MathsCBSE Class 12 · Board exam 2025 (session 2024-25)

CBSE Class 12 Mathematics: 2025 board paper, set 65/7/1

The question paper set in the CBSE Class 12 board exam of 2025, 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/2025/XII/MATHEMATICS.zip (a zip; this paper is the file 65-7-1_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 in 2 questions of Section B, 3 of C, 2 of D and 2 of E; calculator not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQInverse Trigonometric Functions
The given graph illustrates :
1. (A) $y = tan^{- 1} x$
2. (B) $y = cosec^{- 1} x$
3. (C) $y = cot^{- 1} x$
4. (D) $y = sec^{- 1} x$
Figure in the paper: Graph of an increasing S-shaped curve passing through the origin O, approaching the horizontal dashed asymptote y = π/2 (point (0, π/2)) as x increases and y = −π/2 (point (0, −π/2)) as x decreases. See it in the official paper.
Try it first, then show our solutionSolution
1. The curve is defined for all real $x$ , increasing, passes through the origin and has range $(- \frac{π}{2}, \frac{π}{2})$ with asymptotes $y = \pm \frac{π}{2}$ . This is the principal branch of $tan^{- 1} x$ .
  1 mark
Answer: (A) $y = tan^{- 1} x$
Where the marks are
- Identifies the graph as $tan^{- 1} x$ from its domain R and range $(- π /2, π /2)$ (1 mark)
Q2 · 1 mark · MCQInverse Trigonometric Functions
Domain of $f (x) = cos^{- 1} x + sin x$ is :
1. (A)R
2. (B) $(- 1, 1)$
3. (C) $[- 1, 1]$
4. (D) $ϕ$
Try it first, then show our solutionSolution
1. $cos^{- 1} x$ is defined only for $x \in [- 1, 1]$ , and $sin x$ is defined for all R. So the domain is $[- 1, 1] \cap R = [- 1, 1]$ .
  1 mark
Answer: (C) $[- 1, 1]$
Where the marks are
- Takes the intersection of the two domains to get $[- 1, 1]$ (1 mark)
Q3 · 1 mark · MCQMatrices
What is the total number of possible matrices of order $3 \times 3$ with each entry as $2$ or $3$ ?
1. (A)9
2. (B)512
3. (C)615
4. (D)64
Try it first, then show our solutionSolution
1. A $3 \times 3$ matrix has 9 entries, and each entry has 2 choices. Number of matrices $= 2^{9} = 512$ .
  1 mark
Answer: (B) 512
Where the marks are
- Uses $2^{9} = 512$ (1 mark)
Q4 · 1 mark · MCQMatrices
The matrix $A = 300020005$ is a/an :
1. (A)scalar matrix
2. (B)identity matrix
3. (C)null matrix
4. (D)symmetric matrix
Try it first, then show our solutionSolution
1. $A$ is a diagonal matrix with unequal diagonal entries, so it is not scalar, identity or null. Since $A^{T} = A$ , it is symmetric.
  1 mark
Answer: (D) symmetric matrix
Where the marks are
- Notes that $A^{T} = A$ , so A is symmetric (1 mark)
Q5 · 1 mark · MCQDeterminants
If A and B are two square matrices each of order 3 with $∣ A ∣ = 3$ and $∣ B ∣ = 5$ , then $∣2 A B ∣$ is :
1. (A)30
2. (B)120
3. (C)15
4. (D)225
Try it first, then show our solutionSolution
1. For order 3, $∣ k M ∣ = k^{3} ∣ M ∣$ . So $∣2 A B ∣ = 2^{3} ∣ A ∣∣ B ∣ = 8 \times 3 \times 5 = 120$ .
  1 mark
Answer: (B) 120
Where the marks are
- Uses $∣2 A B ∣ = 8∣ A ∣∣ B ∣ = 120$ (1 mark)
Q6 · 1 mark · MCQDeterminants
Let A be a square matrix of order 3. If $∣ A ∣ = 5$ , then $∣ adj A ∣$ is :
1. (A)5
2. (B)125
3. (C)25
4. (D) $- 5$
Try it first, then show our solutionSolution
1. $∣ adj A ∣ = ∣ A ∣^{n - 1} = 5^{2} = 25$ .
  1 mark
Answer: (C) 25
Where the marks are
- Uses $∣ adj A ∣ = ∣ A ∣^{n - 1}$ to get 25 (1 mark)
Q7 · 1 mark · MCQMatrices
If $[2 x - 1 0 3 x y^{2} - 1] = [x + 3 0 1235]$ , then the value of $(x - y)$ is :
1. (A)2 or 10
2. (B) $- 2$ or 10
3. (C)2 or $- 10$
4. (D) $- 2$ or $- 10$
Try it first, then show our solutionSolution
1. Equating entries: $3 x = 12 \Rightarrow x = 4$ , which also satisfies $2 x - 1 = x + 3$ .
2. $y^{2} - 1 = 35 \Rightarrow y = \pm 6$ . So $x - y = 4 - 6 = - 2$ or $4 + 6 = 10$ .
  1 mark
Answer: (B) $- 2$ or 10
Where the marks are
- Finds $x = 4$ , $y = \pm 6$ and hence $x - y = - 2$ or 10 (1 mark)
Q8 · 1 mark · MCQContinuity and Differentiability
If $f (x) = ⎩ ⎨ ⎧ 1, a x + b, 7, if x \leq 3 if 3 < x < 5 if 5 \leq x$ is continuous in R, then the values of a and b are :
1. (A) $a = 3, b = - 8$
2. (B) $a = 3, b = 8$
3. (C) $a = - 3, b = - 8$
4. (D) $a = - 3, b = 8$
Try it first, then show our solutionSolution
1. Continuity at $x = 3$ : $3 a + b = 1$ . Continuity at $x = 5$ : $5 a + b = 7$ .
2. Subtracting gives $2 a = 6$ , so $a = 3$ and $b = 1 - 9 = - 8$ .
  1 mark
Answer: (A) $a = 3, b = - 8$
Where the marks are
- Forms $3 a + b = 1$ and $5 a + b = 7$ and solves to get $a = 3$ , $b = - 8$ (1 mark)
Q9 · 1 mark · MCQContinuity and Differentiability
If $f (x) = - 2 x^{8}$ , then the correct statement is :
1. (A) $f^{'} (\frac{1}{2}) = f^{'} (- \frac{1}{2})$
2. (B) $f^{'} (\frac{1}{2}) = - f^{'} (- \frac{1}{2})$
3. (C) $- f^{'} (\frac{1}{2}) = f (- \frac{1}{2})$
4. (D) $f (\frac{1}{2}) = - f (- \frac{1}{2})$
Try it first, then show our solutionSolution
1. $f^{'} (x) = - 16 x^{7}$ is an odd function, so $f^{'} (1/2) = - \frac{1}{8}$ and $f^{'} (- 1/2) = \frac{1}{8}$ . Hence $f^{'} (1/2) = - f^{'} (- 1/2)$ .
  1 mark
Answer: (B)
Where the marks are
- Computes $f^{'} (x) = - 16 x^{7}$ and compares $f^{'} (\pm \frac{1}{2})$ (1 mark)
Q10 · 1 mark · MCQApplication of Derivatives
A spherical ball has a variable diameter $\frac{5}{2} (3 x + 1)$ . The rate of change of its volume w.r.t. x, when $x = 1$ , is :
1. (A) $225 π$
2. (B) $300 π$
3. (C) $375 π$
4. (D) $125 π$
Try it first, then show our solutionSolution
1. Radius $r = \frac{5}{4} (3 x + 1)$ , so $V = \frac{4}{3} π \cdot \frac{125}{64} (3 x + 1)^{3}$ .
2. $\frac{d V}{d x} = \frac{125 π}{48} \cdot 3 (3 x + 1)^{2} \cdot 3 = \frac{375 π}{16} (3 x + 1)^{2}$ . At $x = 1$ : $\frac{375 π}{16} \times 16 = 375 π$ .
  1 mark
Answer: (C) $375 π$
Where the marks are
- Writes V in terms of x, differentiates and substitutes $x = 1$ to get $375 π$ (1 mark)
Open Q10 on its own page
Q11 · 1 mark · MCQApplication of Derivatives
If $f : R \to R$ is defined as $f (x) = 2 x - sin x$ , then f is :
1. (A)a decreasing function
2. (B)an increasing function
3. (C)maximum at $x = \frac{π}{2}$
4. (D)maximum at $x = 0$
Try it first, then show our solutionSolution
1. $f^{'} (x) = 2 - cos x \geq 1 > 0$ for all $x$ , so f is strictly increasing on R and has no maximum.
  1 mark
Answer: (B) an increasing function
Where the marks are
- Shows $f^{'} (x) = 2 - cos x > 0$ for all x (1 mark)
Q12 · 1 mark · MCQIntegrals
$\int \frac{e ^{9 l o g x} - e ^{8 l o g x}}{e ^{6 l o g x} - e ^{5 l o g x}} d x$ is equal to :
1. (A) $x + C$
2. (B) $\frac{x ^{2}}{2} + C$
3. (C) $\frac{x ^{4}}{4} + C$
4. (D) $\frac{x ^{3}}{3} + C$
Try it first, then show our solutionSolution
1. Since $e^{n l o g x} = x^{n}$ , the integrand is $\frac{x ^{9} - x ^{8}}{x ^{6} - x ^{5}} = \frac{x ^{8} ( x - 1 )}{x ^{5} ( x - 1 )} = x^{3}$ .
2. $\int x^{3} d x = \frac{x ^{4}}{4} + C$ .
  1 mark
Answer: (C) $\frac{x ^{4}}{4} + C$
Where the marks are
- Simplifies the integrand to $x^{3}$ and integrates to $\frac{x ^{4}}{4} + C$ (1 mark)
Q13 · 1 mark · MCQIntegrals
For a function $f (x)$ , which of the following holds true ?
1. (A) $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
2. (B) $\int_{- a}^{a} f (x) d x = 0$ , if f is an even function
3. (C) $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ , if f is an odd function
4. (D) $\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x - \int_{0}^{a} f (2 a + x) d x$
Try it first, then show our solutionSolution
1. (A) is the standard property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ . (B) and (C) have the even/odd conditions interchanged, and (D) is incorrect.
  1 mark
Answer: (A)
Where the marks are
- Identifies the property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ (1 mark)
Q14 · 1 mark · MCQIntegrals
$\int \frac{e ^{x}}{4 - e ^{2 x}} d x$ is equal to :
1. (A) $\frac{1}{2} cos^{- 1} (e^{x}) + C$
2. (B) $\frac{1}{2} sin^{- 1} (e^{x}) + C$
3. (C) $\frac{e ^{x}}{2} + C$
4. (D) $sin^{- 1} (\frac{e ^{x}}{2}) + C$
Try it first, then show our solutionSolution
1. Put $t = e^{x}$ , so $d t = e^{x} d x$ . Then $\int \frac{d t}{4 - t ^{2}} = sin^{- 1} \frac{t}{2} + C = sin^{- 1} \frac{e ^{x}}{2} + C$ .
  1 mark
Answer: (D) $sin^{- 1} (\frac{e ^{x}}{2}) + C$
Where the marks are
- Substitutes $t = e^{x}$ and uses the standard $sin^{- 1}$ form (1 mark)
Q15 · 1 mark · MCQCompetency-basedVector Algebra
A student tries to tie ropes, parallel to each other from one end of the wall to the other. If one rope is along the vector $3 \hat{i} + 15 \hat{j} + 6 \hat{k}$ and the other is along the vector $2 \hat{i} + 10 \hat{j} + λ \hat{k}$ , then the value of $λ$ is :
1. (A)6
2. (B)1
3. (C) $\frac{1}{4}$
4. (D)4
Try it first, then show our solutionSolution
1. For parallel vectors the direction ratios are proportional: $\frac{2}{3} = \frac{10}{15} = \frac{λ}{6}$ , so $λ = 4$ .
  1 mark
Answer: (D) 4
Where the marks are
- Equates the ratios of components to get $λ = 4$ (1 mark)
Q16 · 1 mark · MCQVector Algebra
If $∣ a + b ∣ = ∣ a - b ∣$ for any two vectors, then vectors $a$ and $b$ are :
1. (A)orthogonal vectors
2. (B)parallel to each other
3. (C)unit vectors
4. (D)collinear vectors
Try it first, then show our solutionSolution
1. Squaring both sides: $∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2}$ , so $a \cdot b = 0$ and the vectors are orthogonal.
  1 mark
Answer: (A) orthogonal vectors
Where the marks are
- Squares both sides to get $a \cdot b = 0$ (1 mark)
Q17 · 1 mark · MCQProbability
If $P (A) = \frac{1}{7}$ , $P (B) = \frac{5}{7}$ and $P (A \cap B) = \frac{4}{7}$ , then $P (\overline{A} ∣ B)$ is :
1. (A) $\frac{6}{7}$
2. (B) $\frac{3}{4}$
3. (C) $\frac{4}{5}$
4. (D) $\frac{1}{5}$
Try it first, then show our solutionSolution
1. $P (\overset{ˉ}{A} ∣ B) = 1 - P (A ∣ B) = 1 - \frac{4/7}{5/7} = 1 - \frac{4}{5} = \frac{1}{5}$ .
  1 mark
Answer: (D) $\frac{1}{5}$
Where the marks are
- Uses $P (\overset{ˉ}{A} ∣ B) = 1 - \frac{P ( A \cap B )}{P ( B )}$ to get $\frac{1}{5}$ (1 mark)
Q18 · 1 mark · MCQProbability
A coin is tossed and a card is selected at random from a well shuffled pack of 52 playing cards. The probability of getting head on the coin and a face card from the pack is :
1. (A) $\frac{2}{13}$
2. (B) $\frac{3}{26}$
3. (C) $\frac{19}{26}$
4. (D) $\frac{3}{13}$
Try it first, then show our solutionSolution
1. The events are independent. $P (head) = \frac{1}{2}$ and $P (face card) = \frac{12}{52} = \frac{3}{13}$ , so the required probability is $\frac{1}{2} \times \frac{3}{13} = \frac{3}{26}$ .
  1 mark
Answer: (B) $\frac{3}{26}$
Where the marks are
- Multiplies the independent probabilities $\frac{1}{2} \times \frac{3}{13}$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedContinuity and Differentiability
Assertion (A) : $f (x) = {x sin \frac{1}{x}, 0, x \neq = 0 x = 0$ is continuous at $x = 0$ .
Reason (R) : When $x \to 0$ , $sin \frac{1}{x}$ is a finite value between $- 1$ and 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, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Since $∣ sin \frac{1}{x} ∣ \leq 1$ (R is true), $∣ x sin \frac{1}{x} ∣ \leq ∣ x ∣ \to 0$ . So $lim_{x \to 0} f (x) = 0 = f (0)$ and A is true. R is the reason A holds.
  1 mark
Answer: (A)
Where the marks are
- Uses the boundedness of $sin \frac{1}{x}$ to show the limit is 0 = f(0), so R explains A (1 mark)
Q20 · 1 mark · Assertion-reasonCompetency-basedInverse Trigonometric Functions
Assertion (A) : Set of values of $sec^{- 1} (\frac{3}{2})$ is a null set.
Reason (R) : $sec^{- 1} x$ is defined for $x \in R - (- 1, 1)$ .
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, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. $\frac{3}{2} \approx 0.866 \in (- 1, 1)$ , which lies outside the domain $R - (- 1, 1)$ of $sec^{- 1}$ . So no value exists, the set of values is null, and R explains A.
  1 mark
Answer: (A)
Where the marks are
- Notes that $\frac{3}{2} \in (- 1, 1)$ lies outside the domain of $sec^{- 1}$ (1 mark)

Section B

VSA

Q21 · 2 marks · Very short answerRelations and Functions
Let $f : A \to B$ be defined by $f (x) = \frac{x - 2}{x - 3}$ , where $A = R - {3}$ and $B = R - {1}$ . Discuss the bijectivity of the function.
Try it first, then show our solutionSolution
1. One-one: Let $x_{1}, x_{2} \in A$ with $f (x_{1}) = f (x_{2})$ . Then $\frac{x _{1} - 2}{x _{1} - 3} = \frac{x _{2} - 2}{x _{2} - 3}$ , so $(x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3)$ . Expanding gives $- 3 x_{1} - 2 x_{2} = - 3 x_{2} - 2 x_{1}$ , i.e. $x_{1} = x_{2}$ . Hence $f$ is one-one.
  1 mark
2. Onto: Let $y \in B = R - {1}$ and set $\frac{x - 2}{x - 3} = y$ . Then $x - 2 = x y - 3 y$ , so $x (y - 1) = 3 y - 2$ and $x = \frac{3 y - 2}{y - 1}$ , which is defined because $y \neq = 1$ . Also $x \neq = 3$ , since $\frac{3 y - 2}{y - 1} = 3$ would give $- 2 = - 3$ , which is impossible. So $x \in A$ and $f (x) = y$ . Hence $f$ is onto.
  ½ mark
3. Since $f$ is both one-one and onto, $f$ is a bijection.
  ½ mark
Answer: f is one-one and onto, hence bijective.
Where the marks are
- Shows $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ (1 mark)
- Finds pre-image $x = \frac{3 y - 2}{y - 1} \in A$ for every $y \in B$ , so f is onto (½ mark)
- Concludes f is bijective (½ mark)
Open Q21 on its own page
Q22 · 2 marks · Very short answerMatrices
If $A = [2 - 1 32]$ , then show that $A^{2} - 4 A + 7 I = 0$ .
Try it first, then show our solutionSolution
1. $A^{2} = [2 - 1 32] [2 - 1 32] = [4 - 3 - 2 - 2 6 + 6 - 3 + 4] = [1 - 4 121]$
  1 mark
2. $A^{2} - 4 A + 7 I = [1 - 4 121] - [8 - 4 128] + [7007] = [1 - 8 + 7 - 4 + 4 + 0 12 - 12 + 0 1 - 8 + 7] = [0000] = O$ . Hence proved.
  1 mark
Answer: $A^{2} - 4 A + 7 I = O$ (verified)
Where the marks are
- Correctly computes $A^{2} = [1 - 4 121]$ (1 mark)
- Substitutes and shows the result is the zero matrix (1 mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerContinuity and Differentiability
Differentiate $(\frac{5 ^{x}}{x ^{5}})$ with respect to x.
Try it first, then show our solutionSolution
1. Write $y = 5^{x} \cdot x^{- 5}$ . By the product rule, $\frac{d y}{d x} = (5^{x})^{'} x^{- 5} + 5^{x} (x^{- 5})^{'} = 5^{x} lo g 5 \cdot x^{- 5} + 5^{x} (- 5 x^{- 6})$ .
  1 mark
2. $\frac{d y}{d x} = \frac{5 ^{x} l o g 5}{x ^{5}} - \frac{5 ^{x + 1}}{x ^{6}}$
  1 mark
Answer: $\frac{d y}{d x} = \frac{5 ^{x} l o g 5}{x ^{5}} - \frac{5 ^{x + 1}}{x ^{6}}$
Where the marks are
- Applies the product (or quotient) rule with $\frac{d}{d x} 5^{x} = 5^{x} lo g 5$ (1 mark)
- Correct simplified derivative (1 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerContinuity and Differentiability
If $- 2 x^{2} - 5 x y + y^{3} = 76$ , then find $\frac{d y}{d x}$ .
Try it first, then show our solutionSolution
1. Differentiate both sides with respect to x, using the product rule on $x y$ : $- 4 x - 5 (y + x \frac{d y}{d x}) + 3 y^{2} \frac{d y}{d x} = 0$ , i.e. $- 4 x - 5 y - 5 x \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0$ .
  1½ marks
2. Collect the $\frac{d y}{d x}$ terms: $\frac{d y}{d x} (3 y^{2} - 5 x) = 4 x + 5 y$ , so $\frac{d y}{d x} = \frac{4 x + 5 y}{3 y ^{2} - 5 x}$ .
  ½ mark
Answer: $\frac{d y}{d x} = \frac{4 x + 5 y}{3 y ^{2} - 5 x}$
Where the marks are
- Correct implicit differentiation, including the product rule for $x y$ (1½ marks)
- Rearranges to get $\frac{d y}{d x} = \frac{4 x + 5 y}{3 y ^{2} - 5 x}$ (½ mark)
Open Q23(b) on its own page
Q24 · 2 marks · Very short answerLinear Programming
In a Linear Programming Problem, the objective function $Z = 5 x + 4 y$ needs to be maximised under constraints $3 x + y \leq 6$ , $x \leq 1$ , $x, y \geq 0$ . Express the LPP on the graph and shade the feasible region and mark the corner points.
Try it first, then show our solutionSolution
1. Plot the boundary lines: $3 x + y = 6$ passes through $(2, 0)$ and $(0, 6)$ , and $x = 1$ is a vertical line through $(1, 0)$ . Test with the origin: $0 \leq 6$ and $0 \leq 1$ are both true, so the region lies on the origin side of both lines. With $x, y \geq 0$ the region is in the first quadrant.
  1½ marks
2. Feasible region and corner points: The lines $x = 1$ and $3 x + y = 6$ meet at $(1, 3)$ . The shaded feasible region is the bounded quadrilateral $O C B A$ , between the y-axis and $x = 1$ , above the x-axis and below $3 x + y = 6$ . Its corner points are $O (0, 0)$ , $C (1, 0)$ , $B (1, 3)$ and $A (0, 6)$ .
  ½ mark
Answer: Feasible region OCBA with corner points O(0,0), C(1,0), B(1,3), A(0,6).
Where the marks are
- Correct plotting of the lines $3 x + y = 6$ and $x = 1$ (1½ marks)
- Correct shading of the feasible region and marking of the corner points (½ mark)
Open Q24 on its own page
Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerProbability
10 identical blocks are marked with '0' on two of them, '1' on three of them, '2' on four of them and '3' on one of them and put in a box. If X denotes the number written on the block, then write the probability distribution of X and calculate its mean.
Try it first, then show our solutionSolution
1. X can take the values 0, 1, 2, 3.
  ½ mark
2. Probability distribution:
  X 0 1 2 3
  P(X) $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$ $\frac{1}{10}$
  1 mark
3. Mean $= E (X) = \sum p_{i} x_{i} = 0 \cdot \frac{2}{10} + 1 \cdot \frac{3}{10} + 2 \cdot \frac{4}{10} + 3 \cdot \frac{1}{10} = \frac{14}{10} = \frac{7}{5} = 1.4$
  ½ mark
Answer: P(X=0,1,2,3) = 2/10, 3/10, 4/10, 1/10; Mean = 7/5 = 1.4
Where the marks are
- Correct values of X (½ mark)
- Correct probabilities (1 mark)
- Mean = 7/5 (½ mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerCompetency-basedProbability
In a village of 8000 people, 3000 go out of the village to work and 4000 are women. It is noted that 30% of women go out of the village to work. What is the probability that a randomly chosen individual is either a woman or a person working outside the village ?
Try it first, then show our solutionSolution
1. Let A be the event that the person is a woman and B the event that the person works outside the village. The number of women working outside is $30%$ of $4000 = 1200$ . So $P (A) = \frac{4000}{8000} = \frac{1}{2}$ , $P (B) = \frac{3000}{8000} = \frac{3}{8}$ and $P (A \cap B) = \frac{1200}{8000} = \frac{3}{20}$ .
  1½ marks
2. $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{1}{2} + \frac{3}{8} - \frac{3}{20} = \frac{20 + 15 - 6}{40} = \frac{29}{40}$
  ½ mark
Answer: $\frac{29}{40}$
Where the marks are
- Finds $P (A) = \frac{1}{2}$ , $P (B) = \frac{3}{8}$ , $P (A \cap B) = \frac{3}{20}$ (1½ marks)
- Uses the addition rule to get $\frac{29}{40}$ (½ mark)
Open Q25(b) on its own page

X	0	1	2	3
P(X)	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$	$\frac{1}{10}$

Section C

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerRelations and Functions
Show that the function $f : R \to R$ defined by $f (x) = 4 x^{3} - 5$ , $\forall x \in R$ is one-one and onto.
Try it first, then show our solutionSolution
1. One-one: Let $x_{1}, x_{2} \in R$ with $f (x_{1}) = f (x_{2})$ . Then $4 x_{1}^{3} - 5 = 4 x_{2}^{3} - 5$ , so $x_{1}^{3} = x_{2}^{3}$ , which gives $x_{1} = x_{2}$ (the real cube root is unique). Hence $f$ is one-one.
  1½ marks
2. Onto: Let $y \in R$ (co-domain). Take $x = (\frac{y + 5}{4})^{1/3}$ , which is real for every real y. Then $f (x) = 4 \cdot \frac{y + 5}{4} - 5 = y$ . So every element of the co-domain has a pre-image, and $f$ is onto. Hence $f$ is both one-one and onto.
  1½ marks
Answer: f is one-one and onto (bijective).
Where the marks are
- Proves $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ (1½ marks)
- Shows every real y has the real pre-image $3 (y + 5) /4$ , so range = co-domain (1½ marks)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerRelations and Functions
Let R be a relation defined on a set N of natural numbers such that $R = {(x, y) : x y is a square of a natural number, x, y \in N}$ . Determine if the relation R is an equivalence relation.
Try it first, then show our solutionSolution
1. Reflexive: For any $x \in N$ , $x \cdot x = x^{2}$ is the square of the natural number x. So $(x, x) \in R$ and R is reflexive.
  1 mark
2. Symmetric: If $(x, y) \in R$ , then $x y$ is a perfect square. Since $y x = x y$ , $y x$ is also a perfect square, so $(y, x) \in R$ and R is symmetric.
  1 mark
3. Transitive: Let $(x, y), (y, z) \in R$ , so $x y = a^{2}$ and $y z = b^{2}$ for some $a, b \in N$ . Then $x y \cdot y z = a^{2} b^{2}$ , so $x z = \frac{a ^{2} b ^{2}}{y ^{2}} = (\frac{ab}{y})^{2}$ . Here $x z$ is a natural number and $\frac{ab}{y}$ is a positive rational whose square is a natural number, so $\frac{ab}{y} \in N$ . Hence $(x, z) \in R$ and R is transitive.
  ½ mark
4. Since R is reflexive, symmetric and transitive, R is an equivalence relation.
  ½ mark
Answer: Yes, R is an equivalence relation.
Where the marks are
- Proves reflexivity (1 mark)
- Proves symmetry using $x y = y x$ (1 mark)
- Proves transitivity by showing $x z = (ab / y)^{2}$ (½ mark)
- Concludes R is an equivalence relation (½ mark)
Open Q26(b) on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerCompetency-basedMatrices
Let $2 x + 5 y - 1 = 0$ and $3 x + 2 y - 7 = 0$ represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.
Try it first, then show our solutionSolution
1. The equations are $2 x + 5 y = 1$ and $3 x + 2 y = 7$ . In matrix form $A X = B$ , where $A = [2352]$ , $X = [x y]$ , $B = [17]$ .
  1 mark
2. $∣ A ∣ = 4 - 15 = - 11 \neq = 0$ , so $A^{- 1}$ exists and $adj A = [2 - 3 - 5 2]$ . Then $X = A^{- 1} B = - \frac{1}{11} [2 - 3 - 5 2] [17] = - \frac{1}{11} [- 33 11] = [3 - 1]$
  1½ marks
3. So $x = 3$ , $y = - 1$ . The point common to the paths of the ants is $(3, - 1)$ .
  ½ mark
Answer: (3, −1)
Where the marks are
- Writes the system as $A X = B$ with correct matrices (1 mark)
- Finds $∣ A ∣$ , adj A and $X = A^{- 1} B$ correctly (1½ marks)
- States the common point (3, −1) (½ mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerCompetency-basedMatrices
A shopkeeper sells 50 Chemistry, 60 Physics and 35 Maths books on day I and sells 40 Chemistry, 45 Physics and 50 Maths books on day II. If the selling price for each such subject book is ₹ 150 (Chemistry), ₹ 175 (Physics) and ₹ 180 (Maths), then find his total sale in two days, using matrix method. If cost price of all the books together is ₹ 35,000, what profit did he earn after the sale of two days ?
Try it first, then show our solutionSolution
1. Sales matrix (rows: Day I, Day II; columns: Chemistry, Physics, Maths) $A = [504060453550]$ and price matrix $B = 150175180$ .
  1 mark
2. $A B = [50 (150) + 60 (175) + 35 (180) 40 (150) + 45 (175) + 50 (180)] = [7500 + 10500 + 6300 6000 + 7875 + 9000] = [2430022875]$ . So the sale is ₹ 24,300 on Day I and ₹ 22,875 on Day II.
  1 mark
3. Total sale in two days = ₹ 24,300 + ₹ 22,875 = ₹ 47,175.
  ½ mark
4. Profit = ₹ 47,175 − ₹ 35,000 = ₹ 12,175.
  ½ mark
Answer: Total sale ₹ 47,175; profit ₹ 12,175
Where the marks are
- Forms the sales and price matrices correctly (1 mark)
- Multiplies to get ₹24,300 and ₹22,875 (1 mark)
- Total sale ₹47,175 (½ mark)
- Profit ₹12,175 (½ mark)
Open Q27(b) on its own page
Q28 · 3 marks · Short answerContinuity and Differentiability
Differentiate $y = lo g {sin (\frac{x ^{3}}{3} - 1)}$ with respect to x.
Try it first, then show our solutionSolution
1. Let $u = \frac{x ^{3}}{3} - 1$ . By the chain rule, the outer square root gives $\frac{d y}{d x} = \frac{1}{2 l o g { s i n u }} \cdot \frac{d}{d x} lo g (sin u)$ .
  1 mark
2. $\frac{d}{d x} lo g (sin u) = \frac{1}{s i n u} \cdot cos u \cdot \frac{d u}{d x}$
  1 mark
3. $\frac{d u}{d x} = \frac{3 x ^{2}}{3} = x^{2}$
  ½ mark
4. $\frac{d y}{d x} = \frac{x ^{2} c o t ( \frac{x ^{3}}{3} - 1 )}{2 l o g { s i n ( \frac{x ^{3}}{3} - 1 ) }}$
  ½ mark
Answer: $\frac{d y}{d x} = \frac{x ^{2} c o t ( \frac{x ^{3}}{3} - 1 )}{2 l o g { s i n ( \frac{x ^{3}}{3} - 1 )}}$
Where the marks are
- Derivative of the square root (1 mark)
- Derivative of log(sin u) as $\frac{c o s u}{s i n u}$ (1 mark)
- Derivative of the inner function $x^{2}$ (½ mark)
- Simplified final answer (½ mark)
Open Q28 on its own page
Q29 · 3 marks · Short answerApplication of Derivatives
Amongst all pairs of positive integers with product as 289, find which of the two numbers add up to the least.
Try it first, then show our solutionSolution
1. Let the numbers be x and y with $x y = 289$ , so $y = \frac{289}{x}$ . Their sum is $S = x + \frac{289}{x}$ , $x > 0$ .
  1 mark
2. $\frac{d S}{d x} = 1 - \frac{289}{x ^{2}}$ . Setting $\frac{d S}{d x} = 0$ gives $x^{2} = 289$ , so $x = 17$ (taking the positive value).
  1½ marks
3. $\frac{d ^{2} S}{d x ^{2}} = \frac{578}{x ^{3}} > 0$ at $x = 17$ , so S is minimum there. Then $y = \frac{289}{17} = 17$ . The numbers are 17 and 17, with least sum 34.
  ½ mark
Answer: 17 and 17 (least sum 34)
Where the marks are
- Forms $S = x + \frac{289}{x}$ (1 mark)
- Solves $S^{'} = 0$ to get $x = 17$ (1½ marks)
- Second-derivative test confirms the minimum; numbers are 17, 17 (½ mark)
Open Q29 on its own page
Q30 · 3 marks · Short answerLinear Programming
In the Linear Programming Problem for objective function $Z = 18 x + 10 y$ subject to constraints
$4 x + y \geq 20$
$2 x + 3 y \geq 30$
$x, y \geq 0$
find the minimum value of Z.
Try it first, then show our solutionSolution
1. Graph: $4 x + y = 20$ passes through $(5, 0)$ and $(0, 20)$ , and $2 x + 3 y = 30$ passes through $(15, 0)$ and $(0, 10)$ . Both inequalities are '≥', so the region lies away from the origin. With $x, y \geq 0$ the feasible region is unbounded, lying above both lines in the first quadrant. Solving the two lines: $y = 20 - 4 x$ gives $2 x + 60 - 12 x = 30$ , so $x = 3$ and $y = 8$ . The corner points are $A (0, 20)$ , $B (3, 8)$ and $C (15, 0)$ .
  1½ marks
2. Corner point $Z = 18 x + 10 y$
  A(0, 20) 200
  B(3, 8) 134
  C(15, 0) 270
  1 mark
3. The region is unbounded, so draw the half-plane $18 x + 10 y < 134$ . It has no point in common with the feasible region. Hence the minimum value of Z is 134 at (3, 8).
  ½ mark
Answer: Minimum Z = 134 at (3, 8)
Where the marks are
- Correct graph of the unbounded feasible region with corner points (1½ marks)
- Evaluates Z at A, B, C correctly (1 mark)
- Checks the half-plane $18 x + 10 y < 134$ and concludes min Z = 134 at (3,8) (½ mark)
Open Q30 on its own page
Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerVector Algebra
The scalar product of the vector $a = \hat{i} - \hat{j} + 2 \hat{k}$ with a unit vector along sum of vectors $b = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $c = λ \hat{i} - 2 \hat{j} - 3 \hat{k}$ is equal to 1. Find the value of $λ$ .
Try it first, then show our solutionSolution
1. $d = b + c = (2 + λ) \hat{i} - 6 \hat{j} + 2 \hat{k}$
  1 mark
2. $\hat{d} = \frac{( 2 + λ ) i ^ - 6 j ^ + 2 k ^}{( 2 + λ ) ^{2} + 40}$
  ½ mark
3. $a \cdot \hat{d} = \frac{( 2 + λ ) + 6 + 4}{( 2 + λ ) ^{2} + 40} = 1$ , so $λ + 12 = (2 + λ)^{2} + 40$ .
  1 mark
4. Squaring: $λ^{2} + 24 λ + 144 = λ^{2} + 4 λ + 44$ , so $20 λ = - 100$ and $λ = - 5$ . Check: $7 = 9 + 40$ ✓.
  ½ mark
Answer: λ = −5
Where the marks are
- Finds $b + c$ (1 mark)
- Writes the unit vector along it (½ mark)
- Sets up $a \cdot \hat{d} = 1$ correctly (1 mark)
- Solves to get λ = −5 (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerThree Dimensional Geometry
Find the shortest distance between the lines :
$r = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 3 \hat{k})$
$r = (\hat{i} + 4 \hat{k}) + μ (3 \hat{i} - 6 \hat{j} + 9 \hat{k})$ .
Try it first, then show our solutionSolution
1. $3 \hat{i} - 6 \hat{j} + 9 \hat{k} = 3 (\hat{i} - 2 \hat{j} + 3 \hat{k})$ , so the lines are parallel with $b = \hat{i} - 2 \hat{j} + 3 \hat{k}$ . Here $a_{1} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $a_{2} = \hat{i} + 4 \hat{k}$ , so $a_{2} - a_{1} = - \hat{i} + \hat{j} + \hat{k}$ .
  ½ mark
2. $b \times (a_{2} - a_{1}) = \hat{i} 1 - 1 \hat{j} - 2 1 \hat{k} 31 = \hat{i} (- 2 - 3) - \hat{j} (1 + 3) + \hat{k} (1 - 2) = - 5 \hat{i} - 4 \hat{j} - \hat{k}$
  1½ marks
3. $S D = \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣} = \frac{25 + 16 + 1}{1 + 4 + 9} = \frac{42}{14} = 3$ units
  1 mark
Answer: $3$ units
Where the marks are
- Recognises the lines are parallel and finds $a_{2} - a_{1}$ (½ mark)
- Computes $b \times (a_{2} - a_{1}) = - 5 \hat{i} - 4 \hat{j} - \hat{k}$ (1½ marks)
- Shortest distance $= 3$ (1 mark)
Open Q31(b) on its own page

Corner point	$Z = 18 x + 10 y$
A(0, 20)	200
B(3, 8)	134
C(15, 0)	270

Section D

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerIntegrals
Find : $\int \frac{x ^{2} + 1}{( x ^{2} + 2 ) ( 2 x ^{2} + 1 )} d x$
Try it first, then show our solutionSolution
1. Put $x^{2} = t$ (for the partial fractions only): $\frac{t + 1}{( t + 2 ) ( 2 t + 1 )} = \frac{A}{t + 2} + \frac{B}{2 t + 1}$ , so $t + 1 = A (2 t + 1) + B (t + 2)$ .
2. At $t = - 2$ : $- 1 = - 3 A$ , so $A = \frac{1}{3}$ . At $t = - \frac{1}{2}$ : $\frac{1}{2} = \frac{3}{2} B$ , so $B = \frac{1}{3}$ .
  2 marks
3. $\int \frac{x ^{2} + 1}{( x ^{2} + 2 ) ( 2 x ^{2} + 1 )} d x = \frac{1}{3} \int \frac{d x}{x ^{2} + 2} + \frac{1}{3} \int \frac{d x}{2 x ^{2} + 1}$
  ½ mark
4. $\frac{1}{3} \int \frac{d x}{x ^{2} + ( 2 ) ^{2}} = \frac{1}{3} \cdot \frac{1}{2} tan^{- 1} \frac{x}{2}$
  1 mark
5. $\frac{1}{3} \int \frac{d x}{2 x ^{2} + 1} = \frac{1}{6} \int \frac{d x}{x ^{2} + ( 1/ 2 ) ^{2}} = \frac{1}{6} \cdot 2 tan^{- 1} (2 x) = \frac{1}{3 2} tan^{- 1} (2 x)$
  1 mark
6. $\int \frac{x ^{2} + 1}{( x ^{2} + 2 ) ( 2 x ^{2} + 1 )} d x = \frac{1}{3 2} [tan^{- 1} \frac{x}{2} + tan^{- 1} (2 x)] + C$
  ½ mark
Answer: $\frac{1}{3 2} [tan^{- 1} \frac{x}{2} + tan^{- 1} (2 x)] + C$
Where the marks are
- Partial fractions with $A = B = \frac{1}{3}$ (2 marks)
- Splits the integral correctly (½ mark)
- Integrates $\frac{1}{x ^{2} + 2}$ correctly (1 mark)
- Integrates $\frac{1}{2 x ^{2} + 1}$ correctly (1 mark)
- Final answer with C (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerIntegrals
Evaluate : $\int_{0}^{π} \frac{x t a n x}{s e c x + t a n x} d x$
Try it first, then show our solutionSolution
1. Let $I = \int_{0}^{π} \frac{x t a n x}{s e c x + t a n x} d x$ …(i). Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ with $tan (π - x) = - tan x$ and $sec (π - x) = - sec x$ : $I = \int_{0}^{π} \frac{( π - x ) t a n x}{s e c x + t a n x} d x$ …(ii).
  1½ marks
2. Adding (i) and (ii): $2 I = π \int_{0}^{π} \frac{t a n x}{s e c x + t a n x} d x = π \int_{0}^{π} \frac{t a n x ( s e c x - t a n x )}{s e c ^{2} x - t a n ^{2} x} d x$
  1 mark
3. Since $sec^{2} x - tan^{2} x = 1$ : $2 I = π \int_{0}^{π} (sec x tan x - tan^{2} x) d x = π \int_{0}^{π} (sec x tan x - sec^{2} x + 1) d x$
  1 mark
4. $2 I = π [sec x - tan x + x]_{0}^{π}$
  1 mark
5. $2 I = π [(- 1 - 0 + π) - (1 - 0 + 0)] = π (π - 2)$ , so $I = \frac{π}{2} (π - 2)$ .
  ½ mark
Answer: $I = \frac{π}{2} (π - 2)$
Where the marks are
- Applies the property $x \to π - x$ correctly (1½ marks)
- Adds and rationalises with $sec x - tan x$ (1 mark)
- Rewrites as $sec x tan x - sec^{2} x + 1$ (1 mark)
- Correct antiderivative $sec x - tan x + x$ (1 mark)
- Final value $\frac{π}{2} (π - 2)$ (½ mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerCompetency-basedApplication of Integrals
A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle $\frac{π}{4}$ anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.
Try it first, then show our solutionSolution
1. Figure: Draw a circle of radius 8 centred at the origin O and the line $y = x$ through O. In the first quadrant they meet at A. Drop the perpendicular from A to the x-axis at $x = 42$ . Shade the region bounded by the x-axis, the line OA and the arc from A to C(8,0).
  1 mark
2. Equation of the circular table top: $x^{2} + y^{2} = 64$ .
  ½ mark
3. The scratch passes through the origin at angle $\frac{π}{4}$ , so its slope is $tan \frac{π}{4} = 1$ and its equation is $y = x$ .
  ½ mark
4. Solving, $2 x^{2} = 64$ gives $x = 42$ in the first quadrant, so $A (42, 42)$ .
  ½ mark
5. Required area $= \int_{0}^{42} x d x + \int_{42}^{8} 64 - x^{2} d x$
  1 mark
6. $= [\frac{x ^{2}}{2}]_{0}^{42} + [\frac{x}{2} 64 - x^{2} + 32 sin^{- 1} \frac{x}{8}]_{42}^{8}$
  1 mark
7. $= 16 + [32 \cdot \frac{π}{2} - (22 \cdot 42 + 32 \cdot \frac{π}{4})] = 16 + 16 π - 16 - 8 π = 8 π cm^{2}$
  ½ mark
Answer: $8 π$ cm²
Where the marks are
- Correct figure with region shaded (1 mark)
- Equation of circle $x^{2} + y^{2} = 64$ (½ mark)
- Equation of scratch $y = x$ (½ mark)
- Point of intersection $x = 42$ (½ mark)
- Correct integral set-up (1 mark)
- Correct integration (1 mark)
- Final area $8 π$ cm² (½ mark)
Open Q33 on its own page
Q34 · 5 marks · Long answerDifferential Equations
Solve the differential equation $\frac{d y}{d x} = cos x - 2 y$ .
Try it first, then show our solutionSolution
1. Rewrite the equation in linear form: $\frac{d y}{d x} + 2 y = cos x$ . This is of the type $\frac{d y}{d x} + P y = Q$ with $P = 2$ and $Q = cos x$ .
  ½ mark
2. Integrating factor: $I.F. = e^{\int 2 d x} = e^{2 x}$ .
  1 mark
3. The solution is $y \cdot e^{2 x} = \int e^{2 x} cos x d x + C$ .
  1½ marks
4. Let $I_{1} = \int e^{2 x} cos x d x$ . Integrate by parts, taking $cos x$ as the first function:
  $I_{1} = cos x \cdot \frac{e ^{2 x}}{2} + \frac{1}{2} \int sin x e^{2 x} d x = \frac{e ^{2 x} c o s x}{2} + \frac{1}{2} [sin x \cdot \frac{e ^{2 x}}{2} - \int cos x \cdot \frac{e ^{2 x}}{2} d x]$
  $\Rightarrow I_{1} = \frac{e ^{2 x} c o s x}{2} + \frac{e ^{2 x} s i n x}{4} - \frac{1}{4} I_{1}$
  $\Rightarrow \frac{5}{4} I_{1} = \frac{e ^{2 x}}{4} (2 cos x + sin x) \Rightarrow I_{1} = \frac{e ^{2 x}}{5} (2 cos x + sin x)$ .
  1½ marks
5. Hence $y e^{2 x} = \frac{e ^{2 x}}{5} (2 cos x + sin x) + C$ , i.e. the general solution is $y = \frac{1}{5} (2 cos x + sin x) + C e^{- 2 x}$ .
  ½ mark
Answer: $y = \frac{1}{5} (2 cos x + sin x) + C e^{- 2 x}$
Where the marks are
- Writes the equation as $\frac{d y}{d x} + 2 y = cos x$ and identifies $P = 2$ , $Q = cos x$ (½ mark)
- Finds I.F. $= e^{2 x}$ (1 mark)
- Writes the solution form $y e^{2 x} = \int e^{2 x} cos x d x + C$ (1½ marks)
- Evaluates $\int e^{2 x} cos x d x = \frac{e ^{2 x}}{5} (2 cos x + sin x)$ by parts (1½ marks)
- Writes the general solution $y = \frac{1}{5} (2 cos x + sin x) + C e^{- 2 x}$ (½ mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerThree Dimensional Geometry
Find the point Q on the line $\frac{2 x + 4}{6} = \frac{y + 1}{2} = \frac{- 2 z + 6}{- 4}$ at a distance of $32$ from the point P(1, 2, 3).
Try it first, then show our solutionSolution
1. Write the line in standard form: $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} = λ$ . Any point on it is $Q (3 λ - 2, 2 λ - 1, 2 λ + 3)$ for some $λ \in R$ .
  2 marks
2. Since $P Q = 32$ , we have $P Q^{2} = 18$ :
  $(3 λ - 3)^{2} + (2 λ - 3)^{2} + (2 λ)^{2} = 18$ .
  1 mark
3. Expanding: $9 λ^{2} - 18 λ + 9 + 4 λ^{2} - 12 λ + 9 + 4 λ^{2} = 18 \Rightarrow 17 λ^{2} - 30 λ = 0 \Rightarrow λ = 0$ or $λ = \frac{30}{17}$ .
  1 mark
4. For $λ = 0$ : $Q (- 2, - 1, 3)$ . For $λ = \frac{30}{17}$ : $Q (\frac{90}{17} - 2, \frac{60}{17} - 1, \frac{60}{17} + 3) = Q (\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$ .
  1 mark
Answer: $Q (- 2, - 1, 3)$ or $Q (\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$
Where the marks are
- Converts the line to standard form and writes the general point $(3 λ - 2, 2 λ - 1, 2 λ + 3)$ (2 marks)
- Sets up $P Q^{2} = 18$ correctly (1 mark)
- Solves $17 λ^{2} - 30 λ = 0$ to get $λ = 0$ or $\frac{30}{17}$ (1 mark)
- States both points Q (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerThree Dimensional Geometry
Find the image of the point $(- 1, 5, 2)$ in the line $\frac{2 x - 4}{2} = \frac{y}{2} = \frac{2 - z}{3}$ . Find the length of the line segment joining the points (given point and the image point).
Try it first, then show our solutionSolution
1. Standard form of the line: $\frac{x - 2}{1} = \frac{y}{2} = \frac{z - 2}{- 3}$ , with direction ratios $1, 2, - 3$ . Let $A (- 1, 5, 2)$ , let $A^{'} (a, b, c)$ be its image and let $M$ be the foot of the perpendicular from $A$ to the line. Then $M$ is the mid-point of $A A^{'}$ .
  1 mark
2. Any point on the line is $M (λ + 2, 2 λ, - 3 λ + 2)$ .
  ½ mark
3. Direction ratios of $A M$ : $λ + 3, 2 λ - 5, - 3 λ$ .
  ½ mark
4. Since $A M ⊥$ line: $1 (λ + 3) + 2 (2 λ - 5) - 3 (- 3 λ) = 0 \Rightarrow 14 λ - 7 = 0 \Rightarrow λ = \frac{1}{2}$ . So $M (\frac{5}{2}, 1, \frac{1}{2})$ .
  1 mark
5. $M$ is the mid-point of $A A^{'}$ : $\frac{a - 1}{2} = \frac{5}{2}, \frac{b + 5}{2} = 1, \frac{c + 2}{2} = \frac{1}{2} \Rightarrow a = 6, b = - 3, c = - 1$ .
  1 mark
6. Image of the point: $A^{'} (6, - 3, - 1)$ .
  ½ mark
7. Length $A A^{'} = (6 + 1)^{2} + (- 3 - 5)^{2} + (- 1 - 2)^{2} = 49 + 64 + 9 = 122$ units.
  ½ mark
Answer: Image $(6, - 3, - 1)$ ; length of segment $= 122$ units
Where the marks are
- Writes the line in standard form $\frac{x - 2}{1} = \frac{y}{2} = \frac{z - 2}{- 3}$ (1 mark)
- General point $M (λ + 2, 2 λ, - 3 λ + 2)$ (½ mark)
- Direction ratios of AM (½ mark)
- Uses perpendicularity to get $λ = \frac{1}{2}$ (1 mark)
- Uses mid-point condition to get $a = 6, b = - 3, c = - 1$ (1 mark)
- States image $(6, - 3, - 1)$ (½ mark)
- Finds $A A^{'} = 122$ (½ mark)
Open Q35(b) on its own page

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedVector Algebra
Case Study – 1
Three friends A, B and C move out from the same location O at the same time in three different directions to reach their destinations. They move out on straight paths and decide that A and B after reaching their destinations will meet up with C at his predecided destination, following straight paths from A to C and B to C in such a way that $O A = a$ , $O B = b$ and $O C = 5 a - 2 b$ respectively.
Based upon the above information, answer the following questions :
1. (i)
  Complete the given figure to explain their entire movement plan along the respective vectors.
  1 mark
2. (ii)
  Find vectors $A C$ and $B C$ .
  1 mark
3. (iii)
  (a) If $a \cdot b = 1$ , distance of O to A is 1 km and that from O to B is 2 km, then find the angle between $O A$ and $O B$ . Also, find $∣ a \times b ∣$ .
  2 marks
4. OR
  (iii)
  (b) If $a = 2 \hat{i} - \hat{j} + 4 \hat{k}$ and $b = \hat{j} - \hat{k}$ , then find a unit vector perpendicular to $(a + b)$ and $(a - b)$ .
  2 marks
Figure in the paper: Point O at left with three arrows from it: OA along vector a to point A (horizontal, right); OB along vector b to point B (upper right); OC along vector 5a − 2b to point C (lower right). Segments AC, BC, AB are not drawn (to be completed). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Complete the figure by drawing the straight path from A to C (arrow from A towards C) and from B to C (arrow from B towards C), so A moves $O \to A \to C$ and B moves $O \to B \to C$ , while C moves directly $O \to C$ along $5 a - 2 b$ ; the three meet at C.
  1 mark
2. (ii) $A C = O C - O A = (5 a - 2 b) - a = 4 a - 2 b$ .
  ½ mark
3. (ii) $B C = O C - O B = (5 a - 2 b) - b = 5 a - 3 b$ .
  ½ mark
4. (iii)(a) Given $∣ a ∣ = 1$ , $∣ b ∣ = 2$ , $a \cdot b = 1$ . Angle $θ = cos^{- 1} (\frac{a \cdot b}{∣ a ∣∣ b ∣}) = cos^{- 1} (\frac{1}{1 \times 2}) = \frac{π}{3}$ .
  1 mark
5. (iii)(a) $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ = 1 \times 2 \times \frac{3}{2} = 3$ .
  1 mark
6. (iii)(b) $a + b = 2 \hat{i} + 0 \hat{j} + 3 \hat{k}$ and $a - b = 2 \hat{i} - 2 \hat{j} + 5 \hat{k}$ .
  1 mark
7. (iii)(b) A vector perpendicular to both: $c = (a + b) \times (a - b) = \hat{i} 22 \hat{j} 0 - 2 \hat{k} 35 = (0 + 6) \hat{i} - (10 - 6) \hat{j} + (- 4 - 0) \hat{k} = 6 \hat{i} - 4 \hat{j} - 4 \hat{k}$ , with $∣ c ∣ = 36 + 16 + 16 = 68 = 217$ .
  ½ mark
8. (iii)(b) Required unit vector: $\overset{c}{^} = \frac{1}{2 17} (6 \hat{i} - 4 \hat{j} - 4 \hat{k}) = \frac{1}{17} (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ (its negative is also acceptable).
  ½ mark
Answer: (i) Draw arrows A→C and B→C; (ii) $A C = 4 a - 2 b$ , $B C = 5 a - 3 b$ ; (iii)(a) $θ = \frac{π}{3}$ , $∣ a \times b ∣ = 3$ ; (iii)(b) $\frac{1}{17} (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$
Where the marks are
- (i) Completes figure with paths A→C and B→C (1 mark)
- (ii) $A C = 4 a - 2 b$ (½ mark) and $B C = 5 a - 3 b$ (½ mark)
- (iii)(a) Angle $\frac{π}{3}$ (1 mark); $∣ a \times b ∣ = 3$ (1 mark)
- (iii)(b) Finds $a + b$ and $a - b$ (½ + ½ mark); cross product $6 \hat{i} - 4 \hat{j} - 4 \hat{k}$ with magnitude $68$ (½ mark); unit vector $\frac{1}{17} (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$ (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedDifferential Equations
Case Study – 2
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature. (Cylindrical-shaped Camphor tablets)
A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, $\frac{d V}{d t} = k S$ is the differential equation, where V is the volume, S is the surface area and t is the time in hours.
Based upon the above information, answer the following questions :
1. (i)
  Write the order and degree of the given differential equation.
  1 mark
2. (ii)
  Substituting $V = π r^{3}$ and $S = 2 π r^{2}$ , we get the differential equation $\frac{d r}{d t} = \frac{2}{3} k$ . Solve it, given that $r (0) = 5$ mm.
  1 mark
3. (iii)
  (a) If it is given that $r = 3$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
  2 marks
4. OR
  (iii)
  (b) If it is given that $r = 1$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
  2 marks
Figure in the paper: Photograph of cylindrical-shaped camphor tablets. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In $\frac{d V}{d t} = k S$ the highest-order derivative is $\frac{d V}{d t}$ (first order) and it appears with power 1. Order $= 1$ , Degree $= 1$ .
  1 mark
2. (ii) Separating variables and integrating: $\int d r = \frac{2 k}{3} \int d t \Rightarrow r = \frac{2}{3} k t + C$ .
  ½ mark
3. (ii) Using $r (0) = 5$ : $C = 5$ , so $r = \frac{2}{3} k t + 5$ .
  ½ mark
4. (iii)(a) Put $r = 3$ , $t = 1$ : $3 = \frac{2}{3} k + 5 \Rightarrow \frac{2}{3} k = - 2 \Rightarrow k = - 3$ .
  1 mark
5. (iii)(a) Then $r = - 2 t + 5$ . For $r = 0$ : $t = \frac{5}{2}$ hours $= 2.5$ hours.
  1 mark
6. (iii)(b) Put $r = 1$ , $t = 1$ : $1 = \frac{2}{3} k + 5 \Rightarrow \frac{2}{3} k = - 4 \Rightarrow k = - 6$ .
  1 mark
7. (iii)(b) Then $r = - 4 t + 5$ . For $r = 0$ : $t = \frac{5}{4}$ hours $= 1.25$ hours.
  1 mark
Answer: (i) Order 1, degree 1; (ii) $r = \frac{2}{3} k t + 5$ ; (iii)(a) $k = - 3$ , $t = 2.5$ h; (iii)(b) $k = - 6$ , $t = 1.25$ h
Where the marks are
- (i) Order = 1 (½ mark), Degree = 1 (½ mark)
- (ii) Integrates to $r = \frac{2}{3} k t + C$ (½ mark); uses $r (0) = 5$ to get $r = \frac{2}{3} k t + 5$ (½ mark)
- (iii)(a) Finds $k = - 3$ (1 mark); $r = - 2 t + 5$ and $t = \frac{5}{2}$ h (1 mark)
- (iii)(b) Finds $k = - 6$ (1 mark); $r = - 4 t + 5$ and $t = \frac{5}{4}$ h (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
Case Study – 3
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let $A_{1}$ : People with good health,
$A_{2}$ : People with average health,
and $A_{3}$ : People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category $A_{1}$ , $A_{2}$ and $A_{3}$ are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions :
1. (i)
  A person was tested randomly. What is the probability that he/she has contracted the disease ?
  2 marks
2. (ii)
  Given that the person has not contracted the disease, what is the probability that the person is from category $A_{2}$ ?
  2 marks
Try it first, then show our solutionSolution
1. (i) Let $A$ : the person has contracted the disease. $P (A_{1}) = \frac{7}{10}$ , $P (A_{2}) = \frac{2}{10}$ , $P (A_{3}) = \frac{1}{10}$ ; $P (A ∣ A_{1}) = \frac{25}{100}$ , $P (A ∣ A_{2}) = \frac{35}{100}$ , $P (A ∣ A_{3}) = \frac{50}{100}$ . By the theorem of total probability:
  $P (A) = \frac{7}{10} \cdot \frac{25}{100} + \frac{2}{10} \cdot \frac{35}{100} + \frac{1}{10} \cdot \frac{50}{100}$ .
  1½ marks
2. (i) $P (A) = \frac{175 + 70 + 50}{1000} = \frac{295}{1000} = 0.295$ (or $\frac{59}{200}$ ).
  ½ mark
3. (ii) $P (\overset{ˉ}{A} ∣ A_{1}) = \frac{75}{100}$ , $P (\overset{ˉ}{A} ∣ A_{2}) = \frac{65}{100}$ , $P (\overset{ˉ}{A} ∣ A_{3}) = \frac{50}{100}$ . By Bayes' theorem:
  $P (A_{2} ∣ \overset{ˉ}{A}) = \frac{P ( A _{2} ) P ( A ˉ ∣ A _{2} )}{P ( A _{1} ) P ( A ˉ ∣ A _{1} ) + P ( A _{2} ) P ( A ˉ ∣ A _{2} ) + P ( A _{3} ) P ( A ˉ ∣ A _{3} )} = \frac{\frac{2}{10} \cdot \frac{65}{100}}{\frac{7}{10} \cdot \frac{75}{100} + \frac{2}{10} \cdot \frac{65}{100} + \frac{1}{10} \cdot \frac{50}{100}}$ .
  1½ marks
4. (ii) $= \frac{130}{525 + 130 + 50} = \frac{130}{705} = \frac{26}{141}$ .
  ½ mark
Answer: (i) $0.295$ (= $\frac{59}{200}$ ); (ii) $\frac{26}{141}$
Where the marks are
- (i) Applies total probability with correct priors and conditional probabilities (1½ marks)
- (i) Obtains $P (A) = 0.295$ (½ mark)
- (ii) Applies Bayes' theorem with complementary probabilities 75%, 65%, 50% (1½ marks)
- (ii) Obtains $\frac{26}{141}$ (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.