MathsCBSE Class 12 · Sample paper, session 2025-26

CBSE Class 12 Mathematics: sample paper 2025-26

The sample question paper CBSE published for the 2025-26 session, 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 sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassXII_2025_26/Maths-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them. Questions printed only for visually impaired candidates are left out (7).

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; internal choice in 2 questions of Section B, 3 of Section C, 2 of Section D and one subpart each in 2 questions of Section E. Calculator not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQInverse Trigonometric Functions
Identify the function shown in the graph
1. (A) $sin^{- 1} x$
2. (B) $sin^{- 1} (2 x)$
3. (C) $sin^{- 1} (\frac{x}{2})$
4. (D) $2 sin^{- 1} x$
Figure in the paper: Graph of an inverse trigonometric curve on X'OX, Y'OY axes, with y-values marked -π/2 to π/2 (dashed horizontal lines at ±π/2) and dashed vertical lines at the ends of the domain, which lies between about -0.5 and 0.5 (x=1 marked on axis); the curve is increasing, passing through the origin. See it in the official paper.
Try it first, then show our solutionSolution
1. From the graph, the domain is $[- \frac{1}{2}, \frac{1}{2}]$ and the range is $[- \frac{π}{2}, \frac{π}{2}]$ ; the curve increases through the origin.
2. $sin^{- 1} (2 x)$ is defined when $- 1 \leq 2 x \leq 1$ , i.e. $- \frac{1}{2} \leq x \leq \frac{1}{2}$ , and its range is $[- \frac{π}{2}, \frac{π}{2}]$ . So the graph is of $sin^{- 1} (2 x)$ , option (B).
  1 mark
Answer: $sin^{- 1} (2 x)$ , option (B)
Where the marks are
- Matches the domain $[- \frac{1}{2}, \frac{1}{2}]$ to $sin^{- 1} (2 x)$ (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQMatrices
If for three matrices $A = [a_{ij}]_{m \times 4}$ , $B = [b_{ij}]_{n \times 3}$ and $C = [c_{ij}]_{p \times q}$ products $A B$ and $A C$ both are defined and are square matrices of same order, then value of $m, n, p$ and $q$ are:
1. (A) $m = q = 3$ and $n = p = 4$
2. (B) $m = 2, q = 3$ and $n = p = 4$
3. (C) $m = q = 4$ and $n = p = 3$
4. (D) $m = 4, p = 2$ and $n = q = 3$
Try it first, then show our solutionSolution
1. $A B$ is defined, so the columns of $A$ equal the rows of $B$ : $n = 4$ . $A C$ is defined, so $p = 4$ .
2. $A B$ has order $m \times 3$ and $A C$ has order $m \times q$ . Both are square and of the same order, so $m = 3$ and $q = 3$ . Option (A).
  1 mark
Answer: $m = q = 3$ , $n = p = 4$ , option (A)
Where the marks are
- Uses the conditions for the products to be defined and square to get $n = p = 4$ , $m = q = 3$ (1 mark)
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Q3 · 1 mark · MCQMatrices
If the matrix $A = 03 q r p - 4 - 2 t 0$ is skew-symmetric, then value of $\frac{q + t}{p + r}$ is....
1. (A) $- 2$
2. (B)0
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. For a skew-symmetric matrix, $a_{ij} = - a_{j i}$ and every diagonal entry is 0. So $p = 0$ , $r = - 3$ , $q = 2$ and $t = 4$ .
2. $\frac{q + t}{p + r} = \frac{2 + 4}{0 - 3} = - 2$ . Option (A).
  1 mark
Answer: $- 2$ , option (A)
Where the marks are
- Finds $p, q, r, t$ from $a_{ij} = - a_{j i}$ and evaluates the ratio as $- 2$ (1 mark)
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Q4 · 1 mark · MCQDeterminants
If $A$ is a square matrix of order 4 and $∣ a d j A ∣ = 27$ , then $A (a d j A)$ is equal to
1. (A)3
2. (B)9
3. (C) $3 I$
4. (D) $9 I$
Try it first, then show our solutionSolution
1. For order $n = 4$ , $∣ a d j A ∣ = ∣ A ∣^{n - 1} = ∣ A ∣^{3} = 27$ , so $∣ A ∣ = 3$ .
2. $A (a d j A) = ∣ A ∣ I = 3 I$ . Option (C).
  1 mark
Answer: $3 I$ , option (C)
Where the marks are
- Uses $∣ a d j A ∣ = ∣ A ∣^{n - 1}$ to get $∣ A ∣ = 3$ , then $A (a d j A) = ∣ A ∣ I = 3 I$ (1 mark)
Q5 · 1 mark · MCQMatrices
The inverse of the matrix $300020005$ is...
1. (A) $005020300$
2. (B) $\frac{1}{3} 00 0 \frac{1}{2} 0 00 \frac{1}{5}$
3. (C) $- \frac{1}{3} 00 0 - \frac{1}{2} 0 00 - \frac{1}{5}$
4. (D) $- 3 00 0 - 2 0 00 - 5$
Try it first, then show our solutionSolution
1. The inverse of a diagonal matrix (with non-zero diagonal entries) is the diagonal matrix of the reciprocals.
2. So the inverse is $diag (\frac{1}{3}, \frac{1}{2}, \frac{1}{5})$ . Check: the product with the given matrix is $I$ . Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Takes the reciprocals of the diagonal entries (1 mark)
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Q6 · 1 mark · MCQDeterminants
Value of the determinant $cos 6 7^{\circ} sin 2 3^{\circ} sin 6 7^{\circ} cos 2 3^{\circ}$ is
1. (A)0
2. (B) $\frac{1}{2}$
3. (C) $\frac{3}{2}$
4. (D)1
Try it first, then show our solutionSolution
1. Expanding: $cos 6 7^{\circ} cos 2 3^{\circ} - sin 6 7^{\circ} sin 2 3^{\circ} = cos (6 7^{\circ} + 2 3^{\circ})$ .
2. $= cos 9 0^{\circ} = 0$ . Option (A).
  1 mark
Answer: 0, option (A)
Where the marks are
- Expands the determinant and uses $cos (A + B)$ to get $cos 9 0^{\circ} = 0$ (1 mark)
Q7 · 1 mark · MCQContinuity and Differentiability
If a function defined by $f (x) = {k x + 1, cos x, x \leq π x > π$ is continuous at $x = π$ , then the value of $k$ is
1. (A) $π$
2. (B) $\frac{- 1}{π}$
3. (C)0
4. (D) $\frac{- 2}{π}$
Try it first, then show our solutionSolution
1. For continuity at $x = π$ : LHL $= k π + 1 = f (π)$ and RHL $= cos π = - 1$ .
2. Equating, $k π + 1 = - 1$ , so $k = - \frac{2}{π}$ . Option (D).
  1 mark
Answer: $k = - \frac{2}{π}$ , option (D)
Where the marks are
- Equates the left and right limits and solves $k π + 1 = - 1$ (1 mark)
Q8 · 1 mark · MCQContinuity and Differentiability
If $f (x) = x tan^{- 1} x$ , then $f^{'} (1)$ is equal to
1. (A) $\frac{π}{4} - \frac{1}{2}$
2. (B) $\frac{π}{4} + \frac{1}{2}$
3. (C) $- \frac{π}{4} - \frac{1}{2}$
4. (D) $- \frac{π}{4} + \frac{1}{2}$
Try it first, then show our solutionSolution
1. By the product rule, $f^{'} (x) = tan^{- 1} x + \frac{x}{1 + x ^{2}}$ .
2. $f^{'} (1) = \frac{π}{4} + \frac{1}{2}$ . Option (B).
  1 mark
Answer: $\frac{π}{4} + \frac{1}{2}$ , option (B)
Where the marks are
- Differentiates by the product rule and substitutes $x = 1$ (1 mark)
Q9 · 1 mark · MCQApplication of Derivatives
A function $f (x) = 10 - x - 2 x^{2}$ is increasing on the interval
1. (A) $(- \infty, - \frac{1}{4}]$
2. (B) $(- \infty, \frac{1}{4})$
3. (C) $[- \frac{1}{4}, \infty)$
4. (D) $[- \frac{1}{4}, \frac{1}{4}]$
Try it first, then show our solutionSolution
1. $f^{'} (x) = - 1 - 4 x$ .
2. For $f$ to be increasing, $f^{'} (x) \geq 0$ : $- 1 - 4 x \geq 0 \Rightarrow x \leq - \frac{1}{4}$ . So $f$ is increasing on $(- \infty, - \frac{1}{4}]$ . Option (A).
  1 mark
Answer: $(- \infty, - \frac{1}{4}]$ , option (A)
Where the marks are
- Solves $f^{'} (x) \geq 0$ to get $x \leq - \frac{1}{4}$ (1 mark)
Q10 · 1 mark · MCQDifferential Equations
The solution of the differential equation $x d x + y d y = 0$ represents a family of
1. (A)straight lines
2. (B)parabolas
3. (C)Circles
4. (D)Ellipses
Try it first, then show our solutionSolution
1. Integrating, $\int x d x = - \int y d y \Rightarrow \frac{x ^{2}}{2} = - \frac{y ^{2}}{2} + k$ .
2. So $x^{2} + y^{2} = 2 k$ , which is a family of circles centred at the origin. Option (C).
  1 mark
Answer: Circles, option (C)
Where the marks are
- Separates the variables, integrates to $x^{2} + y^{2} = 2 k$ and identifies circles (1 mark)
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Q11 · 1 mark · MCQIntegrals
If $f (a + b - x) = f (x)$ , then $\int_{a}^{b} x f (x) d x$ is equal to
1. (A) $\frac{a + b}{2} \int_{a}^{b} f (b - x) d x$
2. (B) $\frac{a + b}{2} \int_{a}^{b} f (a - x) d x$
3. (C) $\frac{b - a}{2} \int_{a}^{b} f (x) d x$
4. (D) $\frac{a + b}{2} \int_{a}^{b} f (x) d x$
Try it first, then show our solutionSolution
1. Let $I = \int_{a}^{b} x f (x) d x$ . By the property $\int_{a}^{b} g (x) d x = \int_{a}^{b} g (a + b - x) d x$ , $I = \int_{a}^{b} (a + b - x) f (a + b - x) d x = \int_{a}^{b} (a + b - x) f (x) d x$ [given $f (a + b - x) = f (x)$ ].
2. So $I = (a + b) \int_{a}^{b} f (x) d x - I$ , giving $2 I = (a + b) \int_{a}^{b} f (x) d x$ and $I = \frac{a + b}{2} \int_{a}^{b} f (x) d x$ . Option (D).
  1 mark
Answer: $\frac{a + b}{2} \int_{a}^{b} f (x) d x$ , option (D)
Where the marks are
- Applies the $a + b - x$ property and solves for $I$ (1 mark)
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Q12 · 1 mark · MCQIntegrals
If $\int x^{3} sin^{4} (x^{4}) cos (x^{4}) d x = a sin^{5} (x^{4}) + C$ , then $a$ is equal to
1. (A) $- \frac{1}{10}$
2. (B) $\frac{1}{20}$
3. (C) $\frac{1}{4}$
4. (D) $\frac{1}{5}$
Try it first, then show our solutionSolution
1. Put $sin (x^{4}) = t$ . Then $4 x^{3} cos (x^{4}) d x = d t$ , so $x^{3} cos (x^{4}) d x = \frac{1}{4} d t$ .
2. $\int t^{4} \cdot \frac{1}{4} d t = \frac{t ^{5}}{20} + C = \frac{1}{20} sin^{5} (x^{4}) + C$ , so $a = \frac{1}{20}$ . Option (B).
  1 mark
Answer: $a = \frac{1}{20}$ , option (B)
Where the marks are
- Substitutes $t = sin (x^{4})$ and integrates to get $a = \frac{1}{20}$ (1 mark)
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Q13 · 1 mark · MCQCompetency-basedVector Algebra
A bird flies through a distance in a straight line given by the vector $\hat{i} + 2 \hat{j} + \hat{k}$ . A man standing beside a straight metro rail track given by $r = (3 + λ) \hat{i} + (2 λ - 1) \hat{j} + 3 λ \hat{k}$ is observing the bird. The projected length of its flight on the metro track is
1. (A) $\frac{6}{14}$ units
2. (B) $\frac{14}{6}$ units
3. (C) $\frac{8}{14}$ units
4. (D) $\frac{5}{6}$ units
Try it first, then show our solutionSolution
1. The track is $r = (3 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k})$ , so its direction is $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$ .
2. Projection $= \frac{a \cdot b}{∣ b ∣} = \frac{1 + 4 + 3}{14} = \frac{8}{14}$ units. Option (C).
  1 mark
Answer: $\frac{8}{14}$ units, option (C)
Where the marks are
- Identifies the direction of the track and computes the projection $\frac{8}{14}$ (1 mark)
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Q14 · 1 mark · MCQThree Dimensional Geometry
The distance of the point with position vector $3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ from the y-axis is
1. (A)4 units
2. (B) $34$ units
3. (C)5 units
4. (D) $52$ units
Try it first, then show our solutionSolution
1. The distance of a point $(a, b, c)$ from the y-axis is $a^{2} + c^{2}$ .
2. $= 9 + 25 = 34$ units. Option (B).
  1 mark
Answer: $34$ units, option (B)
Where the marks are
- Uses $x^{2} + z^{2}$ to get $34$ (1 mark)
Q15 · 1 mark · MCQVector Algebra
If $a = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ , $b = \hat{i} + \hat{j} - 3 \hat{k}$ and $c = 6 \hat{i} - \hat{j} + 2 \hat{k}$ are three given vectors, then $(2 a \cdot \hat{i}) \hat{i} - (b \cdot \hat{j}) \hat{j} + (c \cdot \hat{k}) \hat{k}$ is same as the vector
1. (A) $a$
2. (B) $b + c$
3. (C) $a - b$
4. (D) $c$
Try it first, then show our solutionSolution
1. $2 a \cdot \hat{i} = 6$ , $b \cdot \hat{j} = 1$ , $c \cdot \hat{k} = 2$ .
2. The vector is $6 \hat{i} - \hat{j} + 2 \hat{k} = c$ . Option (D).
  1 mark
Answer: $c$ , option (D)
Where the marks are
- Evaluates each dot product and identifies the vector as $c$ (1 mark)
Q16 · 1 mark · MCQLinear Programming
A student of class XII studying Mathematics comes across an incomplete question in a book.
Maximise $Z = 3 x + 2 y + 1$
Subject to the constraints $x \geq 0, y \geq 0, 3 x + 4 y \leq 12$ ,
He/She notices the below shown graph for the said LPP problem, and finds that a constraint is missing in it:
Help him/her choose the required constraint from the graph.
The missing constraint is
1. (A) $x + 2 y \leq 2$
2. (B) $2 x + y \geq 2$
3. (C) $2 x + y \leq 2$
4. (D) $x + 2 y \geq 2$
Figure in the paper: Graph with a shaded (hatched) feasible region in the first quadrant bounded by the axes, the line 3x+4y=12 (intercepts (4,0) and (0,3)) and another line passing through about (1,0) and (0,2) (the missing constraint line); the shaded region lies between the two lines. See it in the official paper.
Try it first, then show our solutionSolution
1. The line through $(1, 0)$ and $(0, 2)$ is $2 x + y = 2$ .
2. The feasible region lies away from the origin (testing $(0, 0)$ gives $0 \geq 2$ , false, and the origin is not shaded), so the constraint is $2 x + y \geq 2$ . Option (B).
  1 mark
Answer: $2 x + y \geq 2$ , option (B)
Where the marks are
- Finds the line $2 x + y = 2$ and picks the side not containing the origin (1 mark)
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Q17 · 1 mark · MCQLinear Programming
The feasible region of a linear programming problem is bounded but the objective function attains its minimum value at more than one point. One of the points is $(5, 0)$ .
Then one of the other possible points at which the objective function attains its minimum value is
1. (A) $(2, 9)$
2. (B) $(6, 6)$
3. (C) $(4, 7)$
4. (D) $(0, 0)$
Figure in the paper: Graph with a shaded bounded feasible region in the first quadrant: a triangle-like region with vertices near (5,0), (10,0) and (0,16), bounded by two lines (one through (5,0) and (0,16), another steeper line) and the x-axis. See it in the official paper.
Try it first, then show our solutionSolution
1. If the minimum occurs at more than one point, it occurs all along a boundary edge. The edge through $(5, 0)$ and $(0, 15)$ is $3 x + y = 15$ .
2. $(2, 9)$ satisfies $3 (2) + 9 = 15$ , so it lies on that edge. Option (A).
  1 mark
Answer: $(2, 9)$ , option (A)
Where the marks are
- Identifies the edge $3 x + y = 15$ through $(5, 0)$ and checks that $(2, 9)$ lies on it (1 mark)
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Q18 · 1 mark · MCQCompetency-basedProbability
A person observed the first 4 digits of your 6-digit PIN. What is the probability that the person can guess your PIN?
1. (A) $\frac{1}{81}$
2. (B) $\frac{1}{100}$
3. (C) $\frac{1}{90}$
4. (D)1
Try it first, then show our solutionSolution
1. Only the last 2 digits are unknown, and each can be any of 10 digits.
2. $P = \frac{1}{10} \times \frac{1}{10} = \frac{1}{100}$ . Option (B).
  1 mark
Answer: $\frac{1}{100}$ , option (B)
Where the marks are
- Counts the 100 equally likely choices for the last two digits (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedInverse Trigonometric Functions
Assertion (A): Value of the expression $sin^{- 1} (\frac{3}{2}) + tan^{- 1} 1 - sec^{- 1} (2)$ is $\frac{π}{4}$ .
Reason (R): Principal value branch of $sin^{- 1} x$ is $[- \frac{π}{2}, \frac{π}{2}]$ and that of $sec^{- 1} x$ is $[0, π] - {\frac{π}{2}}$ .
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (C)(A) is true but (R) is false.
4. (D)(A) is false but (R) is true.
Try it first, then show our solutionSolution
1. $sin^{- 1} \frac{3}{2} + tan^{- 1} 1 - sec^{- 1} 2 = \frac{π}{3} + \frac{π}{4} - \frac{π}{4} = \frac{π}{3} \neq = \frac{π}{4}$ , so (A) is false.
2. The principal value branches stated in (R) are correct, so (R) is true. Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Evaluates the expression as $\frac{π}{3}$ and confirms that (R) is correct (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedVector Algebra
Assertion (A): Given two non-zero vectors $a$ and $b$ . If $r$ is another non-zero vector such that $r \times (a + b) = 0$ . Then $r$ is perpendicular to $a \times b$ .
Reason (R): The vector $(a + b)$ is perpendicular to the plane of $a$ and $b$
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (C)(A) is true but (R) is false.
4. (D)(A) is false but (R) is true.
Try it first, then show our solutionSolution
1. $r \times (a + b) = 0$ means $r ∥ (a + b)$ . Since $a + b$ lies in the plane of $a, b$ and $a \times b$ is normal to that plane, $r ⊥ a \times b$ . So (A) is true.
2. $a + b$ lies in the plane of $a$ and $b$ , so it is not perpendicular to that plane; (R) is false. Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- Shows $r ∥ (a + b)$ , which lies in the plane, so (A) is true and (R) is false (1 mark)
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Section B

VSA

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerInverse Trigonometric Functions
Evaluate $tan (tan^{- 1} (- 1) + \frac{π}{3})$
Try it first, then show our solutionSolution
1. The principal value branch of $tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ , so $tan^{- 1} (- 1) = - \frac{π}{4}$ . The expression becomes $tan (- \frac{π}{4} + \frac{π}{3}) = tan (\frac{π}{3} - \frac{π}{4})$ .
  ½ mark
2. Using $tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ : $tan (\frac{π}{3} - \frac{π}{4}) = \frac{tan \frac{π}{3} - tan \frac{π}{4}}{1 + tan \frac{π}{3} tan \frac{π}{4}} = \frac{3 - 1}{1 + 3}$ .
  1 mark
3. Rationalising: $\frac{( 3 - 1 ) ^{2}}{3 - 1} = \frac{4 - 2 3}{2} = 2 - 3$ .
  ½ mark
Answer: $\frac{3 - 1}{3 + 1} = 2 - 3$
Where the marks are
- Uses the principal value $tan^{- 1} (- 1) = - π /4$ (½ mark)
- Applies the $tan (A - B)$ formula with correct values (1 mark)
- Gets $\frac{3 - 1}{1 + 3}$ or $2 - 3$ (½ mark)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerInverse Trigonometric Functions
Find the domain of $cos^{- 1} (3 x - 2)$
Try it first, then show our solutionSolution
1. $cos^{- 1} t$ is defined only for $- 1 \leq t \leq 1$ , so we need $- 1 \leq 3 x - 2 \leq 1$ .
  ½ mark
2. Adding 2 throughout: $1 \leq 3 x \leq 3$ .
  ½ mark
3. Dividing by 3: $\frac{1}{3} \leq x \leq 1$ .
  ½ mark
4. Hence the domain of $cos^{- 1} (3 x - 2)$ is $[\frac{1}{3}, 1]$ .
  ½ mark
Answer: $[\frac{1}{3}, 1]$
Where the marks are
- Writes the condition $- 1 \leq 3 x - 2 \leq 1$ (½ mark)
- Simplifies to $1 \leq 3 x \leq 3$ (½ mark)
- Gets $\frac{1}{3} \leq x \leq 1$ (½ mark)
- States the domain as $[1/3, 1]$ (½ mark)
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Q22 · 2 marks · Very short answerContinuity and Differentiability
If $y = lo g tan (\frac{π}{4} + \frac{x}{2})$ , then prove that $\frac{d y}{d x} - sec x = 0$
Try it first, then show our solutionSolution
1. Differentiating w.r.t. $x$ by the chain rule: $\frac{d y}{d x} = \frac{1}{tan ( \frac{π}{4} + \frac{x}{2} )} \cdot sec^{2} (\frac{π}{4} + \frac{x}{2}) \cdot \frac{1}{2}$ .
  ½ mark
2. $= \frac{cos ( \frac{π}{4} + \frac{x}{2} )}{sin ( \frac{π}{4} + \frac{x}{2} )} \cdot \frac{1}{cos ^{2} ( \frac{π}{4} + \frac{x}{2} )} \cdot \frac{1}{2} = \frac{1}{2 sin ( \frac{π}{4} + \frac{x}{2} ) cos ( \frac{π}{4} + \frac{x}{2} )} = \frac{1}{sin ( \frac{π}{2} + x )} = \frac{1}{cos x}$ [ $2 sin A cos A = sin 2 A$ ].
  1 mark
3. So $\frac{d y}{d x} = sec x$ , i.e. $\frac{d y}{d x} - sec x = 0$ . Hence proved.
  ½ mark
Answer: $\frac{d y}{d x} = sec x$ , so $\frac{d y}{d x} - sec x = 0$
Where the marks are
- Correct chain-rule derivative (½ mark)
- Simplifies using $2 sin A cos A = sin 2 A$ to $1/ cos x$ (1 mark)
- Concludes $\frac{d y}{d x} - sec x = 0$ (½ mark)
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Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerIntegrals
Find: $\int \frac{( x - 3 )}{( x - 1 ) ^{3}} e^{x} d x$
Try it first, then show our solutionSolution
1. Write $x - 3 = (x - 1) - 2$ : $\int \frac{( x - 3 ) e ^{x}}{( x - 1 ) ^{3}} d x = \int e^{x} [\frac{1}{( x - 1 ) ^{2}} - \frac{2}{( x - 1 ) ^{3}}] d x$ . With $f (x) = \frac{1}{( x - 1 ) ^{2}}$ , $f^{'} (x) = - \frac{2}{( x - 1 ) ^{3}}$ , so the integrand is $e^{x} [f (x) + f^{'} (x)]$ .
  1 mark
2. Since $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ , the integral $= \frac{e ^{x}}{( x - 1 ) ^{2}} + C$ .
  1 mark
Answer: $\frac{e ^{x}}{( x - 1 ) ^{2}} + C$
Where the marks are
- Splits integrand into $e^{x} [f (x) + f^{'} (x)]$ form with $f (x) = 1/ (x - 1)^{2}$ (1 mark)
- Writes $\frac{e ^{x}}{( x - 1 ) ^{2}} + C$ (1 mark)
Open Q23(A) on its own page
OR
Q23(B) · 2 marks · Very short answerApplication of Integrals
Find out the area of shaded region in the enclosed figure.
Figure in the paper: Parabola $x^{2} = y$ with vertex O at origin, and the horizontal line $y = 4$ meeting the parabola at A (left) and B (right); N is the point where y=4 meets the Y-axis. The shaded region is the part between the parabola and the line y=4 on the right of the Y-axis (region ONB). See it in the official paper.
Try it first, then show our solutionSolution
1. The region ONB lies between the Y-axis and the parabola $x^{2} = y$ (right branch $x = y$ ), from $y = 0$ to $y = 4$ . So Area $= \int_{0}^{4} x d y = \int_{0}^{4} y d y$ .
  1 mark
2. $= [\frac{2}{3} y^{3/2}]_{0}^{4} = \frac{2}{3} (8 - 0) = \frac{16}{3}$ sq. units.
  1 mark
Answer: $\frac{16}{3}$ sq. units
Where the marks are
- Sets up area as $\int_{0}^{4} y d y$ (1 mark)
- Evaluates to $16/3$ sq. units (1 mark)
Open Q23(B) on its own page
Q24 · 2 marks · Very short answerContinuity and Differentiability
If $f (x + y) = f (x) f (y)$ for all $x, y \in R$ and $f (5) = 2$ , $f^{'} (0) = 3$ , then using the definition of derivatives, find $f^{'} (5)$ .
Try it first, then show our solutionSolution
1. Put $x = 0, y = 5$ : $f (5) = f (0) f (5) \Rightarrow 2 = 2 f (0) \Rightarrow f (0) = 1$ .
  ½ mark
2. By definition, $f^{'} (5) = lim_{h \to 0} \frac{f ( 5 + h ) - f ( 5 )}{h} = lim_{h \to 0} \frac{f ( 5 ) f ( h ) - f ( 5 )}{h}$ [ $f (x + y) = f (x) f (y)$ ] $= lim_{h \to 0} \frac{2 f ( h ) - 2}{h} = 2 lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h}$ [ $f (0) = 1$ ].
  1 mark
3. $= 2 f^{'} (0) = 2 \times 3 = 6$ .
  ½ mark
Answer: $f^{'} (5) = 6$
Where the marks are
- Finds $f (0) = 1$ (½ mark)
- Uses first-principle definition and functional equation to reach $2 lim \frac{f ( h ) - f ( 0 )}{h}$ (1 mark)
- Gets $f^{'} (5) = 2 f^{'} (0) = 6$ (½ mark)
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Q25 · 2 marks · Very short answerVector Algebra
The two vectors $\hat{i} + \hat{j} + \hat{k}$ and $3 \hat{i} - \hat{j} + 3 \hat{k}$ represent the two sides $O A$ and $O B$ , respectively of a $Δ O A B$ , where $O$ is the origin. The point $P$ lies on $A B$ such that $O P$ is a median. Find the area of the parallelogram formed by the two adjacent sides as $O A$ and $O P$ .
Try it first, then show our solutionSolution
1. P is the midpoint of AB, so $O P = \frac{1}{2} (O A + O B) = \frac{1}{2} (4 \hat{i} + 4 \hat{k}) = 2 \hat{i} + 2 \hat{k}$ .
  ½ mark
2. Area of parallelogram with adjacent sides OA, OP $= ∣ O A \times O P ∣$ .
  ½ mark
3. $O A \times O P = \hat{i} 12 \hat{j} 10 \hat{k} 12 = \hat{i} (2 - 0) - \hat{j} (2 - 2) + \hat{k} (0 - 2) = 2 \hat{i} - 2 \hat{k}$ .
  ½ mark
4. Area $= 4 + 4 = 22$ square units.
  ½ mark
Answer: $22$ sq. units
Where the marks are
- Finds $O P = 2 \hat{i} + 2 \hat{k}$ (½ mark)
- Uses area $= ∣ O A \times O P ∣$ (½ mark)
- Computes cross product $2 \hat{i} - 2 \hat{k}$ (½ mark)
- Area $22$ sq. units (½ mark)
Open Q25 on its own page

Section C

Answer one of these 2 alternatives.
Q26(A) · 3 marks · Short answerContinuity and Differentiability
If $x^{y} = e^{x - y}$ prove that $\frac{d y}{d x} = \frac{l o g x}{( l o g ( x e ) ) ^{2}}$ and hence find its value at $x = e$ .
Try it first, then show our solutionSolution
1. Taking log: $y lo g x = (x - y) lo g e = x - y$ [ $lo g e = 1$ ] $\Rightarrow y (1 + lo g x) = x \Rightarrow y = \frac{x}{1 + lo g x}$ .
  1 mark
2. By quotient rule: $\frac{d y}{d x} = \frac{( 1 + lo g x ) \cdot 1 - x \cdot \frac{1}{x}}{( 1 + lo g x ) ^{2}} = \frac{lo g x}{( 1 + lo g x ) ^{2}} = \frac{lo g x}{( lo g e + lo g x ) ^{2}} = \frac{lo g x}{( lo g ( x e ) ) ^{2}}$ . Hence proved.
  1 mark
3. At $x = e$ : $\frac{d y}{d x} = \frac{lo g e}{( lo g e ^{2} ) ^{2}} = \frac{1}{( 2 ) ^{2}} = \frac{1}{4}$ .
  1 mark
Answer: $\frac{d y}{d x} = \frac{l o g x}{( l o g ( x e ) ) ^{2}}$ ; value at $x = e$ is $\frac{1}{4}$
Where the marks are
- Takes logs and expresses $y = \frac{x}{1 + l o g x}$ (1 mark)
- Differentiates and shows the required form (1 mark)
- Evaluates $\frac{1}{4}$ at $x = e$ (1 mark)
Open Q26(A) on its own page
OR
Q26(B) · 3 marks · Short answerContinuity and Differentiability
If $x = a (θ - sin θ)$ , $y = a (1 - cos θ)$ find $\frac{d ^{2} y}{d x ^{2}}$ .
Try it first, then show our solutionSolution
1. $\frac{d x}{d θ} = a (1 - cos θ)$ , $\frac{d y}{d θ} = a sin θ$ .
  1 mark
2. $\frac{d y}{d x} = \frac{a sin θ}{a ( 1 - cos θ )} = \frac{2 sin \frac{θ}{2} cos \frac{θ}{2}}{2 sin ^{2} \frac{θ}{2}} = cot \frac{θ}{2}$ .
  1 mark
3. $\frac{d ^{2} y}{d x ^{2}} = - \frac{1}{2} csc^{2} \frac{θ}{2} \cdot \frac{d θ}{d x} = - \frac{1}{2} csc^{2} \frac{θ}{2} \cdot \frac{1}{2 a sin ^{2} \frac{θ}{2}} = - \frac{1}{4 a} csc^{4} \frac{θ}{2}$ .
  1 mark
Answer: $\frac{d ^{2} y}{d x ^{2}} = - \frac{1}{4 a} csc^{4} \frac{θ}{2}$
Where the marks are
- Finds $d x / d θ$ and $d y / d θ$ (1 mark)
- Obtains $d y / d x = cot (θ /2)$ (1 mark)
- Differentiates again using $d θ / d x$ to get $- \frac{1}{4 a} csc^{4} (θ /2)$ (1 mark)
Open Q26(B) on its own page
Q27 · 3 marks · Short answerCompetency-basedApplication of Derivatives
A spherical ball of ice melts in such a way that the rate at which its volume decreases at any instant is directly proportional to its surface area. Prove that the radius of the ice ball decreases at a constant rate.
Try it first, then show our solutionSolution
1. Let $r$ be the radius at time $t$ . Volume $V = \frac{4}{3} π r^{3}$ …(1), surface area $S = 4 π r^{2}$ …(2).
  ½ mark
2. Given the volume decreases at a rate proportional to $S$ : $\frac{d V}{d t} = - k S$ , $k > 0$ a constant …(3).
  ½ mark
3. Differentiating (1) w.r.t. $t$ : $\frac{d V}{d t} = \frac{4}{3} π \cdot 3 r^{2} \frac{d r}{d t} = 4 π r^{2} \frac{d r}{d t}$ …(4).
  1 mark
4. From (3) and (4): $- k S = 4 π r^{2} \frac{d r}{d t} = S \frac{d r}{d t}$ [using (2)].
  ½ mark
5. $\Rightarrow \frac{d r}{d t} = - k$ , a constant. So the radius decreases at a constant rate. Hence proved.
  ½ mark
Answer: $\frac{d r}{d t} = - k$ (constant), so the radius decreases at a constant rate
Where the marks are
- Writes $V$ and $S$ in terms of $r$ (½ mark)
- Forms $\frac{d V}{d t} = - k S$ (½ mark)
- Differentiates $V$ to get $4 π r^{2} \frac{d r}{d t}$ (1 mark)
- Equates using $S = 4 π r^{2}$ (½ mark)
- Concludes $\frac{d r}{d t} = - k$ (½ mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(A) · 3 marks · Short answerApplication of Integrals
Sketch the graph $y = ∣ x + 1∣$ . Evaluate $\int_{- 4}^{2} ∣ x + 1∣ d x$ . What does the value of this integral represent on the graph?
Try it first, then show our solutionSolution
1. Graph: $y = ∣ x + 1∣ = - x - 1$ for $x < - 1$ and $x + 1$ for $x \geq - 1$ . It is a V-shape with vertex at $(- 1, 0)$ , passing through $(- 4, 3)$ , $(0, 1)$ and $(2, 3)$ . Draw the lines $x = - 4$ and $x = 2$ and shade the two triangles between the V and the x-axis from $x = - 4$ to $x = 2$ .
  1 mark
2. $\int_{- 4}^{2} ∣ x + 1∣ d x = \int_{- 4}^{- 1} (- x - 1) d x + \int_{- 1}^{2} (x + 1) d x$ .
  ½ mark
3. $= [- \frac{( x + 1 ) ^{2}}{2}]_{- 4}^{- 1} + [\frac{( x + 1 ) ^{2}}{2}]_{- 1}^{2}$ .
  ½ mark
4. $= - (0 - \frac{9}{2}) + (\frac{9}{2} - 0) = 9$ .
  ½ mark
5. The value 9 is the area (in sq. units) of the region bounded by $y = ∣ x + 1∣$ , the x-axis and the lines $x = - 4$ and $x = 2$ (the shaded region).
  ½ mark
Answer: Integral $= 9$ ; it is the area bounded by $y = ∣ x + 1∣$ , the x-axis, $x = - 4$ and $x = 2$
Where the marks are
- Correct sketch of $y = ∣ x + 1∣$ with region shaded (1 mark)
- Splits the integral at $x = - 1$ (½ mark)
- Correct antiderivatives (½ mark)
- Value 9 (½ mark)
- Interprets as area of shaded region (½ mark)
Open Q28(A) on its own page
OR
Q28(B) · 3 marks · Short answerApplication of Integrals
Using integration find the area of the region ${(x, y) : x^{2} - 4 y \leq 0, y - x \leq 0}$
Try it first, then show our solutionSolution
1. The region lies above the parabola $x^{2} = 4 y$ (i.e. $y \geq x^{2} /4$ ) and below the line $y = x$ . Solving $\frac{x ^{2}}{4} = x$ gives $x = 0, 4$ , so they meet at $(0, 0)$ and $(4, 4)$ . Sketch: upward parabola with vertex O and the line $y = x$ ; shade the region between them from $x = 0$ to $x = 4$ .
  1 mark
2. Required area $= \int_{0}^{4} x d x - \int_{0}^{4} \frac{x ^{2}}{4} d x$ .
  1 mark
3. $= [\frac{x ^{2}}{2}]_{0}^{4} - \frac{1}{12} [x^{3}]_{0}^{4}$ .
  ½ mark
4. $= 8 - \frac{64}{12} = 8 - \frac{16}{3} = \frac{8}{3}$ sq. units.
  ½ mark
Answer: $\frac{8}{3}$ sq. units
Where the marks are
- Finds intersection points and sketches/shades region (1 mark)
- Sets up area as $\int_{0}^{4} (x - x^{2} /4) d x$ (1 mark)
- Integrates correctly (½ mark)
- Area $8/3$ sq. units (½ mark)
Open Q28(B) on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerThree Dimensional Geometry
Find the distance of the point $(2, - 1, 3)$ from the line $r = (2 \hat{i} - \hat{j} + 2 \hat{k}) + μ (3 \hat{i} + 6 \hat{j} + 2 \hat{k})$ measured parallel to the z-axis.
Try it first, then show our solutionSolution
1. Line through $(2, - 1, 3)$ parallel to the z-axis: $r = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ \hat{k}$ .
  1 mark
2. Any point on it: $P (2, - 1, 3 + λ)$ .
  ½ mark
3. Any point on the given line: $Q (2 + 3 μ, - 1 + 6 μ, 2 + 2 μ)$ .
  ½ mark
4. For intersection, $P = Q$ : $2 + 3 μ = 2 \Rightarrow μ = 0$ (also $- 1 + 6 μ = - 1$ holds). Point of intersection $(2, - 1, 2)$ .
  ½ mark
5. Required distance $= 0^{2} + 0^{2} + (3 - 2)^{2} = 1$ unit.
  ½ mark
Answer: 1 unit
Where the marks are
- Equation of line through the point parallel to z-axis (1 mark)
- General point on this line (½ mark)
- General point on given line (½ mark)
- Finds $μ = 0$ and point $(2, - 1, 2)$ (½ mark)
- Distance 1 unit (½ mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerThree Dimensional Geometry
Find the point of intersection of the line $r = (3 \hat{i} + \hat{k}) + μ (\hat{i} + \hat{j} + \hat{k})$ and the line through $(2, - 1, 1)$ parallel to the z-axis. How far is this point from the z-axis?
Try it first, then show our solutionSolution
1. Line through $(2, - 1, 1)$ parallel to the z-axis: $r = (2 \hat{i} - \hat{j} + \hat{k}) + λ \hat{k}$ ; any point $P (2, - 1, 1 + λ)$ .
  1 mark
2. Any point on the given line: $A (3 + μ, μ, 1 + μ)$ . Equating: $3 + μ = 2$ and $μ = - 1$ $\Rightarrow μ = - 1$ (consistent).
  1 mark
3. Point of intersection: $(3 - 1, - 1, 1 - 1) = (2, - 1, 0)$ .
  ½ mark
4. Distance of $(2, - 1, 0)$ from the z-axis $= x^{2} + y^{2} = 4 + 1 = 5$ units.
  ½ mark
Answer: Point $(2, - 1, 0)$ ; distance from z-axis $5$ units
Where the marks are
- Equation/general point of line parallel to z-axis (1 mark)
- Equates general points and gets $μ = - 1$ (1 mark)
- Point of intersection $(2, - 1, 0)$ (½ mark)
- Distance $5$ units (½ mark)
Open Q29(B) on its own page
Q30 · 3 marks · Short answerLinear Programming
Solve graphically:
Maximise $Z = 2 x + y$ subject to
$x + y \leq 1200$
$x + y \geq 600$
$y \leq \frac{x}{2}$
$x \geq 0, y \geq 0$ .
Try it first, then show our solutionSolution
1. Graph: draw $x + y = 1200$ (through $(1200, 0), (0, 1200)$ ), $x + y = 600$ (through $(600, 0), (0, 600)$ ) and $y = \frac{x}{2}$ (through $(0, 0), (800, 400)$ ). The feasible region lies between the two parallel lines, below $y = x /2$ , in the first quadrant: the quadrilateral ABCD (bounded) shown shaded. Intersections: $x + y = 1200, y = x /2 \Rightarrow C (800, 400)$ ; $x + y = 600, y = x /2 \Rightarrow D (400, 200)$ .
  1½ marks
2. Corner points: $A (600, 0)$ , $B (1200, 0)$ , $C (800, 400)$ , $D (400, 200)$ .
  ½ mark
3. $Z = 2 x + y$ : $Z_{A} = 1200$ , $Z_{B} = 2400$ , $Z_{C} = 2000$ , $Z_{D} = 1000$ .
  ½ mark
4. Maximum $Z = 2400$ at $x = 1200, y = 0$ .
  ½ mark
Answer: Max $Z = 2400$ at $(1200, 0)$
Where the marks are
- Correct graph with feasible region shaded (1½ marks)
- Correct corner points (½ mark)
- Values of Z at corners (½ mark)
- Maximum 2400 at (1200, 0) (½ mark)
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Q31 · 3 marks · Short answerCompetency-basedProbability
Two students Mehul and Rashi are seeking admission in a college. The probability that Mehul is selected is 0.4 and the probability of selection of exactly one of the them is 0.5. Chances of selection of them is independent of each other. Find the chances of selection of Rashi. Also find the probability of selection of at least one of them.
Try it first, then show our solutionSolution
1. Let A: Mehul is selected, B: Rashi is selected; A, B independent. $P (A) = 0.4$ , $P (B) = x$ . Exactly one selected: $P (A \cap \overset{ˉ}{B}) + P (\overset{ˉ}{A} \cap B) = 0.5$ .
  1 mark
2. $P (A) P (\overset{ˉ}{B}) + P (\overset{ˉ}{A}) P (B) = 0.5 \Rightarrow 0.4 (1 - x) + 0.6 x = 0.5 \Rightarrow 0.2 x = 0.1 \Rightarrow x = 0.5$ . So $P (Rashi selected) = 0.5$ .
  1 mark
3. $P (at least one) = 1 - P (\overset{ˉ}{A} \cap \overset{ˉ}{B}) = 1 - P (\overset{ˉ}{A}) P (\overset{ˉ}{B}) = 1 - 0.6 \times 0.5 = 0.7$ .
  1 mark
Answer: Rashi: 0.5; at least one: 0.7
Where the marks are
- Defines events and writes the 'exactly one' condition (1 mark)
- Uses independence to get $P (B) = 0.5$ (1 mark)
- Probability of at least one $= 0.7$ (1 mark)
Open Q31 on its own page

Section D

Q32 · 5 marks · Long answerMatrices
For two matrices $A = 32 - 2 - 6 - 5 4 - 1 - 1 1$ and $B = 102 - 2 - 1 0 - 1 - 1 3$ , find the product $A B$ and hence solve the system of equations:
$3 x - 6 y - z = 3$
$2 x - 5 y - z + 2 = 0$
$- 2 x + 4 y + z = 5$
Try it first, then show our solutionSolution
1. $A B = 32 - 2 - 6 - 5 4 - 1 - 1 1 102 - 2 - 1 0 - 1 - 1 3 = 3 - 0 - 2 2 - 0 - 2 - 2 + 0 + 2 - 6 + 6 + 0 - 4 + 5 + 0 4 - 4 + 0 - 3 + 6 - 3 - 2 + 5 - 3 2 - 4 + 3 = 100010001 = I$ .
  1 mark
2. Hence $A^{- 1} = B$ (and $B^{- 1} = A$ ).
  ½ mark
3. The system is $3 x - 6 y - z = 3$ , $2 x - 5 y - z = - 2$ , $- 2 x + 4 y + z = 5$ , i.e. $A X = C$ .
  ½ mark
4. where $X = x y z$ , $C = 3 - 2 5$ .
  ½ mark
5. $∣ A ∣ = 3 (- 5 + 4) + 6 (2 - 2) - 1 (8 - 10) = - 3 + 0 + 2 = - 1 \neq = 0$ , so the system has a unique solution.
  ½ mark
6. $X = A^{- 1} C = B C$ .
  ½ mark
7. $B C = 102 - 2 - 1 0 - 1 - 1 3 3 - 2 5 = 3 + 4 - 5 0 + 2 - 5 6 + 0 + 15 = 2 - 3 21$ .
  1 mark
8. Therefore $x = 2, y = - 3, z = 21$ .
  ½ mark
Answer: $A B = I$ ; $x = 2, y = - 3, z = 21$
Where the marks are
- Computes $A B = I$ (1 mark)
- Concludes $A^{- 1} = B$ (½ mark)
- Writes system as $A X = C$ (½ mark)
- Correct $X$ and $C$ (½ mark)
- $∣ A ∣ \neq = 0$ , unique solution (½ mark)
- Uses $X = B C$ (½ mark)
- Computes $B C$ (1 mark)
- States $x = 2, y = - 3, z = 21$ (½ mark)
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Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerIntegrals
Evaluate: $\int_{0}^{1} \frac{l o g ( 1 + x )}{1 + x ^{2}} d x$
Try it first, then show our solutionSolution
1. Put $x = tan θ$ , $d x = sec^{2} θ d θ$ ; $x = 0 \to θ = 0$ , $x = 1 \to θ = \frac{π}{4}$ . $I = \int_{0}^{π /4} \frac{lo g ( 1 + tan θ )}{1 + tan ^{2} θ} sec^{2} θ d θ$ .
  ½ mark
2. $I = \int_{0}^{π /4} lo g (1 + tan θ) d θ = \int_{0}^{π /4} lo g [1 + tan (\frac{π}{4} - θ)] d θ$ [ $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ ].
  1 mark
3. $= \int_{0}^{π /4} lo g [1 + \frac{1 - tan θ}{1 + tan θ}] d θ = \int_{0}^{π /4} lo g \frac{2}{1 + tan θ} d θ$ .
  1 mark
4. $= \int_{0}^{π /4} lo g 2 d θ - \int_{0}^{π /4} lo g (1 + tan θ) d θ = lo g 2 [θ]_{0}^{π /4} - I$ .
  1 mark
5. $\Rightarrow 2 I = \frac{π}{4} lo g 2$ .
  1 mark
6. $\Rightarrow I = \frac{π}{8} lo g 2$ .
  ½ mark
Answer: $\frac{π}{8} lo g 2$
Where the marks are
- Substitution $x = tan θ$ with changed limits (½ mark)
- Applies property $f (a - x)$ (1 mark)
- Simplifies to $lo g \frac{2}{1 + t a n θ}$ (1 mark)
- Splits to $lo g 2 \cdot \frac{π}{4} - I$ (1 mark)
- $2 I = \frac{π}{4} lo g 2$ (1 mark)
- $I = \frac{π}{8} lo g 2$ (½ mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerIntegrals
Find $\int \frac{( 3 s i n θ - 2 ) c o s θ}{5 - c o s ^{2} θ - 4 s i n θ} d θ$
Try it first, then show our solutionSolution
1. $I = \int \frac{( 3 sin θ - 2 ) cos θ}{5 - ( 1 - sin ^{2} θ ) - 4 sin θ} d θ$ .
  ½ mark
2. Put $sin θ = t$ , $cos θ d θ = d t$ : $I = \int \frac{3 t - 2}{t ^{2} - 4 t + 4} d t = \int \frac{3 t - 2}{( t - 2 ) ^{2}} d t$ .
  1 mark
3. Let $\frac{3 t - 2}{( t - 2 ) ^{2}} = \frac{A}{t - 2} + \frac{B}{( t - 2 ) ^{2}} \Rightarrow 3 t - 2 = A (t - 2) + B$ . Comparing coefficients: $A = 3$ ; $- 2 A + B = - 2 \Rightarrow B = 4$ .
  1 mark
4. $I = \int \frac{3}{t - 2} d t + \int \frac{4}{( t - 2 ) ^{2}} d t = 3 lo g ∣ t - 2∣ - \frac{4}{t - 2} + C$ .
  2 marks
5. $I = 3 lo g ∣ sin θ - 2∣ - \frac{4}{sin θ - 2} + C$ .
  ½ mark
Answer: $3 lo g ∣ sin θ - 2∣ - \frac{4}{sin θ - 2} + C$
Where the marks are
- Writes $cos^{2} θ = 1 - sin^{2} θ$ (½ mark)
- Substitutes $t = sin θ$ and gets $\frac{3 t - 2}{( t - 2 ) ^{2}}$ (1 mark)
- Partial fractions with $A = 3$ , $B = 4$ (½ + ½ marks)
- Integrates both terms correctly (1 + 1 marks)
- Final answer in $θ$ (½ mark)
Open Q33(B) on its own page
Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerDifferential Equations
Solve the differential equation: $y + \frac{d}{d x} (x y) = x (sin x + x)$
Try it first, then show our solutionSolution
1. Expand the derivative of the product:
  $y + (x \frac{d y}{d x} + y) = x (sin x + x) \Rightarrow x \frac{d y}{d x} + 2 y = x (sin x + x)$ .
  Divide by $x$ :
  $\frac{d y}{d x} + \frac{2}{x} y = sin x + x$ .
  1 mark
2. This is a linear equation $\frac{d y}{d x} + P y = Q$ with $P = \frac{2}{x}$ and $Q = sin x + x$ .
  Integrating factor:
  $I.F. = e^{\int \frac{2}{x} d x} = e^{2 l o g x} = x^{2}$ .
  1 mark
3. The solution is $y \cdot I.F. = \int Q \cdot I.F. d x$ , so
  $y x^{2} = \int (sin x + x) x^{2} d x = \int x^{2} sin x d x + \int x^{3} d x$ .
  1 mark
4. Integrate $\int x^{2} sin x d x$ by parts:
  $\int x^{2} sin x d x = - x^{2} cos x + 2 \int x cos x d x$ .
  Also $\int x^{3} d x = \frac{x ^{4}}{4}$ .
  1 mark
5. By parts again, $\int x cos x d x = x sin x + cos x$ .
  Hence the general solution is
  $y x^{2} = - x^{2} cos x + 2 (x sin x + cos x) + \frac{x ^{4}}{4} + C .$
  1 mark
Answer: $x^{2} y = - x^{2} cos x + 2 x sin x + 2 cos x + \frac{x ^{4}}{4} + C$
Where the marks are
- Reduces the equation to $\frac{d y}{d x} + \frac{2}{x} y = sin x + x$ (1 mark)
- Identifies $P$ , $Q$ and finds I.F. $= x^{2}$ (1 mark)
- Writes $y x^{2} = \int (sin x + x) x^{2} d x$ (1 mark)
- Applies integration by parts once and integrates $x^{3}$ correctly (1 mark)
- Completes the integration and states the general solution with constant (1 mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerDifferential Equations
Find the particular solution of the differential equation:
$2 y e^{x / y} d x + (y - 2 x e^{x / y}) d y = 0$ given that $y (0) = 1$
Try it first, then show our solutionSolution
1. Rearrange the equation:
  $\frac{d x}{d y} = \frac{2 x e ^{x / y} - y}{2 y e ^{x / y}} = \frac{2 \frac{x}{y} e ^{x / y} - 1}{2 e ^{x / y}} .$
  The right side is a function of $\frac{x}{y}$ only, so the equation is homogeneous.
  1 mark
2. Put $x = v y$ , so that $\frac{d x}{d y} = v + y \frac{d v}{d y}$ .
  Then $v + y \frac{d v}{d y} = \frac{2 v e ^{v} - 1}{2 e ^{v}}$ .
  1 mark
3. Simplify:
  $y \frac{d v}{d y} = \frac{2 v e ^{v} - 1 - 2 v e ^{v}}{2 e ^{v}} = \frac{- 1}{2 e ^{v}}$ .
  Separate the variables:
  $2 e^{v} d v = - \frac{d y}{y}$ .
  1 mark
4. Integrate both sides: $2 e^{v} = - lo g ∣ y ∣ + C$ .
  Substitute back $v = \frac{x}{y}$ :
  $2 e^{x / y} + lo g ∣ y ∣ = C$ .
  1 mark
5. Use $y (0) = 1$ : $2 e^{0} + lo g 1 = C \Rightarrow C = 2$ .
  The particular solution is
  $2 e^{x / y} + lo g ∣ y ∣ = 2.$
  1 mark
Answer: $2 e^{x / y} + lo g ∣ y ∣ = 2$
Where the marks are
- Writes $\frac{d x}{d y}$ as a function of $x / y$ and recognises the equation as homogeneous (1 mark)
- Substitutes $x = v y$ with $\frac{d x}{d y} = v + y \frac{d v}{d y}$ (1 mark)
- Simplifies to the separable form $2 e^{v} d v = - \frac{d y}{y}$ (1 mark)
- Integrates to get $2 e^{x / y} + lo g ∣ y ∣ = C$ (1 mark)
- Uses $y (0) = 1$ to get $C = 2$ and states the particular solution (1 mark)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerThree Dimensional Geometry
The two lines $\frac{x - 1}{3} = - y, z + 1 = 0$ and $\frac{- x}{2} = \frac{y + 1}{2} = z + 2$ intersect at a point whose y-coordinate is 1. Find the co-ordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.
Try it first, then show our solutionSolution
1. Write the first line in symmetric form: $\frac{x - 1}{3} = \frac{y - 0}{- 1} = \frac{z + 1}{0} = λ$ .
  A general point on it is $(3 λ + 1, - λ, - 1)$ .
  ½ mark
2. The point of intersection has $y = 1$ , so $- λ = 1 \Rightarrow λ = - 1$ .
  1 mark
3. The point of intersection is $(3 (- 1) + 1, 1, - 1) = (- 2, 1, - 1)$ .
  Check on the second line: $\frac{- ( - 2 )}{2} = 1$ , $\frac{1 + 1}{2} = 1$ and $- 1 + 2 = 1$ , so it lies on both lines.
  ½ mark
4. Direction of the first line: $m = 3 \hat{i} - \hat{j} + 0 \hat{k}$ .
  1 mark
5. Rewrite the second line as $\frac{x}{- 2} = \frac{y + 1}{2} = \frac{z + 2}{1}$ .
  Its direction is $n = - 2 \hat{i} + 2 \hat{j} + \hat{k}$ .
  ½ mark
6. A line perpendicular to both has direction $m \times n$ :
  $m \times n = \hat{i} 3 - 2 \hat{j} - 1 2 \hat{k} 01 = \hat{i} (- 1 - 0) - \hat{j} (3 - 0) + \hat{k} (6 - 2) = - \hat{i} - 3 \hat{j} + 4 \hat{k} .$
  1 mark
7. The required line passes through $(- 2, 1, - 1)$ with this direction:
  $r = (- 2 \hat{i} + \hat{j} - \hat{k}) + μ (- \hat{i} - 3 \hat{j} + 4 \hat{k}) .$
  ½ mark
Answer: Point of intersection $(- 2, 1, - 1)$ ; line $r = (- 2 \hat{i} + \hat{j} - \hat{k}) + μ (- \hat{i} - 3 \hat{j} + 4 \hat{k})$
Where the marks are
- Writes the general point $(3 λ + 1, - λ, - 1)$ of the first line (½ mark)
- Uses $y = 1$ to get $λ = - 1$ (1 mark)
- Obtains the point $(- 2, 1, - 1)$ (½ mark)
- Direction of the first line $3 \hat{i} - \hat{j}$ (1 mark)
- Direction of the second line $- 2 \hat{i} + 2 \hat{j} + \hat{k}$ (½ mark)
- Computes $m \times n = - \hat{i} - 3 \hat{j} + 4 \hat{k}$ (1 mark)
- States the vector equation of the required line (½ mark)
Open Q35 on its own page

Section E

Case study

Q36 · 4 marks · Case-basedCompetency-basedRelations and Functions
Case Study -1
A city's traffic management department is planning to optimize traffic flow by analyzing the connectivity between various traffic signals. The city has five major spots labelled $A, B, C, D$ and $E$ .
The department has collected the following data regarding one-way traffic flow between spots:
1. Traffic flows from $A$ to $B$ , $A$ to $C$ , and $A$ to $D$ .
2. Traffic flows from $B$ to $C$ and $B$ to $E$ .
3. Traffic flows from $C$ to $E$ .
4. Traffic flows from $D$ to $E$ and $D$ to $C$ .
The department wants to represent and analyze this data using relations and functions. Use the given data to answer the following questions:
1. I
  Is the traffic flow reflexive? Justify.
  1 mark
2. II
  Is the traffic flow transitive? Justify.
  1 mark
3. III A
  Represent the relation describing the traffic flow as a set of ordered pairs. Also state the domain and range of the relation.
  2 marks
4. OR
  III B
  Does the traffic flow represent a function? Justify your answer.
  2 marks
Figure in the paper: Illustration of a road intersection with traffic signals labelled A, C, B, D, E and white arrows painted on the road showing directions of traffic. See it in the official paper.
Try it first, then show our solutionSolution
1. I. The relation is $R$ = "traffic flows from first spot to second".
  It is not reflexive, because $(A, A) \in / R$ ; in fact no spot has traffic flowing to itself.
  1 mark
2. II. It is not transitive, because $(A, B) \in R$ and $(B, E) \in R$ , but $(A, E) \in / R$ .
  1 mark
3. III A. The relation as a set of ordered pairs is
  $R = {(A, B), (A, C), (A, D), (B, C), (B, E), (C, E), (D, E), (D, C)} .$
  1 mark
4. III A (contd.) The first elements give the domain $= {A, B, C, D}$ .
  The second elements give the range $= {B, C, D, E}$ .
  1 mark
5. III B. No, the traffic flow does not represent a function.
  1 mark
6. III B (contd.) Justification: in a function each element of the domain has exactly one image. Here $A$ is related to three spots $B$ , $C$ and $D$ , so $A$ has three images.
  (Also $E$ has no image. Similarly $B$ and $D$ each have two images.)
  1 mark
Answer: I. Not reflexive, since $(A, A) \in / R$ . II. Not transitive, since $(A, B), (B, E) \in R$ but $(A, E) \in / R$ . III A. $R = {(A, B), (A, C), (A, D), (B, C), (B, E), (C, E), (D, E), (D, C)}$ ; domain ${A, B, C, D}$ , range ${B, C, D, E}$ . III B. Not a function, as $A$ has three images.
Where the marks are
- States not reflexive with reason $(A, A) \in / R$ (1 mark)
- States not transitive with the counterexample $(A, B), (B, E) \in R$ but $(A, E) \in / R$ (1 mark)
- III A: writes $R$ correctly as ordered pairs (1 mark)
- III A: domain ${A, B, C, D}$ (½ mark) and range ${B, C, D, E}$ (½ mark)
- III B: says it is not a function (1 mark), justified by $A$ having more than one image (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
Case Study -2
LED bulbs are energy-efficient because they use significantly less electricity than traditional bulbs while producing the same amount of light. They convert more energy into light rather than heat, reducing waste. Additionally, their long lifespan means fewer replacements, saving resources and money over time.
A company manufactures a new type of energy-efficient LED bulb. The cost of production and the revenue generated by selling x bulbs (in an hour) are modelled as $C (x) = 0.5 x^{2} - 10 x + 150$ and $R (x) = - 0.3 x^{2} + 20 x$ respectively, where $C (x)$ and $R (x)$ are both in ₹.
To maximize the profit, the company needs to analyze these functions using calculus. Use the given models to answer the following questions:
1. I
  Derive the profit function $P (x)$
  1 mark
2. II
  Find the critical points of $P (x)$ .
  1 mark
3. III A
  Determine whether the critical points correspond to a maximum or a minimum profit by using the second derivative test.
  2 marks
4. OR
  III B
  Identify the possible practical value of $x$ (i.e., the number of bulbs that can realistically be produced and sold) that can maximize the profit, if the resources available and the expenditure on machines allows to produce minimum 10 but not more than 18 bulbs per hour. Also calculate the maximum profit.
  2 marks
Figure in the paper: Photograph of a factory assembly line where workers in blue uniforms, safety glasses and gloves assemble LED bulbs on conveyor belts. See it in the official paper.
Try it first, then show our solutionSolution
1. I. Profit is revenue minus cost:
  $P (x) = R (x) - C (x) = (- 0.3 x^{2} + 20 x) - (0.5 x^{2} - 10 x + 150)$ .
  So $P (x) = - 0.8 x^{2} + 30 x - 150$ .
  1 mark
2. II. Differentiate: $P^{'} (x) = - 1.6 x + 30$ .
  For critical points, $P^{'} (x) = 0 \Rightarrow x = \frac{30}{1.6} = 18.75$ .
  So $x = 18.75$ is the only critical point.
  1 mark
3. III A. Differentiate again: $P^{''} (x) = - 1.6$ , so $P^{''} (18.75) = - 1.6$ .
  1 mark
4. III A (contd.) Since $P^{''} (18.75) < 0$ , by the second derivative test $x = 18.75$ gives a maximum profit.
  1 mark
5. III B. Production must satisfy $10 \leq x \leq 18$ , and $x$ must be a whole number of bulbs.
  On $[10, 18]$ we have $P^{'} (x) = - 1.6 x + 30 > 0$ , because $x < 18.75$ . So $P$ is increasing there.
  Hence the maximum profit occurs at the largest allowed value, $x = 18$ bulbs.
  1 mark
6. III B (contd.) Maximum profit:
  $P (18) = - 0.8 (18)^{2} + 30 (18) - 150 = - 259.2 + 540 - 150 = 130.80$ .
  So the maximum profit is ₹130.80.
  1 mark
Answer: I. $P (x) = - 0.8 x^{2} + 30 x - 150$ . II. $x = 18.75$ . III A. $P^{''} = - 1.6 < 0$ , so maximum profit. III B. $x = 18$ bulbs, maximum profit ₹130.80.
Where the marks are
- Obtains $P (x) = - 0.8 x^{2} + 30 x - 150$ (1 mark)
- Solves $P^{'} (x) = - 1.6 x + 30 = 0$ to get $x = 18.75$ (1 mark)
- III A: finds $P^{''} (x) = - 1.6$ (1 mark) and concludes maximum since $P^{''} < 0$ (1 mark)
- III B: identifies $x = 18$ as the practical value (1 mark) and computes the maximum profit ₹130.80 (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
Case Study -3
Excessive use of screens can result in vision problems, obesity, sleep disorders, anxiety, low retention problems and can impede social and emotional comprehension and expression. It is essential to be mindful of the amount of time we spend on screens and to reduce our screen-time by taking regular breaks, setting time limits, and engaging in non-screen-based activities.
In a class of students of the age group 14 to 17, the students were categorised into three groups according to a feedback form filled by them. The first group constituted of the students who spent more than 4 hours per day on the mobile screen or the gaming screens, while the second group spent 2 to 4 hours /day on the same activities. The third group spent less than 2 hours /day on the same. The first group with the high screen time is 60% of all the students, whereas the second group with moderate screen time is 30% and the third group with low screen time is only 10% of the total number of students. It was observed that 80% students of first group faced severe anxiety and low retention issues, with 70% of second group, and 30% of third group having the same symptoms.
1. I
  What is the total percentage of students who suffer from anxiety and low retention issues in the class?
  2 marks
2. II
  A student is selected at random, and he is found to suffer from anxiety and low retention issues. What is the probability that he/she spends screen time more than 4 hours per day?
  2 marks
Figure in the paper: Illustration of three children sitting on a sofa using mobile phones, with a television showing cartoon characters behind them. See it in the official paper.
Try it first, then show our solutionSolution
1. Define the events:
  $E_{1}$ : the student spends more than 4 h/day on screens
  $E_{2}$ : the student spends 2–4 h/day on screens
  $E_{3}$ : the student spends less than 2 h/day on screens
  $A$ : the student has anxiety and low retention issues
  Then $P (E_{1}) = 0.6$ , $P (E_{2}) = 0.3$ , $P (E_{3}) = 0.1$ .
  Also $P (A ∣ E_{1}) = 0.8$ , $P (A ∣ E_{2}) = 0.7$ , $P (A ∣ E_{3}) = 0.3$ .
2. I. By the theorem of total probability:
  $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) + P (E_{3}) P (A ∣ E_{3})$ .
  So $P (A) = 0.6 \times 0.8 + 0.3 \times 0.7 + 0.1 \times 0.3$ .
  1 mark
3. I (contd.) $P (A) = 0.48 + 0.21 + 0.03 = 0.72$ .
  Hence 72% of the students suffer from anxiety and low retention issues.
  1 mark
4. II. By Bayes' theorem:
  $P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( A )} = \frac{0.6 \times 0.8}{0.72} .$
  1 mark
5. II (contd.) $P (E_{1} ∣ A) = \frac{0.48}{0.72} = \frac{48}{72} = \frac{2}{3}$ .
  1 mark
Answer: I. 72%. II. $\frac{2}{3}$
Where the marks are
- Applies the total probability formula with the correct values (1 mark)
- Obtains $P (A) = 0.72$ , i.e. 72% (1 mark)
- Applies Bayes' theorem for $P (E_{1} ∣ A)$ (1 mark)
- Gets $P (E_{1} ∣ A) = \frac{2}{3}$ (1 mark)
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Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.