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

CBSE Class 12 Mathematics: 2023 board paper, set 65/5/1

The question paper set in the CBSE Class 12 board exam of 2023, as printed: 38 questions for 80 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2023/XII/MATHEMATICS.zip (a zip; this paper is the file 65-5-1 MATHEMATICS.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

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

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQRelations and Functions
Let $A = {3, 5}$ . Then number of reflexive relations on A is
1. (A)2
2. (B)4
3. (C)0
4. (D)8
Try it first, then show our solutionSolution
1. A reflexive relation must contain $(3, 3)$ and $(5, 5)$ ; the other pairs $(3, 5), (5, 3)$ may each be included or not, giving $2^{n^{2} - n} = 2^{2} = 4$ relations.
  1 mark
Answer: 4, option (B)
Where the marks are
- Uses $2^{n^{2} - n}$ with $n = 2$ to get 4 (1 mark)
Q2 · 1 mark · MCQInverse Trigonometric Functions
$sin [\frac{π}{3} + sin^{- 1} (\frac{1}{2})]$ is equal to
1. (A)1
2. (B) $\frac{1}{2}$
3. (C) $\frac{1}{3}$
4. (D) $\frac{1}{4}$
Try it first, then show our solutionSolution
1. $sin^{- 1} \frac{1}{2} = \frac{π}{6}$ , so the expression is $sin (\frac{π}{3} + \frac{π}{6}) = sin \frac{π}{2} = 1$ .
  1 mark
Answer: 1, option (A)
Where the marks are
- Principal value $π /6$ and result 1 (1 mark)
Q3 · 1 mark · MCQMatrices
If for a square matrix A, $A^{2} - A + I = O$ , then $A^{- 1}$ equals
1. (A)A
2. (B)A + I
3. (C)I − A
4. (D)A − I
Try it first, then show our solutionSolution
1. $A^{2} - A + I = O \Rightarrow A (I - A) = I$ . Hence $A^{- 1} = I - A$ .
  1 mark
Answer: $I - A$ , option (C)
Where the marks are
- Rearranges to $A (I - A) = I$ (1 mark)
Q4 · 1 mark · MCQMatrices
If $A = [1201]$ , $B = [x 1 01]$ and $A = B^{2}$ , then $x$ equals
1. (A) $\pm 1$
2. (B) $- 1$
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. $B^{2} = [x^{2} x + 1 01]$ . Equating with A: $x^{2} = 1$ and $x + 1 = 2$ , so $x = 1$ .
  1 mark
Answer: $x = 1$ , option (C)
Where the marks are
- Computes $B^{2}$ and equates entries giving $x = 1$ (1 mark)
Q5 · 1 mark · MCQDeterminants
If $α 11 324411 = 0$ , then the value of $α$ is
1. (A)1
2. (B)2
3. (C)3
4. (D)4
Try it first, then show our solutionSolution
1. Expanding: $α (2 - 4) - 3 (1 - 1) + 4 (4 - 2) = - 2 α + 8 = 0 \Rightarrow α = 4$ .
  1 mark
Answer: $α = 4$ , option (D)
Where the marks are
- Correct expansion giving $α = 4$ (1 mark)
Q6 · 1 mark · MCQContinuity and Differentiability
The derivative of $x^{2 x}$ w.r.t. $x$ is
1. (A) $x^{2 x - 1}$
2. (B) $2 x^{2 x} lo g x$
3. (C) $2 x^{2 x} (1 + lo g x)$
4. (D) $2 x^{2 x} (1 - lo g x)$
Try it first, then show our solutionSolution
1. Let $y = x^{2 x}$ ; $lo g y = 2 x lo g x$ . Differentiating: $\frac{1}{y} \frac{d y}{d x} = 2 lo g x + 2$ , so $\frac{d y}{d x} = 2 x^{2 x} (1 + lo g x)$ .
  1 mark
Answer: $2 x^{2 x} (1 + lo g x)$ , option (C)
Where the marks are
- Logarithmic differentiation to correct result (1 mark)
Q7 · 1 mark · MCQContinuity and Differentiability
The function $f (x) = [x]$ , where $[x]$ denotes the greatest integer less than or equal to $x$ , is continuous at
1. (A) $x = 1$
2. (B) $x = 1.5$
3. (C) $x = - 2$
4. (D) $x = 4$
Try it first, then show our solutionSolution
1. $[x]$ is discontinuous at every integer; 1.5 is not an integer, and near it $[x] = 1$ constant, so continuous at $x = 1.5$ .
  1 mark
Answer: $x = 1.5$ , option (B)
Where the marks are
- Recognises discontinuity only at integers (1 mark)
Q8 · 1 mark · MCQContinuity and Differentiability
If $x = A cos 4 t + B sin 4 t$ , then $\frac{d ^{2} x}{d t ^{2}}$ is equal to
1. (A) $x$
2. (B) $- x$
3. (C) $16 x$
4. (D) $- 16 x$
Try it first, then show our solutionSolution
1. $\frac{d x}{d t} = - 4 A sin 4 t + 4 B cos 4 t$ , $\frac{d ^{2} x}{d t ^{2}} = - 16 (A cos 4 t + B sin 4 t) = - 16 x$ .
  1 mark
Answer: $- 16 x$ , option (D)
Where the marks are
- Differentiates twice to get $- 16 x$ (1 mark)
Q9 · 1 mark · MCQApplication of Derivatives
The interval in which the function $f (x) = 2 x^{3} + 9 x^{2} + 12 x - 1$ is decreasing, is
1. (A) $(- 1, \infty)$
2. (B) $(- 2, - 1)$
3. (C) $(- \infty, - 2)$
4. (D) $[- 1, 1]$
Try it first, then show our solutionSolution
1. $f^{'} (x) = 6 x^{2} + 18 x + 12 = 6 (x + 1) (x + 2)$ , which is negative for $- 2 < x < - 1$ .
  1 mark
Answer: $(- 2, - 1)$ , option (B)
Where the marks are
- Factorises $f^{'} (x)$ and finds sign (1 mark)
Q10 · 1 mark · MCQIntegrals
$\int \frac{s e c x}{s e c x - t a n x} d x$ equals
1. (A) $sec x - tan x + c$
2. (B) $sec x + tan x + c$
3. (C) $tan x - sec x + c$
4. (D) $- (sec x + tan x) + c$
Try it first, then show our solutionSolution
1. Multiply by $(sec x + tan x)$ : since $sec^{2} x - tan^{2} x = 1$ , integrand $= sec^{2} x + sec x tan x$ .
2. Integral $= tan x + sec x + c$ .
  1 mark
Answer: $sec x + tan x + c$ , option (B)
Where the marks are
- Rationalises and integrates correctly (1 mark)
Q11 · 1 mark · MCQIntegrals
$\int_{- 1}^{1} \frac{∣ x - 2∣}{x - 2} d x$ , $x \neq = 2$ is equal to
1. (A)1
2. (B) $- 1$
3. (C)2
4. (D) $- 2$
Try it first, then show our solutionSolution
1. For $- 1 \leq x \leq 1$ , $x - 2 < 0$ so $∣ x - 2∣ = - (x - 2)$ and integrand $= - 1$ .
2. $\int_{- 1}^{1} (- 1) d x = - 2$ .
  1 mark
Answer: $- 2$ , option (D)
Where the marks are
- Integrand is $- 1$ on interval, value $- 2$ (1 mark)
Q12 · 1 mark · MCQDifferential Equations
The sum of the order and the degree of the differential equation $\frac{d}{d x} ((\frac{d y}{d x})^{3})$ is
1. (A)2
2. (B)3
3. (C)5
4. (D)0
Try it first, then show our solutionSolution
1. The given expression has no '= ' sign, so it is not a differential equation. If read as $\frac{d}{d x} ((\frac{d y}{d x})^{3}) = 0$ , i.e. $3 (\frac{d y}{d x})^{2} \frac{d ^{2} y}{d x ^{2}} = 0$ , order 2, degree 1, sum 3 (option B).
  1 mark
Answer: Question erroneous; on the natural reading the sum is 3
Where the marks are
- 1 mark awarded to all who attempted, due to error in question (1 mark)
Q13 · 1 mark · MCQVector Algebra
Two vectors $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ are collinear if
1. (A) $a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = 0$
2. (B) $\frac{a _{1}}{b _{1}} = \frac{a _{2}}{b _{2}} = \frac{a _{3}}{b _{3}}$
3. (C) $a_{1} = b_{1}, a_{2} = b_{2}, a_{3} = b_{3}$
4. (D) $a_{1} + a_{2} + a_{3} = b_{1} + b_{2} + b_{3}$
Try it first, then show our solutionSolution
1. Collinear means $a = λ b$ , i.e. components proportional: $\frac{a _{1}}{b _{1}} = \frac{a _{2}}{b _{2}} = \frac{a _{3}}{b _{3}}$ .
  1 mark
Answer: option (B)
Where the marks are
- Condition of proportional components (1 mark)
Q14 · 1 mark · MCQVector Algebra
The magnitude of the vector $6 \hat{i} - 2 \hat{j} + 3 \hat{k}$ is
1. (A)1
2. (B)5
3. (C)7
4. (D)12
Try it first, then show our solutionSolution
1. $36 + 4 + 9 = 49 = 7$ .
  1 mark
Answer: 7, option (C)
Where the marks are
- Correct magnitude 7 (1 mark)
Q15 · 1 mark · MCQThree Dimensional Geometry
If a line makes angles of $9 0^{\circ}$ , $13 5^{\circ}$ and $4 5^{\circ}$ with the x, y and z axes respectively, then its direction cosines are
1. (A) $0, - \frac{1}{2}, \frac{1}{2}$
2. (B) $- \frac{1}{2}, 0, \frac{1}{2}$
3. (C) $\frac{1}{2}, 0, - \frac{1}{2}$
4. (D) $0, \frac{1}{2}, \frac{1}{2}$
Try it first, then show our solutionSolution
1. $cos 9 0^{\circ} = 0$ , $cos 13 5^{\circ} = - \frac{1}{2}$ , $cos 4 5^{\circ} = \frac{1}{2}$ .
  1 mark
Answer: $0, - \frac{1}{2}, \frac{1}{2}$ , option (A)
Where the marks are
- Takes cosines of the angles (1 mark)
Q16 · 1 mark · MCQThree Dimensional Geometry
The angle between the lines $2 x = 3 y = - z$ and $6 x = - y = - 4 z$ is
1. (A) $0^{\circ}$
2. (B) $3 0^{\circ}$
3. (C) $4 5^{\circ}$
4. (D) $9 0^{\circ}$
Try it first, then show our solutionSolution
1. Line 1: $\frac{x}{1/2} = \frac{y}{1/3} = \frac{z}{- 1}$ , dr's $3, 2, - 6$ . Line 2: $\frac{x}{1/6} = \frac{y}{- 1} = \frac{z}{- 1/4}$ , dr's $2, - 12, - 3$ .
2. $3 (2) + 2 (- 12) + (- 6) (- 3) = 6 - 24 + 18 = 0$ , so lines are perpendicular: $9 0^{\circ}$ .
  1 mark
Answer: $9 0^{\circ}$ , option (D)
Where the marks are
- Direction ratios and zero dot product (1 mark)
Open Q16 on its own page
Q17 · 1 mark · MCQProbability
If for any two events A and B, $P (A) = \frac{4}{5}$ and $P (A \cap B) = \frac{7}{10}$ , then $P (B / A)$ is equal to
1. (A) $\frac{1}{10}$
2. (B) $\frac{1}{8}$
3. (C) $\frac{7}{8}$
4. (D) $\frac{17}{20}$
Try it first, then show our solutionSolution
1. $P (B / A) = \frac{P ( A \cap B )}{P ( A )} = \frac{7/10}{4/5} = \frac{7}{8}$ .
  1 mark
Answer: $\frac{7}{8}$ , option (C)
Where the marks are
- Conditional probability formula and value (1 mark)
Q18 · 1 mark · MCQProbability
Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is
1. (A) $\frac{27}{32}$
2. (B) $\frac{5}{32}$
3. (C) $\frac{31}{32}$
4. (D) $\frac{1}{32}$
Try it first, then show our solutionSolution
1. $P (no head) = (\frac{1}{2})^{5} = \frac{1}{32}$ , so $P (at least one) = 1 - \frac{1}{32} = \frac{31}{32}$ .
  1 mark
Answer: $\frac{31}{32}$ , option (C)
Where the marks are
- Complement method (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedProbability
Assertion (A) : Two coins are tossed simultaneously. The probability of getting two heads, if it is known that at least one head comes up, is $\frac{1}{3}$ .
Reason (R) : Let E and F be two events with a random experiment, then $P (F / E) = \frac{P ( E \cap F )}{P ( E )}$ .
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 and (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. E = at least one head = {HH,HT,TH}, P(E)=3/4; F = two heads, $P (E \cap F) = 1/4$ .
2. $P (F / E) = \frac{1/4}{3/4} = \frac{1}{3}$ ; A is true, R is the definition used, so R explains A.
  1 mark
Answer: option (A)
Where the marks are
- Uses conditional probability to verify 1/3 (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedIntegrals
Assertion (A) : $\int_{2}^{8} \frac{10 - x}{x + 10 - x} d x = 3$
Reason (R) : $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
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 and (R) is false.
4. (D)(A) is false, but (R) is true.
Try it first, then show our solutionSolution
1. By R with $a + b = 10$ : $I = \int_{2}^{8} \frac{x}{10 - x + x} d x$ . Adding, $2 I = \int_{2}^{8} 1 d x = 6$ , so $I = 3$ .
  1 mark
Answer: option (A)
Where the marks are
- Applies property and gets $I = 3$ (1 mark)

Section B

VSA

Q21 · 2 marks · Very short answerInverse Trigonometric Functions
Write the domain and range (principle value branch) of the following functions :
$f (x) = tan^{- 1} x$
Try it first, then show our solutionSolution
1. $tan x$ restricted to $(- \frac{π}{2}, \frac{π}{2})$ is one-one and onto $R$ . So its inverse $tan^{- 1} x$ is defined for every real $x$ .
  Domain of $f (x) = tan^{- 1} x$ is $R$ (all real numbers).
  1 mark
2. Range (principal value branch) of $tan^{- 1} x$ is $(- \frac{π}{2}, \frac{π}{2})$ . The end points are excluded because $tan x$ is not defined at $\pm \frac{π}{2}$ .
  1 mark
Answer: Domain = $R$ ; Range = $(- \frac{π}{2}, \frac{π}{2})$
Where the marks are
- States domain as $R$ (1 mark)
- States principal value range as the open interval $(- \frac{π}{2}, \frac{π}{2})$ (1 mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerContinuity and Differentiability
If $f (x) = {x^{2}, x, if x \geq 1 if x < 1$ , then show that f is not differentiable at $x = 1$ .
Try it first, then show our solutionSolution
1. Here $f (1) = 1^{2} = 1$ .
  RHD: $lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h} = lim_{h \to 0} \frac{( 1 + h ) ^{2} - 1}{h} = lim_{h \to 0} \frac{2 h + h ^{2}}{h} = lim_{h \to 0} (2 + h) = 2$
  1 mark
2. LHD: $lim_{h \to 0} \frac{f ( 1 - h ) - f ( 1 )}{- h} = lim_{h \to 0} \frac{( 1 - h ) - 1}{- h} = lim_{h \to 0} \frac{- h}{- h} = 1$
  ½ mark
3. Since LHD $= 1 \neq = 2 =$ RHD, $f$ is not differentiable at $x = 1$ .
  ½ mark
Answer: RHD = 2, LHD = 1; since they are unequal, f is not differentiable at x = 1.
Where the marks are
- Correctly finds RHD = 2 and LHD = 1 (1½ marks)
- Concludes non-differentiability because LHD ≠ RHD (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerContinuity and Differentiability
Find the value(s) of ' $λ$ ', if the function
$f (x) = {\frac{s i n ^{2} λ x}{x ^{2}}, 1, if x \neq = 0 if x = 0$ is continuous at $x = 0$ .
Try it first, then show our solutionSolution
1. For $λ \neq = 0$ :
  $lim_{x \to 0} \frac{s i n ^{2} λ x}{x ^{2}} = lim_{x \to 0} (\frac{s i n λ x}{λ x})^{2} \cdot λ^{2} = 1 \cdot λ^{2} = λ^{2}$ [using $lim_{θ \to 0} \frac{s i n θ}{θ} = 1$ ]
  1 mark
2. Since $f$ is continuous at $x = 0$ , $lim_{x \to 0} f (x) = f (0)$ , so $λ^{2} = 1$ .
  ½ mark
3. Hence $λ = \pm 1$ .
  ( $λ = 0$ is not possible: the limit would then be $0 \neq = 1$ .)
  ½ mark
Answer: $λ = \pm 1$
Where the marks are
- Evaluates the limit as $λ^{2}$ (1 mark)
- Equates the limit to $f (0) = 1$ by continuity (½ mark)
- Gets $λ = \pm 1$ (½ mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerApplication of Integrals
Sketch the region bounded by the lines $2 x + y = 8$ , $y = 2$ , $y = 4$ and the y-axis. Hence, obtain its area using integration.
Try it first, then show our solutionSolution
1. Figure: The line $2 x + y = 8$ meets the y-axis at $(0, 8)$ and the x-axis at $(4, 0)$ .
  The horizontal lines $y = 2$ and $y = 4$ cut it at $E (3, 2)$ and $F (2, 4)$ .
  The required region is the trapezium with vertices $C (0, 2)$ , $E (3, 2)$ , $F (2, 4)$ and $B (0, 4)$ , lying between the y-axis and the line. Shade this region.
  ½ mark
2. Using horizontal strips, $x = \frac{8 - y}{2}$ and $y$ runs from $2$ to $4$ :
  Area $= \int_{2}^{4} x d y = \int_{2}^{4} \frac{1}{2} (8 - y) d y$
  ½ mark
3. $= \frac{1}{2} [8 y - \frac{y ^{2}}{2}]_{2}^{4} = \frac{1}{2} [(32 - 8) - (16 - 2)]$
  ½ mark
4. $= \frac{1}{2} (24 - 14) = 5$ sq units
  ½ mark
Answer: Area = 5 sq units
Where the marks are
- Correct sketch of the trapezium region (½ mark)
- Sets up the integral $\int_{2}^{4} \frac{1}{2} (8 - y) d y$ (½ mark)
- Correct integration with limits (½ mark)
- Area = 5 sq units (½ mark)
Open Q23 on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerVector Algebra
If the vectors $a$ and $b$ are such that $∣ a ∣ = 3$ , $∣ b ∣ = \frac{2}{3}$ and $a \times b$ is a unit vector, then find the angle between $a$ and $b$ .
Try it first, then show our solutionSolution
1. Let $θ$ be the angle between $a$ and $b$ .
  Since $a \times b$ is a unit vector, $∣ a \times b ∣ = 1$ .
  So $∣ a ∣∣ b ∣ sin θ = 1$ , i.e. $3 \cdot \frac{2}{3} sin θ = 1$ .
  1 mark
2. $sin θ = \frac{1}{2}$ , so $θ = \frac{π}{6}$ (i.e. $3 0^{\circ}$ ).
  1 mark
Answer: $θ = \frac{π}{6}$ (30°)
Where the marks are
- Uses $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ = 1$ (1 mark)
- Gets $sin θ = \frac{1}{2}$ and $θ = \frac{π}{6}$ (1 mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerVector Algebra
Find the area of a parallelogram whose adjacent sides are determined by the vectors $a = \hat{i} - \hat{j} + 3 \hat{k}$ and $b = 2 \hat{i} - 7 \hat{j} + \hat{k}$ .
Try it first, then show our solutionSolution
1. $a \times b = \hat{i} 12 \hat{j} - 1 - 7 \hat{k} 31$
  $= \hat{i} (- 1 + 21) - \hat{j} (1 - 6) + \hat{k} (- 7 + 2)$
  $= 20 \hat{i} + 5 \hat{j} - 5 \hat{k}$
  1½ marks
2. Area of parallelogram $= ∣ a \times b ∣ = 400 + 25 + 25 = 450 = 152$ sq units
  ½ mark
Answer: $152$ sq units
Where the marks are
- Correct cross product $20 \hat{i} + 5 \hat{j} - 5 \hat{k}$ (1½ marks)
- Area $= ∣ a \times b ∣ = 152$ sq units (½ mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerThree Dimensional Geometry
Find the vector and the cartesian equations of a line that passes through the point A(1, 2, −1) and parallel to the line $5 x - 25 = 14 - 7 y = 35 z$ .
Try it first, then show our solutionSolution
1. Rewrite the given line: $5 (x - 5) = - 7 (y - 2) = 35 z$ .
  Dividing by 35: $\frac{x - 5}{7} = \frac{y - 2}{- 5} = \frac{z}{1}$
  So its direction ratios are $7, - 5, 1$ .
  1 mark
2. The required line passes through $(1, 2, - 1)$ and is parallel to $7 \hat{i} - 5 \hat{j} + \hat{k}$ .
  Vector equation: $r = (\hat{i} + 2 \hat{j} - \hat{k}) + λ (7 \hat{i} - 5 \hat{j} + \hat{k})$
  ½ mark
3. Cartesian equation: $\frac{x - 1}{7} = \frac{y - 2}{- 5} = \frac{z + 1}{1}$
  ½ mark
Answer: $r = (\hat{i} + 2 \hat{j} - \hat{k}) + λ (7 \hat{i} - 5 \hat{j} + \hat{k})$ ; $\frac{x - 1}{7} = \frac{y - 2}{- 5} = \frac{z + 1}{1}$
Where the marks are
- Writes the given line in standard form with direction ratios 7, −5, 1 (1 mark)
- Correct vector equation (½ mark)
- Correct cartesian equation (½ mark)
Open Q25 on its own page

Section C

Q26 · 3 marks · Short answerMatrices
If $A = 134 2 - 2 2 311$ , then show that $A^{3} - 23 A - 40 I = O$ .
Try it first, then show our solutionSolution
1. $A^{2} = A \cdot A = 134 2 - 2 2 311 134 2 - 2 2 311 = 1911441268815$
  1 mark
2. $A^{3} = A^{2} \cdot A = 1911441268815 134 2 - 2 2 311 = 636992 46 - 6 46 692363$
  1 mark
3. $A^{3} - 23 A - 40 I = 636992 46 - 6 46 692363 - 236992 46 - 46 46 692323 - 400004000040$
  ½ mark
4. $= 000000000 = O$ . Hence proved.
  ½ mark
Answer: $A^{3} - 23 A - 40 I = O$ (verified)
Where the marks are
- Correct $A^{2}$ (1 mark)
- Correct $A^{3}$ (1 mark)
- Substitutes $A^{3}$ , $23 A$ and $40 I$ (½ mark)
- Simplifies to the null matrix (½ mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerInverse Trigonometric Functions
Differentiate $sec^{- 1} (\frac{1}{1 - x ^{2}})$ w.r.t. $sin^{- 1} (2 x 1 - x^{2})$ .
Try it first, then show our solutionSolution
1. Put $x = sin θ$ (taking $0 < x < \frac{1}{2}$ , so $0 < θ < \frac{π}{4}$ ).
  $U = sec^{- 1} (\frac{1}{1 - s i n ^{2} θ}) = sec^{- 1} (\frac{1}{c o s θ}) = sec^{- 1} (sec θ) = θ = sin^{- 1} x$
  So $\frac{d U}{d x} = \frac{1}{1 - x ^{2}}$
  1 mark
2. $V = sin^{- 1} (2 sin θ 1 - sin^{2} θ) = sin^{- 1} (2 sin θ cos θ) = sin^{- 1} (sin 2 θ) = 2 θ = 2 sin^{- 1} x$
  So $\frac{d V}{d x} = \frac{2}{1 - x ^{2}}$
  1 mark
3. $\frac{d U}{d V} = \frac{d U / d x}{d V / d x} = \frac{1/ 1 - x ^{2}}{2/ 1 - x ^{2}} = \frac{1}{2}$
  1 mark
Answer: $\frac{d U}{d V} = \frac{1}{2}$
Where the marks are
- Simplifies $U = sin^{- 1} x$ and differentiates (1 mark)
- Simplifies $V = 2 sin^{- 1} x$ and differentiates (1 mark)
- Divides to get $\frac{d U}{d V} = \frac{1}{2}$ (1 mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerContinuity and Differentiability
If $y = tan x + sec x$ , then prove that $\frac{d ^{2} y}{d x ^{2}} = \frac{c o s x}{( 1 - s i n x ) ^{2}}$ .
Try it first, then show our solutionSolution
1. $y = tan x + sec x = \frac{s i n x + 1}{c o s x}$
  By the quotient rule:
  $\frac{d y}{d x} = \frac{c o s x \cdot c o s x - ( 1 + s i n x ) ( - s i n x )}{c o s ^{2} x} = \frac{c o s ^{2} x + s i n ^{2} x + s i n x}{c o s ^{2} x}$
  $= \frac{1 + s i n x}{1 - s i n ^{2} x} = \frac{1 + s i n x}{( 1 - s i n x ) ( 1 + s i n x )} = \frac{1}{1 - s i n x}$
  1½ marks
2. Differentiate again:
  $\frac{d ^{2} y}{d x ^{2}} = \frac{( 1 - s i n x ) \cdot 0 - 1 \cdot ( - c o s x )}{( 1 - s i n x ) ^{2}} = \frac{c o s x}{( 1 - s i n x ) ^{2}}$
  Hence proved.
  1½ marks
Answer: $\frac{d ^{2} y}{d x ^{2}} = \frac{c o s x}{( 1 - s i n x ) ^{2}}$ (proved)
Where the marks are
- Obtains $\frac{d y}{d x} = \frac{1}{1 - s i n x}$ (1½ marks)
- Differentiates again to get $\frac{c o s x}{( 1 - s i n x ) ^{2}}$ (1½ marks)
Open Q27(b) on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntegrals
Evaluate : $\int_{0}^{2 π} \frac{1}{1 + e ^{s i n x}} d x$
Try it first, then show our solutionSolution
1. Let $I = \int_{0}^{2 π} \frac{1}{1 + e ^{s i n x}} d x$ .
  Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ :
  $I = \int_{0}^{2 π} \frac{1}{1 + e ^{s i n (2 π - x)}} d x$
  1 mark
2. Since $sin (2 π - x) = - sin x$ :
  $I = \int_{0}^{2 π} \frac{1}{1 + e ^{- s i n x}} d x = \int_{0}^{2 π} \frac{e ^{s i n x}}{e ^{s i n x} + 1} d x$
  1 mark
3. Adding the two forms of $I$ :
  $2 I = \int_{0}^{2 π} \frac{1 + e ^{s i n x}}{1 + e ^{s i n x}} d x = \int_{0}^{2 π} 1 d x = 2 π$
  ½ mark
4. $I = π$
  ½ mark
Answer: $I = π$
Where the marks are
- Applies the property $f (a - x)$ with $a = 2 π$ (1 mark)
- Simplifies to $\int_{0}^{2 π} \frac{e ^{s i n x}}{1 + e ^{s i n x}} d x$ (1 mark)
- Adds the two forms to get $2 I = 2 π$ (½ mark)
- $I = π$ (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntegrals
Find : $\int \frac{x ^{4}}{( x - 1 ) ( x ^{2} + 1 )} d x$
Try it first, then show our solutionSolution
1. The denominator is $(x - 1) (x^{2} + 1) = x^{3} - x^{2} + x - 1$ .
  By division, $x^{4} = (x + 1) (x^{3} - x^{2} + x - 1) + 1$ . So
  $\frac{x ^{4}}{( x - 1 ) ( x ^{2} + 1 )} = x + 1 + \frac{1}{( x - 1 ) ( x ^{2} + 1 )}$
  1 mark
2. Let $\frac{1}{( x - 1 ) ( x ^{2} + 1 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 1}$ . Then $1 = A (x^{2} + 1) + (B x + C) (x - 1)$ .
  Putting $x = 1$ : $A = \frac{1}{2}$ .
  Comparing $x^{2}$ terms: $A + B = 0$ , so $B = - \frac{1}{2}$ .
  Comparing constants: $A - C = 1$ , so $C = - \frac{1}{2}$ .
  So $\frac{1}{( x - 1 ) ( x ^{2} + 1 )} = \frac{1}{2 ( x - 1 )} - \frac{1}{2} \cdot \frac{x + 1}{x ^{2} + 1}$
  1 mark
3. $I = \int [x + 1 + \frac{1}{2 ( x - 1 )} - \frac{1}{4} \cdot \frac{2 x}{x ^{2} + 1} - \frac{1}{2} \cdot \frac{1}{x ^{2} + 1}] d x$
  ½ mark
4. $= \frac{x ^{2}}{2} + x + \frac{1}{2} lo g ∣ x - 1∣ - \frac{1}{4} lo g (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C$
  ½ mark
Answer: $\frac{x ^{2}}{2} + x + \frac{1}{2} lo g ∣ x - 1∣ - \frac{1}{4} lo g (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C$
Where the marks are
- Divides to get $x + 1 + \frac{1}{( x - 1 ) ( x ^{2} + 1 )}$ (1 mark)
- Correct partial fractions with $A = \frac{1}{2}$ , $B = C = - \frac{1}{2}$ (1 mark)
- Splits the integrand into integrable parts (½ mark)
- Correct final integral with constant (½ mark)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerApplication of Integrals
Find the area of the following region using integration :
${(x, y) : y^{2} \leq 2 x and y \geq x - 4}$
Try it first, then show our solutionSolution
1. Figure: The parabola $y^{2} = 2 x$ has its vertex at O and opens to the right.
  The line $y = x - 4$ passes through $(4, 0)$ and $(0, - 4)$ .
  The region is the part inside the parabola that lies above the line. It is enclosed between the points $(2, - 2)$ and $(8, 4)$ . Shade this region.
  1 mark
2. Solving $y^{2} = 2 x$ and $x = y + 4$ : $y^{2} = 2 y + 8$ , so $y^{2} - 2 y - 8 = 0$ , i.e. $(y - 4) (y + 2) = 0$ .
  So $y = 4$ or $y = - 2$ , giving the points $(8, 4)$ and $(2, - 2)$ .
  ½ mark
3. Using horizontal strips, the line ( $x = y + 4$ ) lies to the right of the parabola ( $x = \frac{y ^{2}}{2}$ ):
  Area $= \int_{- 2}^{4} [(y + 4) - \frac{y ^{2}}{2}] d y = [\frac{y ^{2}}{2} + 4 y - \frac{y ^{3}}{6}]_{- 2}^{4}$
  1 mark
4. $= (8 + 16 - \frac{64}{6}) - (2 - 8 + \frac{8}{6}) = \frac{40}{3} - (- \frac{14}{3}) = 18$ sq units
  ½ mark
Answer: Area = 18 sq units
Where the marks are
- Correct figure with shaded region (1 mark)
- Finds intersection at $y = 4$ and $y = - 2$ (½ mark)
- Sets up and integrates $\int_{- 2}^{4} [(y + 4) - \frac{y ^{2}}{2}] d y$ (1 mark)
- Area = 18 sq units (½ mark)
Open Q29 on its own page
Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerThree Dimensional Geometry
Find the coordinates of the foot of the perpendicular drawn from the point P(0, 2, 3) to the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ .
Try it first, then show our solutionSolution
1. A general point on the line is $M (5 λ - 3, 2 λ + 1, 3 λ - 4)$ .
  Direction ratios of $P M$ are $5 λ - 3, 2 λ - 1, 3 λ - 7$ .
  1 mark
2. Since $P M ⊥$ the line (direction ratios $5, 2, 3$ ):
  $5 (5 λ - 3) + 2 (2 λ - 1) + 3 (3 λ - 7) = 0$
  $38 λ - 38 = 0$ , so $λ = 1$
  1 mark
3. Foot of the perpendicular: $M (2, 3, - 1)$
  1 mark
Answer: (2, 3, −1)
Where the marks are
- Writes the general point and direction ratios of PM (1 mark)
- Uses the perpendicularity condition to get $λ = 1$ (1 mark)
- Foot of perpendicular is $(2, 3, - 1)$ (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerVector Algebra
Three vectors $a$ , $b$ and $c$ satisfy the condition $a + b + c = 0$ . Evaluate the quantity $μ = a \cdot b + b \cdot c + c \cdot a$ , if $∣ a ∣ = 3$ , $∣ b ∣ = 4$ and $∣ c ∣ = 2$ .
Try it first, then show our solutionSolution
1. Since $a + b + c = 0$ , we have $(a + b + c) \cdot (a + b + c) = 0$ .
  1 mark
2. $∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 (a \cdot b + b \cdot c + c \cdot a) = 0$
  $9 + 16 + 4 + 2 μ = 0$
  1 mark
3. $2 μ = - 29$ , so $μ = - \frac{29}{2}$
  1 mark
Answer: $μ = - \frac{29}{2}$
Where the marks are
- Takes the dot product of $a + b + c$ with itself and equates to 0 (1 mark)
- Expands as $∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 μ = 0$ (1 mark)
- $μ = - \frac{29}{2}$ (1 mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerThree Dimensional Geometry
Find the distance between the lines :
$r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ ;
$r = (3 \hat{i} + 3 \hat{j} - 5 \hat{k}) + μ (4 \hat{i} + 6 \hat{j} + 12 \hat{k})$
Try it first, then show our solutionSolution
1. Here $a_{1} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $a_{2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$ .
  The directions are $2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $4 \hat{i} + 6 \hat{j} + 12 \hat{k} = 2 (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ .
  So the lines are parallel, with $b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ .
  ½ mark
2. $a_{2} - a_{1} = 2 \hat{i} + \hat{j} - \hat{k}$
  ½ mark
3. $(a_{2} - a_{1}) \times b = \hat{i} 22 \hat{j} 13 \hat{k} - 1 6$
  $= \hat{i} (6 + 3) - \hat{j} (12 + 2) + \hat{k} (6 - 2) = 9 \hat{i} - 14 \hat{j} + 4 \hat{k}$
  1 mark
4. Distance $d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣} = \frac{81 + 196 + 16}{4 + 9 + 36} = \frac{293}{7}$ units
  1 mark
Answer: $\frac{293}{7}$ units
Where the marks are
- Identifies that the lines are parallel (½ mark)
- Finds $a_{2} - a_{1}$ (½ mark)
- Correct cross product $9 \hat{i} - 14 \hat{j} + 4 \hat{k}$ (1 mark)
- Applies the parallel-lines distance formula to get $\frac{293}{7}$ (1 mark)
Open Q31 on its own page

Section D

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerApplication of Derivatives
The median of an equilateral triangle is increasing at the rate of $23$ cm/s. Find the rate at which its side is increasing.
Try it first, then show our solutionSolution
1. In an equilateral triangle ABC, the median from A is also the altitude. Let $h$ be the length of the median and $x$ the side.
  (Figure: equilateral triangle ABC with the median AD drawn to the midpoint D of BC, perpendicular to BC.)
  1 mark
2. In right $△ A D B$ : $h^{2} = x^{2} - (\frac{x}{2})^{2} = \frac{3 x ^{2}}{4}$ .
  So $h = \frac{3}{2} x$ , i.e. $x = \frac{2 h}{3}$ ...(i)
  2 marks
3. Given $\frac{d h}{d t} = 23$ cm/s. Differentiating (i) with respect to $t$ :
  $\frac{d x}{d t} = \frac{2}{3} \frac{d h}{d t}$
  1 mark
4. $\frac{d x}{d t} = \frac{2}{3} \times 23 = 4$
  The side is increasing at the rate of 4 cm/s.
  1 mark
Answer: 4 cm/s
Where the marks are
- Notes that the median equals the altitude, with figure (1 mark)
- Obtains $h = \frac{3}{2} x$ (2 marks)
- Differentiates with respect to t using $\frac{d h}{d t} = 23$ (1 mark)
- Rate of increase of side = 4 cm/s (1 mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerApplication of Derivatives
Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.
Try it first, then show our solutionSolution
1. Let the numbers be $x$ and $y$ with $x + y = 5$ , so $y = 5 - x$ .
  ½ mark
2. Sum of cubes: $S = x^{3} + y^{3} = x^{3} + (5 - x)^{3}$
  1 mark
3. $\frac{d S}{d x} = 3 x^{2} - 3 (5 - x)^{2} = 3 (x - 5 + x) (x + 5 - x) = 15 (2 x - 5)$
  1 mark
4. Setting $\frac{d S}{d x} = 0$ gives $x = \frac{5}{2}$ .
  ½ mark
5. $\frac{d ^{2} S}{d x ^{2}} = 6 x + 6 (5 - x) = 30 > 0$ , so $S$ is minimum at $x = \frac{5}{2}$ .
  1 mark
6. The numbers are $\frac{5}{2}$ and $\frac{5}{2}$ .
  Sum of squares $= \frac{25}{4} + \frac{25}{4} = \frac{25}{2}$
  1 mark
Answer: Sum of squares = $\frac{25}{2}$ (numbers are $\frac{5}{2}$ and $\frac{5}{2}$ )
Where the marks are
- Expresses $y = 5 - x$ (½ mark)
- Forms $S = x^{3} + (5 - x)^{3}$ (1 mark)
- Gets $\frac{d S}{d x} = 15 (2 x - 5)$ (1 mark)
- Critical point $x = \frac{5}{2}$ (½ mark)
- Second derivative test shows a minimum (1 mark)
- Sum of squares $= \frac{25}{2}$ (1 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerIntegrals
Evaluate : $\int_{0}^{π /2} sin 2 x tan^{- 1} (sin x) d x$
Try it first, then show our solutionSolution
1. $I = \int_{0}^{π /2} sin 2 x tan^{- 1} (sin x) d x = \int_{0}^{π /2} 2 sin x cos x tan^{- 1} (sin x) d x$
  ½ mark
2. Put $sin x = t$ , so $cos x d x = d t$ .
  When $x = 0$ , $t = 0$ ; when $x = \frac{π}{2}$ , $t = 1$ .
  ½ mark
3. $I = 2 \int_{0}^{1} t tan^{- 1} t d t$
  ½ mark
4. Integrating by parts, with $tan^{- 1} t$ as the first function:
  $I = 2 [\frac{t ^{2}}{2} tan^{- 1} t_{0}^{1} - \int_{0}^{1} \frac{1}{1 + t ^{2}} \cdot \frac{t ^{2}}{2} d t]$
  1 mark
5. $= 2 \cdot \frac{1}{2} \cdot \frac{π}{4} - \int_{0}^{1} \frac{t ^{2}}{1 + t ^{2}} d t = \frac{π}{4} - \int_{0}^{1} [1 - \frac{1}{1 + t ^{2}}] d t$
  1 mark
6. $= \frac{π}{4} - [t]_{0}^{1} + [tan^{- 1} t]_{0}^{1} = \frac{π}{4} - 1 + \frac{π}{4}$
  1 mark
7. $I = \frac{π}{2} - 1$
  ½ mark
Answer: $\frac{π}{2} - 1$
Where the marks are
- Writes $sin 2 x = 2 sin x cos x$ (½ mark)
- Substitution $sin x = t$ with new limits (½ mark)
- Reduces to $2 \int_{0}^{1} t tan^{- 1} t d t$ (½ mark)
- Correct integration by parts (1 mark)
- Splits $\frac{t ^{2}}{1 + t ^{2}} = 1 - \frac{1}{1 + t ^{2}}$ (1 mark)
- Correct evaluation of both integrals (1 mark)
- Final answer $\frac{π}{2} - 1$ (½ mark)
Open Q33 on its own page
Q34 · 5 marks · Long answerLinear Programming
Solve the following Linear Programming Problem graphically :
Maximize : $P = 70 x + 40 y$
subject to : $3 x + 2 y \leq 9$ , $3 x + y \leq 9$ , $x \geq 0, y \geq 0$
Try it first, then show our solutionSolution
1. Graph of the constraints. Draw each constraint line and shade the side that satisfies the inequality.
  Line $3 x + 2 y = 9$ passes through $(3, 0)$ and $(0, 4.5)$ . The point $(0, 0)$ gives $0 \leq 9$ , so we shade the origin side.
  Line $3 x + y = 9$ passes through $(3, 0)$ and $(0, 9)$ . Again we shade the origin side.
  $x \geq 0, y \geq 0$ restricts the region to the first quadrant.
  Both lines meet the $x$ -axis at the same point $C (3, 0)$ : subtracting the equations gives $y = 0$ , so $x = 3$ .
  For $y \geq 0$ we have $3 x + y \leq 3 x + 2 y \leq 9$ . So the second constraint is automatically satisfied.
  The feasible region is therefore the bounded triangle $O B C$ with vertices $O (0, 0)$ , $B (0, 4.5)$ and $C (3, 0)$ . It is shaded in the first quadrant below the line $3 x + 2 y = 9$ .
  3 marks
2. Corner point method. Evaluate $P = 70 x + 40 y$ at each vertex of the feasible region.
  Corner point $P = 70 x + 40 y$
  $O (0, 0)$ $0$
  $B (0, 4.5)$ $40 \times 4.5 = 180$
  $C (3, 0)$ $70 \times 3 = 210$ (maximum)
  1½ marks
3. The feasible region is bounded, so the maximum occurs at a corner point.
  Hence the maximum value of $P$ is $210$ , attained at $x = 3, y = 0$ .
  ½ mark
Answer: Maximum P = 210 at x = 3, y = 0
Where the marks are
- Correct graph of both lines with the feasible triangle O(0,0), B(0,4.5), C(3,0) correctly shaded (3 marks)
- Table of corner points with values of P: 0, 180, 210 (1½ marks)
- States maximum P = 210 at (3, 0) (½ mark)
Open Q34 on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerCompetency-basedProbability
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{5}$ be the probability that he knows the answer and $\frac{2}{5}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{3}$ . What is the probability that the student knows the answer, given that he answered it correctly ?
Try it first, then show our solutionSolution
1. Define the events:
  $A$ : the student knows the answer
  $B$ : the student guesses the answer
  $E$ : the student answers correctly
  1 mark
2. Prior probabilities: $P (A) = \frac{3}{5}$ and $P (B) = \frac{2}{5}$ .
  ½ mark
3. Conditional probabilities:
  A student who knows the answer is certainly correct, so $P (E ∣ A) = 1$ .
  A student who guesses is correct with probability $\frac{1}{3}$ , so $P (E ∣ B) = \frac{1}{3}$ .
  1 mark
4. By Bayes' theorem:
  $P (A ∣ E) = \frac{P ( A ) P ( E ∣ A )}{P ( A ) P ( E ∣ A ) + P ( B ) P ( E ∣ B )} = \frac{\frac{3}{5} \times 1}{\frac{3}{5} \times 1 + \frac{2}{5} \times \frac{1}{3}}$
  1½ marks
5. Simplify:
  $P (A ∣ E) = \frac{\frac{3}{5}}{\frac{3}{5} + \frac{2}{15}} = \frac{\frac{9}{15}}{\frac{11}{15}} = \frac{9}{11}$
  So the probability that the student knew the answer, given that he answered correctly, is $\frac{9}{11}$ .
  1 mark
Answer: 9/11
Where the marks are
- Defines the events (knows, guesses, answers correctly) (1 mark)
- Writes P(A) = 3/5 and P(B) = 2/5 (½ mark)
- Writes P(E|A) = 1 and P(E|B) = 1/3 (1 mark)
- Applies Bayes' theorem with correct substitution (1½ marks)
- Obtains 9/11 (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerCompetency-basedProbability
A box contains 10 tickets, 2 of which carry a prize of ₹ 8 each, 5 of which carry a prize of ₹ 4 each, and remaining 3 carry a prize of ₹ 2 each. If one ticket is drawn at random, find the mean value of the prize.
Try it first, then show our solutionSolution
1. Let $X$ be the prize value (in ₹) on the ticket drawn.
  Then $X$ can take the values $8$ , $4$ and $2$ .
  1 mark
2. Probabilities, out of 10 tickets:
  $P (X = 8) = \frac{2}{10} = \frac{1}{5}$
  $P (X = 4) = \frac{5}{10} = \frac{1}{2}$
  $P (X = 2) = \frac{3}{10}$
  $X$ 8 4 2
  $P (X)$ $\frac{1}{5}$ $\frac{1}{2}$ $\frac{3}{10}$
  $X P (X)$ $\frac{8}{5}$ $2$ $\frac{6}{10}$
  3 marks
3. Mean:
  $E (X) = \sum X P (X) = \frac{8}{5} + 2 + \frac{6}{10} = \frac{16 + 20 + 6}{10} = \frac{42}{10} = 4.2$
  So the mean value of the prize is ₹ 4.20.
  1 mark
Answer: Mean prize = 42/10 = ₹ 4.20
Where the marks are
- Identifies values of X as 8, 4, 2 (1 mark)
- Correct probability distribution 1/5, 1/2, 3/10 with XP(X) column (3 marks)
- Mean = 42/10 = ₹ 4.20 (1 mark)
Open Q35(b) on its own page

Corner point	$P = 70 x + 40 y$
$O (0, 0)$	$0$
$B (0, 4.5)$	$40 \times 4.5 = 180$
$C (3, 0)$	$70 \times 3 = 210$ (maximum)

$X$	8	4	2
$P (X)$	$\frac{1}{5}$	$\frac{1}{2}$	$\frac{3}{10}$
$X P (X)$	$\frac{8}{5}$	$2$	$\frac{6}{10}$

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedRelations and Functions
Case Study-I
An organization conducted bike race under two different categories – Boys and Girls. There were 28 participants in all. Among all of them, finally three from category 1 and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project.
Let $B = {b_{1}, b_{2}, b_{3}}$ and $G = {g_{1}, g_{2}}$ , where B represents the set of Boys selected and G the set of Girls selected for the final race.
Based on the above information, answer the following questions :
1. (I)
  How many relations are possible from B to G ?
  1 mark
2. (II)
  Among all the possible relations from B to G, how many functions can be formed from B to G ?
  1 mark
3. (III)
  Let $R : B \to B$ be defined by $R = {(x, y) : x and y are students of the same sex}$ . Check if R is an equivalence relation.
  2 marks
4. OR
  (III)
  A function $f : B \to G$ be defined by $f = {(b_{1}, g_{1}), (b_{2}, g_{2}), (b_{3}, g_{1})}$ . Check if f is bijective. Justify your answer.
  2 marks
Figure in the paper: Photograph of two riders on racing bikes, one bike ahead tilted. See it in the official paper.
Try it first, then show our solutionSolution
1. (I) We have $n (B \times G) = 3 \times 2 = 6$ .
  Every subset of $B \times G$ is a relation from $B$ to $G$ . So the number of relations is $2^{6} = 64$ .
  1 mark
2. (II) In a function, each of the 3 elements of $B$ must have exactly one image in $G$ . Each has 2 choices.
  So the number of functions is $2 \times 2 \times 2 = 2^{3} = 8$ .
  1 mark
3. (III)(a) All elements of $B$ are boys, so any two of them are of the same sex. Hence $R = B \times B$ .
  Reflexive: $(b_{i}, b_{i}) \in R$ for every $b_{i}$ .
  Symmetric: if $(x, y) \in R$ , then $(y, x) \in R$ , since they are of the same sex.
  Transitive: if $(x, y), (y, z) \in R$ , then $x$ and $z$ are of the same sex, so $(x, z) \in R$ .
  Hence $R$ is an equivalence relation.
  2 marks
4. (III)(b) We have $f (b_{1}) = g_{1}$ and $f (b_{3}) = g_{1}$ , but $b_{1} \neq = b_{3}$ .
  So $f$ is not one-one. (It is onto, since both $g_{1}$ and $g_{2}$ have pre-images.)
  Since $f$ is not one-one, $f$ is not bijective.
  2 marks
Answer: (I) 64; (II) 8; (III)(a) R is an equivalence relation; (III)(b) f is not bijective (not one-one, since f(b₁) = f(b₃) = g₁)
Where the marks are
- (I) Number of relations = 2^6 = 64 (1 mark)
- (II) Number of functions = 2^3 = 8 (1 mark)
- (III)(a) Shows R is reflexive, symmetric and transitive, hence equivalence (2 marks)
- (III)(b) Shows f is not one-one as f(b₁) = f(b₃) (1 mark); concludes f is not bijective (1 mark)
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Q37 · 4 marks · Case-basedCompetency-basedMatrices
Case Study-II
Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹ 160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹ 190. Also Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹ 250.
Based on the above information, answer the following questions :
1. (I)
  Convert the given above situation into a matrix equation of the form $A X = B$ .
  1 mark
2. (II)
  Find $∣ A ∣$ .
  1 mark
3. (III)
  Find $A^{- 1}$ .
  2 marks
4. OR
  (III)
  Determine $P = A^{2} - 5 A$ .
  2 marks
Try it first, then show our solutionSolution
1. (I) Let the prices (in ₹) of a pen, a bag and an instrument box be $x, y, z$ . Then
  $5 x + 3 y + z = 160$
  $2 x + y + 3 z = 190$
  $x + 2 y + 4 z = 250$
  In matrix form, $A X = B$ with
  $A = 521312134, X = x y z, B = 160190250$
  1 mark
2. (II) Expanding along the first row:
  $∣ A ∣ = 5 (4 - 6) - 3 (8 - 3) + 1 (4 - 1) = - 10 - 15 + 3 = - 22$
  1 mark
3. (III)(a) Cofactors:
  $C_{11} = - 2, C_{12} = - 5, C_{13} = 3$
  $C_{21} = - 10, C_{22} = 19, C_{23} = - 7$
  $C_{31} = 8, C_{32} = - 13, C_{33} = - 1$
  The adjoint is the transpose of the cofactor matrix:
  $adj A = - 2 - 5 3 - 10 19 - 7 8 - 13 - 1$
  1 mark
4. Since $∣ A ∣ = - 22 \neq = 0$ , $A^{- 1}$ exists and
  $A^{- 1} = \frac{1}{∣ A ∣} adj A = - \frac{1}{22} - 2 - 5 3 - 10 19 - 7 8 - 13 - 1$
  1 mark
5. (III)(b) Compute $A^{2}$ and $5 A$ :
  $A^{2} = 321513201313181723, 5 A = 251051551051520$
  1½ marks
6. Subtract:
  $P = A^{2} - 5 A = 7585831323$
  ½ mark
Answer: (I) A = [[5,3,1],[2,1,3],[1,2,4]], X = [x,y,z]ᵀ, B = [160,190,250]ᵀ; (II) |A| = −22; (III)(a) A⁻¹ = −(1/22)[[−2,−10,8],[−5,19,−13],[3,−7,−1]]; (III)(b) P = [[7,5,13],[5,8,2],[8,3,3]]
Where the marks are
- (I) Correct matrix equation AX = B (1 mark)
- (II) |A| = −22 (1 mark)
- (III)(a) Correct adj A (1 mark); A⁻¹ = (1/−22) adj A (1 mark)
- (III)(b) Correct A² and 5A (1½ marks); P = [[7,5,13],[5,8,2],[8,3,3]] (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedDifferential Equations
Case Study-III
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form $\frac{d y}{d x} = F (x, y)$ is said to be homogeneous if $F (x, y)$ is a homogeneous function of degree zero, whereas a function $F (x, y)$ is a homogenous function of degree n if $F (λ x, λ y) = λ^{n} F (x, y)$ . To solve a homogeneous differential equation of the type $\frac{d y}{d x} = F (x, y) = g (\frac{y}{x})$ , we make the substitution $y = v x$ and then separate the variables.
Based on the above, answer the following questions :
1. (I)
  Show that $(x^{2} - y^{2}) d x + 2 x y d y = 0$ is a differential equation of the type $\frac{d y}{d x} = g (\frac{y}{x})$ .
  2 marks
2. (II)
  Solve the above equation to find its general solution.
  2 marks
Try it first, then show our solutionSolution
1. (I) From $(x^{2} - y^{2}) d x + 2 x y d y = 0$ :
  $\frac{d y}{d x} = - \frac{x ^{2} - y ^{2}}{2 x y} = \frac{y ^{2} - x ^{2}}{2 x y}$
  1 mark
2. Divide the numerator and denominator by $x^{2}$ :
  $\frac{d y}{d x} = \frac{( \frac{y}{x} ) ^{2} - 1}{2 ( \frac{y}{x} )} = g (\frac{y}{x})$
  So the equation is of the type $\frac{d y}{d x} = g (\frac{y}{x})$ . It is homogeneous, with $F (λ x, λ y) = λ^{0} F (x, y)$ .
  1 mark
3. (II) Put $y = v x$ , so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Then
  $v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} \Rightarrow x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v = \frac{- 1 - v ^{2}}{2 v}$
  ½ mark
4. Separate the variables:
  $\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{d x}{x}$
  1 mark
5. Integrate:
  $lo g ∣1 + v^{2} ∣ + lo g ∣ x ∣ = lo g C \Rightarrow x (1 + \frac{y ^{2}}{x ^{2}}) = C$
  Hence the general solution is $x^{2} + y^{2} = C x$ .
  ½ mark
Answer: (I) dy/dx = ((y/x)² − 1)/(2(y/x)) = g(y/x); (II) x² + y² = Cx
Where the marks are
- (I) Writes dy/dx = (y² − x²)/(2xy) (1 mark)
- (I) Expresses it as a function of y/x (1 mark)
- (II) Substitutes y = vx, dy/dx = v + x dv/dx (½ mark)
- (II) Separates variables: 2v/(1+v²) dv = −dx/x (1 mark)
- (II) Integrates to get x² + y² = Cx (½ mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.