MathsCBSE Class 12 · Board exam 2023 (session 2022-23)

CBSE Class 12 Mathematics: 2023 board paper, set 65/1/1

The question paper set in the CBSE Class 12 board exam of 2023, 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/2023/XII/MATHEMATICS.zip (a zip; this paper is the file 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

Instructions in brief: All questions compulsory; no overall choice; internal choice in 2 questions of Section B, 3 of Section C, 2 of Section D and 2 of Section E; calculators not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQMatrices
If for a square matrix A, $A^{2} - 3 A + I = O$ and $A^{- 1} = x A + y I$ , then the value of $x + y$ is :
1. (A) $- 2$
2. (B) $2$
3. (C) $3$
4. (D) $- 3$
Try it first, then show our solutionSolution
1. Pre-multiply $A^{2} - 3 A + I = O$ by $A^{- 1}$ : $A - 3 I + A^{- 1} = O$ , so $A^{- 1} = - A + 3 I$ .
  ½ mark
2. Comparing with $x A + y I$ : $x = - 1, y = 3$ , so $x + y = 2$ . Option (B).
  ½ mark
Answer: $x + y = 2$ , option (B)
Where the marks are
- Obtains $A^{- 1} = 3 I - A$ and $x + y = 2$ (1 mark)
Q2 · 1 mark · MCQDeterminants
If $∣ A ∣ = 2$ , where A is a $2 \times 2$ matrix, then $∣4 A^{- 1} ∣$ equals :
1. (A) $4$
2. (B) $2$
3. (C) $8$
4. (D) $\frac{1}{32}$
Try it first, then show our solutionSolution
1. For a $2 \times 2$ matrix, $∣ k B ∣ = k^{2} ∣ B ∣$ and $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ .
  ½ mark
2. $∣4 A^{- 1} ∣ = 16 \cdot \frac{1}{2} = 8$ . Option (C).
  ½ mark
Answer: 8, option (C)
Where the marks are
- Uses $∣ k A ∣ = k^{n} ∣ A ∣$ and $∣ A^{- 1} ∣ = 1/∣ A ∣$ to get 8 (1 mark)
Q3 · 1 mark · MCQDeterminants
Let A be a $3 \times 3$ matrix such that $∣ adj A ∣ = 64$ . Then $∣ A ∣$ is equal to :
1. (A) $8$ only
2. (B) $- 8$ only
3. (C) $64$
4. (D) $8$ or $- 8$
Try it first, then show our solutionSolution
1. For order 3, $∣ adj A ∣ = ∣ A ∣^{2}$ , so $∣ A ∣^{2} = 64$ .
  ½ mark
2. $∣ A ∣ = \pm 8$ . Option (D).
  ½ mark
Answer: $8$ or $- 8$ , option (D)
Where the marks are
- Uses $∣ adj A ∣ = ∣ A ∣^{n - 1}$ and gets $\pm 8$ (1 mark)
Q4 · 1 mark · MCQMatrices
If $A = [3542]$ and $2 A + B$ is a null matrix, then B is equal to :
1. (A) $[61084]$
2. (B) $[- 6 - 10 - 8 - 4]$
3. (C) $[51083]$
4. (D) $[- 5 - 10 - 8 - 3]$
Try it first, then show our solutionSolution
1. $B = - 2 A = [- 6 - 10 - 8 - 4]$ . Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- $B = - 2 A$ computed correctly (1 mark)
Q5 · 1 mark · MCQIntegrals
If $\frac{d}{d x} (f (x)) = lo g x$ , then $f (x)$ equals :
1. (A) $- \frac{1}{x} + C$
2. (B) $x (lo g x - 1) + C$
3. (C) $x (lo g x + x) + C$
4. (D) $\frac{1}{x} + C$
Try it first, then show our solutionSolution
1. $f (x) = \int lo g x d x$ ; by parts with $1$ as second function: $x lo g x - \int x \cdot \frac{1}{x} d x = x lo g x - x + C$ .
  ½ mark
2. $= x (lo g x - 1) + C$ . Option (B).
  ½ mark
Answer: $x (lo g x - 1) + C$ , option (B)
Where the marks are
- Integrates $lo g x$ by parts correctly (1 mark)
Q6 · 1 mark · MCQIntegrals
$\int_{0}^{π /6} sec^{2} (x - \frac{π}{6}) d x$ is equal to :
1. (A) $\frac{1}{3}$
2. (B) $- \frac{1}{3}$
3. (C) $3$
4. (D) $- 3$
Try it first, then show our solutionSolution
1. $\int_{0}^{π /6} sec^{2} (x - \frac{π}{6}) d x = [tan (x - \frac{π}{6})]_{0}^{π /6} = 0 - tan (- \frac{π}{6})$ .
  ½ mark
2. $= \frac{1}{3}$ . Option (A).
  ½ mark
Answer: $\frac{1}{3}$ , option (A)
Where the marks are
- Antiderivative $tan (x - π /6)$ and correct limits (1 mark)
Q7 · 1 mark · MCQDifferential Equations
The sum of the order and the degree of the differential equation $\frac{d ^{2} y}{d x ^{2}} + (\frac{d y}{d x})^{3} = sin y$ is :
1. (A) $5$
2. (B) $2$
3. (C) $3$
4. (D) $4$
Try it first, then show our solutionSolution
1. Highest derivative is $\frac{d ^{2} y}{d x ^{2}}$ : order 2; its power is 1: degree 1.
  ½ mark
2. Sum $= 3$ . Option (C).
  ½ mark
Answer: 3, option (C)
Where the marks are
- Order 2, degree 1, sum 3 (1 mark)
Q8 · 1 mark · MCQVector Algebra
The value of p for which the vectors $2 \hat{i} + p \hat{j} + \hat{k}$ and $- 4 \hat{i} - 6 \hat{j} + 26 \hat{k}$ are perpendicular to each other, is :
1. (A) $3$
2. (B) $- 3$
3. (C) $- \frac{17}{3}$
4. (D) $\frac{17}{3}$
Try it first, then show our solutionSolution
1. Perpendicular ⇒ dot product zero: $- 8 - 6 p + 26 = 0$ .
  ½ mark
2. $p = 3$ . Option (A).
  ½ mark
Answer: $p = 3$ , option (A)
Where the marks are
- Sets dot product to 0 and gets $p = 3$ (1 mark)
Q9 · 1 mark · MCQVector Algebra
The value of $(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$ is :
1. (A) $2$
2. (B) $0$
3. (C) $1$
4. (D) $- 1$
Try it first, then show our solutionSolution
1. $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{k} \cdot \hat{j} = 0$ ; $\hat{j} \times \hat{i} = - \hat{k}$ , $- \hat{k} \cdot \hat{k} = - 1$ .
  ½ mark
2. Sum $= - 1$ . Option (D).
  ½ mark
Answer: $- 1$ , option (D)
Where the marks are
- Correct cross products and dot products giving −1 (1 mark)
Q10 · 1 mark · MCQVector Algebra
If $a + b = \hat{i}$ and $a = 2 \hat{i} - 2 \hat{j} + 2 \hat{k}$ , then $∣ b ∣$ equals :
1. (A) $14$
2. (B) $3$
3. (C) $12$
4. (D) $17$
Try it first, then show our solutionSolution
1. $b = \hat{i} - a = - \hat{i} + 2 \hat{j} - 2 \hat{k}$ .
  ½ mark
2. $∣ b ∣ = 1 + 4 + 4 = 3$ . Option (B).
  ½ mark
Answer: 3, option (B)
Where the marks are
- Finds $b$ and its magnitude 3 (1 mark)
Q11 · 1 mark · MCQThree Dimensional Geometry
Direction cosines of the line $\frac{x - 1}{2} = \frac{1 - y}{3} = \frac{2 z - 1}{12}$ are :
1. (A) $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$
2. (B) $\frac{2}{157}, - \frac{3}{157}, \frac{12}{157}$
3. (C) $\frac{2}{7}, - \frac{3}{7}, - \frac{6}{7}$
4. (D) $\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}$
Try it first, then show our solutionSolution
1. Standard form: $\frac{x - 1}{2} = \frac{y - 1}{- 3} = \frac{z - 1/2}{6}$ , so d.r.s $2, - 3, 6$ .
  ½ mark
2. Magnitude $4 + 9 + 36 = 7$ ; d.c.s $\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}$ . Option (D).
  ½ mark
Answer: $\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}$ , option (D)
Where the marks are
- Converts to standard form and normalises d.r.s (1 mark)
Q12 · 1 mark · MCQProbability
If $P (\frac{A}{B}) = 0.3$ , $P (A) = 0.4$ and $P (B) = 0.8$ , then $P (\frac{B}{A})$ is equal to :
1. (A) $0.6$
2. (B) $0.3$
3. (C) $0.06$
4. (D) $0.4$
Try it first, then show our solutionSolution
1. $P (A \cap B) = P (A ∣ B) P (B) = 0.3 \times 0.8 = 0.24$ .
  ½ mark
2. $P (B ∣ A) = \frac{0.24}{0.4} = 0.6$ . Option (A).
  ½ mark
Answer: 0.6, option (A)
Where the marks are
- Finds $P (A \cap B) = 0.24$ and $P (B ∣ A) = 0.6$ (1 mark)
Q13 · 1 mark · MCQContinuity and Differentiability
The value of k for which $f (x) = {3 x + 5, k x^{2}, x \geq 2 x < 2$ is a continuous function, is :
1. (A) $- \frac{11}{4}$
2. (B) $\frac{4}{11}$
3. (C) $11$
4. (D) $\frac{11}{4}$
Try it first, then show our solutionSolution
1. Continuity at $x = 2$ : LHL $= 4 k$ , $f (2) = 11$ .
  ½ mark
2. $4 k = 11 \Rightarrow k = \frac{11}{4}$ . Option (D).
  ½ mark
Answer: $k = \frac{11}{4}$ , option (D)
Where the marks are
- Equates LHL and value at 2 to get $k = 11/4$ (1 mark)
Q14 · 1 mark · MCQMatrices
If $A = [0 - 1 10]$ and $(3 I + 4 A) (3 I - 4 A) = x^{2} I$ , then the value(s) x is/are :
1. (A) $\pm 7$
2. (B) $0$
3. (C) $\pm 5$
4. (D) $25$
Try it first, then show our solutionSolution
1. $(3 I + 4 A) (3 I - 4 A) = 9 I - 16 A^{2}$ (A commutes with I); $A^{2} = - I$ .
  ½ mark
2. So $25 I = x^{2} I \Rightarrow x = \pm 5$ . Option (C).
  ½ mark
Answer: $x = \pm 5$ , option (C)
Where the marks are
- Uses $A^{2} = - I$ to get $25 I$ , so $x = \pm 5$ (1 mark)
Q15 · 1 mark · MCQDifferential Equations
The general solution of the differential equation $x d y - (1 + x^{2}) d x = d x$ is :
1. (A) $y = 2 x + \frac{x ^{3}}{3} + C$
2. (B) $y = 2 lo g x + \frac{x ^{3}}{3} + C$
3. (C) $y = \frac{x ^{2}}{2} + C$
4. (D) $y = 2 lo g x + \frac{x ^{2}}{2} + C$
Try it first, then show our solutionSolution
1. $x d y = (2 + x^{2}) d x \Rightarrow d y = (\frac{2}{x} + x) d x$ .
  ½ mark
2. $y = 2 lo g x + \frac{x ^{2}}{2} + C$ . Option (D).
  ½ mark
Answer: Option (D)
Where the marks are
- Separates variables and integrates (1 mark)
Q16 · 1 mark · MCQApplication of Derivatives
If $f (x) = a (x - cos x)$ is strictly decreasing in $R$ , then 'a' belongs to
1. (A) ${0}$
2. (B) $(0, \infty)$
3. (C) $(- \infty, 0)$
4. (D) $(- \infty, \infty)$
Try it first, then show our solutionSolution
1. $f^{'} (x) = a (1 + sin x)$ , and $1 + sin x \geq 0$ (zero only at isolated points).
  ½ mark
2. $f^{'} \leq 0$ with isolated zeros needs $a < 0$ . Option (C).
  ½ mark
Answer: $a \in (- \infty, 0)$ , option (C)
Where the marks are
- Differentiates and deduces $a < 0$ (1 mark)
Q17 · 1 mark · MCQLinear Programming
The corner points of the feasible region in the graphical representation of a linear programming problem are $(2, 72)$ , $(15, 20)$ and $(40, 15)$ . If $z = 18 x + 9 y$ be the objective function, then :
1. (A)z is maximum at $(2, 72)$ , minimum at $(15, 20)$
2. (B)z is maximum at $(15, 20)$ , minimum at $(40, 15)$
3. (C)z is maximum at $(40, 15)$ , minimum at $(15, 20)$
4. (D)z is maximum at $(40, 15)$ , minimum at $(2, 72)$
Try it first, then show our solutionSolution
1. z: (2,72)→36+648=684; (15,20)→270+180=450; (40,15)→720+135=855.
  ½ mark
2. Max at (40,15), min at (15,20). Option (C).
  ½ mark
Answer: Option (C)
Where the marks are
- Evaluates z at corners and identifies max/min (1 mark)
Q18 · 1 mark · MCQLinear Programming
The number of corner points of the feasible region determined by the constraints $x - y \geq 0$ , $2 y \leq x + 2$ , $x \geq 0$ , $y \geq 0$ is :
1. (A) $2$
2. (B) $3$
3. (C) $4$
4. (D) $5$
Try it first, then show our solutionSolution
1. Lines $y = x$ and $2 y = x + 2$ meet at (2,2); region with $y \leq x$ , $2 y \leq x + 2$ , $y \geq 0$ is unbounded, bounded by $y = 0$ from (0,0), and edges from (0,0) to (2,2) and beyond along $2 y = x + 2$ .
  ½ mark
2. Corner points: (0,0) and (2,2), i.e. 2. Option (A).
  ½ mark
Answer: 2, option (A)
Where the marks are
- Identifies corners (0,0),(2,2) (1 mark)
Open Q18 on its own page
Q19 · 1 mark · Assertion-reasonCompetency-basedInverse Trigonometric Functions
Assertion (A) : The range of the function $f (x) = 2 sin^{- 1} x + \frac{3 π}{2}$ , where $x \in [- 1, 1]$ , is $[\frac{π}{2}, \frac{5 π}{2}]$ .
Reason (R) : The range of the principal value branch of $sin^{- 1} (x)$ is $[0, π]$ .
1. (A)Both Assertion (A) and Reason (R) are true and 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. $sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}]$ , so $2 sin^{- 1} x + \frac{3 π}{2} \in [\frac{π}{2}, \frac{5 π}{2}]$ : A true.
  ½ mark
2. Principal range of $sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ , not $[0, π]$ : R false. Option (C).
  ½ mark
Answer: Option (C)
Where the marks are
- A true, R false with reason (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedThree Dimensional Geometry
Assertion (A) : Equation of a line passing through the points $(1, 2, 3)$ and $(3, - 1, 3)$ is $\frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 3}{0}$ .
Reason (R) : Equation of a line passing through points $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ is given by $\frac{x - x _{1}}{x _{2} - x _{1}} = \frac{y - y _{1}}{y _{2} - y _{1}} = \frac{z - z _{1}}{z _{2} - z _{1}}$ .
1. (A)Both Assertion (A) and Reason (R) are true and 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. R is the standard two-point form: true.
  ½ mark
2. D.r.s are $2, - 3, 0$ , so the line is $\frac{x - 3}{2} = \frac{y + 1}{- 3} = \frac{z - 3}{0}$ ; given denominator 3 is wrong, A false. Option (D).
  ½ mark
Answer: Option (D)
Where the marks are
- Finds d.r.s $2, - 3, 0$ so A false, 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 answerRelations and Functions
A function $f : A \to B$ defined as $f (x) = 2 x$ is both one-one and onto. If $A = {1, 2, 3, 4}$ , then find the set B.
Try it first, then show our solutionSolution
1. Find the image of each element of $A$ under $f (x) = 2 x$ : $f (1) = 2, f (2) = 4, f (3) = 6, f (4) = 8$ .
  1½ marks
2. Since $f$ is onto, every element of $B$ is an image. So $B$ is exactly the range: $B = {2, 4, 6, 8}$ .
  ½ mark
Answer: $B = {2, 4, 6, 8}$
Where the marks are
- Computes $f (1), f (2), f (3), f (4)$ correctly (1½ marks)
- Uses ontoness to conclude $B = {2, 4, 6, 8}$ (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerInverse Trigonometric Functions
Evaluate : $sin^{- 1} (sin \frac{3 π}{4}) + cos^{- 1} (cos \frac{3 π}{4}) + tan^{- 1} (1)$
Try it first, then show our solutionSolution
1. $\frac{3 π}{4} \in / [- \frac{π}{2}, \frac{π}{2}]$ . Since $sin \frac{3 π}{4} = sin \frac{π}{4}$ , we get $sin^{- 1} (sin \frac{3 π}{4}) = \frac{π}{4}$ .
  $\frac{3 π}{4} \in [0, π]$ , so $cos^{- 1} (cos \frac{3 π}{4}) = \frac{3 π}{4}$ .
  Also $tan^{- 1} (1) = \frac{π}{4}$ .
  Required value $= \frac{π}{4} + \frac{3 π}{4} + \frac{π}{4}$ .
  1½ marks
2. $= \frac{5 π}{4}$
  ½ mark
Answer: $\frac{5 π}{4}$
Where the marks are
- Writes the three principal values $\frac{π}{4}, \frac{3 π}{4}, \frac{π}{4}$ (1½ marks)
- Final sum $\frac{5 π}{4}$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerVector Algebra
Find all the vectors of magnitude $33$ which are collinear to vector $\hat{i} + \hat{j} + \hat{k}$ .
Try it first, then show our solutionSolution
1. $∣ \hat{i} + \hat{j} + \hat{k} ∣ = 3$ , so the unit vector along it is $\frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}$ .
  1 mark
2. A collinear vector of magnitude $33$ is $\pm 33 \times$ (unit vector). Taking the $+$ sign gives $3 \hat{i} + 3 \hat{j} + 3 \hat{k}$ .
  ½ mark
3. Taking the $-$ sign gives $- 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$ .
  ½ mark
Answer: $3 \hat{i} + 3 \hat{j} + 3 \hat{k}$ and $- 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$
Where the marks are
- Unit vector $\frac{1}{3} (\hat{i} + \hat{j} + \hat{k})$ (1 mark)
- Vector $3 \hat{i} + 3 \hat{j} + 3 \hat{k}$ (½ mark)
- Vector $- 3 \hat{i} - 3 \hat{j} - 3 \hat{k}$ (½ mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerVector Algebra
Position vectors of the points A, B and C as shown in the figure below are $a$ , $b$ and $c$ respectively. If $A C = \frac{5}{4} A B$ , express $c$ in terms of $a$ and $b$ .
Figure in the paper: A horizontal line with three points in order from left to right: A( $a$ ), B( $b$ ), C( $c$ ), with C beyond B. See it in the official paper.
Try it first, then show our solutionSolution
1. $A C = c - a$ and $A B = b - a$ . So $c - a = \frac{5}{4} (b - a)$ .
  1½ marks
2. $c = a + \frac{5}{4} b - \frac{5}{4} a = \frac{5 b}{4} - \frac{a}{4}$ , i.e. $c = \frac{5 b - a}{4}$ .
  ½ mark
Answer: $c = \frac{5 b - a}{4}$
Where the marks are
- Writes $c - a = \frac{5}{4} (b - a)$ (1½ marks)
- Simplifies to $c = \frac{5 b}{4} - \frac{a}{4}$ (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerThree Dimensional Geometry
Check whether the lines given by equations $x = 2 λ + 2$ , $y = 7 λ + 1$ , $z = - 3 λ - 3$ and $x = - μ - 2$ , $y = 2 μ + 8$ , $z = 4 μ + 5$ are perpendicular to each other or not.
Try it first, then show our solutionSolution
1. From the parametric forms, the coefficients of the parameters give the direction ratios: line 1 has $⟨ 2, 7, - 3 ⟩$ and line 2 has $⟨ - 1, 2, 4 ⟩$ .
  1 mark
2. $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 (- 1) + 7 (2) + (- 3) (4) = - 2 + 14 - 12 = 0$ .
  Hence the given lines are perpendicular to each other.
  1 mark
Answer: Yes, the lines are perpendicular.
Where the marks are
- Correct direction ratios of both lines (1 mark)
- Shows $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ and concludes the lines are perpendicular (1 mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerContinuity and Differentiability
If $y = (x + x^{2} - 1)^{2}$ , then show that $(x^{2} - 1) (\frac{d y}{d x})^{2} = 4 y^{2}$ .
Try it first, then show our solutionSolution
1. By the chain rule,
  $\frac{d y}{d x} = 2 (x + x^{2} - 1) (1 + \frac{x}{x ^{2} - 1})$
  $= 2 (x + x^{2} - 1) \cdot \frac{x ^{2} - 1 + x}{x ^{2} - 1}$
  $= \frac{2 ( x + x ^{2} - 1 ) ^{2}}{x ^{2} - 1}$
  1½ marks
2. Since $(x + x^{2} - 1)^{2} = y$ , we get $x^{2} - 1 \frac{d y}{d x} = 2 y$ .
  Squaring both sides: $(x^{2} - 1) (\frac{d y}{d x})^{2} = 4 y^{2}$ . Hence proved.
  ½ mark
Answer: $(x^{2} - 1) (\frac{d y}{d x})^{2} = 4 y^{2}$ (proved)
Where the marks are
- Correct derivative simplified to $\frac{2 ( x + x ^{2} - 1 ) ^{2}}{x ^{2} - 1}$ (1½ marks)
- Writes $x^{2} - 1 y^{'} = 2 y$ and squares to get the result (½ mark)
Open Q24 on its own page
Q25 · 2 marks · Very short answerApplication of Derivatives
Show that the function $f (x) = \frac{16 s i n x}{4 + c o s x} - x$ , is strictly decreasing in $(\frac{π}{2}, π)$ .
Try it first, then show our solutionSolution
1. Using the quotient rule,
  $f^{'} (x) = \frac{16 c o s x ( 4 + c o s x ) + 16 s i n ^{2} x}{( 4 + c o s x ) ^{2}} - 1$
  1 mark
2. The numerator becomes $64 cos x + 16 cos^{2} x + 16 sin^{2} x = 64 cos x + 16$ . Hence
  $f^{'} (x) = \frac{64 c o s x + 16 - ( 16 + 8 c o s x + c o s ^{2} x )}{( 4 + c o s x ) ^{2}} = \frac{c o s x ( 56 - c o s x )}{( 4 + c o s x ) ^{2}}$
  ½ mark
3. In $(\frac{π}{2}, π)$ we have $cos x < 0$ , while $56 - cos x > 0$ and $(4 + cos x)^{2} > 0$ . So $f^{'} (x) < 0$ .
  Hence $f$ is strictly decreasing in $(\frac{π}{2}, π)$ .
  ½ mark
Answer: $f^{'} (x) = \frac{c o s x ( 56 - c o s x )}{( 4 + c o s x ) ^{2}} < 0$ in $(π /2, π)$ , so $f$ is strictly decreasing there.
Where the marks are
- Correct $f^{'} (x)$ using the quotient rule (1 mark)
- Simplifies to $\frac{c o s x ( 56 - c o s x )}{( 4 + c o s x ) ^{2}}$ (½ mark)
- Sign argument with $cos x < 0$ and conclusion (½ mark)
Open Q25 on its own page

Section C

Q26 · 3 marks · Short answerIntegrals
Evaluate :
$\int_{0}^{π /2} [lo g (sin x) - lo g (2 cos x)] d x$
Try it first, then show our solutionSolution
1. $I = \int_{0}^{π /2} lo g \frac{s i n x}{2 c o s x} d x = \int_{0}^{π /2} lo g (\frac{t a n x}{2}) d x$ ... (1)
  ½ mark
2. Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , and $tan (\frac{π}{2} - x) = cot x$ :
  $I = \int_{0}^{π /2} lo g (\frac{c o t x}{2}) d x$ ... (2)
  1 mark
3. Adding (1) and (2):
  $2 I = \int_{0}^{π /2} lo g (\frac{t a n x}{2} \cdot \frac{c o t x}{2}) d x = \int_{0}^{π /2} lo g \frac{1}{4} d x$
  ½ mark
4. $2 I = lo g \frac{1}{4} [x]_{0}^{π /2} = \frac{π}{2} lo g \frac{1}{4}$
  ½ mark
5. $I = \frac{π}{4} lo g \frac{1}{4} = - \frac{π}{2} lo g 2$
  ½ mark
Answer: $\frac{π}{4} lo g \frac{1}{4} = - \frac{π}{2} lo g 2$
Where the marks are
- Combines the logs into $lo g \frac{t a n x}{2}$ (½ mark)
- Applies the property $\int_{0}^{a} f (a - x) d x$ to get $lo g \frac{c o t x}{2}$ (1 mark)
- Adds the two forms to get $lo g \frac{1}{4}$ (½ mark)
- Integrates the constant over $[0, π /2]$ (½ mark)
- Final answer $- \frac{π}{2} lo g 2$ (½ mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerIntegrals
Find :
$\int \frac{1}{x ( x + 1 ) ( x + 2 )} d x$
Try it first, then show our solutionSolution
1. Put $x = t$ , so $\frac{1}{2 x} d x = d t$ . Then $I = 2 \int \frac{d t}{( t + 1 ) ( t + 2 )}$ .
  ½ mark
2. By partial fractions, $\frac{1}{( t + 1 ) ( t + 2 )} = \frac{1}{t + 1} - \frac{1}{t + 2}$ . So $I = 2 \int (\frac{1}{t + 1} - \frac{1}{t + 2}) d t$ .
  1 mark
3. $I = 2 [lo g ∣ t + 1∣ - lo g ∣ t + 2∣] + C$
  1 mark
4. $I = 2 lo g (\frac{x + 1}{x + 2}) + C$
  ½ mark
Answer: $2 lo g (\frac{x + 1}{x + 2}) + C$
Where the marks are
- Substitution $x = t$ giving $2 \int \frac{d t}{( t + 1 ) ( t + 2 )}$ (½ mark)
- Correct partial fractions (1 mark)
- Integrates to logarithms (1 mark)
- Substitutes back in terms of $x$ (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerDifferential Equations
Find the particular solution of the differential equation $\frac{d y}{d x} + sec^{2} x \cdot y = tan x \cdot sec^{2} x$ , given that $y (0) = 0$ .
Try it first, then show our solutionSolution
1. This is a linear equation with $P = sec^{2} x$ and $Q = tan x sec^{2} x$ . So I.F. $= e^{\int s e c^{2} x d x} = e^{t a n x}$ .
  ½ mark
2. The solution is $y e^{t a n x} = \int tan x sec^{2} x e^{t a n x} d x + C$ .
  ½ mark
3. Put $tan x = t$ , so $sec^{2} x d x = d t$ .
  ½ mark
4. Integrating by parts: $\int t e^{t} d t = t e^{t} - e^{t} = e^{t} (t - 1)$ .
  ½ mark
5. Hence $y e^{t a n x} = e^{t a n x} (tan x - 1) + C$ .
  ½ mark
6. Using $y (0) = 0$ : $0 = 1 \cdot (0 - 1) + C$ , so $C = 1$ .
  The particular solution is $y e^{t a n x} = e^{t a n x} (tan x - 1) + 1$ , i.e. $y = tan x - 1 + e^{- t a n x}$ .
  ½ mark
Answer: $y = tan x - 1 + e^{- t a n x}$
Where the marks are
- I.F. $= e^{t a n x}$ (½ mark)
- Writes the solution form $y \cdot$ I.F. $= \int Q \cdot$ I.F. $d x$ (½ mark)
- Substitution $tan x = t$ (½ mark)
- $\int t e^{t} d t = e^{t} (t - 1)$ (½ mark)
- General solution (½ mark)
- Finds $C = 1$ and writes the particular solution (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerDifferential Equations
Solve the differential equation given by $x d y - y d x - x^{2} + y^{2} d x = 0$ .
Try it first, then show our solutionSolution
1. Rewrite the equation as $\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x} = \frac{y}{x} + 1 + (\frac{y}{x})^{2}$ ... (i). This is a homogeneous equation (taking $x > 0$ ).
  ½ mark
2. Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Substituting in (i): $v + x \frac{d v}{d x} = v + 1 + v^{2}$ .
  ½ mark
3. Separating variables: $\frac{d v}{1 + v ^{2}} = \frac{d x}{x}$ .
  ½ mark
4. Integrating both sides: $lo g v + 1 + v^{2} = lo g ∣ x ∣ + lo g C$ .
  1 mark
5. So $\frac{y}{x} + \frac{x ^{2} + y ^{2}}{x} = C x$ , which gives $y + x^{2} + y^{2} = C x^{2}$ .
  ½ mark
Answer: $y + x^{2} + y^{2} = C x^{2}$
Where the marks are
- Expresses the equation in homogeneous form (½ mark)
- Substitutes $y = v x$ (½ mark)
- Separates the variables (½ mark)
- Integrates correctly (1 mark)
- Final general solution $y + x^{2} + y^{2} = C x^{2}$ (½ mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerLinear Programming
Solve graphically the following linear programming problem :
Maximise $z = 6 x + 3 y$ ,
subject to the constraints
$4 x + y \geq 80$ ,
$3 x + 2 y \leq 150$ ,
$x + 5 y \geq 115$ ,
$x \geq 0, y \geq 0$ .
Try it first, then show our solutionSolution
1. Graph: draw the lines
  $4 x + y = 80$ through $(20, 0)$ and $(0, 80)$ ;
  $3 x + 2 y = 150$ through $(50, 0)$ and $(0, 75)$ ;
  $x + 5 y = 115$ through $(115, 0)$ and $(0, 23)$ .
  The feasible region lies above $4 x + y = 80$ , below $3 x + 2 y = 150$ and above $x + 5 y = 115$ , in the first quadrant. It is the bounded triangle with corner points:
  $(2, 72)$ , from $4 x + y = 80$ and $3 x + 2 y = 150$ ;
  $(15, 20)$ , from $4 x + y = 80$ and $x + 5 y = 115$ ;
  $(40, 15)$ , from $3 x + 2 y = 150$ and $x + 5 y = 115$ .
  2 marks
2. Values of $z = 6 x + 3 y$ at the corner points:
  Corner point $z = 6 x + 3 y$
  $(2, 72)$ $12 + 216 = 228$
  $(15, 20)$ $90 + 60 = 150$
  $(40, 15)$ $240 + 45 = 285$ (maximum)
  Hence the maximum value of $z$ is 285, at $x = 40, y = 15$ .
  1 mark
Answer: Maximum $z = 285$ at $(40, 15)$
Where the marks are
- Correct graph with the bounded feasible region and corner points $(2, 72), (15, 20), (40, 15)$ (2 marks)
- Evaluates $z$ at the corners and states the maximum 285 at $(40, 15)$ (1 mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerProbability
The probability distribution of a random variable X is given below :
X 1 2 3
P(X) $\frac{k}{2}$ $\frac{k}{3}$ $\frac{k}{6}$
1. (i)
  Find the value of k.
  0 marks
2. (ii)
  Find $P (1 \leq X < 3)$ .
  0 marks
3. (iii)
  Find $E (X)$ , the mean of X.
  0 marks
Try it first, then show our solutionSolution
1. (i) The probabilities must add to 1: $\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$ .
  1 mark
2. (i) This simplifies to $k \cdot \frac{3 + 2 + 1}{6} = k = 1$ .
  ½ mark
3. (ii) $P (1 \leq X < 3) = P (X = 1) + P (X = 2) = \frac{k}{2} + \frac{k}{3} = \frac{5 k}{6} = \frac{5}{6}$ .
  ½ mark
4. (iii) $E (X) = \sum p_{i} x_{i} = 1 \cdot \frac{k}{2} + 2 \cdot \frac{k}{3} + 3 \cdot \frac{k}{6} = \frac{k}{2} + \frac{2 k}{3} + \frac{k}{2} = \frac{5 k}{3}$ .
  ½ mark
5. (iii) With $k = 1$ : $E (X) = \frac{5}{3}$ .
  ½ mark
Answer: (i) $k = 1$ ; (ii) $\frac{5}{6}$ ; (iii) $E (X) = \frac{5}{3}$
Where the marks are
- Sets the sum of probabilities equal to 1 (1 mark)
- Gets $k = 1$ (½ mark)
- $P (1 \leq X < 3) = \frac{5}{6}$ (½ mark)
- Sets up $\sum p_{i} x_{i} = \frac{5 k}{3}$ (½ mark)
- $E (X) = \frac{5}{3}$ (½ mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerProbability
A and B are independent events such that $P (A \cap \overset{ˉ}{B}) = \frac{1}{4}$ and $P (\overset{ˉ}{A} \cap B) = \frac{1}{6}$ . Find $P (A)$ and $P (B)$ .
Try it first, then show our solutionSolution
1. Since A and B are independent, so are A, $\overset{ˉ}{B}$ and $\overset{ˉ}{A}$ , B. Hence $P (A) P (\overset{ˉ}{B}) = \frac{1}{4}$ and $P (\overset{ˉ}{A}) P (B) = \frac{1}{6}$ .
  ½ mark
2. Let $P (A) = x$ and $P (B) = y$ . Then $x (1 - y) = \frac{1}{4}$ and $(1 - x) y = \frac{1}{6}$ .
  Subtracting: $x - y = \frac{1}{12}$ , so $y = x - \frac{1}{12}$ .
  1 mark
3. Substituting into the first equation: $x (\frac{13}{12} - x) = \frac{1}{4}$ , which gives $12 x^{2} - 13 x + 3 = 0$ .
  Factorising: $(3 x - 1) (4 x - 3) = 0$ , so $x = \frac{1}{3}$ or $\frac{3}{4}$ .
  1 mark
4. If $P (A) = \frac{1}{3}$ , then $P (B) = \frac{1}{4}$ .
  If $P (A) = \frac{3}{4}$ , then $P (B) = \frac{2}{3}$ .
  ½ mark
Answer: $P (A) = \frac{1}{3}, P (B) = \frac{1}{4}$ or $P (A) = \frac{3}{4}, P (B) = \frac{2}{3}$
Where the marks are
- Uses independence to write the product equations (½ mark)
- Forms the equations and gets $x - y = \frac{1}{12}$ (1 mark)
- Gets the quadratic $12 x^{2} - 13 x + 3 = 0$ and solves it (1 mark)
- Both pairs of values of $P (A), P (B)$ (½ mark)
Open Q30(b) on its own page
Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerIntegrals
Evaluate :
$\int_{0}^{π /2} e^{x} sin x d x$
Try it first, then show our solutionSolution
1. Let $I = \int e^{x} sin x d x$ . Taking $sin x$ as the first function: $I = e^{x} sin x - \int e^{x} cos x d x$ .
  1 mark
2. Integrating by parts again: $I = e^{x} sin x - [e^{x} cos x + \int e^{x} sin x d x] = e^{x} sin x - e^{x} cos x - I$ .
  ½ mark
3. So $2 I = e^{x} (sin x - cos x)$ , i.e. $I = \frac{1}{2} e^{x} (sin x - cos x)$ .
  ½ mark
4. $\int_{0}^{π /2} e^{x} sin x d x = \frac{1}{2} [e^{π /2} (1 - 0) - e^{0} (0 - 1)] = \frac{1}{2} (e^{π /2} + 1)$
  1 mark
Answer: $\frac{1}{2} (e^{π /2} + 1)$
Where the marks are
- First integration by parts (1 mark)
- Second integration by parts giving $- I$ (½ mark)
- Solves for $I = \frac{1}{2} e^{x} (sin x - cos x)$ (½ mark)
- Applies the limits to get $\frac{1}{2} (e^{π /2} + 1)$ (1 mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerIntegrals
Find :
$\int \frac{1}{cos ( x - a ) cos ( x - b )} d x$
Try it first, then show our solutionSolution
1. Write $sin (a - b) = sin [(x - b) - (x - a)]$ . Then
  $I = \frac{1}{s i n ( a - b )} \int \frac{s i n [( x - b ) - ( x - a )]}{c o s ( x - a ) c o s ( x - b )} d x$
  1 mark
2. Expanding the sine:
  $I = \frac{1}{s i n ( a - b )} \int [\frac{s i n ( x - b ) c o s ( x - a )}{c o s ( x - a ) c o s ( x - b )} - \frac{c o s ( x - b ) s i n ( x - a )}{c o s ( x - a ) c o s ( x - b )}] d x$
  ½ mark
3. $I = \frac{1}{s i n ( a - b )} \int [tan (x - b) - tan (x - a)] d x$
  ½ mark
4. $I = \frac{1}{s i n ( a - b )} [lo g ∣ sec (x - b) ∣ - lo g ∣ sec (x - a) ∣] + C = \frac{1}{s i n ( a - b )} lo g \frac{c o s ( x - a )}{c o s ( x - b )} + C$
  1 mark
Answer: $\frac{1}{sin ( a - b )} lo g \frac{cos ( x - a )}{cos ( x - b )} + C$
Where the marks are
- Multiplies and divides by $sin (a - b)$ written as $sin [(x - b) - (x - a)]$ (1 mark)
- Expands the sine and splits the fraction (½ mark)
- Reduces to $tan (x - b) - tan (x - a)$ (½ mark)
- Integrates to log sec terms (1 mark)
Open Q31(b) on its own page

Corner point	$z = 6 x + 3 y$
$(2, 72)$	$12 + 216 = 228$
$(15, 20)$	$90 + 60 = 150$
$(40, 15)$	$240 + 45 = 285$ (maximum)

X	1	2	3
P(X)	$\frac{k}{2}$	$\frac{k}{3}$	$\frac{k}{6}$

Section D

Q32 · 5 marks · Long answerRelations and Functions
A relation R is defined on a set of real numbers $R$ as
$R = {(x, y) : x \cdot y is an irrational number}$ .
Check whether R is reflexive, symmetric and transitive or not.
Try it first, then show our solutionSolution
1. Reflexive: take $1 \in R$ . Then $1 \cdot 1 = 1$ , which is rational, so $(1, 1) \in / R$ .
  Hence R is not reflexive.
  1½ marks
2. Symmetric: let $(x, y) \in R$ , so $x y$ is irrational. Since $y x = x y$ , $y x$ is also irrational, so $(y, x) \in R$ .
  Hence R is symmetric.
  1½ marks
3. Transitive: $(1, 2) \in R$ since $2$ is irrational, and $(2, 2) \in R$ since $22$ is irrational.
  But $1 \cdot 2 = 2$ is rational, so $(1, 2) \in / R$ .
  Hence R is not transitive.
  2 marks
Answer: R is symmetric, but neither reflexive nor transitive.
Where the marks are
- Counter-example showing R is not reflexive (1½ marks)
- General proof that R is symmetric (1½ marks)
- Counter-example showing R is not transitive (2 marks)
Open Q32 on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerMatrices
If $A = 1 - 1 0 23 - 2 - 2 01$ and $B^{- 1} = 3 - 15 5 - 1 6 - 2 1 - 5 2$ , find $(A B)^{- 1}$ .
Try it first, then show our solutionSolution
1. Use the reversal law: $(A B)^{- 1} = B^{- 1} A^{- 1}$ .
  ½ mark
2. $∣ A ∣ = 1 (3 - 0) - 2 (- 1 - 0) + (- 2) (2 - 0) = 3 + 2 - 4 = 1 \neq = 0$ , so $A^{- 1}$ exists.
  1 mark
3. Cofactors of $A$ :
  $A_{11} = 3, A_{12} = 1, A_{13} = 2$
  $A_{21} = 2, A_{22} = 1, A_{23} = 2$
  $A_{31} = 6, A_{32} = 2, A_{33} = 5$
  So $adj A = 312212625$ .
  2 marks
4. $A^{- 1} = \frac{1}{∣ A ∣} adj A = 312212625$
  ½ mark
5. $(A B)^{- 1} = B^{- 1} A^{- 1} = 3 - 15 5 - 1 6 - 2 1 - 5 2 312212625 = 10 - 49 17 7 - 34 12 21 - 103 36$
  1 mark
Answer: $(A B)^{- 1} = 10 - 49 17 7 - 34 12 21 - 103 36$
Where the marks are
- States $(A B)^{- 1} = B^{- 1} A^{- 1}$ (½ mark)
- $∣ A ∣ = 1$ (1 mark)
- Correct $adj A$ (2 marks)
- Correct $A^{- 1}$ (½ mark)
- Correct product $B^{- 1} A^{- 1}$ (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerMatrices
Solve the following system of equations by matrix method :
$x + 2 y + 3 z = 6$
$2 x - y + z = 2$
$3 x + 2 y - 2 z = 3$
Try it first, then show our solutionSolution
1. Write the system as $A X = B$ , where $A = 123 2 - 1 2 31 - 2$ , $X = x y z$ , $B = 623$ . Then $X = A^{- 1} B$ .
  $∣ A ∣ = 1 (2 - 2) - 2 (- 4 - 3) + 3 (4 + 3) = 0 + 14 + 21 = 35 \neq = 0$ , so $A^{- 1}$ exists.
  ½ mark
2. Cofactors (first two rows):
  $A_{11} = 0, A_{12} = 7, A_{13} = 7$
  $A_{21} = 10, A_{22} = - 11, A_{23} = 4$
  1 mark
3. Cofactors (third row): $A_{31} = 5, A_{32} = 5, A_{33} = - 5$ .
  1½ marks
4. $A^{- 1} = \frac{1}{35} 077 10 - 11 4 55 - 5$
  1 mark
5. $X = \frac{1}{35} 077 10 - 11 4 55 - 5 623 = \frac{1}{35} 353535 = 111$
  Hence $x = 1, y = 1, z = 1$ .
  1 mark
Answer: $x = 1, y = 1, z = 1$
Where the marks are
- Matrix form and $∣ A ∣ = 35 \neq = 0$ (½ mark)
- Correct cofactors (1 + 1½ marks)
- Correct $A^{- 1}$ (1 mark)
- $X = A^{- 1} B$ giving $x = y = z = 1$ (1 mark)
Open Q33(b) on its own page
Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerThree Dimensional Geometry
Find the vector and the Cartesian equations of a line passing through the point $(1, 2, - 4)$ and parallel to the line joining the points $A (3, 3, - 5)$ and $B (1, 0, - 11)$ . Hence, find the distance between the two lines.
Try it first, then show our solutionSolution
1. Direction of $A B$ : $A B = (1 - 3) \hat{i} + (0 - 3) \hat{j} + (- 11 + 5) \hat{k} = - 2 \hat{i} - 3 \hat{j} - 6 \hat{k}$ , so we may take $b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ as the direction. The required line passes through $(1, 2, - 4)$ , so its vector equation is $r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ .
  1 mark
2. Cartesian equation: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ .
  1 mark
3. Line $A B$ through $A (3, 3, - 5)$ with the same direction: $r = (3 \hat{i} + 3 \hat{j} - 5 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ . The two lines are parallel, so the distance between them is $d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$ .
  ½ mark
4. Here $a_{1} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $a_{2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$ , so $a_{2} - a_{1} = 2 \hat{i} + \hat{j} - \hat{k}$ .
  ½ mark
5. $(a_{2} - a_{1}) \times b = \hat{i} 22 \hat{j} 13 \hat{k} - 1 6 = \hat{i} (6 + 3) - \hat{j} (12 + 2) + \hat{k} (6 - 2) = 9 \hat{i} - 14 \hat{j} + 4 \hat{k}$ .
  1 mark
6. $∣ (a_{2} - a_{1}) \times b ∣ = 81 + 196 + 16 = 293$ and $∣ b ∣ = 4 + 9 + 36 = 7$ . Therefore $d = \frac{293}{7}$ units.
  1 mark
Answer: $r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ ; $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ ; distance $= \frac{293}{7}$ units
Where the marks are
- Correct vector equation of the required line (1 mark)
- Correct Cartesian equation (1 mark)
- Equation of line AB and the parallel-lines distance formula (½ mark)
- Finds $a_{2} - a_{1} = 2 \hat{i} + \hat{j} - \hat{k}$ (½ mark)
- Cross product $9 \hat{i} - 14 \hat{j} + 4 \hat{k}$ (1 mark)
- Distance $293 /7$ (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerThree Dimensional Geometry
Find the equations of the line passing through the points $A (1, 2, 3)$ and $B (3, 5, 9)$ . Hence, find the coordinates of the points on this line which are at a distance of 14 units from point B.
Try it first, then show our solutionSolution
1. Direction ratios of $A B$ : $3 - 1, 5 - 2, 9 - 3 = 2, 3, 6$ . The equation of line $A B$ is $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ , or in vector form $r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ .
  1 mark
2. Any point on this line is $P (2 λ + 1, 3 λ + 2, 6 λ + 3)$ for some $λ$ .
  1 mark
3. Given $P B = 14$ with $B (3, 5, 9)$ : $(2 λ - 2)^{2} + (3 λ - 3)^{2} + (6 λ - 6)^{2} = 1 4^{2}$ , so $49 (λ - 1)^{2} = 196$ , which gives $λ^{2} - 2 λ - 3 = 0$ .
  1 mark
4. $(λ - 3) (λ + 1) = 0$ , so $λ = 3$ or $λ = - 1$ .
  1 mark
5. $λ = 3$ gives $(7, 11, 21)$ and $λ = - 1$ gives $(- 1, - 1, - 3)$ . These are the required points.
  1 mark
Answer: Line: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}$ ; points $(7, 11, 21)$ and $(- 1, - 1, - 3)$
Where the marks are
- Equation of line AB (1 mark)
- General point $(2 λ + 1, 3 λ + 2, 6 λ + 3)$ (1 mark)
- Distance condition leading to $λ^{2} - 2 λ - 3 = 0$ (1 mark)
- $λ = 3, - 1$ (1 mark)
- Both points $(7, 11, 21)$ and $(- 1, - 1, - 3)$ (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerApplication of Integrals
Find the area of the region bounded by the curves $x^{2} = y$ , $y = x + 2$ and x-axis, using integration.
Try it first, then show our solutionSolution
1. Figure: the upward parabola $y = x^{2}$ has its vertex at $O$ . The line $y = x + 2$ meets the x-axis at $(- 2, 0)$ . The region bounded by the line, the parabola and the x-axis lies above the x-axis between $x = - 2$ and $x = 0$ . It is bounded by the line from $(- 2, 0)$ to $(- 1, 1)$ and by the parabola from $(- 1, 1)$ to $(0, 0)$ . The vertical line $x = - 1$ splits it into two parts.
  1½ marks
2. Intersection of the curves: $x^{2} = x + 2 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow x = - 1, 2$ . The relevant corner is $(- 1, 1)$ .
  ½ mark
3. Required area $= \int_{- 2}^{- 1} (x + 2) d x + \int_{- 1}^{0} x^{2} d x$ .
  1½ marks
4. $= [\frac{( x + 2 ) ^{2}}{2}]_{- 2}^{- 1} + [\frac{x ^{3}}{3}]_{- 1}^{0} = (\frac{1}{2} - 0) + (0 + \frac{1}{3})$
  1 mark
5. $= \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ square units.
  ½ mark
Answer: $\frac{5}{6}$ square units
Where the marks are
- Correct graph with the region shaded (1½ marks)
- Points of intersection $x = - 1, 2$ (½ mark)
- Correct integral set-up $\int_{- 2}^{- 1} (x + 2) d x + \int_{- 1}^{0} x^{2} d x$ (1½ marks)
- Correct integration and substitution of limits (1 mark)
- Area $= 5/6$ (½ mark)
Open Q35 on its own page

Section E

Case study

Q36 · 4 marks · Case-basedCompetency-basedProbability
There are different types of Yoga which involve the usage of different poses of Yoga Asanas, Meditation and Pranayam as shown in the figure below. The Venn diagram below represents the probabilities of three different types of Yoga, A, B and C performed by the people of a society. Further, it is given that probability of a member performing type C Yoga is 0·44.
On the basis of the above information, answer the following questions :
1. (i)
  Find the value of x.
  1 mark
2. (ii)
  Find the value of y.
  1 mark
3. (iii)(a)
  Find $P (\frac{C}{B})$ .
  2 marks
4. OR
  (iii)(b)
  Find the probability that a randomly selected person of the society does Yoga of type A or B but not C.
  2 marks
Figure in the paper: First figure: picture listing Types of Yoga: Anusara, Kundalini, Vinyasa, Bikram, Hatha Yoga. Second figure: Venn diagram in sample space S with three overlapping ovals A, B, C in a row. Regions: A only 0·32, A∩B 0·09, B only y, B∩C x, C only 0·21; outside all sets 0·11. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $P (C) = x + 0.21 = 0.44$ , so $x = 0.23$ .
  1 mark
2. (ii) The probabilities of all regions add to 1: $0.32 + 0.09 + y + 0.23 + 0.21 + 0.11 = 1$ , so $0.96 + y = 1$ and $y = 0.04$ .
  1 mark
3. (iii)(a) $P (C ∣ B) = \frac{P ( C \cap B )}{P ( B )}$ , with $P (B) = 0.09 + 0.04 + 0.23 = 0.36$ .
  1 mark
4. (iii)(a) $P (C \cap B) = x = 0.23$ , so $P (C ∣ B) = \frac{0.23}{0.36} = \frac{23}{36}$ .
  1 mark
5. (iii)(b) The person is in A or B but not in C. These regions are A only, A∩B and B only. So $P = 0.32 + 0.09 + 0.04$ .
  1½ marks
6. (iii)(b) $= 0.45$ .
  ½ mark
Answer: (i) x = 0.23; (ii) y = 0.04; (iii)(a) 23/36; (iii)(b) 0.45
Where the marks are
- (i) $x = 0.44 - 0.21 = 0.23$ (1 mark)
- (ii) Total probability 1 gives $y = 0.04$ (1 mark)
- (iii)(a) $P (B) = 0.36$ (1 mark)
- (iii)(a) $P (C ∣ B) = 23/36$ (1 mark)
- (iii)(b) Adds the regions $0.32 + 0.09 + 0.04$ (1½ marks)
- (iii)(b) Answer 0.45 (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank. A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of $2 cm^{3} / s$ . The semi-vertical angle of the conical tank is $4 5^{\circ}$ .
On the basis of given information, answer the following questions :
1. (i)
  Find the volume of water in the tank in terms of its radius r.
  1 mark
2. (ii)
  Find rate of change of radius at an instant when $r = 22$ cm.
  1 mark
3. (iii)(a)
  Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius $r = 22$ cm.
  2 marks
4. OR
  (iii)(b)
  Find the rate of change of height 'h' at an instant when slant height is 4 cm.
  2 marks
Figure in the paper: Photo of a storage tank: a cylindrical upper part with a conical bottom, supported on a metal frame with legs. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Semi-vertical angle $4 5^{\circ}$ gives $tan 4 5^{\circ} = \frac{r}{h} = 1$ , so $h = r$ . Then $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{3}$ .
  1 mark
2. (ii) $\frac{d V}{d t} = π r^{2} \frac{d r}{d t}$ , and $\frac{d V}{d t} = - 2$ cm³/s because the volume is decreasing.
  ½ mark
3. (ii) At $r = 22$ : $- 2 = π (8) \frac{d r}{d t}$ , so $\frac{d r}{d t} = - \frac{1}{4 π}$ cm/s. The radius decreases at $\frac{1}{4 π}$ cm/s.
  ½ mark
4. (iii)(a) Slant height $l = r^{2} + h^{2} = 2 r$ . The wet (curved) surface is $C = π r l = 2 π r^{2}$ .
  1 mark
5. (iii)(a) $\frac{d C}{d t} = 22 π r \frac{d r}{d t}$ .
  ½ mark
6. (iii)(a) At $r = 22$ : $\frac{d C}{d t} = 22 π (22) (- \frac{1}{4 π}) = - 2$ cm²/s. The wet surface is decreasing at $2$ cm²/s.
  ½ mark
7. (iii)(b) $l^{2} = h^{2} + r^{2} = 2 h^{2}$ since $r = h$ . When $l = 4$ : $h^{2} = 8$ , so $h = r = 22$ cm.
  1 mark
8. (iii)(b) Since $h = r$ at all times, $\frac{d h}{d t} = \frac{d r}{d t}$ . By (ii), at $r = 22$ this is $- \frac{1}{4 π}$ cm/s. The height decreases at $\frac{1}{4 π}$ cm/s.
  1 mark
Answer: (i) $V = \frac{1}{3} π r^{3}$ ; (ii) $- \frac{1}{4 π}$ cm/s; (iii)(a) decreasing at 2 cm²/s; (iii)(b) $- \frac{1}{4 π}$ cm/s
Where the marks are
- (i) Uses $h = r$ to get $V = \frac{1}{3} π r^{3}$ (1 mark)
- (ii) Differentiates: $d V / d t = π r^{2} d r / d t$ (½ mark)
- (ii) $d r / d t = - 1/ (4 π)$ cm/s (½ mark)
- (iii)(a) $C = 2 π r^{2}$ (1 mark)
- (iii)(a) $d C / d t = 22 π r d r / d t$ (½ mark)
- (iii)(a) $- 2$ cm²/s (½ mark)
- (iii)(b) Finds $r = h = 22$ when $l = 4$ (1 mark)
- (iii)(b) $d h / d t = d r / d t = - 1/ (4 π)$ cm/s (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedContinuity and Differentiability
The equation of the path traced by a roller-coaster is given by the polynomial $f (x) = a (x + 9) (x + 1) (x - 3)$ . If the roller-coaster crosses y-axis at a point $(0, - 1)$ , answer the following :
1. (i)
  Find the value of 'a'.
  2 marks
2. (ii)
  Find $f^{''} (x)$ at $x = 1$ .
  2 marks
Figure in the paper: Picture of a roller-coaster track drawn over coordinate axes, with a large arch-shaped hill on the left of the y-axis and smaller undulations to the right; y-axis marked with dots. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The curve passes through $(0, - 1)$ , so $f (0) = - 1$ , i.e. $a (0 + 9) (0 + 1) (0 - 3) = - 1$ .
  1 mark
2. (i) $- 27 a = - 1$ , so $a = \frac{1}{27}$ .
  1 mark
3. (ii) $f (x) = \frac{1}{27} (x + 9) (x + 1) (x - 3) = \frac{1}{27} (x^{3} + 7 x^{2} - 21 x - 27)$ .
  ½ mark
4. (ii) $f^{'} (x) = \frac{1}{27} (3 x^{2} + 14 x - 21)$ .
  ½ mark
5. (ii) $f^{''} (x) = \frac{6 x + 14}{27}$ .
  ½ mark
6. (ii) $f^{''} (1) = \frac{6 + 14}{27} = \frac{20}{27}$ .
  ½ mark
Answer: (i) $a = \frac{1}{27}$ ; (ii) $f^{''} (1) = \frac{20}{27}$
Where the marks are
- (i) Substitutes $(0, - 1)$ into f(x) (1 mark)
- (i) $a = 1/27$ (1 mark)
- (ii) Expands f(x) to $\frac{1}{27} (x^{3} + 7 x^{2} - 21 x - 27)$ (½ mark)
- (ii) Correct $f^{'} (x)$ (½ mark)
- (ii) Correct $f^{''} (x) = (6 x + 14) /27$ (½ mark)
- (ii) $f^{''} (1) = 20/27$ (½ mark)
Open Q38 on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.