frac{d y}{d x} + y)$
  1 mark
2. Collect the $\frac{d y}{d x}$ terms: $\frac{d y}{d x} (23 y - 4 x) = 4 y - 23 x$ , so
  $\frac{d y}{d x} = \frac{2 y - 3 x}{3 y - 2 x}$
  ½ mark
3. At $(\frac{1}{2}, \frac{3}{2})$ : numerator $= 3 - \frac{3}{2} = \frac{3}{2}$ , denominator $= \frac{3}{2} - 1 = \frac{1}{2}$ .
  $∴ \frac{d y}{d x} = \frac{3 /2}{1/2} = 3$
  ½ mark
Answer: $\frac{d y}{d x} = 3$
Where the marks are
- Correct implicit differentiation of both sides (1 mark)
- Expression $\frac{d y}{d x} = \frac{2 y - 3 x}{3 y - 2 x}$ (½ mark)
- Substitutes the point to get $3$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerCompetency-basedApplication of Derivatives
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 mm $^{3}$ /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 $\frac{π}{6}$ .
Figure in the paper: Left: a room freshener bottle shaped as an inverted cone spraying perfume. Right: an inverted cone with shaded perfume; the semi-vertical angle is marked 30° and the perfume level height is marked 10 mm. See it in the official paper.
Try it first, then show our solutionSolution
1. Let $V$ , $r$ , $h$ be the volume, radius and height of perfume at time $t$ . Given $\frac{d V}{d t} = - 1$ mm³/min.
  From the semi-vertical angle: $\frac{r}{h} = tan \frac{π}{6} = \frac{1}{3} \Rightarrow r = \frac{h}{3}$
  ½ mark
2. $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \cdot \frac{h ^{2}}{3} \cdot h = \frac{π h ^{3}}{9}$
  ½ mark
3. Differentiating w.r.t. $t$ : $\frac{d V}{d t} = \frac{π}{9} \cdot 3 h^{2} \frac{d h}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$
  ½ mark
4. So $\frac{d h}{d t} = - \frac{3}{π h ^{2}}$ . At $h = 10$ mm: $\frac{d h}{d t} = - \frac{3}{100 π}$ .
  The level of perfume is dropping at $\frac{3}{100 π}$ mm/min.
  ½ mark
Answer: Level drops at $\frac{3}{100 π}$ mm/min
Where the marks are
- Uses $tan \frac{π}{6}$ to get $r = \frac{h}{3}$ (½ mark)
- Writes $V = \frac{π h ^{3}}{9}$ (½ mark)
- Differentiates: $\frac{d V}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$ (½ mark)
- Gets rate $\frac{3}{100 π}$ mm/min at $h = 10$ (½ mark)
Open Q22 on its own page
Q23 · 2 marks · Very short answerVector Algebra
Find the vector of magnitude 14 in the direction of $QP$ , where P and Q are the points (1, 3, 2) and (–1, 0, 8) respectively.
Try it first, then show our solutionSolution
1. $QP = (1 - (- 1)) \hat{i} + (3 - 0) \hat{j} + (2 - 8) \hat{k} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k}$
  ½ mark
2. $∣ QP ∣ = 4 + 9 + 36 = 7$ , so the unit vector is $\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}$
  ½ mark
3. Required vector $= 14 (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) = 4 \hat{i} + 6 \hat{j} - 12 \hat{k}$
  1 mark
Answer: $4 \hat{i} + 6 \hat{j} - 12 \hat{k}$
Where the marks are
- Finds $QP = 2 \hat{i} + 3 \hat{j} - 6 \hat{k}$ (½ mark)
- Finds magnitude 7 / unit vector (½ mark)
- Multiplies by 14 to get $4 \hat{i} + 6 \hat{j} - 12 \hat{k}$ (1 mark)
Open Q23 on its own page
Q24 · 2 marks · Very short answerVector Algebra
Vectors $a = 3 \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{k}$ represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths.
Try it first, then show our solutionSolution
1. One diagonal: $a + b = (3 + 1) \hat{i} + (- 2 + 0) \hat{j} + (2 + 2) \hat{k} = 4 \hat{i} - 2 \hat{j} + 4 \hat{k}$ (or its negative)
  ½ mark
2. Other diagonal: $a - b = 2 \hat{i} - 2 \hat{j} + 0 \hat{k} = 2 \hat{i} - 2 \hat{j}$ (or its negative)
  ½ mark
3. Lengths: $∣ a + b ∣ = 16 + 4 + 16 = 6$ and $∣ a - b ∣ = 4 + 4 = 22$
  1 mark
Answer: Diagonals $4 \hat{i} - 2 \hat{j} + 4 \hat{k}$ and $2 \hat{i} - 2 \hat{j}$ ; lengths 6 and $22$
Where the marks are
- Diagonal $a + b = 4 \hat{i} - 2 \hat{j} + 4 \hat{k}$ (½ mark)
- Diagonal $a - b = 2 \hat{i} - 2 \hat{j}$ (½ mark)
- Lengths 6 and $22$ (1 mark)
Open Q24 on its own page
Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerInverse Trigonometric Functions
Simplify : $tan^{- 1} (\frac{c o s 2 x - s i n 2 x}{c o s 2 x + s i n 2 x})$ , $0 < x < \frac{π}{4}$ .
Try it first, then show our solutionSolution
1. Divide numerator and denominator by $cos 2 x$ ( $\neq = 0$ since $0 < 2 x < \frac{π}{2}$ ):
  $tan^{- 1} (\frac{1 - tan 2 x}{1 + tan 2 x})$
  1 mark
2. $= tan^{- 1} [tan (\frac{π}{4} - 2 x)] = \frac{π}{4} - 2 x$ , since $- \frac{π}{4} < \frac{π}{4} - 2 x < \frac{π}{4}$ lies in $(- \frac{π}{2}, \frac{π}{2})$ .
  1 mark
Answer: $\frac{π}{4} - 2 x$
Where the marks are
- Converts to $\frac{1 - t a n 2 x}{1 + t a n 2 x}$ (1 mark)
- Writes as $tan (\frac{π}{4} - 2 x)$ and simplifies to $\frac{π}{4} - 2 x$ (1 mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerInverse Trigonometric Functions
Evaluate : $tan (sin^{- 1} 1 - cos^{- 1} (- \frac{1}{2}))$
Try it first, then show our solutionSolution
1. $sin^{- 1} 1 = \frac{π}{2}$ and $cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ , so the expression $= tan (\frac{π}{2} - \frac{2 π}{3})$
  1 mark
2. $= tan (- \frac{π}{6}) = - \frac{1}{3}$
  1 mark
Answer: $- \frac{1}{3}$
Where the marks are
- Principal values $\frac{π}{2}$ and $\frac{2 π}{3}$ (1 mark)
- Simplifies to $tan (- \frac{π}{6}) = - \frac{1}{3}$ (1 mark)
Open Q25(b) on its own page

Section C

Q26 · 3 marks · Short answerIntegrals
Evaluate : $\int_{0}^{1} x tan^{- 1} x d x$ .
Try it first, then show our solutionSolution
1. Integrate by parts taking $tan^{- 1} x$ as first function:
  $\int_{0}^{1} x tan^{- 1} x d x = [tan^{- 1} x \cdot \frac{x ^{2}}{2}]_{0}^{1} - \int_{0}^{1} \frac{1}{1 + x ^{2}} \cdot \frac{x ^{2}}{2} d x$
  1 mark
2. $= \frac{π}{8} - \frac{1}{2} \int_{0}^{1} \frac{( 1 + x ^{2} ) - 1}{1 + x ^{2}} d x = \frac{π}{8} - \frac{1}{2} [x - tan^{- 1} x]_{0}^{1}$
  1 mark
3. $= \frac{π}{8} - \frac{1}{2} (1 - \frac{π}{4}) = \frac{π}{8} - \frac{1}{2} + \frac{π}{8} = \frac{π}{4} - \frac{1}{2}$
  1 mark
Answer: $\frac{π}{4} - \frac{1}{2}$
Where the marks are
- Correct application of integration by parts (1 mark)
- Splits $\frac{x ^{2}}{1 + x ^{2}} = 1 - \frac{1}{1 + x ^{2}}$ and integrates (1 mark)
- Final value $\frac{π}{4} - \frac{1}{2}$ (1 mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerIntegrals
Find $\int \frac{x + 2}{x - 2} d x$
Try it first, then show our solutionSolution
1. Rationalise: $I = \int \frac{x + 2}{x - 2} \cdot \frac{x + 2}{x + 2} d x = \int \frac{x + 2}{x ^{2} - 4} d x$
  1 mark
2. $I = \frac{1}{2} \int \frac{2 x}{x ^{2} - 4} d x + 2 \int \frac{1}{x ^{2} - 4} d x = I_{1} + I_{2}$
  1 mark
3. For $I_{1}$ put $x^{2} - 4 = t \Rightarrow 2 x d x = d t$ : $I_{1} = \frac{1}{2} \int \frac{d t}{t} = t = x^{2} - 4$
  ½ mark
4. $I_{2} = 2 lo g x + x^{2} - 4$
  $∴ I = x^{2} - 4 + 2 lo g x + x^{2} - 4 + C$
  ½ mark
Answer: $x^{2} - 4 + 2 lo g x + x^{2} - 4 + C$
Where the marks are
- Rationalises to $\int \frac{x + 2}{x ^{2} - 4} d x$ (1 mark)
- Splits into two standard integrals (1 mark)
- Evaluates $I_{1} = x^{2} - 4$ (½ mark)
- Evaluates $I_{2} = 2 lo g ∣ x + x^{2} - 4 ∣$ and writes final answer with $C$ (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerIntegrals
Find : $\int \frac{x ^{2}}{( x ^{2} + 9 ) ( x ^{2} + 16 )} d x$
Try it first, then show our solutionSolution
1. Put $x^{2} = t$ (only for partial fractions): $\frac{t}{( t + 9 ) ( t + 16 )} = \frac{A}{t + 9} + \frac{B}{t + 16}$ , i.e. $t = A (t + 16) + B (t + 9)$
  ½ mark
2. $t = - 9$ : $- 9 = 7 A \Rightarrow A = - \frac{9}{7}$ ; $t = - 16$ : $- 16 = - 7 B \Rightarrow B = \frac{16}{7}$
  1 mark
3. Given integral $= - \frac{9}{7} \int \frac{d x}{x ^{2} + 9} + \frac{16}{7} \int \frac{d x}{x ^{2} + 16}$
  ½ mark
4. $= - \frac{9}{7} \cdot \frac{1}{3} tan^{- 1} \frac{x}{3} + \frac{16}{7} \cdot \frac{1}{4} tan^{- 1} \frac{x}{4} + C = - \frac{3}{7} tan^{- 1} \frac{x}{3} + \frac{4}{7} tan^{- 1} \frac{x}{4} + C$
  1 mark
Answer: $- \frac{3}{7} tan^{- 1} \frac{x}{3} + \frac{4}{7} tan^{- 1} \frac{x}{4} + C$
Where the marks are
- Sets up partial fractions with $x^{2} = t$ (½ mark)
- Finds $A = - \frac{9}{7}$ , $B = \frac{16}{7}$ (1 mark)
- Rewrites the integral in $x$ (½ mark)
- Integrates to the correct $tan^{- 1}$ expression (1 mark)
Open Q27(b) on its own page
Q28 · 3 marks · Short answerIntegrals
If $I_{1} = \int_{- π /4}^{π /4} \frac{d x}{1 + c o s 2 x}$ and $I_{2} = \int_{- 1/2}^{1/2} ∣ x ∣ d x$ , then show that $I_{1} - 4 I_{2} = 0$ .
Try it first, then show our solutionSolution
1. $\frac{1}{1 + cos 2 x}$ is an even function, so
  $I_{1} = 2 \int_{0}^{π /4} \frac{d x}{1 + cos 2 x} = 2 \int_{0}^{π /4} \frac{d x}{2 cos ^{2} x} = \int_{0}^{π /4} sec^{2} x d x = [tan x]_{0}^{π /4} = 1$
  1½ marks
2. $∣ x ∣$ is an even function, so $I_{2} = 2 \int_{0}^{1/2} ∣ x ∣ d x = 2 \int_{0}^{1/2} x d x$
  ½ mark
3. $= [x^{2}]_{0}^{1/2} = \frac{1}{4}$
  ½ mark
4. $I_{1} - 4 I_{2} = 1 - 4 \cdot \frac{1}{4} = 0$ . Hence proved.
  ½ mark
Answer: $I_{1} = 1$ , $I_{2} = \frac{1}{4}$ , so $I_{1} - 4 I_{2} = 0$
Where the marks are
- Uses even-function property and $1 + cos 2 x = 2 cos^{2} x$ to get $I_{1} = 1$ (1½ marks)
- Uses even property for $∣ x ∣$ (½ mark)
- Evaluates $I_{2} = \frac{1}{4}$ (½ mark)
- Shows $I_{1} - 4 I_{2} = 0$ (½ mark)
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Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerDifferential Equations
Find the general solution of the following differential equation :
$x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2}$
Try it first, then show our solutionSolution
1. Dividing by $x^{2}$ : $\frac{d y}{d x} = 1 + \frac{y}{x} + (\frac{y}{x})^{2}$ , a homogeneous equation.
  ½ mark
2. Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
  ½ mark
3. $v + x \frac{d v}{d x} = 1 + v + v^{2} \Rightarrow x \frac{d v}{d x} = 1 + v^{2} \Rightarrow \frac{d v}{1 + v ^{2}} = \frac{d x}{x}$
  ½ mark
4. Integrating: $\int \frac{d v}{1 + v ^{2}} = \int \frac{d x}{x} \Rightarrow tan^{- 1} v = lo g ∣ x ∣ + C$
  1 mark
5. Replacing $v = \frac{y}{x}$ : $tan^{- 1} (\frac{y}{x}) = lo g ∣ x ∣ + C$
  ½ mark
Answer: $tan^{- 1} (\frac{y}{x}) = lo g ∣ x ∣ + C$
Where the marks are
- Writes equation in homogeneous form (½ mark)
- Substitutes $y = v x$ (½ mark)
- Separates variables $\frac{d v}{1 + v ^{2}} = \frac{d x}{x}$ (½ mark)
- Integrates to $tan^{- 1} v = lo g ∣ x ∣ + C$ (1 mark)
- Back-substitutes $v = \frac{y}{x}$ (½ mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerDifferential Equations
Find the particular solution of the differential equation
$x y \frac{d y}{d x} = (x + 2) (y + 2)$ , given that $y (1) = - 1$ .
Try it first, then show our solutionSolution
1. Separating variables: $\frac{y}{y + 2} d y = \frac{x + 2}{x} d x \Rightarrow \int (1 - \frac{2}{y + 2}) d y = \int (1 + \frac{2}{x}) d x$
  1 mark
2. $y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ + C$ (general solution)
  1½ marks
3. Using $y (1) = - 1$ : $- 1 - 2 lo g 1 = 1 + 2 lo g 1 + C \Rightarrow C = - 2$ .
  Particular solution: $y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ - 2$ , i.e. $y = x + 2 lo g ∣ x (y + 2) ∣ - 2$
  ½ mark
Answer: $y - 2 lo g ∣ y + 2∣ = x + 2 lo g ∣ x ∣ - 2$
Where the marks are
- Separates variables correctly (1 mark)
- Integrates to the general solution (1½ marks)
- Finds $C = - 2$ and writes the particular solution (½ mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerLinear Programming
Solve the following linear programming problem graphically :
Minimize $Z = 13 x - 15 y$
Subject to constraints
$x + y \leq 7$ ,
$2 x - 3 y + 6 \geq 0$ ,
$x \geq 0, y \geq 0$
Try it first, then show our solutionSolution
1. Draw the lines $x + y = 7$ (through $(7, 0)$ , $(0, 7)$ ) and $2 x - 3 y + 6 = 0$ (through $(- 3, 0)$ , $(0, 2)$ ) in the first quadrant. $(0, 0)$ satisfies both inequalities, so shade the side containing the origin. They intersect where $x = 7 - y$ , $2 (7 - y) - 3 y = - 6 \Rightarrow y = 4, x = 3$ . The bounded feasible region is the quadrilateral with corners $O (0, 0)$ , $(7, 0)$ , $(3, 4)$ , $(0, 2)$ .
  1½ marks
2. Corner point $Z = 13 x - 15 y$
  (0, 0) 0
  (7, 0) 91
  (3, 4) 39 − 60 = −21
  (0, 2) −30
  1 mark
3. Minimum value of $Z$ is $- 30$ at $x = 0, y = 2$ .
  ½ mark
Answer: Minimum Z = −30 at (0, 2)
Where the marks are
- Correct graph and shading of bounded feasible region with corners (0,0), (7,0), (3,4), (0,2) (1½ marks)
- Evaluates Z at all corner points (1 mark)
- States minimum −30 at (0, 2) (½ mark)
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Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerProbability
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.
Try it first, then show our solutionSolution
1. Let $E_{1}$ : die shows a number less than 3 (bag I chosen); $E_{2}$ : die shows 3 or more (bag II chosen); $A$ : red ball is drawn.
  ½ mark
2. $P (E_{1}) = \frac{2}{6}$ , $P (A ∣ E_{1}) = \frac{3}{7}$ , $P (E_{2}) = \frac{4}{6}$ , $P (A ∣ E_{2}) = \frac{8}{14}$
  1 mark
3. By the theorem of total probability:
  $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) = \frac{2}{6} \cdot \frac{3}{7} + \frac{4}{6} \cdot \frac{8}{14}$
  1 mark
4. $= \frac{1}{7} + \frac{8}{21} = \frac{3 + 8}{21} = \frac{11}{21}$
  ½ mark
Answer: $\frac{11}{21}$
Where the marks are
- Defines events $E_{1}$ , $E_{2}$ , $A$ (½ mark)
- Correct probabilities $P (E_{1}), P (E_{2}), P (A ∣ E_{1}), P (A ∣ E_{2})$ (1 mark)
- Applies total probability theorem (1 mark)
- Final answer $\frac{11}{21}$ (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerProbability
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 $P (X^{'}) + P (Y^{'}) = 2 - 2 a + b$ .
Try it first, then show our solutionSolution
1. At least one of X, Y occurs: $P (X \cup Y) = a$
  ½ mark
2. Exactly one occurs: $P (X \cup Y) - P (X \cap Y) = b \Rightarrow P (X \cap Y) = a - b$
  1 mark
3. $P (X^{'}) + P (Y^{'}) = [1 - P (X)] + [1 - P (Y)] = 2 - [P (X) + P (Y)] = 2 - [P (X \cup Y) + P (X \cap Y)]$
  1 mark
4. $= 2 - [a + (a - b)] = 2 - 2 a + b$ . Hence proved.
  ½ mark
Answer: $P (X^{'}) + P (Y^{'}) = 2 - 2 a + b$ (proved)
Where the marks are
- Writes $P (X \cup Y) = a$ (½ mark)
- Obtains $P (X \cap Y) = a - b$ (1 mark)
- Uses complements and addition theorem (1 mark)
- Simplifies to $2 - 2 a + b$ (½ mark)
Open Q31(b) on its own page

Corner point	$Z = 13 x - 15 y$
(0, 0)	0
(7, 0)	91
(3, 4)	39 − 60 = −21
(0, 2)	−30

Section D

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerRelations and Functions
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.
Try it first, then show our solutionSolution
1. Reflexive: For any $x \in Z$ , $∣ x - x ∣ = 0$ , which is divisible by $p$ ( $0 = 0 \cdot p$ ). So $(x, x) \in R$ and R is reflexive.
  1½ marks
2. Symmetric: Let $(x, y) \in R$ , so $∣ x - y ∣$ is divisible by $p$ . Since $∣ y - x ∣ = ∣ x - y ∣$ , $∣ y - x ∣$ is divisible by $p$ , so $(y, x) \in R$ . R is symmetric.
  1½ marks
3. Transitive: Let $(x, y), (y, z) \in R$ . Then $x - y = k p$ and $y - z = m p$ for some integers $k, m$ . Adding, $x - z = (k + m) p$ , so $∣ x - z ∣$ is divisible by $p$ and $(x, z) \in R$ . R is transitive.
  1½ marks
4. Since R is reflexive, symmetric and transitive, R is an equivalence relation.
  ½ mark
Answer: R is an equivalence relation
Where the marks are
- Proves reflexivity using $∣ x - x ∣ = 0$ (1½ marks)
- Proves symmetry using $∣ x - y ∣ = ∣ y - x ∣$ (1½ marks)
- Proves transitivity by adding $x - y$ and $y - z$ (1½ marks)
- Concludes equivalence relation (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerRelations and Functions
A function $f : R - {\frac{3}{5}} ⟶ R - {\frac{3}{5}}$ is defined as $f (x) = \frac{3 x + 2}{5 x - 3}$ . Show that f is one-one and onto.
Try it first, then show our solutionSolution
1. One-one: Let $x_{1}, x_{2} \in R - {\frac{3}{5}}$ with $f (x_{1}) = f (x_{2})$ :
  $\frac{3 x _{1} + 2}{5 x _{1} - 3} = \frac{3 x _{2} + 2}{5 x _{2} - 3}$
  $\Rightarrow 15 x_{1} x_{2} - 9 x_{1} + 10 x_{2} - 6 = 15 x_{1} x_{2} + 10 x_{1} - 9 x_{2} - 6$
  $\Rightarrow 19 x_{2} = 19 x_{1} \Rightarrow x_{1} = x_{2}$ . Hence $f$ is one-one.
  2½ marks
2. Onto: Let $y \in R - {\frac{3}{5}}$ (codomain). Setting $\frac{3 x + 2}{5 x - 3} = y$ : $5 x y - 3 y = 3 x + 2 \Rightarrow x (5 y - 3) = 3 y + 2 \Rightarrow x = \frac{3 y + 2}{5 y - 3}$ , defined as $y \neq = \frac{3}{5}$ . Also $x \neq = \frac{3}{5}$ , since $\frac{3 y + 2}{5 y - 3} = \frac{3}{5}$ gives $15 y + 10 = 15 y - 9$ , impossible. So $x$ is in the domain, and
  $f (x) = \frac{3 ( 3 y + 2 ) + 2 ( 5 y - 3 )}{5 ( 3 y + 2 ) - 3 ( 5 y - 3 )} = \frac{19 y}{19} = y$ .
  Every element of the codomain has a pre-image, so $f$ is onto.
  2½ marks
Answer: f is one-one and onto (pre-image $x = \frac{3 y + 2}{5 y - 3}$ )
Where the marks are
- Shows $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ (2½ marks)
- Finds $x = \frac{3 y + 2}{5 y - 3}$ in the domain with $f (x) = y$ , concluding onto (2½ marks)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerMatrices
If $A = 0 - 2 1 2 - 1 - 1 1 - 2 0$ , find $A^{- 1}$ and use it to solve the following system of equations :
$- 2 y + z = 7$ , $2 x - y - z = 8$ , $x - 2 y = 10$
Try it first, then show our solutionSolution
1. $∣ A ∣ = 0 (0 - 2) - 2 (0 + 2) + 1 (2 + 1) = - 4 + 3 = - 1 \neq = 0$ , so $A^{- 1}$ exists.
  1 mark
2. Cofactors: $C_{11} = - 2, C_{12} = - 2, C_{13} = 3; C_{21} = - 1, C_{22} = - 1, C_{23} = 2; C_{31} = - 3, C_{32} = - 2, C_{33} = 4$ .
  $adj A = - 2 - 2 3 - 1 - 1 2 - 3 - 2 4$
  1½ marks
3. $A^{- 1} = \frac{1}{∣ A ∣} adj A = 22 - 3 11 - 2 32 - 4$
  ½ mark
4. The coefficient matrix of the system is $021 - 2 - 1 - 2 1 - 1 0 = A^{T}$ . So the system is $A^{T} X = B$ with $X = x y z$ , $B = 7810$ , giving $X = (A^{T})^{- 1} B = (A^{- 1})^{T} B$ .
  ½ mark
5. $x y z = 213212 - 3 - 2 - 4 7810 = 14 + 16 - 30 7 + 8 - 20 21 + 16 - 40 = 0 - 5 - 3$
  $∴ x = 0, y = - 5, z = - 3$
  1½ marks
Answer: $A^{- 1} = 22 - 3 11 - 2 32 - 4$ ; $x = 0, y = - 5, z = - 3$
Where the marks are
- Finds $∣ A ∣ = - 1$ (1 mark)
- Correct adj A (1½ marks)
- Correct $A^{- 1}$ (½ mark)
- Recognises system as $A^{T} X = B$ so $X = (A^{- 1})^{T} B$ (½ mark)
- Solves to get $x = 0, y = - 5, z = - 3$ (1½ marks)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerDeterminants
If $3 - 7 - 11 - 1 47 sin 3 x cos 2 x 2$ is a singular matrix, then find all values of $x$ where $x \in [0, \frac{π}{2}]$ .
Try it first, then show our solutionSolution
1. A singular matrix has zero determinant, so
  $3 - 7 - 11 - 1 47 sin 3 x cos 2 x 2 = 0$ .
  1 mark
2. Expand along $C_{3}$ :
  $sin 3 x (- 49 + 44) - cos 2 x (21 - 11) + 2 (12 - 7) = 0$
  $\Rightarrow - 5 sin 3 x - 10 cos 2 x + 10 = 0 \Rightarrow sin 3 x + 2 cos 2 x = 2$ .
  1 mark
3. Use $sin 3 x = 3 sin x - 4 sin^{3} x$ and $cos 2 x = 1 - 2 sin^{2} x$ :
  $3 sin x - 4 sin^{3} x + 2 - 4 sin^{2} x = 2$
  $\Rightarrow 4 sin^{3} x + 4 sin^{2} x - 3 sin x = 0$ .
  1 mark
4. Factorise:
  $sin x (4 sin^{2} x + 4 sin x - 3) = 0 \Rightarrow sin x (2 sin x - 1) (2 sin x + 3) = 0$ .
  So $sin x = 0$ , $sin x = \frac{1}{2}$ or $sin x = - \frac{3}{2}$ .
  1 mark
5. $sin x = - \frac{3}{2}$ is impossible because $∣ sin x ∣ \leq 1$ .
  For $x \in [0, \frac{π}{2}]$ : $sin x = 0 \Rightarrow x = 0$ , and $sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}$ .
  1 mark
Answer: $x = 0$ or $x = \frac{π}{6}$
Where the marks are
- Sets the determinant equal to 0 because the matrix is singular (1 mark)
- Expands the determinant to get $sin 3 x + 2 cos 2 x = 2$ (1 mark)
- Converts to the cubic $4 sin^{3} x + 4 sin^{2} x - 3 sin x = 0$ (1 mark)
- Factorises as $sin x (2 sin x - 1) (2 sin x + 3) = 0$ (1 mark)
- Rejects $sin x = - 3/2$ and gets $x = 0, π /6$ (1 mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerContinuity and Differentiability
If $x = cos t$ , $y = cos m t$ , prove that $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ .
Try it first, then show our solutionSolution
1. Differentiate both equations w.r.t. $t$ :
  $\frac{d x}{d t} = - sin t$ and $\frac{d y}{d t} = - m sin m t$ .
  1 mark
2. Divide:
  $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{m s i n m t}{s i n t}$ , so $sin t \frac{d y}{d x} = m sin m t$ .
  1 mark
3. Differentiate again w.r.t. $x$ using the chain rule:
  $sin t \frac{d ^{2} y}{d x ^{2}} + cos t \frac{d t}{d x} \frac{d y}{d x} = m^{2} cos m t \frac{d t}{d x}$ .
  Multiply by $sin t$ and use $sin t \frac{d t}{d x} = \frac{s i n t}{- s i n t} = - 1$ :
  $sin^{2} t \frac{d ^{2} y}{d x ^{2}} - cos t \frac{d y}{d x} = - m^{2} cos m t$ .
  2 marks
4. Substitute $sin^{2} t = 1 - cos^{2} t = 1 - x^{2}$ , $cos t = x$ and $cos m t = y$ :
  $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ . Hence proved.
  1 mark
Answer: $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ (proved)
Where the marks are
- Finds $d x / d t = - sin t$ and $d y / d t = - m sin m t$ (1 mark)
- Gets $sin t \frac{d y}{d x} = m sin m t$ (1 mark)
- Differentiates again w.r.t. $x$ correctly, using $d t / d x = - 1/ sin t$ (2 marks)
- Substitutes $x = cos t$ , $y = cos m t$ to reach the result (1 mark)
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Q35 · 5 marks · Long answerThree Dimensional Geometry
Check whether the lines given by $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ are parallel or not. If parallel, find the distance between them, otherwise find their point of intersection, if the lines are intersecting.
Try it first, then show our solutionSolution
1. The direction ratios of the first line are $⟨ 2, 3, 4 ⟩$ . Writing the second line as $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1}$ , its direction ratios are $⟨ 5, 2, 1 ⟩$ .
  1 mark
2. Since $\frac{2}{5} \neq = \frac{3}{2} \neq = \frac{4}{1}$ , the direction ratios are not proportional. Hence the lines are not parallel.
  1 mark
3. A general point on line 1 is $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ .
  A general point on line 2 is $(5 μ + 4, 2 μ + 1, μ)$ .
  1 mark
4. For the lines to intersect, these points must coincide for some $λ, μ$ :
  $2 λ + 1 = 5 μ + 4$ , $3 λ + 2 = 2 μ + 1$ , $4 λ + 3 = μ$ .
  The first two give $2 λ - 5 μ = 3$ and $3 λ - 2 μ = - 1$ , so $λ = - 1$ , $μ = - 1$ .
  Check in the third equation: $4 (- 1) + 3 = - 1 = μ$ . It is satisfied, so the lines intersect.
  1 mark
5. Put $λ = - 1$ in the point on line 1: $(2 (- 1) + 1, 3 (- 1) + 2, 4 (- 1) + 3)$ .
  The point of intersection is $(- 1, - 1, - 1)$ .
  1 mark
Answer: The lines are not parallel; they intersect at $(- 1, - 1, - 1)$ .
Where the marks are
- Writes the direction ratios $⟨ 2, 3, 4 ⟩$ and $⟨ 5, 2, 1 ⟩$ (1 mark)
- Concludes the lines are not parallel because the direction ratios are not proportional (1 mark)
- Writes general points on both lines (1 mark)
- Solves to get $λ = μ = - 1$ and verifies the third equation (1 mark)
- States the point of intersection $(- 1, - 1, - 1)$ (1 mark)
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Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ $x$ .
Based on the given information, answer the following questions :
1. (i)
  How many subscribers will discontinue after an increase of ₹ $x$ in annual fee ?
  1 mark
2. (ii)
  If R( $x$ ) denotes the total revenue collected after the increase of ₹ $x$ in subscription fee, express R( $x$ ) as a function of $x$ .
  1 mark
3. (iii)(a)
  Find the value of $x$ for which R( $x$ ) is maximum.
  2 marks
4. OR
  (iii)(b)
  Find the sub-intervals of (0, 5000) in which R( $x$ ) is increasing and decreasing.
  2 marks
Figure in the paper: Illustration of an online delivery app and a delivery person, alongside a graph with 'Subscription Fee' on x-axis and 'No. of Subscribers' on y-axis showing a straight line decreasing. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Each ₹1 increase loses 10 subscribers, so an increase of ₹ $x$ makes $10 x$ subscribers discontinue.
  1 mark
2. (ii) Subscribers left $= 5000 - 10 x$ and the new fee $= ₹ (300 + x)$ .
  So $R (x) = (5000 - 10 x) (300 + x) = 1500000 + 2000 x - 10 x^{2}$ .
  1 mark
3. (iii)(a) Differentiate: $R^{'} (x) = 2000 - 20 x$ .
  ½ mark
4. (iii)(a) Set $R^{'} (x) = 0$ : $2000 - 20 x = 0 \Rightarrow x = 100$ .
  ½ mark
5. (iii)(a) Second derivative: $R^{''} (x) = - 20 < 0$ at $x = 100$ .
  ½ mark
6. (iii)(a) So $R (x)$ is maximum at $x = 100$ , i.e. when the fee is raised by ₹100. The maximum revenue is $R (100) = 4000 \times 400 = ₹16, 00, 000$ .
  ½ mark
7. (iii)(b) $R^{'} (x) = 2000 - 20 x$ .
  ½ mark
8. (iii)(b) $R^{'} (x) = 0 \Rightarrow x = 100$ , which splits $(0, 5000)$ into $(0, 100)$ and $(100, 5000)$ .
  ½ mark
9. (iii)(b) For $0 < x < 100$ , $R^{'} (x) > 0$ , so $R (x)$ is increasing on $(0, 100)$ (or $(0, 100]$ ).
  ½ mark
10. (iii)(b) For $100 < x < 5000$ , $R^{'} (x) < 0$ , so $R (x)$ is decreasing on $(100, 5000)$ (or $[100, 5000)$ ).
  ½ mark
Answer: (i) $10 x$ ; (ii) $R (x) = (5000 - 10 x) (300 + x)$ ; (iii)(a) $x = 100$ ; (iii)(b) increasing on $(0, 100)$ , decreasing on $(100, 5000)$
Where the marks are
- (i) States $10 x$ subscribers discontinue (1 mark)
- (ii) Writes $R (x) = (5000 - 10 x) (300 + x)$ (1 mark)
- (iii)(a) Finds $R^{'} (x) = 2000 - 20 x$ (½ mark)
- (iii)(a) Solves $R^{'} (x) = 0$ to get $x = 100$ (½ mark)
- (iii)(a) Shows $R^{''} (x) = - 20 < 0$ (½ mark)
- (iii)(a) Concludes $R$ is maximum at $x = 100$ (½ mark)
- (iii)(b) Finds $R^{'} (x)$ (½ mark)
- (iii)(b) Gets the critical point $x = 100$ (½ mark)
- (iii)(b) States increasing on $(0, 100)$ with reason $R^{'} > 0$ (½ mark)
- (iii)(b) States decreasing on $(100, 5000)$ with reason $R^{'} < 0$ (½ mark)
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Q37 · 4 marks · Case-basedCompetency-basedProbability
In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each.
A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold there by raising a fund of ₹ 1,00,00,000.
Rohan bought one ticket.
Based on given information, answer the following questions :
1. (i)
  What are the possible amounts, the person can win ?
  1 mark
2. (ii)(a)
  What is the probability that the person wins atleast ₹ 2,00,000 ?
  2 marks
3. OR
  (ii)(b)
  What is the probability that the person does not win any amount ?
  2 marks
4. (iii)
  In another jackpot, Rohan also bought a ticket having a prize money of ₹ 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
Figure in the paper: Banner image of a lucky ticket booth with text 'Lucky Ticket', 'WIN', 'PRIZE ₹ 200,000 ₹ 100,000'. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Rohan can win ₹3,00,000 (first prize), ₹2,00,000 (second prize) or ₹50,000 (third prize).
  1 mark
2. (ii)(a) Winning at least ₹2,00,000 means winning the first prize or a second prize. There are $1 + 2 = 3$ such tickets out of $1, 00, 000$ .
  So $P = \frac{3}{100000}$ .
  2 marks
3. (ii)(b) There are $1 + 2 + 3 = 6$ winning tickets in all.
  $P (no prize) = 1 - \frac{6}{100000} = \frac{99994}{100000} = \frac{49997}{50000}$ .
  2 marks
4. (iii) In the first jackpot, $P (win) = \frac{6}{100000}$ and $P (lose) = \frac{99994}{100000}$ . In the second jackpot, $P (win) = \frac{1}{100000}$ and $P (lose) = \frac{99999}{100000}$ . The two draws are independent.
  $P (exactly one win) = \frac{6}{1 0 ^{5}} \cdot \frac{99999}{1 0 ^{5}} + \frac{99994}{1 0 ^{5}} \cdot \frac{1}{1 0 ^{5}}$
  $= \frac{599994 + 99994}{1 0 ^{10}} = \frac{699988}{1 0 ^{10}} = \frac{174997}{2500000000}$ .
  1 mark
Answer: (i) ₹3,00,000, ₹2,00,000, ₹50,000; (ii)(a) $\frac{3}{100000}$ ; (ii)(b) $\frac{49997}{50000}$ ; (iii) $\frac{699988}{1 0 ^{10}} = \frac{174997}{2500000000}$
Where the marks are
- (i) Lists the three prize amounts (1 mark)
- (ii)(a) Counts 3 favourable tickets and gets $3/100000$ (2 marks)
- (ii)(b) Gets $1 - 6/100000 = 49997/50000$ (2 marks)
- (iii) Uses (win first, lose second) + (lose first, win second) to get $699988/1 0^{10}$ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedApplication of Integrals
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 $C_{2} : x^{2} + y^{2} = 4$ .
Based on the given information, answer the following questions :
1. (i)
  Represent the given equations $C_{1}$ and $C_{2}$ with the help of a diagram.
  1 mark
2. (ii)
  Express y as a function of $x$ , ( $y = f (x)$ ), for both $C_{1}$ and $C_{2}$ .
  1 mark
3. (iii)(a)
  Using integration find the area of region covered by the roundabout.
  2 marks
4. OR
  (iii)(b)
  Using integration, find the area of region covered by circular pond.
  2 marks
Figure in the paper: Aerial picture of a roundabout with a circular pond and fountain at its centre, surrounded by trees and roads. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Diagram: draw the $x$ - and $y$ -axes with origin O.
  Draw $C_{1}$ , a circle centred at O with radius 8, passing through $(\pm 8, 0)$ and $(0, \pm 8)$ .
  Draw $C_{2}$ , a smaller concentric circle with radius 2, passing through $(\pm 2, 0)$ and $(0, \pm 2)$ .
  Label both circles.
  1 mark
2. (ii) For $C_{1}$ : $y = 64 - x^{2}$ (upper half) or $y = - 64 - x^{2}$ (lower half), with $- 8 \leq x \leq 8$ .
  ½ mark
3. (ii) For $C_{2}$ : $y = 4 - x^{2}$ or $y = - 4 - x^{2}$ , with $- 2 \leq x \leq 2$ .
  ½ mark
4. (iii)(a) The circle is symmetric about both axes, so the area is 4 times the area in the first quadrant:
  Area $= 4 \int_{0}^{8} 64 - x^{2} d x$ .
  1 mark
5. (iii)(a) Use $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ :
  $= 4 [\frac{x}{2} 64 - x^{2} + 32 sin^{- 1} \frac{x}{8}]_{0}^{8} = 4 [0 + 32 \cdot \frac{π}{2} - 0] = 64 π$ sq. units.
  1 mark
6. (iii)(b) By symmetry, Area of pond $= 4 \int_{0}^{2} 4 - x^{2} d x$ .
  1 mark
7. (iii)(b) $= 4 [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{0}^{2} = 4 [0 + 2 \cdot \frac{π}{2} - 0] = 4 π$ sq. units.
  1 mark
Answer: (ii) $C_{1}$ : $y = \pm 64 - x^{2}$ ; $C_{2}$ : $y = \pm 4 - x^{2}$ ; (iii)(a) $64 π$ sq. units; (iii)(b) $4 π$ sq. units
Where the marks are
- (i) Draws two concentric circles centred at O with radii 8 and 2, correctly labelled (1 mark)
- (ii) Writes $y = \pm 64 - x^{2}$ for $C_{1}$ (½ mark)
- (ii) Writes $y = \pm 4 - x^{2}$ for $C_{2}$ (½ mark)
- (iii)(a) Sets up Area $= 4 \int_{0}^{8} 64 - x^{2} d x$ (1 mark)
- (iii)(a) Evaluates to $64 π$ sq. units (1 mark)
- (iii)(b) Sets up Area $= 4 \int_{0}^{2} 4 - x^{2} d x$ (1 mark)
- (iii)(b) Evaluates to $4 π$ sq. units (1 mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.