MathsCBSE Class 12 · Sample paper, session 2026-27

CBSE Class 12 Mathematics: sample paper 2026-27

The sample question paper CBSE published for the 2026-27 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_2026_27/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 (6).

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 Section C, 2 of Section D and one sub-part each in 2 questions of Section E. Time 3 hours.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQInverse Trigonometric Functions
Identify the function whose graph is given below:
1. (A) $cos^{- 1} (\frac{x}{2})$
2. (B) $cos^{- 1} (2 x)$
3. (C) $π - cos^{- 1} x$
4. (D) $π + cos^{- 1} x$
Figure in the paper: Graph on X-Y axes (x from -2 to 2, y marked -π, π, 2π, 3π) showing a decreasing curve starting near x=-1 at y about 2π... ending near x=1 at y about π, passing through about (0, 3π/2). See it in the official paper.
Try it first, then show our solutionSolution
1. The curve is defined only for $x \in [- 1, 1]$ and decreases, like $cos^{- 1} x$ , whose range is $[0, π]$ .
2. Its values run from $2 π$ at $x = - 1$ down to $π$ at $x = 1$ , so it is $cos^{- 1} x$ shifted up by $π$ . Check: at $x = 0$ , $π + \frac{π}{2} = \frac{3 π}{2}$ . Hence the function is $π + cos^{- 1} x$ .
  1 mark
Answer: $π + cos^{- 1} x$ , option (D)
Where the marks are
- Identifies the graph as $cos^{- 1} x$ shifted up by $π$ (1 mark)
Open Q1 on its own page
Q2 · 1 mark · MCQInverse Trigonometric Functions
Domain of the function: $sec^{- 1} (2 x)$ is
1. (A) $(- \infty, \infty)$
2. (B) $(- \frac{1}{2}, \frac{1}{2})$
3. (C) $(- \infty, - \frac{1}{2}) \cup (\frac{1}{2}, \infty)$
4. (D) $(- \infty, - \frac{1}{2}] \cup [\frac{1}{2}, \infty)$
Try it first, then show our solutionSolution
1. $sec^{- 1} t$ is defined only for $∣ t ∣ \geq 1$ . So we need $∣2 x ∣ \geq 1$ , which gives $∣ x ∣ \geq \frac{1}{2}$ .
  1 mark
2. Domain $= (- \infty, - \frac{1}{2}] \cup [\frac{1}{2}, \infty)$ .
Answer: Option (D)
Where the marks are
- Uses $∣2 x ∣ \geq 1$ to get the domain (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQInverse Trigonometric Functions
Value of $cot^{- 1} (2 cos (\frac{1}{2} sin^{- 1} (\frac{3}{2})))$
1. (A) $\frac{π}{6}$
2. (B) $\frac{π}{4}$
3. (C) $\frac{π}{3}$
4. (D) $\frac{π}{2}$
Try it first, then show our solutionSolution
1. $sin^{- 1} \frac{3}{2} = \frac{π}{3}$ , so $cos \frac{π}{6} = \frac{3}{2}$ .
2. The expression becomes $cot^{- 1} (3) = \frac{π}{6}$ .
  1 mark
Answer: $\frac{π}{6}$ , option (A)
Where the marks are
- Evaluates the expression step by step to get $π /6$ (1 mark)
Q4 · 1 mark · MCQMatrices
If $A = 2 3 a b 10 2 c - 2 41$ is a symmetric matrix, then value of $6 a + 2 b + 3 c$ is:
1. (A) $- 2$
2. (B)2
3. (C) $\pm 2$
4. (D)4
Try it first, then show our solutionSolution
1. For a symmetric matrix $a_{ij} = a_{j i}$ . This gives $3 a = 1$ , $b = - 2$ and $2 c = 4$ .
2. $6 a + 2 b + 3 c = 2 - 4 + 6 = 4$ , and $4 = 2$ (the principal root).
  1 mark
Answer: 2, option (B)
Where the marks are
- Equates the symmetric entries and evaluates the root (1 mark)
Q5 · 1 mark · MCQMatrices
If $P$ and $Q$ are matrices of same order such that $P = [2011]$ , $∣ Q ∣ = 3$ , then find $∣ a d j (P Q) ∣$
1. (A)-3
2. (B)1
3. (C)6
4. (D)36
Try it first, then show our solutionSolution
1. $∣ P ∣ = 2$ , so $∣ P Q ∣ = ∣ P ∣∣ Q ∣ = 2 \times 3 = 6$ .
2. For a matrix of order 2, $∣ adj M ∣ = ∣ M ∣^{2 - 1} = ∣ M ∣$ . Hence $∣ adj (P Q) ∣ = 6$ .
  1 mark
Answer: 6, option (C)
Where the marks are
- Uses $∣ adj M ∣ = ∣ M ∣^{n - 1}$ with $n = 2$ (1 mark)
Q6 · 1 mark · MCQMatrices
If $[1 x] [1002] [4 - 1] = O$ , then value of $x$ is
1. (A) $- 2$
2. (B) $- 1$
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. $[1 x] [1002] = [12 x]$ .
2. $[12 x] [4 - 1] = 4 - 2 x = 0$ , so $x = 2$ .
  1 mark
Answer: x = 2, option (D)
Where the marks are
- Multiplies the matrices and solves $4 - 2 x = 0$ (1 mark)
Q7 · 1 mark · MCQContinuity and Differentiability
The derivative of $cos^{- 1} (2 x^{2} - 1)$ with respect to $cos^{- 1} (x)$ is
1. (A)2
2. (B) $\pm 1$
3. (C) $\frac{1}{2}$
4. (D) $- 3$
Try it first, then show our solutionSolution
1. Let $u = cos^{- 1} x$ , so $x = cos u$ . Then $2 x^{2} - 1 = cos 2 u$ .
2. So $y = cos^{- 1} (cos 2 u) = 2 u$ and $\frac{d y}{d u} = 2$ .
  1 mark
Answer: 2, option (A)
Where the marks are
- Substitutes $x = cos u$ to get $y = 2 u$ (1 mark)
Q8 · 1 mark · MCQContinuity and Differentiability
If $a, b, c$ are arbitrary constants such that the function defined by $f (x) = {a x^{2} + b, b \neq = 0, x \leq 1 b x^{2} + a x + c, x > 1$ is differentiable at $x = 1$ , then which of the following could be a triplet for $(a, b, c)$ ?
1. (A) $(3, 2, 0)$
2. (B) $(0, 0, 3)$
3. (C) $(\frac{3}{2}, 2, 0)$
4. (D) $(3, \frac{3}{2}, 0)$
Try it first, then show our solutionSolution
1. Continuity at $x = 1$ : $a + b = b + a + c$ , so $c = 0$ .
2. Equal one-sided derivatives at $x = 1$ : $2 a = 2 b + a$ , so $a = 2 b$ . The triplet is $(2 b, b, 0)$ with $b \neq = 0$ , which matches $(3, \frac{3}{2}, 0)$ .
  1 mark
Answer: Option (D)
Where the marks are
- Gets $c = 0$ and $a = 2 b$ and picks the matching triplet (1 mark)
Open Q8 on its own page
Q9 · 1 mark · MCQApplication of Derivatives
If $y = a {x + 2 a sin (\frac{x}{a})}$ , $a > 0$ , $x \in [0, π a]$ ; is a decreasing function, then the interval in which $x$ lies is
1. (A) $(\frac{5 π}{6}, π]$
2. (B) $(\frac{2 π}{3}, π]$
3. (C) $(\frac{5 π a}{- 6}, π a]$
4. (D) $(\frac{2 π a}{3}, π a]$
Try it first, then show our solutionSolution
1. $y^{'} = a (1 + 2 cos \frac{x}{a})$ . For $y$ to decrease, $y^{'} < 0$ , so $cos \frac{x}{a} < - \frac{1}{2}$ .
2. Since $\frac{x}{a} \in [0, π]$ , we need $\frac{x}{a} \in (\frac{2 π}{3}, π]$ , i.e. $x \in (\frac{2 π a}{3}, π a]$ .
  1 mark
Answer: Option (D)
Where the marks are
- Solves $y^{'} < 0$ to get the interval (1 mark)
Open Q9 on its own page
Q10 · 1 mark · MCQApplication of Derivatives
The values of $x$ for which the derivative of the function $y = x^{4}$ is greater than the value of the function are
1. (A) $0 < x < 4$
2. (B) $0 < x \leq 4$
3. (C) $x < 0$ or $x > 4$
4. (D) $x < 0$ or $x \geq 4$
Try it first, then show our solutionSolution
1. We need $4 x^{3} > x^{4}$ , i.e. $x^{3} (4 - x) > 0$ .
2. This holds when $x^{3}$ and $4 - x$ have the same sign, which happens only for $0 < x < 4$ .
  1 mark
Answer: 0 < x < 4, option (A)
Where the marks are
- Solves $x^{3} (4 - x) > 0$ (1 mark)
Q11 · 1 mark · MCQDifferential Equations
The degree and order of the differential equation $y - x (\frac{d y}{d x})^{3} = a (y + \frac{d ^{2} y}{d x ^{2}})^{2}$ are respectively
1. (A)3, 1
2. (B)2, 2
3. (C)3, 2
4. (D)4, 2
Try it first, then show our solutionSolution
1. The highest derivative present is $\frac{d ^{2} y}{d x ^{2}}$ , so the order is 2.
2. The equation is a polynomial in derivatives, and $\frac{d ^{2} y}{d x ^{2}}$ appears with highest power 2. So the degree is 2, giving (degree, order) $= (2, 2)$ .
  1 mark
Answer: 2, 2; option (B)
Where the marks are
- Finds order 2 and degree 2 (1 mark)
Open Q11 on its own page
Q12 · 1 mark · MCQIntegrals
If $\int \frac{1}{1 + e ^{2 x}} d x = x - P lo g (1 + e^{2 x}) + C$ , then 'P' is equal to
1. (A) $- \frac{1}{2}$
2. (B) $- \frac{1}{3}$
3. (C) $\frac{1}{3}$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. $\frac{1}{1 + e ^{2 x}} = 1 - \frac{e ^{2 x}}{1 + e ^{2 x}}$ .
2. Integrating gives $x - \frac{1}{2} lo g (1 + e^{2 x}) + C$ , so $P = \frac{1}{2}$ .
  1 mark
Answer: P = 1/2, option (D)
Where the marks are
- Splits the integrand and integrates to find $P$ (1 mark)
Q13 · 1 mark · MCQVector Algebra
If vector $2 \hat{i} - u \hat{j} + 3 \hat{k}$ has magnitude 4, then value of 'u' is
1. (A)1
2. (B) $2$
3. (C) $\pm 3$
4. (D) $\pm 2$
Try it first, then show our solutionSolution
1. $4 + u^{2} + 9 = 4$ , so $u^{2} = 16 - 13 = 3$ and $u = \pm 3$ .
  1 mark
Answer: Option (C)
Where the marks are
- Uses the magnitude formula to solve for $u$ (1 mark)
Q14 · 1 mark · MCQVector Algebra
Angle made by the vector $\hat{i} + \hat{j} + 2 \hat{k}$ with positive direction of z-axis is
1. (A) $\frac{π}{6}$
2. (B) $\frac{π}{4}$
3. (C) $\frac{π}{3}$
4. (D) $\frac{π}{2}$
Try it first, then show our solutionSolution
1. The magnitude is $1 + 1 + 2 = 2$ , so $cos γ = \frac{2}{2} = \frac{1}{2}$ and $γ = \frac{π}{4}$ .
  1 mark
Answer: $π /4$ , option (B)
Where the marks are
- Finds the direction cosine along the z-axis (1 mark)
Q15 · 1 mark · MCQVector Algebra
A vector perpendicular to the vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$ is
1. (A) $- 2 \hat{j}$
2. (B) $3 \hat{k}$
3. (C) $2 \hat{i}$
4. (D) $2 \hat{i} + \hat{k}$
Try it first, then show our solutionSolution
1. $(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = - 2 \hat{k}$ , so any vector perpendicular to both is parallel to $\hat{k}$ .
2. Among the options, only $3 \hat{k}$ is parallel to $\hat{k}$ .
  1 mark
Answer: Option (B)
Where the marks are
- Recognises that the perpendicular vector is along $\hat{k}$ (1 mark)
Open Q15 on its own page
Q16 · 1 mark · MCQProbability
The probability that it rains on a particular day is $\frac{1}{5}$ . Assuming weather conditions of any two days are independent, find the probability that it rains on three consecutive days.
1. (A) $\frac{4}{5}$
2. (B) $\frac{1}{5}$
3. (C) $\frac{1}{25}$
4. (D) $\frac{1}{125}$
Try it first, then show our solutionSolution
1. The days are independent, so $P = (\frac{1}{5})^{3} = \frac{1}{125}$ .
  1 mark
Answer: 1/125, option (D)
Where the marks are
- Multiplies the probabilities of the independent events (1 mark)
Q17 · 1 mark · MCQProbability
A die is thrown twice. What is the probability that at least one number is 4, given the sum of the numbers appearing is a multiple of 5?
1. (A) $\frac{3}{5}$
2. (B) $\frac{4}{5}$
3. (C) $\frac{3}{7}$
4. (D) $\frac{4}{7}$
Try it first, then show our solutionSolution
1. Let $A$ be 'sum is a multiple of 5'. Then $A = {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}$ , which has 7 outcomes.
2. Let $B$ be 'at least one 4'. Then $A \cap B = {(1, 4), (4, 1), (4, 6), (6, 4)}$ , which has 4 outcomes. So $P (B ∣ A) = \frac{4}{7}$ .
  1 mark
Answer: 4/7, option (D)
Where the marks are
- Lists the outcomes and computes the conditional probability (1 mark)
Open Q17 on its own page
Q18 · 1 mark · MCQProbability
$A$ and $B$ are two independent events such that $P (A) = \frac{1}{6}$ , $P (B) = k$ . If the probability of occurrence of exactly one of them is $\frac{1}{3}$ , then find the value of $k$ .
1. (A) $\frac{1}{4}$
2. (B) $\frac{2}{3}$
3. (C) $\frac{3}{4}$
4. (D) $\frac{5}{6}$
Try it first, then show our solutionSolution
1. $P (exactly one) = P (A) P (B^{'}) + P (A^{'}) P (B)$ , so $\frac{1}{6} (1 - k) + \frac{5}{6} k = \frac{1}{3}$ .
2. This gives $\frac{1}{6} + \frac{4}{6} k = \frac{1}{3}$ , so $k = \frac{1}{4}$ .
  1 mark
Answer: k = 1/4, option (A)
Where the marks are
- Sets up the 'exactly one' equation and solves for $k$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedIntegrals
Assertion: The value of the integral $\int_{- 1}^{1} ∣ x - 2∣ d x$ is 4
Reason: If $f (- x) = f (x)$ then $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of 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. On $[- 1, 1]$ we have $x - 2 < 0$ , so $∣ x - 2∣ = 2 - x$ . Then $\int_{- 1}^{1} (2 - x) d x = 4 - 0 = 4$ , so the Assertion is true.
2. The Reason is a true property of even functions. However, $∣ x - 2∣$ is not even, so the Reason does not explain the Assertion.
  1 mark
Answer: Option (B)
Where the marks are
- Evaluates the integral and judges the Reason correctly (1 mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedDifferential Equations
Assertion: The general solution of a differential equation represents a family of curves.
Reason: The derivative of an arbitrary constant is zero.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of 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. The general solution contains arbitrary constants, so it represents a family of curves.
2. Differentiating eliminates these constants because the derivative of a constant is zero. This is why one equation describes the whole family, so the Reason explains the Assertion.
  1 mark
Answer: Option (A)
Where the marks are
- Recognises both statements as true and linked (1 mark)
Open Q20 on its own page

Section B

VSA

Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerIntegrals
Evaluate $\int \frac{1}{x x ^{6} - 1} d x$
Try it first, then show our solutionSolution
1. Multiply the numerator and denominator by $6 x^{5}$ : $I = \int \frac{d x}{x x ^{6} - 1} = \frac{1}{6} \int \frac{6 x ^{5}}{x ^{6} x ^{6} - 1} d x$ . Put $x^{6} - 1 = t^{2}$ , so $6 x^{5} d x = 2 t d t$ and $x^{6} = t^{2} + 1$ .
  ½ mark
2. $I = \frac{1}{6} \int \frac{2 t d t}{( t ^{2} + 1 ) t} = \frac{1}{3} \int \frac{d t}{t ^{2} + 1} = \frac{1}{3} tan^{- 1} t + C = \frac{1}{3} tan^{- 1} (x^{6} - 1) + C$
  1½ marks
Answer: $\frac{1}{3} tan^{- 1} x^{6} - 1 + C$
Where the marks are
- Rewrites the integrand with $6 x^{5}$ and substitutes $x^{6} - 1 = t^{2}$ (½ mark)
- Reduces the integral to $\frac{1}{3} \int \frac{d t}{1 + t ^{2}}$ and writes $\frac{1}{3} tan^{- 1} x^{6} - 1 + C$ (1½ marks)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerIntegrals
Evaluate $\int \frac{c o s x - s i n x}{s i n ( 2 x )} d x$
Try it first, then show our solutionSolution
1. Write $sin 2 x = (1 + sin 2 x) - 1 = (sin x + cos x)^{2} - 1$ , so $I = \int \frac{c o s x - s i n x}{( s i n x + c o s x ) ^{2} - 1} d x$ .
  ½ mark
2. Put $sin x + cos x = t$ . Then $(cos x - sin x) d x = d t$ .
  ½ mark
3. $I = \int \frac{d t}{t ^{2} - 1} = lo g t + t^{2} - 1 + C$
  ½ mark
4. $I = lo g sin x + cos x + sin 2 x + C$
  ½ mark
Answer: $lo g sin x + cos x + sin 2 x + C$
Where the marks are
- Expresses $sin 2 x$ as $(sin x + cos x)^{2} - 1$ (½ mark)
- Substitutes $t = sin x + cos x$ with $d t = (cos x - sin x) d x$ (½ mark)
- Integrates to $lo g ∣ t + t^{2} - 1 ∣$ (½ mark)
- Writes the answer back in terms of $x$ (½ mark)
Open Q21(B) on its own page
Q22 · 2 marks · Very short answerContinuity and Differentiability
If $y = sin (x + y)$ and $\frac{d y}{d x} = 0$ , then prove that $y = \pm 1$ .
Try it first, then show our solutionSolution
1. Differentiate $y = sin (x + y)$ with respect to $x$ : $\frac{d y}{d x} = cos (x + y) (1 + \frac{d y}{d x})$ , so $\frac{d y}{d x} (1 - cos (x + y)) = cos (x + y)$ .
  1 mark
2. Since $\frac{d y}{d x} = 0$ , the left side is $0$ , so $cos (x + y) = 0$ . Then $sin^{2} (x + y) = 1$ , so $sin (x + y) = \pm 1$ . Because $y = sin (x + y)$ , we get $y = \pm 1$ . Hence proved.
  1 mark
Answer: $y = \pm 1$ (proved)
Where the marks are
- Differentiates implicitly and collects the $\frac{d y}{d x}$ terms (1 mark)
- Uses $\frac{d y}{d x} = 0$ to get $cos (x + y) = 0$ , hence $y = sin (x + y) = \pm 1$ (1 mark)
Open Q22 on its own page
Answer one of these 2 alternatives.
Q23(A) · 2 marks · Very short answerThree Dimensional Geometry
A line passing through the points $(2, 6, 5)$ , and $(3, p, 7)$ is parallel to the line joining the points $(- 4, 3, 4)$ and $(- 1, 0, q)$ . Find the value of $\frac{q}{p}$ .
Try it first, then show our solutionSolution
1. Direction ratios of the line through $(2, 6, 5)$ and $(3, p, 7)$ : $⟨ 1, p - 6, 2 ⟩$ . Direction ratios of the line through $(- 4, 3, 4)$ and $(- 1, 0, q)$ : $⟨ 3, - 3, q - 4 ⟩$ . Parallel lines have proportional direction ratios: $\frac{1}{3} = \frac{p - 6}{- 3} = \frac{2}{q - 4}$ .
  1 mark
2. From $\frac{p - 6}{- 3} = \frac{1}{3}$ : $p - 6 = - 1$ , so $p = 5$ . From $\frac{2}{q - 4} = \frac{1}{3}$ : $q - 4 = 6$ , so $q = 10$ .
  ½ mark
3. $\frac{q}{p} = \frac{10}{5} = 2$
  ½ mark
Answer: $\frac{q}{p} = 2$ (with $p = 5, q = 10$ )
Where the marks are
- Finds both sets of direction ratios and sets them proportional (1 mark)
- Obtains $p = 5$ and $q = 10$ (½ mark)
- Gets $\frac{q}{p} = 2$ (½ mark)
Open Q23(A) on its own page
OR
Q23(B) · 2 marks · Very short answerVector Algebra
A rectangle ABCD is such that $A C = u$ , and $B D = v$ , then find the vectors representing four sides of the rectangle (in a cyclic order starting from $A B$ ).
Try it first, then show our solutionSolution
1. Let $A B = a$ , $B C = b$ , $C D = c$ , $D A = d$ . In a rectangle, $c = - a$ and $d = - b$ . Then $u = A C = a + b$ …(i) and $v = B D = b + c = b - a$ …(ii).
  ½ mark
2. Adding (i) and (ii): $b = \frac{1}{2} (u + v)$ . Subtracting (ii) from (i): $a = \frac{1}{2} (u - v)$ .
  1 mark
3. So $A B = \frac{1}{2} (u - v)$ , $B C = \frac{1}{2} (u + v)$ , $C D = \frac{1}{2} (v - u)$ , $D A = - \frac{1}{2} (u + v)$ .
  ½ mark
Answer: $A B = \frac{1}{2} (u - v)$ , $B C = \frac{1}{2} (u + v)$ , $C D = \frac{1}{2} (v - u)$ , $D A = - \frac{1}{2} (u + v)$
Where the marks are
- Writes $u = a + b$ and $v = b - a$ using the triangle law (½ mark)
- Solves to get $a$ and $b$ (1 mark)
- States $C D$ and $D A$ as negatives of the opposite sides (½ mark)
Open Q23(B) on its own page
Q24 · 2 marks · Very short answerVector Algebra
For two non-zero vectors $a$ and $b$ , $∣ a ∣ = 6$ , $a \cdot b = 1$ and $b \times a = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$ . Find $∣ b ∣$ .
Try it first, then show our solutionSolution
1. $∣ a \times b ∣ = ∣ b \times a ∣ = 3^{2} + 4^{2} + (- 2)^{2} = 29$
  ½ mark
2. Use the identity $∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ [since $sin^{2} θ + cos^{2} θ = 1$ ].
  ½ mark
3. $29 + 1 = 6∣ b ∣^{2} \Rightarrow ∣ b ∣^{2} = 5 \Rightarrow ∣ b ∣ = 5$
  1 mark
Answer: $∣ b ∣ = 5$
Where the marks are
- Finds $∣ a \times b ∣ = 29$ (½ mark)
- Uses Lagrange's identity (½ mark)
- Obtains $∣ b ∣ = 5$ (1 mark)
Open Q24 on its own page
Q25 · 2 marks · Very short answerLinear Programming
The feasible region of a linear programming problem is shown in the given figure:
Write all the constraints on the decision variables involved
Figure in the paper: Graph with X and Y axes, origin O. Lines $x + 2 y = 6$ and $3 x + 2 y = 10$ are drawn. The shaded feasible region is the triangle ABC lying above the x-axis: A is the x-intercept of $3 x + 2 y = 10$ on the x-axis, B is the x-intercept of $x + 2 y = 6$ on the x-axis, and C is the intersection of the two lines. Region is bounded by the x-axis and the two lines, to the right of line 3x+2y=10 and below line x+2y=6. See it in the official paper.
Try it first, then show our solutionSolution
1. The region lies on the side of $3 x + 2 y = 10$ away from the origin, so $3 x + 2 y \geq 10$ . It lies on the origin side of $x + 2 y = 6$ , so $x + 2 y \leq 6$ . (Check with the interior point $(4, 0.5)$ : $13 \geq 10$ and $5 \leq 6$ .)
  1½ marks
2. The region lies above the x-axis, so $y \geq 0$ . ( $x \geq 0$ is then automatically satisfied.)
  ½ mark
Answer: $3 x + 2 y \geq 10, x + 2 y \leq 6, y \geq 0$
Where the marks are
- Writes $3 x + 2 y \geq 10$ and $x + 2 y \leq 6$ (1½ marks)
- Writes $y \geq 0$ (½ mark)
Open Q25 on its own page

Section C

Q26 · 3 marks · Short answerMatrices
For two matrices, $P = [200102]$ and $Q = 101020$ , find $P Q$ and $QP$ , whichever is defined. Also, find the inverse of the product matrix, if it exists.
Try it first, then show our solutionSolution
1. $P$ is $2 \times 3$ and $Q$ is $3 \times 2$ , so both products are defined. $P Q = [200102] 101020 = [2202]$
  1 mark
2. $QP = 101020 [200102] = 202020040$
  1 mark
3. $∣ P Q ∣ = 4 - 0 = 4 \neq = 0$ , so $P Q$ is non-singular and its inverse exists. $(P Q)^{- 1} = \frac{1}{∣ P Q ∣} adj (P Q) = \frac{1}{4} [2 - 2 02] = [\frac{1}{2} - \frac{1}{2} 0 \frac{1}{2}]$
  ½ mark
4. $∣ QP ∣ = 0$ because rows 1 and 3 are identical, so $QP$ is singular and its inverse does not exist.
  ½ mark
Answer: $P Q = [2202]$ , $(P Q)^{- 1} = [\frac{1}{2} - \frac{1}{2} 0 \frac{1}{2}]$ ; $QP = 202020040$ is singular, so its inverse does not exist
Where the marks are
- Computes $P Q$ correctly (1 mark)
- Computes $QP$ correctly (1 mark)
- Finds $(P Q)^{- 1}$ using $∣ P Q ∣ = 4$ (½ mark)
- Shows $∣ QP ∣ = 0$ , so its inverse does not exist (½ mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerApplication of Derivatives
The circular area of vision on the ground by a surveillance drone fitted with camera increases at the rate of $44 m^{2}$ per metre as it gains vertical height. The radius of the circle on the ground increases at the rate of 6 m/s. Find the rate at which the height of the drone increases with time, at the instant when the radius of the circular area on the ground is 14 m. (Use $π = \frac{22}{7}$ )
Figure in the paper: Picture of a drone with camera hovering above the ground, a downward arrow indicating its vertical height, and a dashed ellipse/circle on the ground showing the area of vision. See it in the official paper.
Try it first, then show our solutionSolution
1. Let $A$ be the area of the circle of vision, $r$ its radius and $h$ the height of the drone at time $t$ . Given $\frac{d A}{d h} = 44$ m²/m and $\frac{d r}{d t} = 6$ m/s. We need $\frac{d h}{d t}$ when $r = 14$ m.
  ½ mark
2. $A = π r^{2} \Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t}$ . By the chain rule, $\frac{d A}{d t} = \frac{d A}{d h} \cdot \frac{d h}{d t}$ , so $\frac{d A}{d h} \cdot \frac{d h}{d t} = 2 π r \frac{d r}{d t}$ .
  1½ marks
3. $44 \frac{d h}{d t} = 2 \times \frac{22}{7} \times 14 \times 6 = 528 \Rightarrow \frac{d h}{d t} = 12$ m/s. So the height of the drone increases at 12 m/s at that instant.
  1 mark
Answer: 12 m/s
Where the marks are
- Defines the variables and writes the given rates (½ mark)
- Differentiates $A = π r^{2}$ and links the rates by the chain rule (1½ marks)
- Substitutes $r = 14$ to get $\frac{d h}{d t} = 12$ m/s (1 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(A) · 3 marks · Short answerContinuity and Differentiability
If $x = sin^{3} t$ , $y = cos^{3} t$ , Evaluate $\frac{d ^{2} y}{d x ^{2}}$ at $(\frac{1}{2 2}, \frac{1}{2 2})$ .
Try it first, then show our solutionSolution
1. $\frac{d x}{d t} = 3 sin^{2} t cos t$ and $\frac{d y}{d t} = - 3 cos^{2} t sin t$ , so $\frac{d y}{d x} = \frac{- 3 c o s ^{2} t s i n t}{3 s i n ^{2} t c o s t} = - cot t$
  1 mark
2. $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (- cot t) \cdot \frac{d t}{d x} = csc^{2} t \cdot \frac{1}{3 s i n ^{2} t c o s t} = \frac{1}{3} sec t csc^{4} t$
  1 mark
3. At $(\frac{1}{2 2}, \frac{1}{2 2})$ : $sin^{3} t = cos^{3} t = \frac{1}{2 2} \Rightarrow t = \frac{π}{4}$ . Then $\frac{d ^{2} y}{d x ^{2}} = \frac{1}{3} \cdot 2 \cdot (2)^{4} = \frac{4 2}{3}$
  1 mark
Answer: $\frac{d ^{2} y}{d x ^{2}} = \frac{4 2}{3}$
Where the marks are
- Finds $\frac{d y}{d x} = - cot t$ (1 mark)
- Finds $\frac{d ^{2} y}{d x ^{2}} = \frac{1}{3} sec t csc^{4} t$ (1 mark)
- Identifies $t = \frac{π}{4}$ and evaluates $\frac{4 2}{3}$ (1 mark)
Open Q28(A) on its own page
OR
Q28(B) · 3 marks · Short answerContinuity and Differentiability
If $y = x^{l o g x} + (lo g x)^{x}$ ; $x > 1$ , then evaluate $\frac{d y}{d x}$
Try it first, then show our solutionSolution
1. Let $u = x^{l o g x}$ and $v = (lo g x)^{x}$ , so $y = u + v$ . Taking logs: $lo g u = (lo g x)^{2}$ and $lo g v = x lo g (lo g x)$ . Differentiating: $\frac{1}{u} \frac{d u}{d x} = \frac{2 l o g x}{x}$ and $\frac{1}{v} \frac{d v}{d x} = lo g (lo g x) + x \cdot \frac{1}{l o g x} \cdot \frac{1}{x} = lo g (lo g x) + \frac{1}{l o g x}$
  1 mark
2. $\frac{d u}{d x} = x^{l o g x} \cdot \frac{2 l o g x}{x}$ and $\frac{d v}{d x} = (lo g x)^{x} (lo g (lo g x) + \frac{1}{l o g x})$
  1 mark
3. $\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x} = \frac{2 l o g x}{x} x^{l o g x} + (lo g x)^{x} (lo g (lo g x) + \frac{1}{l o g x})$
  1 mark
Answer: $\frac{d y}{d x} = \frac{2 l o g x}{x} x^{l o g x} + (lo g x)^{x} [lo g (lo g x) + \frac{1}{l o g x}]$
Where the marks are
- Takes logarithms of each part and differentiates (1 mark)
- Gets $\frac{d u}{d x}$ and $\frac{d v}{d x}$ (1 mark)
- Adds to obtain $\frac{d y}{d x}$ (1 mark)
Open Q28(B) on its own page
Answer one of these 2 alternatives.
Q29(A) · 3 marks · Short answerDifferential Equations
Find the particular solution of the differential equation $x d y + (y - x^{2} e^{x}) d x = 0$ , given that $y (1) = 0$
Try it first, then show our solutionSolution
1. Dividing by $x d x$ : $\frac{d y}{d x} + \frac{1}{x} y = x e^{x}$ , a linear differential equation with $P = \frac{1}{x}$ and $Q = x e^{x}$ .
  ½ mark
2. I.F. $= e^{\int \frac{1}{x} d x} = e^{l o g x} = x$
  ½ mark
3. The solution is $y \cdot x = \int x \cdot x e^{x} d x = \int x^{2} e^{x} d x$ . By parts: $x y = x^{2} e^{x} - \int 2 x e^{x} d x$
  ½ mark
4. By parts again: $x y = x^{2} e^{x} - 2 (x e^{x} - e^{x}) + C$
  ½ mark
5. Using $y (1) = 0$ : $0 = e - 2 (e - e) + C \Rightarrow C = - e$
  ½ mark
6. The particular solution is $x y = e^{x} (x^{2} - 2 x + 2) - e$
  ½ mark
Answer: $x y = e^{x} (x^{2} - 2 x + 2) - e$
Where the marks are
- Writes the equation in linear form (½ mark)
- Finds I.F. $= x$ (½ mark)
- Sets up $x y = \int x^{2} e^{x} d x$ and applies integration by parts once (½ mark)
- Completes the integration (½ mark)
- Finds $C = - e$ (½ mark)
- States the particular solution (½ mark)
Open Q29(A) on its own page
OR
Q29(B) · 3 marks · Short answerDifferential Equations
Find the particular solution of the differential equation $(x^{2} + y^{2}) d y = x y d x$ , given that $y (0) = 1$
Try it first, then show our solutionSolution
1. $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ is homogeneous. Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Then $v + x \frac{d v}{d x} = \frac{v}{1 + v ^{2}} \Rightarrow x \frac{d v}{d x} = \frac{v - v - v ^{3}}{1 + v ^{2}} = \frac{- v ^{3}}{1 + v ^{2}}$
  1 mark
2. Separating variables: $\int (\frac{1}{v ^{3}} + \frac{1}{v}) d v = - \int \frac{d x}{x} \Rightarrow - \frac{1}{2 v ^{2}} + lo g ∣ v ∣ = - lo g ∣ x ∣ + lo g C$ . So $lo g \frac{v x}{C} = \frac{1}{2 v ^{2}}$ , i.e. $lo g \frac{y}{C} = \frac{x ^{2}}{2 y ^{2}}$ , or $y = C e^{\frac{x ^{2}}{2 y ^{2}}}$
  1 mark
3. Using $y (0) = 1$ : $1 = C e^{0} \Rightarrow C = 1$ . The particular solution is $y = e^{\frac{x ^{2}}{2 y ^{2}}}$
  1 mark
Answer: $y = e^{\frac{x ^{2}}{2 y ^{2}}}$
Where the marks are
- Substitutes $y = v x$ and reduces to $x \frac{d v}{d x} = \frac{- v ^{3}}{1 + v ^{2}}$ (1 mark)
- Separates, integrates and gets the general solution $y = C e^{x^{2} /2 y^{2}}$ (1 mark)
- Uses $y (0) = 1$ to get $C = 1$ and the particular solution (1 mark)
Open Q29(B) on its own page
Answer one of these 2 alternatives.
Q30(A) · 3 marks · Short answerThree Dimensional Geometry
Find the foot of the perpendicular drawn from the point $(2, 1, 4)$ on the line $x - 2 = \frac{2 y + 3}{4} = 2 - z$ . Also find the equation of the line passing through this point and the point $(2, 1, 4)$ .
Try it first, then show our solutionSolution
1. Let $x - 2 = \frac{2 y + 3}{4} = 2 - z = k$ . A general point on the line is $P (k + 2, \frac{4 k - 3}{2}, 2 - k)$ . The line's direction ratios are $⟨ 1, 2, - 1 ⟩$ .
  ½ mark
2. Direction ratios of $A P$ , where $A (2, 1, 4)$ : $⟨ k, \frac{4 k - 5}{2}, - k - 2 ⟩$
  ½ mark
3. $A P ⊥$ the line: $1 (k) + 2 (\frac{4 k - 5}{2}) + (- 1) (- k - 2) = 0 \Rightarrow 6 k - 3 = 0 \Rightarrow k = \frac{1}{2}$
  ½ mark
4. Foot of the perpendicular: $P (\frac{5}{2}, - \frac{1}{2}, \frac{3}{2})$
  ½ mark
5. Direction ratios of $A P$ : $⟨ \frac{1}{2}, - \frac{3}{2}, - \frac{5}{2} ⟩$ , i.e. $⟨ 1, - 3, - 5 ⟩$
  ½ mark
6. Equation of the line through $(2, 1, 4)$ and the foot: $\frac{x - 2}{1} = \frac{y - 1}{- 3} = \frac{z - 4}{- 5}$
  ½ mark
Answer: Foot $(\frac{5}{2}, - \frac{1}{2}, \frac{3}{2})$ ; line $\frac{x - 2}{1} = \frac{y - 1}{- 3} = \frac{z - 4}{- 5}$
Where the marks are
- Writes a general point on the line (½ mark)
- Finds the direction ratios of $A P$ (½ mark)
- Uses perpendicularity to get $k = \frac{1}{2}$ (½ mark)
- Finds the foot $(\frac{5}{2}, - \frac{1}{2}, \frac{3}{2})$ (½ mark)
- Gets the direction ratios $⟨ 1, - 3, - 5 ⟩$ (½ mark)
- Writes the equation of the line (½ mark)
Open Q30(A) on its own page
OR
Q30(B) · 3 marks · Short answerThree Dimensional Geometry
Find the angle between the lines: $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}$ and $\frac{x - 5}{2} = y - \frac{1}{2} = \frac{3 - z}{2}$ . Also find the cartesian equation of a line passing through the point $(1, 1, 1)$ and perpendicular to both lines.
Try it first, then show our solutionSolution
1. The second line in standard form is $\frac{x - 5}{2} = \frac{y - \frac{1}{2}}{1} = \frac{z - 3}{- 2}$ . Direction ratios: $b_{1} = ⟨ 2, 2, 1 ⟩$ and $b_{2} = ⟨ 2, 1, - 2 ⟩$ .
  ½ mark
2. $cos θ = \frac{2 ( 2 ) + 2 ( 1 ) + 1 ( - 2 )}{9 9} = \frac{4}{9} \Rightarrow θ = cos^{- 1} (\frac{4}{9})$
  1 mark
3. The required line is parallel to $b_{1} \times b_{2} = \hat{i} 22 \hat{j} 21 \hat{k} 1 - 2 = - 5 \hat{i} + 6 \hat{j} - 2 \hat{k}$
  1 mark
4. Required line: $\frac{x - 1}{- 5} = \frac{y - 1}{6} = \frac{z - 1}{- 2}$ , or $\frac{x - 1}{5} = \frac{y - 1}{- 6} = \frac{z - 1}{2}$
  ½ mark
Answer: $θ = cos^{- 1} \frac{4}{9}$ ; $\frac{x - 1}{- 5} = \frac{y - 1}{6} = \frac{z - 1}{- 2}$
Where the marks are
- Rewrites the second line in standard form with direction ratios $⟨ 2, 1, - 2 ⟩$ (½ mark)
- Finds $cos θ = \frac{4}{9}$ (1 mark)
- Computes the cross product $- 5 \hat{i} + 6 \hat{j} - 2 \hat{k}$ (1 mark)
- Writes the cartesian equation of the line (½ mark)
Open Q30(B) on its own page
Q31 · 3 marks · Short answerLinear Programming
Solve the linear programming problem graphically:
Maximize $Z = 40 x + 30 y$
Subject to constraints: $x + y \leq 6$ , $2 x \geq y$ , $y \geq 2$
Try it first, then show our solutionSolution
1. Graph: draw the lines $x + y = 6$ (through $(6, 0)$ and $(0, 6)$ ), $y = 2 x$ (through the origin and $(2, 4)$ ) and $y = 2$ . The feasible region is on the origin side of $x + y = 6$ , below or to the right of $y = 2 x$ , and above $y = 2$ . It is the bounded triangle with vertices $A (1, 2)$ [from $y = 2$ , $y = 2 x$ ], $B (4, 2)$ [from $y = 2$ , $x + y = 6$ ] and $C (2, 4)$ [from $y = 2 x$ , $x + y = 6$ ].
  1½ marks
2. Corner point values: $Z_{A} = 40 (1) + 30 (2) = 100$ ; $Z_{B} = 40 (4) + 30 (2) = 220$ ; $Z_{C} = 40 (2) + 30 (4) = 200$
  1 mark
3. Maximum $Z = 220$ at $B (4, 2)$ , i.e. $x = 4, y = 2$ .
  ½ mark
Answer: Maximum $Z = 220$ at $(4, 2)$
Where the marks are
- Draws the graph and shades the correct bounded feasible region with vertices $(1, 2), (4, 2), (2, 4)$ (1½ marks)
- Evaluates $Z$ at every corner point (1 mark)
- States the maximum value 220 at $(4, 2)$ (½ mark)
Open Q31 on its own page

Section D

Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerRelations and Functions
Let $A = N \times N$ , where $N$ is the set of natural numbers and a relation $R$ is defined on $A$ such that $(a, b) R (c, d)$ if and only if $a^{2} + d^{2} = b^{2} + c^{2}$ . Check whether $R$ is an equivalence relation or not?
Try it first, then show our solutionSolution
1. Reflexive: For any $(p, q) \in N \times N$ , $p^{2} + q^{2} = q^{2} + p^{2}$ , so $(p, q) R (p, q)$ . Hence $R$ is reflexive.
  1½ marks
2. Symmetric: Let $(p, q) R (r, s)$ . Then $p^{2} + s^{2} = q^{2} + r^{2}$ , which can be rewritten as $r^{2} + q^{2} = s^{2} + p^{2}$ . So $(r, s) R (p, q)$ , and $R$ is symmetric.
  1½ marks
3. Transitive: Let $(p, q) R (r, s)$ and $(r, s) R (u, v)$ . Then $p^{2} + s^{2} = q^{2} + r^{2}$ …(i) and $r^{2} + v^{2} = s^{2} + u^{2}$ …(ii). Adding (i) and (ii) and cancelling $r^{2} + s^{2}$ gives $p^{2} + v^{2} = q^{2} + u^{2}$ . So $(p, q) R (u, v)$ , and $R$ is transitive.
  1½ marks
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 (1½ marks)
- Proves symmetry (1½ marks)
- Proves transitivity by adding the two conditions (1½ marks)
- Concludes that $R$ is an equivalence relation (½ mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerRelations and Functions
A relation $R : (- \infty, \infty) \to [0, \infty)$ is defined as $R = {(x, y) : y = x^{2} + 5}$
Draw its graph and check whether it is a function or not.
If the given relation is a function, then examine whether the function is injective, and whether it is surjective or not.
Try it first, then show our solutionSolution
1. Graph: $y = x^{2} + 5$ is an upward-opening parabola, symmetric about the y-axis, with vertex $(0, 5)$ . It passes through $(\pm 1, 6)$ , $(\pm 2, 9)$ and $(\pm 3, 14)$ and lies entirely above the line $y = 5$ .
  1 mark
2. Every vertical line $x = a$ meets the graph in exactly one point, so every element of $(- \infty, \infty)$ has a unique image $x^{2} + 5 \in [0, \infty)$ . Hence $R$ is a function; call it $f (x) = x^{2} + 5$ .
  1 mark
3. Injective: $f (- 1) = (- 1)^{2} + 5 = 6$ and $f (1) = 1^{2} + 5 = 6$ . Since $- 1 \neq = 1$ but $f (- 1) = f (1)$ , $f$ is many-one, so it is not injective.
  1½ marks
4. Surjective: The range of $f$ is $[5, \infty)$ , which is not equal to the co-domain $[0, \infty)$ . For example, $y = 1$ would need $x^{2} = - 4$ , which has no real solution. So $f$ is not surjective.
  1½ marks
Answer: $R$ is a function; it is neither injective nor surjective.
Where the marks are
- Draws the correct graph of $y = x^{2} + 5$ (1 mark)
- Justifies that $R$ is a function (1 mark)
- Shows it is not injective, e.g. $f (- 1) = f (1)$ (1½ marks)
- Shows it is not surjective, since the range $[5, \infty)$ is not the co-domain $[0, \infty)$ (1½ marks)
Open Q32(B) on its own page
Q33 · 5 marks · Long answerIntegrals
Evaluate $\int_{- a}^{a} \frac{a - x}{a + x} d x$
Try it first, then show our solutionSolution
1. Multiply the numerator and denominator inside the root by $(a - x)$ : $I = \int_{- a}^{a} \frac{( a - x ) ^{2}}{a ^{2} - x ^{2}} d x = \int_{- a}^{a} \frac{a - x}{a ^{2} - x ^{2}} d x$ [since $a - x \geq 0$ on $[- a, a]$ ].
  ½ mark
2. Split the integral: $I = \int_{- a}^{a} \frac{a}{a ^{2} - x ^{2}} d x - \int_{- a}^{a} \frac{x}{a ^{2} - x ^{2}} d x = I_{1} - I_{2}$
  1 mark
3. $\frac{a}{a ^{2} - x ^{2}}$ is an even function, so $I_{1} = 2 a \int_{0}^{a} \frac{d x}{a ^{2} - x ^{2}}$ [using $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ when $f (- x) = f (x)$ ].
  1 mark
4. $I_{1} = 2 a [sin^{- 1} \frac{x}{a}]_{0}^{a} = 2 a \cdot \frac{π}{2} = π a$
  1 mark
5. $\frac{x}{a ^{2} - x ^{2}}$ is an odd function, so $I_{2} = 0$ [using $\int_{- a}^{a} f (x) d x = 0$ when $f (- x) = - f (x)$ ].
  1 mark
6. $I = I_{1} - I_{2} = π a$
  ½ mark
Answer: $π a$
Where the marks are
- Rationalises to $\frac{a - x}{a ^{2} - x ^{2}}$ (½ mark)
- Splits into $I_{1} - I_{2}$ (1 mark)
- Uses the even-function property for $I_{1}$ (1 mark)
- Evaluates $I_{1} = π a$ (1 mark)
- Uses the odd-function property to get $I_{2} = 0$ (1 mark)
- Concludes $I = π a$ (½ mark)
Open Q33 on its own page
Answer one of these 2 alternatives.
Q34(A) · 5 marks · Long answerApplication of Integrals
Using Integration, find the area of the region ${(x, y) : 0 \leq y \leq 4 - x^{2}, x \geq 2 - y}$ .
Try it first, then show our solutionSolution
1. Figure (described): $y = 4 - x^{2}$ is the upper half of the circle $x^{2} + y^{2} = 4$ , with centre $(0, 0)$ and radius $2$ . The line $x + y = 2$ cuts both axes at $2$ . The condition $x \geq 2 - y$ means the point lies on or above the line, i.e. on the side away from the origin. The required region is the small segment in the first quadrant between the line and the arc, from $x = 0$ to $x = 2$ .
  1 mark
2. Points of intersection: substitute $y = 2 - x$ in $x^{2} + y^{2} = 4$ to get $x^{2} + (2 - x)^{2} = 4 \Rightarrow 2 x^{2} - 4 x = 0 \Rightarrow x = 0$ or $x = 2$ . So the curves meet at $(0, 2)$ and $(2, 0)$ .
  1 mark
3. Required area $= \int_{0}^{2} (4 - x^{2} - (2 - x)) d x$ , because the arc lies above the line on $[0, 2]$ .
  1 mark
4. Integrating, the area is $[\frac{x}{2} 4 - x^{2} + \frac{4}{2} sin^{- 1} \frac{x}{2} - 2 x + \frac{x ^{2}}{2}]_{0}^{2}$ .
  1 mark
5. Substituting the limits gives $(0 + 2 \cdot \frac{π}{2} - 4 + 2) - 0 = π - 2$ . So the required area is $(π - 2)$ sq. units.
  1 mark
Answer: $(π - 2)$ sq. units
Where the marks are
- Correct figure showing the segment between the arc and the line (1 mark)
- Finds the intersection points $(0, 2)$ and $(2, 0)$ (1 mark)
- Sets up the area as $\int_{0}^{2} (4 - x^{2} - (2 - x)) d x$ (1 mark)
- Correct antiderivative (1 mark)
- Evaluates the limits to get $π - 2$ sq. units (1 mark)
Open Q34(A) on its own page
OR
Q34(B) · 5 marks · Long answerApplication of Integrals
Using Integration, find the area of the region bounded by the curve $y^{2} = 2 x$ and the line $x - 2 y = 0$ .
Try it first, then show our solutionSolution
1. Figure (described): $y^{2} = 2 x$ is a right-opening parabola with vertex at the origin. The line $x = 2 y$ passes through the origin. The enclosed region lies in the first quadrant, between the upper branch $y = 2 x$ (above) and the line $y = \frac{x}{2}$ (below), from $x = 0$ to $x = 8$ .
  1 mark
2. Points of intersection: put $x = 2 y$ in $y^{2} = 2 x$ to get $y^{2} = 4 y \Rightarrow y = 0$ or $y = 4$ . This gives the points $(0, 0)$ and $(8, 4)$ .
  1 mark
3. Required area $= \int_{0}^{8} (2 x - \frac{x}{2}) d x$ .
  1 mark
4. Integrating, the area is $[2 \cdot \frac{2}{3} x^{3/2} - \frac{x ^{2}}{4}]_{0}^{8}$ .
  1 mark
5. Now $8^{3/2} = 162$ , so the area is $2 \cdot \frac{2}{3} \cdot 162 - \frac{64}{4} = \frac{64}{3} - 16 = \frac{16}{3}$ sq. units.
  1 mark
Answer: $\frac{16}{3}$ sq. units
Where the marks are
- Correct figure with the region between the parabola and the line shaded (1 mark)
- Finds the intersection points $(0, 0)$ and $(8, 4)$ (1 mark)
- Sets up $\int_{0}^{8} (2 x - x /2) d x$ (1 mark)
- Correct integration (1 mark)
- Final area $16/3$ sq. units (1 mark)
Open Q34(B) on its own page
Q35 · 5 marks · Long answerCompetency-basedProbability
A fitness app syncs with four types of smartwatches: Alpha, Beta, Gamma and Delta. 45% of the users use Alpha, 30% use Beta, 15% use Gamma and 10% use Delta.
Due to device errors, the probabilities of getting a false heart-rate alert from the Alpha, Beta, Gamma and Delta smartwatches are respectively 0.01, 0.02, 0.05, and 0.01.
Based on the given information, answer the following questions:
1. (i)
  Find the probability that a randomly selected user gets a false alert.
  0 marks
2. (ii)
  If a false alert occurs, what is the probability that it came from a Beta smartwatch?
  0 marks
Try it first, then show our solutionSolution
1. Define the events: $E_{1}, E_{2}, E_{3}, E_{4}$ are 'the user uses Alpha, Beta, Gamma, Delta' respectively, and $A$ is 'the user gets a false alert'. The events $E_{1}, \dots, E_{4}$ form a partition of the sample space, with $P (E_{1}) = 0.45, P (E_{2}) = 0.30, P (E_{3}) = 0.15, P (E_{4}) = 0.10$ .
  1 mark
2. The conditional probabilities of a false alert are $P (A ∣ E_{1}) = 0.01, P (A ∣ E_{2}) = 0.02, P (A ∣ E_{3}) = 0.05, P (A ∣ E_{4}) = 0.01$ .
  1 mark
3. (i) By the theorem of total probability, $P (A) = \sum P (E_{i}) P (A ∣ E_{i}) = 0.45 (0.01) + 0.30 (0.02) + 0.15 (0.05) + 0.10 (0.01)$
  $= 0.0045 + 0.006 + 0.0075 + 0.001 = 0.019$ .
  1½ marks
4. (ii) By Bayes' theorem, $P (E_{2} ∣ A) = \frac{P ( E _{2} ) P ( A ∣ E _{2} )}{P ( A )} = \frac{0.30 \times 0.02}{0.019} = \frac{0.006}{0.019} = \frac{6}{19}$ .
  1½ marks
Answer: (i) 0.019; (ii) $\frac{6}{19}$
Where the marks are
- Defines events and writes the prior probabilities (1 mark)
- Writes the conditional probabilities of a false alert (1 mark)
- Applies total probability to get $P (A) = 0.019$ (1½ marks)
- Applies Bayes' theorem to get $6/19$ (1½ marks)
Open Q35 on its own page

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedThree Dimensional Geometry
Two security sensors in a multi-storey mall emit laser beams for movement detection. Three of these beams travel along the following lines:
$l_{1} : r = (λ + 1) \hat{i} + (3 λ + 2) \hat{j} + (3 - λ) \hat{k}$ and
$l_{2} : r = 2 (μ + 1) \hat{i} + (3 μ - 1) \hat{j} + (μ + 1) \hat{k}$
$l_{3} : r = (γ + 3) \hat{i} + (3 γ - 1) \hat{j} + (1 - γ) \hat{k}$
Based on the given information, answer the following questions:
1. (i)
  Show that the lines $l_{1}$ and $l_{2}$ are not parallel to each other.
  1 mark
2. (ii)
  Show that the lines $l_{1}$ and $l_{3}$ are parallel to each other.
  1 mark
3. (iii)(a)
  Find the shortest distance between the lines $l_{1}$ and $l_{2}$ .
  2 marks
4. OR
  (iii)(b)
  Find the shortest distance between the lines $l_{1}$ and $l_{3}$ .
  2 marks
Figure in the paper: Photo of a multi-storey mall interior with two sensors mounted on pillars emitting crossing red laser beams across the atrium. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Rewrite the lines in vector form:
  $l_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + 3 \hat{j} - \hat{k})$ , so $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $b = \hat{i} + 3 \hat{j} - \hat{k}$ .
  $l_{2} : r = (2 \hat{i} - \hat{j} + \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ , so $c = 2 \hat{i} - \hat{j} + \hat{k}$ and $d = 2 \hat{i} + 3 \hat{j} + \hat{k}$ .
  The ratios $\frac{1}{2}, \frac{3}{3}, \frac{- 1}{1}$ are not equal, so $b \neq = k d$ for any scalar $k$ . Hence $l_{1}$ and $l_{2}$ are not parallel.
  1 mark
2. (ii) Similarly, $l_{3} : r = (3 \hat{i} - \hat{j} + \hat{k}) + γ (\hat{i} + 3 \hat{j} - \hat{k})$ , so $e = 3 \hat{i} - \hat{j} + \hat{k}$ and $f = \hat{i} + 3 \hat{j} - \hat{k}$ . Since $f = b$ , the lines $l_{1}$ and $l_{3}$ are parallel.
  1 mark
3. (iii)(a) Find the two vectors needed:
  $c - a = \hat{i} - 3 \hat{j} - 2 \hat{k}$ .
  $b \times d = \hat{i} 12 \hat{j} 33 \hat{k} - 1 1 = 6 \hat{i} - 3 \hat{j} - 3 \hat{k}$ .
  Then $(c - a) \cdot (b \times d) = 6 + 9 + 6 = 21 \neq = 0$ , so the lines are skew.
  1 mark
4. (iii)(a) Here $∣ b \times d ∣ = 36 + 9 + 9 = 36$ . So $S D = \frac{( c - a ) \cdot ( b \times d )}{∣ b \times d ∣} = \frac{21}{3 6} = \frac{7}{6} = \frac{7 6}{6}$ units.
  1 mark
5. (iii)(b) The lines are parallel with common direction $b$ , so $S D = \frac{∣ b \times ( e - a ) ∣}{∣ b ∣}$ , where $e - a = 2 \hat{i} - 3 \hat{j} - 2 \hat{k}$ .
  1 mark
6. (iii)(b) $b \times (e - a) = \hat{i} 12 \hat{j} 3 - 3 \hat{k} - 1 - 2 = - 9 \hat{i} + 0 \hat{j} - 9 \hat{k}$ , so its magnitude is $92$ . Also $∣ b ∣ = 11$ . Hence $S D = \frac{9 2}{11} = \frac{9 22}{11}$ units.
  1 mark
Answer: (i) Direction vectors not proportional ⇒ not parallel; (ii) same direction vector ⇒ parallel; (iii)(a) $\frac{7 6}{6}$ units; (iii)(b) $\frac{9 22}{11}$ units
Where the marks are
- Writes $l_{1}, l_{2}$ in the form $a + λ b$ and shows the direction vectors are not proportional (1 mark)
- Shows the direction vectors of $l_{1}$ and $l_{3}$ are equal (1 mark)
- (iii)(a) Computes $c - a$ , $b \times d$ and the scalar triple product 21 (1 mark)
- (iii)(a) Shortest distance $76 /6$ units (1 mark)
- (iii)(b) Uses the parallel-lines formula $∣ b \times (e - a) ∣/∣ b ∣$ (1 mark)
- (iii)(b) Shortest distance $922 /11$ units (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
Water is essential for life, and rivers are vital water sources across the world. The Yamuna Action Plan (YAP), one of India's largest river restoration projects, was launched to clean the Yamuna, but its effectiveness (WQI) the Water Quality Index must be measured through cost-benefit, economic and environmental analysis. Therefore, comprehensive pollutant treatment, prevention of further contamination, and community awareness are crucial for sustainable river management.
Implementing the same concept to the river Yamuna, if an initial base line WQI is 35, a quadratic expression is formed to measure the impact of investment of ₹ $x$ crore on the WQI i.e., $W (x)$ as
$W (x) = 35 + 1.6 x - 0.02 x^{2}$ , where $x$ is in Rupees in crores
Use the given $W (x)$ to answer the following questions:
1. (i)
  Find the value of $x$ at the stationary point for the function $W (x)$ .
  1 mark
2. (ii)
  Find the maximum value of the WQI using the second derivative test?
  1 mark
3. (iii)(a)
  If the investment is ₹ 35 crores, find by how much the WQI is short of its maximum achievable value.
  2 marks
4. OR
  (iii)(b)
  Evaluate the WQI when the investment is ₹60 crore. Determine whether this indicates an enhancement or a diminishing return in our objective when compared with the optimum WQI.
  2 marks
Figure in the paper: Photo of the polluted Yamuna river with a railway bridge in the background and garbage on the bank. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $W^{'} (x) = 1.6 - 0.04 x$ . Setting $W^{'} (x) = 0$ gives $x = \frac{1.6}{0.04} = 40$ . So the stationary point is at $x = 40$ (₹40 crore).
  1 mark
2. (ii) $W^{''} (x) = - 0.04 < 0$ , so $W$ has a maximum at $x = 40$ by the second derivative test. Maximum WQI $= W (40) = 35 + 64 - 32 = 67$ .
  1 mark
3. (iii)(a) $W (35) = 35 + 1.6 (35) - 0.02 (35)^{2} = 35 + 56 - 24.5 = 66.5$ .
  1 mark
4. (iii)(a) Shortfall from the maximum $= 67 - 66.5 = 0.5$ . So the WQI is short by $0.5$ units.
  1 mark
5. (iii)(b) $W (60) = 35 + 1.6 (60) - 0.02 (3600) = 35 + 96 - 72 = 59$ .
  1 mark
6. (iii)(b) Since $59 < 67$ , raising the investment beyond ₹40 crore to ₹60 crore actually lowers the WQI. This indicates a diminishing return.
  1 mark
Answer: (i) $x = 40$ crore; (ii) max WQI $= 67$ ; (iii)(a) short by 0.5 units; (iii)(b) $W (60) = 59$ , diminishing return
Where the marks are
- Differentiates and solves $W^{'} (x) = 0$ to get $x = 40$ (1 mark)
- Shows $W^{''} < 0$ and finds the maximum $W (40) = 67$ (1 mark)
- (iii)(a) Computes $W (35) = 66.5$ (1 mark)
- (iii)(a) States the shortfall of 0.5 (1 mark)
- (iii)(b) Computes $W (60) = 59$ (1 mark)
- (iii)(b) Compares with 67 and concludes diminishing return (1 mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedMatrices
To celebrate their child's birthday, a family decided to distribute fruits among children in three orphanages. The fruit vendor packed three different types of baskets, according to the quantity required at each orphanage, details are given as follows:
Basket A contains 5 kg Apples, 10 kg Oranges and 5 kg Pomegranate
Basket B contains 10 kg Apples, 10 kg Oranges and 5kg Pomegranate
Basket C contains 8 kg Apples, 15 kg Oranges and 7kg Pomegranate
Cost of basket A, B and C are respectively ₹2750, ₹3500, and ₹4100.
Based on the given information, answer the following questions:
1. (i)
  Write the matrix form of the equations involved in this situation and find the adjoint of the coefficient matrix involved.
  2 marks
2. (ii)
  Using inverse of the matrix, calculate the price of each fruit per kg.
  2 marks
Figure in the paper: Photo of a family buying fruits from a fruit vendor with baskets of fruits on a cart. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let the prices per kg of apples, oranges and pomegranates be ₹ $x$ , ₹ $y$ , ₹ $z$ . Then:
  $5 x + 10 y + 5 z = 2750$
  $10 x + 10 y + 5 z = 3500$
  $8 x + 15 y + 7 z = 4100$
  In matrix form this is $A X = B$ , where $A = 5108101015557$ , $X = x y z$ , $B = 275035004100$ .
  1 mark
2. (i) Cofactors of $A$ :
  $C_{11} = - 5, C_{12} = - 30, C_{13} = 70$
  $C_{21} = 5, C_{22} = - 5, C_{23} = 5$
  $C_{31} = 0, C_{32} = 25, C_{33} = - 50$
  Taking the transpose, $adj A = - 5 - 30 70 5 - 5 5 025 - 50$ .
  1 mark
3. (ii) $∣ A ∣ = 5 (70 - 75) - 10 (70 - 40) + 5 (150 - 80) = - 25 - 300 + 350 = 25 \neq = 0$ , so $A^{- 1}$ exists and the system has a unique solution. Then $X = A^{- 1} B = \frac{1}{25} - 5 - 30 70 5 - 5 5 025 - 50 275035004100$ .
  1 mark
4. (ii) Multiplying out:
  $x = \frac{- 13750 + 17500}{25} = 150$
  $y = \frac{- 82500 - 17500 + 102500}{25} = 100$
  $z = \frac{192500 + 17500 - 205000}{25} = 200$
  So apples cost ₹150/kg, oranges ₹100/kg and pomegranates ₹200/kg.
  1 mark
Answer: (i) $A X = B$ with $A = 5108101015557$ , $adj A = - 5 - 30 70 5 - 5 5 025 - 50$ ; (ii) Apples ₹150/kg, Oranges ₹100/kg, Pomegranates ₹200/kg
Where the marks are
- Forms the equations and writes them as $A X = B$ (1 mark)
- Correct adjoint of $A$ (1 mark)
- Finds $∣ A ∣ = 25 \neq = 0$ and writes $X = A^{- 1} B$ (1 mark)
- Obtains $x = 150, y = 100, z = 200$ with the conclusion (1 mark)
Open Q38 on its own page

Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.