MathsCBSE Class 12 · Sample paper, session 2023-24

CBSE Class 12 Mathematics: sample paper 2023-24

The sample question paper CBSE published for the 2023-24 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_2023_24/Maths-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

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

Instructions in brief: All sections compulsory; internal choices in some questions.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQMatrices
If $A = [a_{ij}]$ is a square matrix of order 2 such that $a_{ij} = {1, 0, when i \neq = j when i = j$ , then $A^{2}$ is
1. (A) $[1100]_{2 \times 2}$
2. (B) $[1010]_{2 \times 2}$
3. (C) $[1110]_{2 \times 2}$
4. (D) $[1001]_{2 \times 2}$
Try it first, then show our solutionSolution
1. Here $a_{11} = a_{22} = 0$ and $a_{12} = a_{21} = 1$ , so $A = [0110]$ .
2. $A^{2} = [0 \cdot 0 + 1 \cdot 1 1 \cdot 0 + 0 \cdot 1 0 \cdot 1 + 1 \cdot 0 1 \cdot 1 + 0 \cdot 0] = [1001]$ , which is option (D).
  1 mark
Answer: $A^{2} = I$ , option (D)
Where the marks are
- Writes $A$ correctly and squares it to get $I$ (1 mark)
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Q2 · 1 mark · MCQMatrices
If $A$ and $B$ are invertible square matrices of the same order, then which of the following is not correct?
1. (A) $∣ A B^{- 1} ∣ = \frac{∣ A ∣}{∣ B ∣}$
2. (B) $∣ (A B)^{- 1} ∣ = \frac{1}{∣ A ∣∣ B ∣}$
3. (C) $(A B)^{- 1} = B^{- 1} A^{- 1}$
4. (D) $(A + B)^{- 1} = B^{- 1} + A^{- 1}$
Try it first, then show our solutionSolution
1. (A), (B) and (C) follow from $∣ A B ∣ = ∣ A ∣∣ B ∣$ , $∣ B^{- 1} ∣ = \frac{1}{∣ B ∣}$ and the reversal law $(A B)^{- 1} = B^{- 1} A^{- 1}$ , so they are true.
2. There is no such rule for the inverse of a sum. For example, with $A = B = I$ : $(2 I)^{- 1} = \frac{1}{2} I$ , whereas $I + I = 2 I$ . So (D) is not correct.
  1 mark
Answer: Option (D)
Where the marks are
- Identifies $(A + B)^{- 1} = B^{- 1} + A^{- 1}$ as false (1 mark)
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Q3 · 1 mark · MCQDeterminants
If the area of the triangle with vertices $(- 3, 0)$ , $(3, 0)$ and $(0, k)$ is 9 sq units, then the value/s of $k$ will be
1. (A)9
2. (B) $\pm 3$
3. (C)-9
4. (D)6
Try it first, then show our solutionSolution
1. Area $= \frac{1}{2} - 3 30 00 k 111$ . Expanding along $C_{2}$ gives the determinant $= - k (- 3 - 3) = 6 k$ , so Area $= \frac{1}{2} ∣6 k ∣ = 3∣ k ∣$ .
2. Setting $3∣ k ∣ = 9$ gives $k = \pm 3$ , which is option (B).
  1 mark
Answer: $k = \pm 3$ , option (B)
Where the marks are
- Uses the determinant area formula and solves $3∣ k ∣ = 9$ (1 mark)
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Q4 · 1 mark · MCQContinuity and Differentiability
If $f (x) = {\frac{k x}{∣ x ∣}, 3, if x < 0 if x \geq 0$ is continuous at $x = 0$ , then the value of $k$ is
1. (A) $- 3$
2. (B)0
3. (C)3
4. (D)any real number
Try it first, then show our solutionSolution
1. For $x < 0$ , $∣ x ∣ = - x$ , so $f (x) = \frac{k x}{- x} = - k$ . Hence LHL $= - k$ .
2. RHL $= f (0) = 3$ . Continuity at $x = 0$ requires $- k = 3$ , so $k = - 3$ , which is option (A).
  1 mark
Answer: $k = - 3$ , option (A)
Where the marks are
- Equates LHL $= - k$ with $f (0) = 3$ to get $k = - 3$ (1 mark)
Q5 · 1 mark · MCQThree Dimensional Geometry
The lines $r = \hat{i} + \hat{j} - \hat{k} + λ (2 \hat{i} + 3 \hat{j} - 6 \hat{k})$ and $r = 2 \hat{i} - \hat{j} - \hat{k} + μ (6 \hat{i} + 9 \hat{j} - 18 \hat{k})$ ; (where $λ$ & $μ$ are scalars) are
1. (A)coincident
2. (B)skew
3. (C)intersecting
4. (D)parallel
Try it first, then show our solutionSolution
1. $6 \hat{i} + 9 \hat{j} - 18 \hat{k} = 3 (2 \hat{i} + 3 \hat{j} - 6 \hat{k})$ , so the two lines have parallel directions.
2. Check whether the point $(1, 1, - 1)$ lies on the second line: $1 = 2 + 6 μ$ gives $μ = - \frac{1}{6}$ , but $1 = - 1 + 9 μ$ gives $μ = \frac{2}{9}$ . The values do not agree, so the point does not lie on it. The lines are therefore parallel, not coincident: option (D).
  1 mark
Answer: Parallel, option (D)
Where the marks are
- Shows the directions are proportional and the lines are not coincident (1 mark)
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Q6 · 1 mark · MCQDifferential Equations
The degree of the differential equation $[1 + (\frac{d y}{d x})^{2}]^{3} = (\frac{d ^{2} y}{d x ^{2}})^{2}$ is
1. (A)4
2. (B) $\frac{3}{2}$
3. (C)2
4. (D)Not defined
Try it first, then show our solutionSolution
1. The equation is polynomial in its derivatives. The highest order derivative is $\frac{d ^{2} y}{d x ^{2}}$ , and it appears with power 2, so the degree is 2: option (C).
  1 mark
Answer: 2, option (C)
Where the marks are
- Degree = power of the highest-order derivative = 2 (1 mark)
Q7 · 1 mark · MCQLinear Programming
The corner points of the bounded feasible region determined by a system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$ . Let $Z = p x + q y$ , where $p, q > 0$ . The condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is
1. (A) $p = 2 q$
2. (B) $p = \frac{q}{2}$
3. (C) $p = 3 q$
4. (D) $p = q$
Try it first, then show our solutionSolution
1. At $(3, 0)$ , $Z = 3 p$ ; at $(1, 1)$ , $Z = p + q$ .
2. For the minimum to occur at both points, $3 p = p + q$ , so $q = 2 p$ , i.e. $p = \frac{q}{2}$ : option (B).
  1 mark
Answer: $p = \frac{q}{2}$ , option (B)
Where the marks are
- Equates $Z$ at both points to get $q = 2 p$ (1 mark)
Q8 · 1 mark · MCQVector Algebra
$A B C D$ is a rhombus whose diagonals intersect at $E$ . Then $E A + E B + E C + E D$ equals to
1. (A) $0$
2. (B) $A D$
3. (C) $2 B D$
4. (D) $2 A D$
Try it first, then show our solutionSolution
1. The diagonals of a rhombus bisect each other at $E$ , so $E A = - E C$ and $E B = - E D$ .
2. Adding the four vectors, the pairs cancel and the sum is $0$ : option (A).
  1 mark
Answer: $0$ , option (A)
Where the marks are
- Uses the bisection of diagonals to cancel opposite vectors (1 mark)
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Q9 · 1 mark · MCQIntegrals
For any integer $n$ , the value of $\int_{- π}^{π} e^{c o s^{2} x} sin^{3} (2 n + 1) x d x$ is
1. (A)-1
2. (B)0
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. Let $f (x) = e^{c o s^{2} x} sin^{3} (2 n + 1) x$ . Then $f (- x) = e^{c o s^{2} x} (- sin (2 n + 1) x)^{3} = - f (x)$ , so $f$ is odd.
2. The integral of an odd function over $[- π, π]$ is 0: option (B).
  1 mark
Answer: 0, option (B)
Where the marks are
- Shows the integrand is odd, so the integral is 0 (1 mark)
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Q10 · 1 mark · MCQDeterminants
The value of $∣ A ∣$ , if $A = 0 1 - 2 x - x 2 x - 1 0 - 2 x x 2 x 0$ , where $x \in R^{+}$ , is
1. (A) $(2 x + 1)^{2}$
2. (B)0
3. (C) $(2 x + 1)^{3}$
4. (D) $(2 x - 1)^{2}$
Try it first, then show our solutionSolution
1. Each entry satisfies $a_{j i} = - a_{ij}$ and the diagonal is zero, so $A^{T} = - A$ : $A$ is skew-symmetric.
2. $∣ A ∣ = ∣ A^{T} ∣ = ∣ - A ∣ = (- 1)^{3} ∣ A ∣$ , so $∣ A ∣ = - ∣ A ∣$ and hence $∣ A ∣ = 0$ : option (B).
  1 mark
Answer: 0, option (B)
Where the marks are
- Recognises a skew-symmetric matrix of odd order, so $∣ A ∣ = 0$ (1 mark)
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Q11 · 1 mark · MCQLinear Programming
The feasible region corresponding to the linear constraints of a Linear Programming Problem is given below. Which of the following is not a constraint to the given Linear Programming Problem?
1. (A) $x + y \geq 2$
2. (B) $x + 2 y \leq 10$
3. (C) $x - y \geq 1$
4. (D) $x - y \leq 1$
Figure in the paper: Graph with shaded feasible region bounded by lines x+y=2 (through (2,0) and (0,2)), x-y=1 (through (1,0)), and x+2y=10 (through (10,0) and (0,5)); the shaded region has corner points near (0,5) and (0,2) on the y-axis, and lies to the left of the line x-y=1. See it in the official paper.
Try it first, then show our solutionSolution
1. The shaded region contains points near the origin side, such as $(0, 3)$ . Test $(0, 3)$ in $x - y \geq 1$ : $0 - 3 = - 3 \neq \geq 1$ . So that half-plane does not contain the region.
2. Hence $x - y \geq 1$ is not a constraint; the region in fact satisfies $x - y \leq 1$ . The answer is option (C).
  1 mark
Answer: $x - y \geq 1$ , option (C)
Where the marks are
- Tests a point of the region against $x - y \geq 1$ and rejects it (1 mark)
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Q12 · 1 mark · MCQVector Algebra
If $a = 4 \hat{i} + 6 \hat{j}$ and $b = 3 \hat{j} + 4 \hat{k}$ , then the vector form of the component of $a$ along $b$ is
1. (A) $\frac{18}{5} (3 \hat{i} + 4 \hat{k})$
2. (B) $\frac{18}{25} (3 \hat{j} + 4 \hat{k})$
3. (C) $\frac{18}{5} (3 \hat{i} + 4 \hat{k})$
4. (D) $\frac{18}{25} (4 \hat{i} + 6 \hat{j})$
Try it first, then show our solutionSolution
1. $a \cdot b = 0 + 18 + 0 = 18$ and $∣ b ∣^{2} = 9 + 16 = 25$ .
2. Vector component of $a$ along $b$ $= \frac{a \cdot b}{∣ b ∣ ^{2}} b = \frac{18}{25} (3 \hat{j} + 4 \hat{k})$ : option (B).
  1 mark
Answer: $\frac{18}{25} (3 \hat{j} + 4 \hat{k})$ , option (B)
Where the marks are
- Uses $\frac{a \cdot b}{∣ b ∣ ^{2}} b$ correctly (1 mark)
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Q13 · 1 mark · MCQMatrices
Given that $A$ is a square matrix of order 3 and $∣ A ∣ = - 2$ , then $∣ a d j (2 A) ∣$ is equal to
1. (A) $- 2^{6}$
2. (B) $+ 4$
3. (C) $- 2^{8}$
4. (D) $2^{8}$
Try it first, then show our solutionSolution
1. For order 3, $∣ a d j M ∣ = ∣ M ∣^{2}$ , and $∣2 A ∣ = 2^{3} ∣ A ∣ = - 16$ .
2. So $∣ a d j (2 A) ∣ = (- 16)^{2} = 256 = 2^{8}$ : option (D).
  1 mark
Answer: $2^{8}$ , option (D)
Where the marks are
- Uses $∣ a d j M ∣ = ∣ M ∣^{n - 1}$ and $∣ k A ∣ = k^{n} ∣ A ∣$ (1 mark)
Q14 · 1 mark · MCQProbability
A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is
1. (A) $\frac{1}{4}$
2. (B) $\frac{1}{3}$
3. (C) $\frac{1}{2}$
4. (D) $\frac{3}{4}$
Try it first, then show our solutionSolution
1. The events are independent, so P(none solves) $= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$ .
2. P(solved) $= 1 - \frac{1}{4} = \frac{3}{4}$ : option (D).
  1 mark
Answer: $\frac{3}{4}$ , option (D)
Where the marks are
- Uses $1 - P (\overset{ˉ}{A}) P (\overset{ˉ}{B}) P (\overset{ˉ}{C})$ (1 mark)
Q15 · 1 mark · MCQDifferential Equations
The general solution of the differential equation $y d x - x d y = 0$ ; (Given $x, y > 0$ ), is of the form (Where 'c' is an arbitrary positive constant of integration)
1. (A) $x y = c$
2. (B) $x = c y^{2}$
3. (C) $y = c x$
4. (D) $y = c x^{2}$
Try it first, then show our solutionSolution
1. Separate the variables: $\frac{d y}{y} = \frac{d x}{x}$ . Integrating gives $lo g y = lo g x + lo g c$ .
2. So $y = c x$ : option (C).
  1 mark
Answer: $y = c x$ , option (C)
Where the marks are
- Separates variables and integrates to $y = c x$ (1 mark)
Q16 · 1 mark · MCQVector Algebra
The value of $λ$ for which two vectors $2 \hat{i} - \hat{j} + 2 \hat{k}$ and $3 \hat{i} + λ \hat{j} + \hat{k}$ are perpendicular is
1. (A)2
2. (B)4
3. (C)6
4. (D)8
Try it first, then show our solutionSolution
1. Perpendicular vectors have zero dot product: $2 (3) + (- 1) λ + 2 (1) = 0$ , so $8 - λ = 0$ and $λ = 8$ : option (D).
  1 mark
Answer: $λ = 8$ , option (D)
Where the marks are
- Sets the dot product to 0 and solves (1 mark)
Q17 · 1 mark · MCQContinuity and Differentiability
The set of all points where the function $f (x) = x + ∣ x ∣$ is differentiable, is
1. (A) $(0, \infty)$
2. (B) $(- \infty, 0)$
3. (C) $(- \infty, 0) \cup (0, \infty)$
4. (D) $(- \infty, \infty)$
Try it first, then show our solutionSolution
1. $f (x) = 2 x$ for $x \geq 0$ and $f (x) = 0$ for $x < 0$ . At $x = 0$ , $L f^{'} (0) = 0$ and $R f^{'} (0) = 2$ , so $f$ is not differentiable there.
2. On each side of 0 the function is linear or constant, so it is differentiable everywhere else: $(- \infty, 0) \cup (0, \infty)$ , option (C).
  1 mark
Answer: $(- \infty, 0) \cup (0, \infty)$ , option (C)
Where the marks are
- Shows $L f^{'} (0) \neq = R f^{'} (0)$ and differentiability elsewhere (1 mark)
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Q18 · 1 mark · MCQThree Dimensional Geometry
If the direction cosines of a line are $< \frac{1}{c}, \frac{1}{c}, \frac{1}{c} >$ , then
1. (A) $0 < c < 1$
2. (B) $c > 2$
3. (C) $c = \pm 2$
4. (D) $c = \pm 3$
Try it first, then show our solutionSolution
1. Direction cosines satisfy $l^{2} + m^{2} + n^{2} = 1$ , so $\frac{3}{c ^{2}} = 1$ , giving $c^{2} = 3$ and $c = \pm 3$ : option (D).
  1 mark
Answer: $c = \pm 3$ , option (D)
Where the marks are
- Uses $l^{2} + m^{2} + n^{2} = 1$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedApplication of Derivatives
Let $f (x)$ be a polynomial function of degree 6 such that $\frac{d}{d x} (f (x)) = (x - 1)^{3} (x - 3)^{2}$ , then
ASSERTION (A): $f (x)$ has a minimum at $x = 1$ .
REASON (R): When $\frac{d}{d x} (f (x)) < 0, \forall x \in (a - h, a)$ and $\frac{d}{d x} (f (x)) > 0, \forall x \in (a, a + h)$ ; where 'h' is an infinitesimally small positive quantity, then $f (x)$ has a minimum at $x = a$ , provided $f (x)$ is continuous at $x = a$ .
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. Near $x = 1$ the factor $(x - 3)^{2}$ is positive, so the sign of $f^{'} (x)$ is the sign of $(x - 1)^{3}$ . Thus $f^{'} < 0$ just left of 1 and $f^{'} > 0$ just right of 1.
2. By the first derivative test, which is exactly R, $f$ has a minimum at $x = 1$ . Both statements are true and R explains A: option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Shows the sign change of $f^{'}$ from $-$ to $+$ at $x = 1$ using R (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedRelations and Functions
ASSERTION (A): The relation $f : {1, 2, 3, 4} \to {x, y, z, p}$ defined by $f = {(1, x), (2, y), (3, z)}$ is a bijective function.
REASON (R): The function $f : {1, 2, 3} \to {x, y, z, p}$ such that $f = {(1, x), (2, y), (3, z)}$ is one-one.
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. A: the element 4 has no image, so $f$ is not even a function. A is false.
2. R: 1, 2 and 3 have distinct images $x$ , $y$ , $z$ , so $f$ is one-one. R is true. The answer is option (D).
  1 mark
Answer: Option (D)
Where the marks are
- A false (4 has no image); R true (distinct images) (1 mark)
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Section B

VSA

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerInverse Trigonometric Functions
Find the value of $sin^{- 1} (cos (\frac{33 π}{5}))$ .
Try it first, then show our solutionSolution
1. Since $cos$ has period $2 π$ : $cos \frac{33 π}{5} = cos (6 π + \frac{3 π}{5}) = cos \frac{3 π}{5} = sin (\frac{π}{2} - \frac{3 π}{5}) = sin (- \frac{π}{10})$ .
  1 mark
2. $- \frac{π}{10} \in [- \frac{π}{2}, \frac{π}{2}]$ , the principal value branch of $sin^{- 1}$ . So $sin^{- 1} (cos \frac{33 π}{5}) = sin^{- 1} (sin (- \frac{π}{10})) = - \frac{π}{10}$ .
  1 mark
Answer: $- \frac{π}{10}$
Where the marks are
- Reduces $cos \frac{33 π}{5}$ to $sin (\frac{π}{2} - \frac{3 π}{5})$ (1 mark)
- Uses the principal branch to get $- \frac{π}{10}$ (1 mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerInverse Trigonometric Functions
Find the domain of $sin^{- 1} (x^{2} - 4)$ .
Try it first, then show our solutionSolution
1. $sin^{- 1} t$ is defined only for $- 1 \leq t \leq 1$ . So $- 1 \leq x^{2} - 4 \leq 1 \Rightarrow 3 \leq x^{2} \leq 5 \Rightarrow 3 \leq ∣ x ∣ \leq 5$ .
  1 mark
2. So $x \in [- 5, - 3] \cup [3, 5]$ , and this is the domain.
  1 mark
Answer: $[- 5, - 3] \cup [3, 5]$
Where the marks are
- Sets $- 1 \leq x^{2} - 4 \leq 1$ and gets $3 \leq ∣ x ∣ \leq 5$ (1 mark)
- Writes the domain as a union of intervals (1 mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerApplication of Derivatives
Find the interval/s in which the function $f : R \to R$ defined by $f (x) = x e^{x}$ , is increasing.
Try it first, then show our solutionSolution
1. By the product rule, $f^{'} (x) = x e^{x} + e^{x} = e^{x} (x + 1)$ .
  1 mark
2. Since $e^{x} > 0$ for all $x \in R$ , $f^{'} (x) \geq 0 ⟺ x + 1 \geq 0 ⟺ x \geq - 1$ . Hence $f$ is increasing on $[- 1, \infty)$ .
  1 mark
Answer: $[- 1, \infty)$
Where the marks are
- Finds $f^{'} (x) = e^{x} (x + 1)$ (1 mark)
- Uses $e^{x} > 0$ to conclude $f$ is increasing on $[- 1, \infty)$ (1 mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerApplication of Derivatives
If $f (x) = \frac{1}{4 x ^{2} + 2 x + 1}$ ; $x \in R$ , then find the maximum value of $f (x)$ .
Try it first, then show our solutionSolution
1. Let $g (x) = 4 x^{2} + 2 x + 1 = 4 (x^{2} + \frac{x}{2} + \frac{1}{16}) + 1 - \frac{1}{4} = 4 (x + \frac{1}{4})^{2} + \frac{3}{4} \geq \frac{3}{4}$ , with equality at $x = - \frac{1}{4}$ .
  1½ marks
2. $f (x) = \frac{1}{g ( x )}$ is largest when $g (x)$ is smallest. So the maximum value of $f (x)$ is $\frac{1}{3/4} = \frac{4}{3}$ , attained at $x = - \frac{1}{4}$ .
  ½ mark
Answer: Maximum value $= \frac{4}{3}$ (at $x = - \frac{1}{4}$ )
Where the marks are
- Completes the square to show $4 x^{2} + 2 x + 1 \geq \frac{3}{4}$ (1½ marks)
- Concludes the maximum of $f$ is $\frac{4}{3}$ (½ mark)
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OR
Q23(b) · 2 marks · Very short answerCompetency-basedApplication of Derivatives
Find the maximum profit that a company can make, if the profit function is given by $P (x) = 72 + 42 x - x^{2}$ , where $x$ is the number of units and $P$ is the profit in rupees.
Try it first, then show our solutionSolution
1. $P^{'} (x) = 42 - 2 x$ . For a maximum or minimum, $P^{'} (x) = 0 \Rightarrow x = 21$ .
  ½ mark
2. $P^{''} (x) = - 2 < 0$ , so $P$ is maximum at $x = 21$ .
  ½ mark
3. Maximum profit $= P (21) = 72 + 42 (21) - (21)^{2} = 72 + 882 - 441 = ₹513$ .
  1 mark
Answer: Maximum profit = ₹513 (at $x = 21$ units)
Where the marks are
- Sets $P^{'} (x) = 0$ and gets $x = 21$ (½ mark)
- Shows $P^{''} (21) < 0$ , so a maximum (½ mark)
- Computes the maximum profit ₹513 (1 mark)
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Q24 · 2 marks · Very short answerIntegrals
Evaluate : $\int_{- 1}^{1} lo g (\frac{2 - x}{2 + x}) d x$ .
Try it first, then show our solutionSolution
1. Let $f (x) = lo g (\frac{2 - x}{2 + x})$ . Then $f (- x) = lo g (\frac{2 + x}{2 - x}) = - lo g (\frac{2 - x}{2 + x}) = - f (x)$ , so $f$ is an odd function.
  1 mark
2. For an odd function, $\int_{- a}^{a} f (x) d x = 0$ . Hence $\int_{- 1}^{1} lo g (\frac{2 - x}{2 + x}) d x = 0$ .
  1 mark
Answer: $0$
Where the marks are
- Shows $f (- x) = - f (x)$ (1 mark)
- Uses the odd-function property to get 0 (1 mark)
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Q25 · 2 marks · Very short answerApplication of Derivatives
Check whether the function $f : R \to R$ defined by $f (x) = x^{3} + x$ , has any critical point/s or not ? If yes, then find the point/s.
Try it first, then show our solutionSolution
1. $f^{'} (x) = 3 x^{2} + 1$ . Since $x^{2} \geq 0$ for all $x \in R$ , $f^{'} (x) \geq 1 > 0$ , so $f^{'} (x) \neq = 0$ for every real $x$ . Also $f$ is a polynomial, so $f^{'}$ exists everywhere.
  1½ marks
2. Hence $f$ has no critical point.
  ½ mark
Answer: No critical point exists.
Where the marks are
- Finds $f^{'} (x) = 3 x^{2} + 1 > 0$ for all real $x$ (1½ marks)
- Concludes that there is no critical point (½ mark)
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Section C

Q26 · 3 marks · Short answerIntegrals
Find : $\int \frac{2 x ^{2} + 3}{x ^{2} ( x ^{2} + 9 )} d x$ ; $x \neq = 0$ .
Try it first, then show our solutionSolution
1. Put $x^{2} = t$ (for the partial fractions only): $\frac{2 t + 3}{t ( t + 9 )} = \frac{A}{t} + \frac{B}{t + 9}$ .
  ½ mark
2. Then $2 t + 3 = A (t + 9) + B t$ . At $t = 0$ : $A = \frac{1}{3}$ . At $t = - 9$ : $- 15 = - 9 B$ , so $B = \frac{5}{3}$ .
  1 mark
3. So $\int \frac{2 x ^{2} + 3}{x ^{2} ( x ^{2} + 9 )} d x = \frac{1}{3} \int \frac{d x}{x ^{2}} + \frac{5}{3} \int \frac{d x}{x ^{2} + 3 ^{2}}$ .
  ½ mark
4. $= - \frac{1}{3 x} + \frac{5}{3} \cdot \frac{1}{3} tan^{- 1} \frac{x}{3} + C = - \frac{1}{3 x} + \frac{5}{9} tan^{- 1} (\frac{x}{3}) + C$ .
  1 mark
Answer: $- \frac{1}{3 x} + \frac{5}{9} tan^{- 1} (\frac{x}{3}) + C$
Where the marks are
- Substitutes $x^{2} = t$ and sets up partial fractions (½ mark)
- Finds $A = \frac{1}{3}$ , $B = \frac{5}{3}$ (1 mark)
- Splits the integral correctly (½ mark)
- Integrates to $- \frac{1}{3 x} + \frac{5}{9} tan^{- 1} \frac{x}{3} + C$ (1 mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerProbability
The random variable $X$ has a probability distribution $P (X)$ of the following form, where 'k' is some real number:
$P (X) = ⎩ ⎨ ⎧ k, 2 k, 3 k, 0, if x = 0 if x = 1 if x = 2 otherwise$
1. (i)
  Determine the value of $k$ .
  0 marks
2. (ii)
  Find $P (X < 2)$ .
  0 marks
3. (iii)
  Find $P (X > 2)$ .
  0 marks
Try it first, then show our solutionSolution
1. (i) The probabilities add to 1: $k + 2 k + 3 k = 1 \Rightarrow 6 k = 1 \Rightarrow k = \frac{1}{6}$ .
  1 mark
2. (ii) $P (X < 2) = P (X = 0) + P (X = 1) = k + 2 k = 3 k = 3 \times \frac{1}{6} = \frac{1}{2}$ .
  1 mark
3. (iii) $X$ takes no value greater than 2, so $P (X > 2) = 0$ .
  1 mark
Answer: (i) $k = \frac{1}{6}$ ; (ii) $\frac{1}{2}$ ; (iii) $0$
Where the marks are
- Uses $\sum P (X) = 1$ to get $k = \frac{1}{6}$ (1 mark)
- $P (X < 2) = 3 k = \frac{1}{2}$ (1 mark)
- $P (X > 2) = 0$ (1 mark)
Open Q27 on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntegrals
Find : $\int \frac{x}{1 - x ^{3}} d x$ ; $x \in (0, 1)$ .
Try it first, then show our solutionSolution
1. Let $x^{3/2} = t \Rightarrow d t = \frac{3}{2} x^{1/2} d x \Rightarrow x d x = \frac{2}{3} d t$ .
  ½ mark
2. $\int \frac{x}{1 - x ^{3}} d x = \frac{2}{3} \int \frac{d t}{1 - t ^{2}}$ .
  ½ mark
3. $= \frac{2}{3} sin^{- 1} t + C$ .
  1 mark
4. $= \frac{2}{3} sin^{- 1} (x^{3/2}) + C$ .
  1 mark
Answer: $\frac{2}{3} sin^{- 1} (x^{3/2}) + C$
Where the marks are
- Chooses the substitution $t = x^{3/2}$ (½ mark)
- Converts to $\frac{2}{3} \int \frac{d t}{1 - t ^{2}}$ (½ mark)
- Integrates to $\frac{2}{3} sin^{- 1} t$ (1 mark)
- Back-substitutes to give the answer in $x$ (1 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntegrals
Evaluate: $\int_{0}^{π /4} lo g (1 + tan x) d x$ .
Try it first, then show our solutionSolution
1. Let $I = \int_{0}^{π /4} lo g (1 + tan x) d x$ ...(i). Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{π /4} lo g (1 + tan (\frac{π}{4} - x)) d x$ .
  1 mark
2. $1 + tan (\frac{π}{4} - x) = 1 + \frac{1 - t a n x}{1 + t a n x} = \frac{2}{1 + t a n x}$ , so $I = \int_{0}^{π /4} lo g 2 d x - \int_{0}^{π /4} lo g (1 + tan x) d x = \int_{0}^{π /4} lo g 2 d x - I$ [by (i)].
  1 mark
3. $2 I = \frac{π}{4} lo g 2 \Rightarrow I = \frac{π}{8} lo g 2$ .
  1 mark
Answer: $\frac{π}{8} lo g 2$
Where the marks are
- Applies the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ (1 mark)
- Simplifies to $lo g 2 - lo g (1 + tan x)$ and obtains $2 I$ (1 mark)
- Gets $I = \frac{π}{8} lo g 2$ (1 mark)
Open Q28(b) on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerDifferential Equations
Solve the differential equation: $y e^{x / y} d x = (x e^{x / y} + y^{2}) d y$ , $(y \neq = 0)$ .
Try it first, then show our solutionSolution
1. Rearrange: $e^{x / y} (y d x - x d y) = y^{2} d y \Rightarrow e^{x / y} (\frac{y d x - x d y}{y ^{2}}) = d y$ .
  1 mark
2. Since $d (\frac{x}{y}) = \frac{y d x - x d y}{y ^{2}}$ , this becomes $e^{x / y} d (\frac{x}{y}) = d y$ .
  1 mark
3. Integrating both sides: $\int e^{x / y} d (\frac{x}{y}) = \int d y \Rightarrow e^{x / y} = y + C$ , where $C$ is an arbitrary constant.
  1 mark
Answer: $e^{x / y} = y + C$
Where the marks are
- Rearranges to $e^{x / y} \frac{y d x - x d y}{y ^{2}} = d y$ (1 mark)
- Recognises $d (x / y)$ (1 mark)
- Integrates to $e^{x / y} = y + C$ (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerDifferential Equations
Solve the differential equation: $(cos^{2} x) \frac{d y}{d x} + y = tan x$ ; $(0 \leq x < \frac{π}{2})$ .
Try it first, then show our solutionSolution
1. Divide by $cos^{2} x$ : $\frac{d y}{d x} + y sec^{2} x = tan x sec^{2} x$ . This is linear, with $P = sec^{2} x$ and $Q = tan x sec^{2} x$ .
  ½ mark
2. I.F. $= e^{\int s e c^{2} x d x} = e^{t a n x}$ .
  ½ mark
3. Solution: $y e^{t a n x} = \int e^{t a n x} tan x sec^{2} x d x$ . Put $t = tan x$ , $d t = sec^{2} x d x$ : $\int t e^{t} d t = t e^{t} - e^{t} + C$ .
  1 mark
4. So $y e^{t a n x} = (tan x - 1) e^{t a n x} + C$ , i.e. $y = tan x - 1 + C e^{- t a n x}$ .
  1 mark
Answer: $y = tan x - 1 + C e^{- t a n x}$
Where the marks are
- Writes the equation in linear form and identifies $P, Q$ (½ mark)
- Finds I.F. $= e^{t a n x}$ (½ mark)
- Sets up $y \cdot$ I.F. $= \int Q \cdot$ I.F. $d x$ with the substitution $t = tan x$ (1 mark)
- Integrates by parts to get the general solution (1 mark)
Open Q29(b) on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerLinear Programming
Solve the following Linear Programming Problem graphically:
Minimize: $z = x + 2 y$ ,
subject to the constraints: $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x, y \geq 0$ .
Try it first, then show our solutionSolution
1. Draw the lines $x + 2 y = 100$ (through $(100, 0), (0, 50)$ ), $2 x - y = 0$ (through $(0, 0), (50, 100)$ ) and $2 x + y = 200$ (through $(100, 0), (0, 200)$ ) in the first quadrant. The feasible region lies above $x + 2 y = 100$ , on the side of $2 x - y = 0$ towards the $y$ -axis, and below $2 x + y = 200$ . It is the bounded quadrilateral with vertices $A (0, 50)$ , $B (20, 40)$ , $C (50, 100)$ , $D (0, 200)$ . Here $B$ comes from solving $x + 2 y = 100, y = 2 x$ , and $C$ from solving $y = 2 x, 2 x + y = 200$ .
  1½ marks
2. Values of $Z = x + 2 y$ at the corners:
  Corner point $Z = x + 2 y$
  $A (0, 50)$ 100 (minimum)
  $B (20, 40)$ 100 (minimum)
  $C (50, 100)$ 250
  $D (0, 200)$ 400
  1 mark
3. The minimum value of $Z$ is 100. It occurs at every point on the segment joining $(0, 50)$ and $(20, 40)$ .
  ½ mark
Answer: Minimum $Z = 100$ at all points on the segment joining $(0, 50)$ and $(20, 40)$
Where the marks are
- Correct graph and feasible region with corner points (1½ marks)
- Evaluates $Z$ at the corner points (1 mark)
- States the minimum 100 on segment AB (½ mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerLinear Programming
Solve the following Linear Programming Problem graphically:
Maximize: $z = - x + 2 y$ ,
subject to the constraints: $x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0$ .
Try it first, then show our solutionSolution
1. Draw $x = 3$ , $x + y = 5$ and $x + 2 y = 6$ with $y \geq 0$ . The feasible region lies to the right of $x = 3$ and above both lines. It is unbounded, with corner points $A (3, 2)$ (from $x = 3$ , $x + y = 5$ ), $B (4, 1)$ (from $x + y = 5$ , $x + 2 y = 6$ ) and $C (6, 0)$ .
  1 mark
2. Values of $Z = - x + 2 y$ :
  Corner point $Z = - x + 2 y$
  $A (3, 2)$ 1
  $B (4, 1)$ $- 2$
  $C (6, 0)$ $- 6$
  ½ mark
3. Because the region is unbounded, $Z = 1$ may not be the maximum. Draw the open half-plane $- x + 2 y > 1$ . It has points in common with the feasible region; for example $(3, 5)$ is feasible and gives $Z = 7 > 1$ . Hence $Z$ has no maximum value.
  1½ marks
Answer: $Z$ has no maximum value (the feasible region is unbounded).
Where the marks are
- Correct graph of the unbounded feasible region (1 mark)
- Corner points and values of $Z$ (½ mark)
- Checks the half-plane $- x + 2 y > 1$ and concludes there is no maximum (1½ marks)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerDifferential Equations
If $(a + b x) e^{y / x} = x$ then prove that $x \frac{d ^{2} y}{d x ^{2}} = (\frac{a}{a + b x})^{2}$ .
Try it first, then show our solutionSolution
1. $e^{y / x} = \frac{x}{a + b x} \Rightarrow \frac{y}{x} = lo g x - lo g (a + b x)$ .
  ½ mark
2. Differentiate with respect to $x$ : $\frac{x \frac{d y}{d x} - y}{x ^{2}} = \frac{1}{x} - \frac{b}{a + b x}$ .
  1 mark
3. $\Rightarrow x \frac{d y}{d x} - y = x^{2} \cdot \frac{a + b x - b x}{x ( a + b x )} = \frac{a x}{a + b x}$ .
  ½ mark
4. Differentiate again: $x \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - \frac{d y}{d x} = \frac{( a + b x ) a - a x \cdot b}{( a + b x ) ^{2}}$ .
  ½ mark
5. $\Rightarrow x \frac{d ^{2} y}{d x ^{2}} = \frac{a ^{2}}{( a + b x ) ^{2}} = (\frac{a}{a + b x})^{2}$ . Hence proved.
  ½ mark
Answer: $x \frac{d ^{2} y}{d x ^{2}} = (\frac{a}{a + b x})^{2}$ (proved)
Where the marks are
- Takes logs to get $\frac{y}{x} = lo g x - lo g (a + b x)$ (½ mark)
- Differentiates correctly once (1 mark)
- Simplifies to $x y^{'} - y = \frac{a x}{a + b x}$ (½ mark)
- Differentiates a second time (½ mark)
- Simplifies to the required result (½ mark)
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Corner point	$Z = x + 2 y$
$A (0, 50)$	100 (minimum)
$B (20, 40)$	100 (minimum)
$C (50, 100)$	250
$D (0, 200)$	400

Corner point	$Z = - x + 2 y$
$A (3, 2)$	1
$B (4, 1)$	$- 2$
$C (6, 0)$	$- 6$

Section D

Q32 · 5 marks · Long answerApplication of Integrals
Make a rough sketch of the region ${(x, y) : 0 \leq y \leq x^{2} + 1, 0 \leq y \leq x + 1, 0 \leq x \leq 2}$ and find the area of the region, using the method of integration.
Try it first, then show our solutionSolution
1. Sketch: draw the parabola $y = x^{2} + 1$ (vertex $(0, 1)$ ) and the line $y = x + 1$ for $0 \leq x \leq 2$ . The region lies above the $x$ -axis and below the lower of the two curves. It is bounded by $P (0, 1)$ , the parabola up to $Q (1, 2)$ , the line from $Q (1, 2)$ to $R (2, 3)$ , the line $x = 2$ down to $T (2, 0)$ , and the $x$ -axis back to $O$ .
  1 mark
2. Intersection: $x^{2} + 1 = x + 1 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0, 1$ , giving $P (0, 1)$ and $Q (1, 2)$ . On $[0, 1]$ , $x^{2} + 1 \leq x + 1$ ; on $[1, 2]$ , $x + 1 \leq x^{2} + 1$ .
  1 mark
3. Required area $= \int_{0}^{1} (x^{2} + 1) d x + \int_{1}^{2} (x + 1) d x$ .
  1 mark
4. $= [\frac{x ^{3}}{3} + x]_{0}^{1} + [\frac{x ^{2}}{2} + x]_{1}^{2}$ .
  ½ mark
5. $= (\frac{1}{3} + 1) + [(2 + 2) - (\frac{1}{2} + 1)] = \frac{4}{3} + \frac{5}{2}$ .
  ½ mark
6. $= \frac{23}{6}$ sq units.
  1 mark
Answer: $\frac{23}{6}$ sq units
Where the marks are
- Correct rough sketch of the region (1 mark)
- Finds the points of intersection $(0, 1)$ and $(1, 2)$ (1 mark)
- Sets up the area as the sum of two integrals (1 mark)
- Integrates correctly (½ mark)
- Substitutes the limits correctly (½ mark)
- Final area $\frac{23}{6}$ sq units (1 mark)
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Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerRelations and Functions
Let $N$ be the set of all natural numbers and $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d) \Leftrightarrow a d = b c$ for all $(a, b), (c, d) \in N \times N$ . Show that $R$ is an equivalence relation on $N \times N$ . Also, find the equivalence class of $(2, 6)$ , i.e., $[(2, 6)]$ .
Try it first, then show our solutionSolution
1. Reflexive: For any $(a, b) \in N \times N$ , $ab = ba$ because multiplication is commutative. So $(a, b) R (a, b)$ , and $R$ is reflexive.
  1 mark
2. Symmetric: Suppose $(a, b) R (c, d)$ . Then $a d = b c$ , so $c b = d a$ . This means $(c, d) R (a, b)$ , and $R$ is symmetric.
  1 mark
3. Transitive: Suppose $(a, b) R (c, d)$ and $(c, d) R (e, f)$ . Then $a d = b c$ and $c f = d e$ . Multiplying, $(a d) (c f) = (b c) (d e)$ , so $a f \cdot c d = b e \cdot c d$ . Since $c, d \in N$ , $c d \neq = 0$ , which gives $a f = b e$ . So $(a, b) R (e, f)$ , and $R$ is transitive.
  1 mark
4. Since $R$ is reflexive, symmetric and transitive, $R$ is an equivalence relation on $N \times N$ .
  ½ mark
5. $[(2, 6)] = {(x, y) \in N \times N : (x, y) R (2, 6)} = {(x, y) : 6 x = 2 y} = {(x, y) : y = 3 x}$ .
  ½ mark
6. $[(2, 6)] = {(x, 3 x) : x \in N} = {(1, 3), (2, 6), (3, 9), \dots}$ .
  1 mark
Answer: $R$ is an equivalence relation; $[(2, 6)] = {(x, 3 x) : x \in N} = {(1, 3), (2, 6), (3, 9), \dots}$
Where the marks are
- Proves reflexivity (1 mark)
- Proves symmetry (1 mark)
- Proves transitivity, cancelling $c d \neq = 0$ (1 mark)
- Concludes that $R$ is an equivalence relation (½ mark)
- Sets up the condition $6 x = 2 y$ for the class (½ mark)
- Writes $[(2, 6)] = {(x, 3 x) : x \in N}$ (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerRelations and Functions
Show that the function $f : R \to {x \in R : - 1 < x < 1}$ defined by $f (x) = \frac{x}{1 + ∣ x ∣}$ , $x \in R$ is one-one and onto function.
Try it first, then show our solutionSolution
1. $f (x) = {\frac{x}{1 + x}, \frac{x}{1 - x}, x \geq 0 x < 0$
  **Case 1 ( $x \geq 0$ ), injectivity:** If $x, y \geq 0$ and $\frac{x}{1 + x} = \frac{y}{1 + y}$ , then $x + x y = y + x y$ , so $x = y$ . Hence $f$ is one-one on $[0, \infty)$ .
  1 mark
2. Case 1, surjectivity: For $x \geq 0$ , $f (x) \geq 0$ and $f (x) = 1 - \frac{1}{1 + x} < 1$ , so $f (x) \in [0, 1)$ . For any $y \in [0, 1)$ , take $x = \frac{y}{1 - y} \geq 0$ . Then $f (x) = \frac{y / ( 1 - y )}{1 + y / ( 1 - y )} = y$ . So $f$ maps $[0, \infty)$ onto $[0, 1)$ .
  1 mark
3. **Case 2 ( $x < 0$ ), injectivity:** If $x, y < 0$ and $\frac{x}{1 - x} = \frac{y}{1 - y}$ , then $x - x y = y - x y$ , so $x = y$ . Hence $f$ is one-one on $(- \infty, 0)$ .
  1 mark
4. Case 2, surjectivity: For $x < 0$ , $f (x) < 0$ and $f (x) = - 1 + \frac{1}{1 - x} > - 1$ , so $f (x) \in (- 1, 0)$ . For any $y \in (- 1, 0)$ , take $x = \frac{y}{1 + y} < 0$ . Then $f (x) = \frac{y / ( 1 + y )}{1 - y / ( 1 + y )} = y$ . So $f$ maps $(- \infty, 0)$ onto $(- 1, 0)$ .
  1 mark
5. **Case 3 ( $x \geq 0$ , $y < 0$ ):** $f (x) \geq 0$ while $f (y) < 0$ , so $f (x) \neq = f (y)$ . (Equivalently, $\frac{x}{1 + x} = \frac{y}{1 - y}$ gives $x - y = 2 x y$ , where the LHS is $> 0$ and the RHS is $\leq 0$ , which is impossible.) So $f$ is one-one on all of $R$ . Since its range is $[0, 1) \cup (- 1, 0) = (- 1, 1)$ , the whole codomain, $f$ is also onto. Hence $f$ is one-one and onto.
  1 mark
Answer: $f$ is one-one and onto from $R$ to $(- 1, 1)$ .
Where the marks are
- Injectivity for $x \geq 0$ (1 mark)
- Surjectivity onto $[0, 1)$ for $x \geq 0$ , with the pre-image $\frac{y}{1 - y}$ (1 mark)
- Injectivity for $x < 0$ (1 mark)
- Surjectivity onto $(- 1, 0)$ for $x < 0$ , with the pre-image $\frac{y}{1 + y}$ (1 mark)
- Mixed-sign case and the final conclusion (1 mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerMatrices
Using the matrix method, solve the following system of linear equations :
$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$ .
Try it first, then show our solutionSolution
1. Let $u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z}$ . The system becomes $A X = B$ , where $A = 246 3 - 6 9 105 - 20$ , $X = 1/ x 1/ y 1/ z$ and $B = 412$ .
  ½ mark
2. $∣ A ∣ = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) = 150 + 330 + 720 = 1200 \neq = 0$ , so $A^{- 1}$ exists.
  ½ mark
3. Cofactors: $A_{11} = 75, A_{12} = 110, A_{13} = 72$ ; $A_{21} = 150, A_{22} = - 100, A_{23} = 0$ ; $A_{31} = 75, A_{32} = 30, A_{33} = - 24$ .
  So $adj A = 7511072 150 - 100 0 7530 - 24$ .
  1½ marks
4. $A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{1200} 7511072 150 - 100 0 7530 - 24$ .
  ½ mark
5. $X = A^{- 1} B = \frac{1}{1200} 7511072 150 - 100 0 7530 - 24 412$ .
  ½ mark
6. $= \frac{1}{1200} 300 + 150 + 150 440 - 100 + 60 288 + 0 - 48 = \frac{1}{1200} 600400240 = 1/2 1/3 1/5$ .
  ½ mark
7. So $\frac{1}{x} = \frac{1}{2}, \frac{1}{y} = \frac{1}{3}, \frac{1}{z} = \frac{1}{5}$ . Hence $x = 2, y = 3, z = 5$ .
  1 mark
Answer: x = 2, y = 3, z = 5
Where the marks are
- Writes the system as $A X = B$ with $X$ in terms of $1/ x, 1/ y, 1/ z$ (½ mark)
- Finds $∣ A ∣ = 1200 \neq = 0$ (½ mark)
- Finds the correct adjoint of $A$ (1½ marks)
- Writes $A^{- 1} = \frac{1}{1200} adj A$ (½ mark)
- Uses $X = A^{- 1} B$ (½ mark)
- Multiplies correctly to get $1/2, 1/3, 1/5$ (½ mark)
- States $x = 2, y = 3, z = 5$ (1 mark)
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Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerThree Dimensional Geometry
Find the coordinates of the image of the point $(1, 6, 3)$ with respect to the line $r = (\hat{j} + 2 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 3 \hat{k})$ ; where ' $λ$ ' is a scalar. Also, find the distance of the image from the $y$ -axis.
Try it first, then show our solutionSolution
1. The line in Cartesian form is $\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = λ$ . Any point on it is $(λ, 2 λ + 1, 3 λ + 2)$ . Let the foot of the perpendicular from $P (1, 6, 3)$ be $L (λ, 2 λ + 1, 3 λ + 2)$ .
  ½ mark
2. Direction ratios of $P L$ : $λ - 1, 2 λ - 5, 3 λ - 1$ .
  ½ mark
3. $P L ⊥$ line (d.r. $1, 2, 3$ ): $(λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0 \Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1$ . So $L = (1, 3, 5)$ .
  1 mark
4. Let $Q (x_{1}, y_{1}, z_{1})$ be the image of $P$ . Then $L$ is the mid-point of $P Q$ : $\frac{x _{1} + 1}{2} = 1, \frac{y _{1} + 6}{2} = 3, \frac{z _{1} + 3}{2} = 5$ .
  1 mark
5. $x_{1} = 1, y_{1} = 0, z_{1} = 7$ . The image is $Q (1, 0, 7)$ .
  1 mark
6. Distance of $(1, 0, 7)$ from the $y$ -axis $= x_{1}^{2} + z_{1}^{2} = 1^{2} + 7^{2} = 50 = 52$ units.
  1 mark
Answer: Image $(1, 0, 7)$ ; distance from y-axis $50 = 52$ units
Where the marks are
- Writes the general point $(λ, 2 λ + 1, 3 λ + 2)$ on the line (½ mark)
- Finds direction ratios of $P L$ (½ mark)
- Uses perpendicularity to get $λ = 1$ and foot $L (1, 3, 5)$ (1 mark)
- Uses $L$ as mid-point of $P$ and its image (1 mark)
- Obtains the image $(1, 0, 7)$ (1 mark)
- Distance from y-axis $50$ units (1 mark)
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OR
Q35(b) · 5 marks · Long answerCompetency-basedThree Dimensional Geometry
An aeroplane is flying along the line $r = λ (\hat{i} - \hat{j} + \hat{k})$ ; where ' $λ$ ' is a scalar and another aeroplane is flying along the line $r = \hat{i} - \hat{j} + μ (- 2 \hat{j} + \hat{k})$ ; where ' $μ$ ' is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest? Find the shortest possible distance between them.
Try it first, then show our solutionSolution
1. The direction vectors $\hat{i} - \hat{j} + \hat{k}$ and $- 2 \hat{j} + \hat{k}$ are not parallel, so the lines are non-parallel. A point $P$ on line 1 is $P (λ, - λ, λ)$ and a point $Q$ on line 2 is $Q (1, - 1 - 2 μ, μ)$ .
  ½ mark
2. $P Q = (1 - λ) \hat{i} + (- 1 - 2 μ + λ) \hat{j} + (μ - λ) \hat{k}$ . For the shortest distance $P Q$ must be perpendicular to both lines.
  ½ mark
3. $P Q \cdot (\hat{i} - \hat{j} + \hat{k}) = 0$ : $(1 - λ) - (- 1 - 2 μ + λ) + (μ - λ) = 0 \Rightarrow 2 + 3 μ - 3 λ = 0$ .
  ½ mark
4. $P Q \cdot (- 2 \hat{j} + \hat{k}) = 0$ : $- 2 (- 1 - 2 μ + λ) + (μ - λ) = 0 \Rightarrow 2 + 5 μ - 3 λ = 0$ .
  ½ mark
5. Subtracting: $2 μ = 0 \Rightarrow μ = 0$ , and then $3 λ = 2 \Rightarrow λ = \frac{2}{3}$ .
  1 mark
6. The aeroplanes should be at $P (\frac{2}{3}, - \frac{2}{3}, \frac{2}{3})$ , i.e. $\frac{2}{3} (\hat{i} - \hat{j} + \hat{k})$ , and at $Q (1, - 1, 0)$ , i.e. $\hat{i} - \hat{j}$ .
  1 mark
7. $P Q = \frac{1}{3} \hat{i} - \frac{1}{3} \hat{j} - \frac{2}{3} \hat{k}$ , so $∣ P Q ∣ = \frac{1}{9} + \frac{1}{9} + \frac{4}{9} = \frac{2}{3}$ units. This is the shortest distance.
  1 mark
Answer: Points $(\frac{2}{3}, - \frac{2}{3}, \frac{2}{3})$ and $(1, - 1, 0)$ ; shortest distance $2/3$ units
Where the marks are
- Writes general points on both lines (½ mark)
- Forms vector $P Q$ (½ mark)
- Perpendicularity condition with first line (½ mark)
- Perpendicularity condition with second line (½ mark)
- Solves to get $μ = 0, λ = 2/3$ (1 mark)
- States the two points (1 mark)
- Finds shortest distance $2/3$ units (1 mark)
Open Q35(b) on its own page

Section E

case based

Q36 · 4 marks · Case-basedCompetency-basedProbability
In an Office three employees Jayant, Sonia and Oliver process incoming copies of a certain form. Jayant processes 50% of the forms, Sonia processes 20% and Oliver the remaining 30% of the forms. Jayant has an error rate of 0.06, Sonia has an error rate of 0.04 and Oliver has an error rate of 0.03.
Based on the above information, answer the following questions.
1. (i)
  Find the probability that Sonia processed the form and committed an error.
  1 mark
2. (ii)
  Find the total probability of committing an error in processing the form.
  1 mark
3. (iii)
  The manager of the Company wants to do a quality check. During inspection, he selects a form at random from the days output of processed form. If the form selected at random has an error, find the probability that the form is not processed by Jayant.
  2 marks
4. OR
  (iii)
  Let $E$ be the event of committing an error in processing the form and let $E_{1}, E_{2}$ and $E_{3}$ be the events that Jayant, Sonia and Oliver processed the form. Find the value of $\sum_{i = 1}^{3} P (E_{i} ∣ E)$ .
  2 marks
Figure in the paper: Photograph of three office employees looking at papers and a laptop. See it in the official paper.
Try it first, then show our solutionSolution
1. Let $E_{1}, E_{2}, E_{3}$ be the events that Jayant, Sonia, Oliver processed the form, and $E$ the event of an error. $P (E_{1}) = \frac{5}{10}, P (E_{2}) = \frac{1}{5}, P (E_{3}) = \frac{3}{10}$ ; $P (E ∣ E_{1}) = 0.06, P (E ∣ E_{2}) = 0.04, P (E ∣ E_{3}) = 0.03$ .
2. (i) $P (E \cap E_{2}) = P (E_{2}) P (E ∣ E_{2}) = \frac{1}{5} \times 0.04 = 0.008$ .
  1 mark
3. (ii) By the theorem of total probability: $P (E) = 0.5 \times 0.06 + 0.2 \times 0.04 + 0.3 \times 0.03 = 0.03 + 0.008 + 0.009 = 0.047$ .
  1 mark
4. (iii)(a) By Bayes' theorem: $P (E_{1} ∣ E) = \frac{P ( E _{1} ) P ( E ∣ E _{1} )}{P ( E )} = \frac{0.03}{0.047} = \frac{30}{47}$ .
  1 mark
5. (iii)(a) Required probability $P (\overset{ˉ}{E}_{1} ∣ E) = 1 - \frac{30}{47} = \frac{17}{47}$ .
  1 mark
6. (iii)(b) $\sum_{i = 1}^{3} P (E_{i} ∣ E) = \frac{P ( E \cap E _{1} ) + P ( E \cap E _{2} ) + P ( E \cap E _{3} )}{P ( E )} = \frac{P ( E \cap ( E _{1} \cup E _{2} \cup E _{3} ))}{P ( E )}$ , since $E_{1}, E_{2}, E_{3}$ are mutually exclusive.
  1 mark
7. (iii)(b) As $E_{1} \cup E_{2} \cup E_{3} = S$ (exhaustive), this equals $\frac{P ( E )}{P ( E )} = 1$ . (Check: $\frac{30}{47} + \frac{8}{47} + \frac{9}{47} = 1$ .)
  1 mark
Answer: (i) 0.008; (ii) 0.047; (iii)(a) 17/47; (iii)(b) 1
Where the marks are
- (i) $P (E_{2}) P (E ∣ E_{2}) = 0.008$ (1 mark)
- (ii) Total probability $P (E) = 0.047$ (1 mark)
- (iii)(a) Bayes' theorem gives $P (E_{1} ∣ E) = 30/47$ (1 mark)
- (iii)(a) Not Jayant: $1 - 30/47 = 17/47$ (1 mark)
- (iii)(b) Uses mutually exclusive and exhaustive events to combine the sum (1 mark)
- (iii)(b) Sum of posterior probabilities $= 1$ (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedVector Algebra
Teams $A, B, C$ went for playing a tug of war game. Teams $A, B, C$ have attached a rope to a metal ring and is trying to pull the ring into their own area.
Team $A$ pulls with force $F_{1} = 6 \hat{i} + 0 \hat{j}$ kN,
Team $B$ pulls with force $F_{2} = - 4 \hat{i} + 4 \hat{j}$ kN,
Team $C$ pulls with force $F_{3} = - 3 \hat{i} - 3 \hat{j}$ kN,
1. (i)
  What is the magnitude of the force of Team $A$ ?
  1 mark
2. (ii)
  Which team will win the game?
  1 mark
3. (iii)
  Find the magnitude of the resultant force exerted by the teams.
  2 marks
4. OR
  (iii)
  In what direction is the ring getting pulled?
  2 marks
Figure in the paper: Ring at centre with three force vectors: a=6i+0j pointing right towards A; b=-4i+4j pointing up-left towards C-side direction labelled B (lower left, arrow towards B); c=-3i-3j pointing towards C (upper left); dashed reference lines through the ring. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) $∣ F_{1} ∣ = 6^{2} + 0^{2} = 6$ kN.
  1 mark
2. (ii) $F_{1} + F_{3} = 3 \hat{i} - 3 \hat{j} = 3 (\hat{i} - \hat{j})$ and $F_{2} = - 4 (\hat{i} - \hat{j})$ , so they are opposite (unlike) vectors. $∣ F_{2} ∣ = 42 > ∣ F_{1} + F_{3} ∣ = 32$ , so the net pull is towards B. Team B will win.
  1 mark
3. (iii)(a) $F = F_{1} + F_{2} + F_{3} = (6 - 4 - 3) \hat{i} + (0 + 4 - 3) \hat{j} = - \hat{i} + \hat{j}$ .
  1 mark
4. (iii)(a) $∣ F ∣ = (- 1)^{2} + 1^{2} = 2$ kN.
  1 mark
5. (iii)(b) Resultant $F = - \hat{i} + \hat{j}$ , which lies in the second quadrant.
  1 mark
6. (iii)(b) Angle with the positive $x$ -axis: $θ = π - tan^{- 1} (\frac{1}{1}) = π - \frac{π}{4} = \frac{3 π}{4}$ . The ring is pulled along $- \hat{i} + \hat{j}$ , i.e. at $13 5^{\circ}$ to the positive $x$ -axis (towards B).
  1 mark
Answer: (i) 6 kN; (ii) Team B; (iii)(a) $2$ kN; (iii)(b) along $- \hat{i} + \hat{j}$ , at $3 π /4$ with positive x-axis
Where the marks are
- (i) Magnitude of $F_{1}$ is 6 kN (1 mark)
- (ii) Compares $F_{2}$ with $F_{1} + F_{3}$ and concludes B wins (1 mark)
- (iii)(a) Resultant $- \hat{i} + \hat{j}$ (1 mark)
- (iii)(a) Magnitude $2$ kN (1 mark)
- (iii)(b) Resultant $- \hat{i} + \hat{j}$ (1 mark)
- (iii)(b) Direction $3 π /4$ with positive x-axis (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
The relation between the height of the plant ('y' in cm) with respect to its exposure to the sunlight is governed by the following equation $y = 4 x - \frac{1}{2} x^{2}$ , where 'x' is the number of days exposed to the sunlight, for $x \leq 3$ .
1. (i)
  Find the rate of growth of the plant with respect to the number of days exposed to the sunlight.
  2 marks
2. (ii)
  Does the rate of growth of the plant increase or decrease in the first three days? What will be the height of the plant after 2 days?
  2 marks
Figure in the paper: Photo of a seedling plant bending towards the bright sun. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Rate of growth $= \frac{d y}{d x} = \frac{d}{d x} (4 x - \frac{1}{2} x^{2}) = 4 - x$ cm per day.
  2 marks
2. (ii) Let $g (x) = \frac{d y}{d x} = 4 - x$ . Then $g^{'} (x) = - 1 < 0$ , so $g (x)$ is decreasing: the rate of growth decreases in the first three days.
  1 mark
3. (ii) Height after 2 days: $y = 4 (2) - \frac{1}{2} (2)^{2} = 8 - 2 = 6$ cm.
  1 mark
Answer: (i) $\frac{d y}{d x} = 4 - x$ ; (ii) rate of growth decreases; height after 2 days = 6 cm
Where the marks are
- (i) Differentiates to get $d y / d x = 4 - x$ (2 marks)
- (ii) Shows $\frac{d}{d x} (d y / d x) = - 1 < 0$ , so rate decreases (1 mark)
- (ii) Height at $x = 2$ is 6 cm (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.