MathsCBSE Class 12 · Board exam 2024 (session 2023-24)

CBSE Class 12 Mathematics: 2024 board paper, set 65/1/1

The question paper set in the CBSE Class 12 board exam of 2024, 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/2024/XII/MATHEMATICS.zip (a zip; this paper is the file 65_1_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 Section C, 2 in Section D and 2 in Section E; calculators not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQRelations and Functions
A function $f : R_{+} \to R$ (where $R_{+}$ is the set of all non-negative real numbers) defined by $f (x) = 4 x + 3$ is :
1. (A)one-one but not onto
2. (B)onto but not one-one
3. (C)both one-one and onto
4. (D)neither one-one nor onto
Try it first, then show our solutionSolution
1. $f (x_{1}) = f (x_{2}) \Rightarrow 4 x_{1} + 3 = 4 x_{2} + 3 \Rightarrow x_{1} = x_{2}$ , so $f$ is one-one.
  ½ mark
2. For $x \geq 0$ , $f (x) \geq 3$ , so values such as $0$ in the codomain $R$ have no pre-image. Hence $f$ is not onto.
  ½ mark
Answer: (A) one-one but not onto
Where the marks are
- Identifies one-one but not onto (1 mark)
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Q2 · 1 mark · MCQMatrices
If a matrix has 36 elements, the number of possible orders it can have, is :
1. (A)13
2. (B)3
3. (C)5
4. (D)9
Try it first, then show our solutionSolution
1. An order $m \times n$ needs $mn = 36$ . The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
  ½ mark
2. Each divisor $m$ gives exactly one order $m \times \frac{36}{m}$ , so there are 9 possible orders.
  ½ mark
Answer: (D) 9
Where the marks are
- Counts the divisors of 36 as the number of orders, 9 (1 mark)
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Q3 · 1 mark · MCQContinuity and Differentiability
Which of the following statements is true for the function $f (x) = {x^{2} + 3, 1, x \neq = 0 x = 0$ ?
1. (A)f(x) is continuous and differentiable $\forall x \in R$
2. (B)f(x) is continuous $\forall x \in R$
3. (C)f(x) is continuous and differentiable $\forall x \in R - {0}$
4. (D)f(x) is discontinuous at infinitely many points
Try it first, then show our solutionSolution
1. $lim_{x \to 0} f (x) = 0 + 3 = 3$ , but $f (0) = 1$ , so $f$ is discontinuous at $x = 0$ and therefore not differentiable there.
  ½ mark
2. For $x \neq = 0$ , $f (x) = x^{2} + 3$ is a polynomial, so it is continuous and differentiable. Hence (C) is true.
  ½ mark
Answer: (C)
Where the marks are
- Shows the discontinuity at 0 and continuity elsewhere, giving (C) (1 mark)
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Q4 · 1 mark · MCQApplication of Derivatives
Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Then, this function f(x) is strictly increasing in (a, b) if
1. (A) $f^{'} (x) < 0, \forall x \in (a, b)$
2. (B) $f^{'} (x) > 0, \forall x \in (a, b)$
3. (C) $f^{'} (x) = 0, \forall x \in (a, b)$
4. (D) $f (x) > 0, \forall x \in (a, b)$
Try it first, then show our solutionSolution
1. By the NCERT theorem, if $f^{'} (x) > 0$ for every $x \in (a, b)$ , then $f$ is strictly increasing on that interval.
  1 mark
Answer: (B)
Where the marks are
- Uses the condition $f^{'} > 0$ (1 mark)
Q5 · 1 mark · MCQMatrices
If $[x + y 5 2 x y] = [6528]$ , then the value of $(\frac{24}{x} + \frac{24}{y})$ is :
1. (A)7
2. (B)6
3. (C)8
4. (D)18
Try it first, then show our solutionSolution
1. Equating corresponding elements gives $x + y = 6$ and $x y = 8$ .
  ½ mark
2. $\frac{24}{x} + \frac{24}{y} = \frac{24 ( x + y )}{x y} = \frac{24 \times 6}{8} = 18$ .
  ½ mark
Answer: (D) 18
Where the marks are
- Equates elements and simplifies to 18 (1 mark)
Q6 · 1 mark · MCQIntegrals
$\int_{a}^{b} f (x) d x$ is equal to :
1. (A) $\int_{a}^{b} f (a - x) d x$
2. (B) $\int_{a}^{b} f (a + b - x) d x$
3. (C) $\int_{a}^{b} f (x - (a + b)) d x$
4. (D) $\int_{a}^{b} f ((a - x) + (b - x)) d x$
Try it first, then show our solutionSolution
1. Substitute $x = a + b - t$ . Then $d x = - d t$ and the limits $a, b$ become $b, a$ .
  ½ mark
2. So $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - t) d t$ , which is option (B).
  ½ mark
Answer: (B)
Where the marks are
- Applies the property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ (1 mark)
Q7 · 1 mark · MCQVector Algebra
Let $θ$ be the angle between two unit vectors $\overset{a}{^}$ and $\hat{b}$ such that $sin θ = \frac{3}{5}$ . Then, $\overset{a}{^} \cdot \hat{b}$ is equal to :
1. (A) $\pm \frac{3}{5}$
2. (B) $\pm \frac{3}{4}$
3. (C) $\pm \frac{4}{5}$
4. (D) $\pm \frac{4}{3}$
Try it first, then show our solutionSolution
1. $cos θ = \pm 1 - \frac{9}{25} = \pm \frac{4}{5}$ .
  ½ mark
2. For unit vectors, $\overset{a}{^} \cdot \hat{b} = ∣ \overset{a}{^} ∣∣ \hat{b} ∣ cos θ = cos θ = \pm \frac{4}{5}$ .
  ½ mark
Answer: (C) $\pm \frac{4}{5}$
Where the marks are
- Finds $cos θ = \pm \frac{4}{5}$ and uses it as the dot product (1 mark)
Q8 · 1 mark · MCQDifferential Equations
The integrating factor of the differential equation $(1 - x^{2}) \frac{d y}{d x} + x y = a x$ , $- 1 < x < 1$ , is :
1. (A) $\frac{1}{x ^{2} - 1}$
2. (B) $\frac{1}{x ^{2} - 1}$
3. (C) $\frac{1}{1 - x ^{2}}$
4. (D) $\frac{1}{1 - x ^{2}}$
Try it first, then show our solutionSolution
1. Dividing by $1 - x^{2}$ gives the linear form $\frac{d y}{d x} + \frac{x}{1 - x ^{2}} y = \frac{a x}{1 - x ^{2}}$ , so $P = \frac{x}{1 - x ^{2}}$ .
  ½ mark
2. $\int P d x = - \frac{1}{2} lo g (1 - x^{2})$ , so $IF = e^{- \frac{1}{2} l o g (1 - x^{2})} = \frac{1}{1 - x ^{2}}$ .
  ½ mark
Answer: (D) $\frac{1}{1 - x ^{2}}$
Where the marks are
- Finds $P = \frac{x}{1 - x ^{2}}$ and $IF = \frac{1}{1 - x ^{2}}$ (1 mark)
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Q9 · 1 mark · MCQThree Dimensional Geometry
If the direction cosines of a line are $3 k, 3 k, 3 k$ , then the value of k is :
1. (A) $\pm 1$
2. (B) $\pm 3$
3. (C) $\pm 3$
4. (D) $\pm \frac{1}{3}$
Try it first, then show our solutionSolution
1. Direction cosines satisfy $l^{2} + m^{2} + n^{2} = 1$ , so $3 k^{2} + 3 k^{2} + 3 k^{2} = 1$ .
  ½ mark
2. $9 k^{2} = 1 \Rightarrow k = \pm \frac{1}{3}$ .
  ½ mark
Answer: (D) $\pm \frac{1}{3}$
Where the marks are
- Uses $l^{2} + m^{2} + n^{2} = 1$ to get $k = \pm \frac{1}{3}$ (1 mark)
Q10 · 1 mark · MCQLinear Programming
A linear programming problem deals with the optimization of a/an :
1. (A)logarithmic function
2. (B)linear function
3. (C)quadratic function
4. (D)exponential function
Try it first, then show our solutionSolution
1. In an LPP, the objective function $Z = a x + b y$ is linear and the constraints are linear inequalities, so it optimises a linear function.
  1 mark
Answer: (B) linear function
Where the marks are
- Identifies that the objective function is linear (1 mark)
Q11 · 1 mark · MCQProbability
If $P (A ∣ B) = P (A^{'} ∣ B)$ , then which of the following statements is true ?
1. (A) $P (A) = P (A^{'})$
2. (B) $P (A) = 2 P (B)$
3. (C) $P (A \cap B) = \frac{1}{2} P (B)$
4. (D) $P (A \cap B) = 2 P (B)$
Try it first, then show our solutionSolution
1. $P (A ∣ B) + P (A^{'} ∣ B) = 1$ , and the two are equal, so $P (A ∣ B) = \frac{1}{2}$ .
  ½ mark
2. $\frac{P ( A \cap B )}{P ( B )} = \frac{1}{2} \Rightarrow P (A \cap B) = \frac{1}{2} P (B)$ .
  ½ mark
Answer: (C)
Where the marks are
- Uses $P (A ∣ B) + P (A^{'} ∣ B) = 1$ to get $P (A \cap B) = \frac{1}{2} P (B)$ (1 mark)
Q12 · 1 mark · MCQDeterminants
$x + 1 x^{2} + x + 1 x - 1 x^{2} - x + 1$ is equal to :
1. (A) $2 x^{3}$
2. (B)2
3. (C)0
4. (D) $2 x^{3} - 2$
Try it first, then show our solutionSolution
1. Value $= (x + 1) (x^{2} - x + 1) - (x - 1) (x^{2} + x + 1)$ .
  ½ mark
2. $= (x^{3} + 1) - (x^{3} - 1) = 2$ .
  ½ mark
Answer: (B) 2
Where the marks are
- Expands the determinant using the sum and difference of cubes to get 2 (1 mark)
Q13 · 1 mark · MCQContinuity and Differentiability
The derivative of $sin (x^{2})$ w.r.t. x, at $x = π$ is :
1. (A)1
2. (B) $- 1$
3. (C) $- 2 π$
4. (D) $2 π$
Try it first, then show our solutionSolution
1. By the chain rule, $\frac{d}{d x} sin (x^{2}) = 2 x cos (x^{2})$ .
  ½ mark
2. At $x = π$ : $2 π cos π = - 2 π$ .
  ½ mark
Answer: (C) $- 2 π$
Where the marks are
- Applies the chain rule and substitutes $x = π$ (1 mark)
Q14 · 1 mark · MCQDifferential Equations
The order and degree of the differential equation $[1 + (\frac{d y}{d x})^{2}]^{3} = \frac{d ^{2} y}{d x ^{2}}$ respectively are :
1. (A)1, 2
2. (B)2, 3
3. (C)2, 1
4. (D)2, 6
Try it first, then show our solutionSolution
1. The highest-order derivative is $\frac{d ^{2} y}{d x ^{2}}$ , so the order is 2.
  ½ mark
2. The equation is polynomial in its derivatives and $\frac{d ^{2} y}{d x ^{2}}$ appears to power 1, so the degree is 1.
  ½ mark
Answer: (C) 2, 1
Where the marks are
- Order 2 and degree 1 (1 mark)
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Q15 · 1 mark · MCQVector Algebra
The vector with terminal point A (2, – 3, 5) and initial point B (3, – 4, 7) is :
1. (A) $\hat{i} - \hat{j} + 2 \hat{k}$
2. (B) $\hat{i} + \hat{j} + 2 \hat{k}$
3. (C) $- \hat{i} - \hat{j} - 2 \hat{k}$
4. (D) $- \hat{i} + \hat{j} - 2 \hat{k}$
Try it first, then show our solutionSolution
1. $B A = terminal - initial = (2 - 3) \hat{i} + (- 3 + 4) \hat{j} + (5 - 7) \hat{k} = - \hat{i} + \hat{j} - 2 \hat{k}$ .
  1 mark
Answer: (D)
Where the marks are
- Computes terminal minus initial point (1 mark)
Q16 · 1 mark · MCQThree Dimensional Geometry
The distance of point P(a, b, c) from y-axis is :
1. (A)b
2. (B) $b^{2}$
3. (C) $a^{2} + c^{2}$
4. (D) $a^{2} + c^{2}$
Try it first, then show our solutionSolution
1. The foot of the perpendicular from P to the y-axis is $(0, b, 0)$ .
  ½ mark
2. Distance $= a^{2} + 0 + c^{2} = a^{2} + c^{2}$ .
  ½ mark
Answer: (C) $a^{2} + c^{2}$
Where the marks are
- Uses the foot $(0, b, 0)$ to get $a^{2} + c^{2}$ (1 mark)
Q17 · 1 mark · MCQLinear Programming
The number of corner points of the feasible region determined by constraints $x \geq 0, y \geq 0, x + y \geq 4$ is :
1. (A)0
2. (B)1
3. (C)2
4. (D)3
Try it first, then show our solutionSolution
1. The feasible region is the unbounded part of the first quadrant lying on or beyond the line $x + y = 4$ .
  ½ mark
2. Its corner points are $(4, 0)$ and $(0, 4)$ , so there are 2 corner points.
  ½ mark
Answer: (C) 2
Where the marks are
- Identifies the corners $(4, 0)$ and $(0, 4)$ (1 mark)
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Q18 · 1 mark · MCQMatrices
If A and B are two non-zero square matrices of same order such that $(A + B)^{2} = A^{2} + B^{2}$ , then :
1. (A) $A B = O$
2. (B) $A B = - B A$
3. (C) $B A = O$
4. (D) $A B = B A$
Try it first, then show our solutionSolution
1. $(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , since matrix multiplication is not commutative.
  ½ mark
2. Equating with $A^{2} + B^{2}$ gives $A B + B A = O$ , so $A B = - B A$ .
  ½ mark
Answer: (B) $A B = - B A$
Where the marks are
- Expands without assuming $A B = B A$ and gets $A B = - B A$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedDeterminants
Assertion (A) : For matrix $A = 1 - cos θ - 1 cos θ 1 - cos θ 1 cos θ 1$ , where $θ \in [0, 2 π]$ , $∣ A ∣ \in [2, 4]$ .
Reason (R) : $cos θ \in [- 1, 1], \forall θ \in [0, 2 π]$ .
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Expanding along $R_{1}$ : $∣ A ∣ = 1 (1 + cos^{2} θ) - cos θ (- cos θ + cos θ) + 1 (cos^{2} θ + 1) = 2 + 2 cos^{2} θ$ .
  ½ mark
2. Since $cos θ \in [- 1, 1]$ , $cos^{2} θ \in [0, 1]$ , so $∣ A ∣ \in [2, 4]$ . Both statements are true, and R is what gives A.
  ½ mark
Answer: (A)
Where the marks are
- Finds $∣ A ∣ = 2 + 2 cos^{2} θ$ and uses the range of $cos θ$ (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedThree Dimensional Geometry
Assertion (A) : A line in space cannot be drawn perpendicular to x, y and z axes simultaneously.
Reason (R) : For any line making angles, $α, β, γ$ with the positive directions of x, y and z axes respectively, $cos^{2} α + cos^{2} β + cos^{2} γ = 1$ .
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. If a line were perpendicular to all three axes, then $α = β = γ = 9 0^{\circ}$ , giving $cos^{2} α + cos^{2} β + cos^{2} γ = 0 \neq = 1$ .
  ½ mark
2. This contradicts R, which is a true identity, so A is true and R correctly explains it.
  ½ mark
Answer: (A)
Where the marks are
- Uses the identity in R to show that A holds (1 mark)
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Section B

VSA

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerContinuity and Differentiability
Check whether the function $f (x) = x^{2} ∣ x ∣$ is differentiable at $x = 0$ or not.
Try it first, then show our solutionSolution
1. Remove the modulus: $f (x) = {x^{3}, - x^{3}, x \geq 0 x < 0$ , so $f (0) = 0$ .
  ½ mark
2. RHD at $0$ : $lim_{h \to 0} \frac{f ( 0 + h ) - f ( 0 )}{h} = lim_{h \to 0} \frac{h ^{3} - 0}{h} = lim_{h \to 0} h^{2} = 0$
  ½ mark
3. LHD at $0$ : $lim_{h \to 0} \frac{f ( 0 - h ) - f ( 0 )}{- h} = lim_{h \to 0} \frac{- ( - h ) ^{3} - 0}{- h} = lim_{h \to 0} \frac{h ^{3}}{- h} = lim_{h \to 0} (- h^{2}) = 0$
  ½ mark
4. Since LHD = RHD = 0, $f (x)$ is differentiable at $x = 0$ , with $f^{'} (0) = 0$ .
  ½ mark
Answer: f is differentiable at x = 0 (f'(0) = 0)
Where the marks are
- Writes f(x) piecewise without modulus (0.5 mark)
- Finds RHD = 0 (0.5 mark)
- Finds LHD = 0 (0.5 mark)
- Concludes differentiable since LHD = RHD (0.5 mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerContinuity and Differentiability
If $y = tan x$ , prove that $x \frac{d y}{d x} = \frac{1 + y ^{4}}{4 y}$ .
Try it first, then show our solutionSolution
1. Differentiate $y = tan x$ by the chain rule: $\frac{d y}{d x} = \frac{1}{2 tan x} \cdot sec^{2} x \cdot \frac{1}{2 x}$
  1 mark
2. So $x \frac{d y}{d x} = \frac{sec ^{2} x}{4 tan x} = \frac{1 + tan ^{2} x}{4 tan x}$ . Since $y^{2} = tan x$ , $tan^{2} x = y^{4}$ and $tan x = y$ , giving $x \frac{d y}{d x} = \frac{1 + y ^{4}}{4 y}$ . Hence proved.
  1 mark
Answer: $x \frac{d y}{d x} = \frac{1 + y ^{4}}{4 y}$ (proved)
Where the marks are
- Correct chain-rule derivative of $tan x$ (1 mark)
- Uses $sec^{2} = 1 + tan^{2}$ and $y^{2} = tan x$ to reach the result (1 mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerApplication of Derivatives
Show that the function $f (x) = 4 x^{3} - 18 x^{2} + 27 x - 7$ has neither maxima nor minima.
Try it first, then show our solutionSolution
1. $f^{'} (x) = 12 x^{2} - 36 x + 27$
  ½ mark
2. $f^{'} (x) = 3 (4 x^{2} - 12 x + 9) = 3 (2 x - 3)^{2} \geq 0$ for all $x \in R$
  ½ mark
3. $f^{'} (x) = 0$ only at $x = \frac{3}{2}$ and $f^{'}$ does not change sign there, so $f$ is increasing on $R$ .
  ½ mark
4. An increasing function has no point where $f^{'}$ changes sign, hence $f (x)$ has neither a maximum nor a minimum.
  ½ mark
Answer: $f^{'} (x) = 3 (2 x - 3)^{2} \geq 0$ , so f is increasing on R and has neither maxima nor minima.
Where the marks are
- Finds $f^{'} (x) = 12 x^{2} - 36 x + 27$ (0.5 mark)
- Writes it as $3 (2 x - 3)^{2} \geq 0$ (0.5 mark)
- States f is increasing on R (0.5 mark)
- Concludes no maxima/minima (0.5 mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerIntegrals
Find : $\int x 1 + 2 x d x$
Try it first, then show our solutionSolution
1. Put $1 + 2 x = t^{2}$ , so $2 d x = 2 t d t \Rightarrow d x = t d t$ and $x = \frac{t ^{2} - 1}{2}$ .
  ½ mark
2. $\int x 1 + 2 x d x = \int \frac{t ^{2} - 1}{2} \cdot t \cdot t d t = \frac{1}{2} \int (t^{4} - t^{2}) d t = \frac{1}{2} [\frac{t ^{5}}{5} - \frac{t ^{3}}{3}] + C$
  1 mark
3. $= \frac{( 1 + 2 x ) ^{5/2}}{10} - \frac{( 1 + 2 x ) ^{3/2}}{6} + C$
  ½ mark
Answer: $\frac{( 1 + 2 x ) ^{5/2}}{10} - \frac{( 1 + 2 x ) ^{3/2}}{6} + C$
Where the marks are
- Correct substitution $1 + 2 x = t^{2}$ with $d x = t d t$ (0.5 mark)
- Reduces to $\frac{1}{2} \int (t^{4} - t^{2}) d t$ and integrates (1 mark)
- Back-substitutes correctly (0.5 mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerIntegrals
Evaluate : $\int_{0}^{\frac{π ^{2}}{4}} \frac{s i n x}{x} d x$
Try it first, then show our solutionSolution
1. Put $x = t \Rightarrow x = t^{2}, d x = 2 t d t$ . When $x = 0, t = 0$ ; when $x = \frac{π ^{2}}{4}, t = \frac{π}{2}$ .
  ½ mark
2. $I = \int_{0}^{π /2} \frac{sin t}{t} \cdot 2 t d t = 2 \int_{0}^{π /2} sin t d t = 2 [- cos t]_{0}^{π /2}$
  1 mark
3. $= 2 (- cos \frac{π}{2} + cos 0) = 2 (0 + 1) = 2$
  ½ mark
Answer: 2
Where the marks are
- Substitution $x = t$ , $d x = 2 t d t$ with changed limits (0.5 mark)
- Reduces to $2 \int_{0}^{π /2} sin t d t$ and integrates (1 mark)
- Correct value 2 (0.5 mark)
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Q24 · 2 marks · Very short answerVector Algebra
If $a$ and $b$ are two non-zero vectors such that $(a + b) ⊥ a$ and $(2 a + b) ⊥ b$ , then prove that $∣ b ∣ = 2 ∣ a ∣$ .
Try it first, then show our solutionSolution
1. $(a + b) ⊥ a \Rightarrow (a + b) \cdot a = 0 \Rightarrow ∣ a ∣^{2} + a \cdot b = 0$ ...(1)
  ½ mark
2. $(2 a + b) ⊥ b \Rightarrow (2 a + b) \cdot b = 0 \Rightarrow 2 a \cdot b + ∣ b ∣^{2} = 0$ ...(2)
  ½ mark
3. From (1), $a \cdot b = - ∣ a ∣^{2}$ ; substituting in (2): $- 2∣ a ∣^{2} + ∣ b ∣^{2} = 0$
  ½ mark
4. $∣ b ∣^{2} = 2∣ a ∣^{2} \Rightarrow ∣ b ∣ = 2 ∣ a ∣$ (magnitudes are positive). Hence proved.
  ½ mark
Answer: $∣ b ∣ = 2 ∣ a ∣$ (proved)
Where the marks are
- Writes dot-product condition (1) (0.5 mark)
- Writes dot-product condition (2) (0.5 mark)
- Eliminates $a \cdot b$ (0.5 mark)
- Concludes $∣ b ∣ = 2 ∣ a ∣$ (0.5 mark)
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Q25 · 2 marks · Very short answerVector Algebra
In the given figure, ABCD is a parallelogram. If $A B = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $D B = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$ , then find $A D$ and hence find the area of parallelogram ABCD.
Figure in the paper: Parallelogram ABCD with A top-left, B top-right, C bottom-right, D bottom-left. Arrows mark vector AB along the top side (A to B), vector AD along the left side (D to A direction arrow), and diagonal DB from D to B. See it in the official paper.
Try it first, then show our solutionSolution
1. By triangle law, $A D + D B = A B \Rightarrow A D = A B - D B = (2 \hat{i} - 4 \hat{j} + 5 \hat{k}) - (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = - \hat{i} + 2 \hat{j} + 3 \hat{k}$
  ½ mark
2. $A D \times A B = \hat{i} - 1 2 \hat{j} 2 - 4 \hat{k} 35 = \hat{i} (10 + 12) - \hat{j} (- 5 - 6) + \hat{k} (4 - 4) = 22 \hat{i} + 11 \hat{j}$
  1 mark
3. Area of parallelogram $= ∣ A D \times A B ∣ = 2 2^{2} + 1 1^{2} = 605 = 115$ sq units
  ½ mark
Answer: $A D = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ ; Area $= 115$ (i.e. $605$ ) sq units
Where the marks are
- Finds $A D = - \hat{i} + 2 \hat{j} + 3 \hat{k}$ (0.5 mark)
- Computes cross product $22 \hat{i} + 11 \hat{j}$ (1 mark)
- Area $= 115$ sq units (0.5 mark)
Open Q25 on its own page

Section C

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerRelations and Functions
A relation R on set $A = {1, 2, 3, 4, 5}$ is defined as $R = {(x, y) : ∣ x^{2} - y^{2} ∣ < 8}$ . Check whether the relation R is reflexive, symmetric and transitive.
Try it first, then show our solutionSolution
1. Reflexive: For every $x \in A$ , $∣ x^{2} - x^{2} ∣ = 0 < 8$ , so $(x, x) \in R$ . Hence R is reflexive.
  ½ mark
2. Symmetric: Let $(x, y) \in R$ . Then $∣ x^{2} - y^{2} ∣ < 8 \Rightarrow ∣ y^{2} - x^{2} ∣ = ∣ x^{2} - y^{2} ∣ < 8 \Rightarrow (y, x) \in R$ . Hence R is symmetric.
  1 mark
3. Transitive: $(1, 2) \in R$ since $∣1 - 4∣ = 3 < 8$ , and $(2, 3) \in R$ since $∣4 - 9∣ = 5 < 8$ . But $∣1 - 9∣ = 8 \neq < 8$ , so $(1, 3) \in / R$ . Hence R is not transitive.
  1½ marks
Answer: R is reflexive and symmetric but not transitive.
Where the marks are
- Proves reflexive (0.5 mark)
- Proves symmetric using $∣ x^{2} - y^{2} ∣ = ∣ y^{2} - x^{2} ∣$ (1 mark)
- Counter-example (1,2),(2,3) ∈ R but (1,3) ∉ R, so not transitive (1.5 marks)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerRelations and Functions
A function f is defined from $R \to R$ as $f (x) = a x + b$ , such that $f (1) = 1$ and $f (2) = 3$ . Find function f(x). Hence, check whether function f(x) is one-one and onto or not.
Try it first, then show our solutionSolution
1. $f (1) = 1 \Rightarrow a + b = 1$ ; $f (2) = 3 \Rightarrow 2 a + b = 3$ . Subtracting, $a = 2$ , $b = - 1$ . So $f (x) = 2 x - 1$ .
  1 mark
2. One-one: Let $f (x_{1}) = f (x_{2})$ for $x_{1}, x_{2} \in R$ . Then $2 x_{1} - 1 = 2 x_{2} - 1 \Rightarrow x_{1} = x_{2}$ . Hence f is one-one.
  1 mark
3. Onto: Let $y \in R$ (codomain) and $y = 2 x - 1 \Rightarrow x = \frac{y + 1}{2} \in R$ (domain). Then $f (\frac{y + 1}{2}) = 2 \cdot \frac{y + 1}{2} - 1 = y$ . So every $y$ has a pre-image; f is onto.
  1 mark
Answer: f(x) = 2x − 1; f is both one-one and onto.
Where the marks are
- Solves for a = 2, b = −1 to get f(x) = 2x − 1 (1 mark)
- Shows one-one (1 mark)
- Shows onto by finding pre-image $(y + 1) /2$ (1 mark)
Open Q26(b) on its own page
Answer one of these 2 alternatives.
Q27(a) · 3 marks · Short answerContinuity and Differentiability
If $1 - x^{2} + 1 - y^{2} = a (x - y)$ , prove that $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}}$ .
Try it first, then show our solutionSolution
1. Put $x = sin θ$ , $y = sin ϕ$ . Then the equation becomes $cos θ + cos ϕ = a (sin θ - sin ϕ)$ .
  ½ mark
2. $2 cos \frac{θ + ϕ}{2} cos \frac{θ - ϕ}{2} = 2 a cos \frac{θ + ϕ}{2} sin \frac{θ - ϕ}{2} \Rightarrow cot \frac{θ - ϕ}{2} = a$
  ½ mark
3. $θ - ϕ = 2 cot^{- 1} a \Rightarrow sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$
  ½ mark
4. Differentiating w.r.t. $x$ (RHS is constant): $\frac{1}{1 - x ^{2}} - \frac{1}{1 - y ^{2}} \frac{d y}{d x} = 0$
  1 mark
5. $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}} = \frac{1 - y ^{2}}{1 - x ^{2}}$ . Hence proved.
  ½ mark
Answer: $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}}$ (proved)
Where the marks are
- Substitution $x = sin θ$ , $y = sin ϕ$ (0.5 mark)
- Sum-to-product to get $cot \frac{θ - ϕ}{2} = a$ (0.5 mark)
- Writes $sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$ (0.5 mark)
- Differentiates correctly (1 mark)
- Final result (0.5 mark)
Open Q27(a) on its own page
OR
Q27(b) · 3 marks · Short answerContinuity and Differentiability
If $y = (tan x)^{x}$ , then find $\frac{d y}{d x}$ .
Try it first, then show our solutionSolution
1. Taking log on both sides: $lo g y = x lo g (tan x)$
  ½ mark
2. Differentiating w.r.t. $x$ (product rule): $\frac{1}{y} \frac{d y}{d x} = x \cdot \frac{sec ^{2} x}{tan x} + lo g (tan x) \cdot 1$
  2 marks
3. $\frac{d y}{d x} = (tan x)^{x} [\frac{x sec ^{2} x}{tan x} + lo g (tan x)]$ (equivalently $(tan x)^{x} [2 x cosec 2 x + lo g tan x]$ )
  ½ mark
Answer: $\frac{d y}{d x} = (tan x)^{x} [\frac{x s e c ^{2} x}{t a n x} + lo g (tan x)]$
Where the marks are
- Takes logarithm: $lo g y = x lo g tan x$ (0.5 mark)
- Correct differentiation using product and chain rules (2 marks)
- Final expression for dy/dx (0.5 mark)
Open Q27(b) on its own page
Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntegrals
Find : $\int \frac{x ^{2}}{( x ^{2} + 4 ) ( x ^{2} + 9 )} d x$
Try it first, then show our solutionSolution
1. Put $x^{2} = t$ (only for partial fractions): $\frac{t}{( t + 4 ) ( t + 9 )} = \frac{A}{t + 4} + \frac{B}{t + 9}$ , i.e. $t = A (t + 9) + B (t + 4)$ .
  ½ mark
2. $t = - 4$ : $- 4 = 5 A \Rightarrow A = - \frac{4}{5}$ ; $t = - 9$ : $- 9 = - 5 B \Rightarrow B = \frac{9}{5}$ . So $I = - \frac{4}{5} \int \frac{d x}{2 ^{2} + x ^{2}} + \frac{9}{5} \int \frac{d x}{3 ^{2} + x ^{2}}$
  1½ marks
3. $I = - \frac{4}{5} \cdot \frac{1}{2} tan^{- 1} \frac{x}{2} + \frac{9}{5} \cdot \frac{1}{3} tan^{- 1} \frac{x}{3} + C = - \frac{2}{5} tan^{- 1} \frac{x}{2} + \frac{3}{5} tan^{- 1} \frac{x}{3} + C$
  1 mark
Answer: $- \frac{2}{5} tan^{- 1} \frac{x}{2} + \frac{3}{5} tan^{- 1} \frac{x}{3} + C$
Where the marks are
- Puts $x^{2} = t$ and sets up partial fractions (0.5 mark)
- Finds A = −4/5, B = 9/5 and splits the integral (1.5 marks)
- Integrates to tan⁻¹ terms correctly (1 mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntegrals
Evaluate : $\int_{1}^{3} (∣ x - 1∣ + ∣ x - 2∣ + ∣ x - 3∣) d x$
Try it first, then show our solutionSolution
1. On $[1, 3]$ : $∣ x - 1∣ = x - 1$ , $∣ x - 3∣ = 3 - x$ ; $∣ x - 2∣ = 2 - x$ on $[1, 2]$ and $x - 2$ on $[2, 3]$ . So $I = \int_{1}^{3} (x - 1) d x + \int_{1}^{2} (2 - x) d x + \int_{2}^{3} (x - 2) d x + \int_{1}^{3} (3 - x) d x = \int_{1}^{3} 2 d x + \int_{1}^{2} (2 - x) d x + \int_{2}^{3} (x - 2) d x$
  1½ marks
2. $= [2 x]_{1}^{3} + [\frac{( 2 - x ) ^{2}}{- 2}]_{1}^{2} + [\frac{( x - 2 ) ^{2}}{2}]_{2}^{3} = 4 + \frac{1}{2} + \frac{1}{2} = 5$
  1½ marks
Answer: 5
Where the marks are
- Removes modulus correctly on the sub-intervals (1.5 marks)
- Evaluates the integrals to get 5 (1.5 marks)
Open Q28(b) on its own page
Q29 · 3 marks · Short answerDifferential Equations
Find the particular solution of the differential equation given by $x^{2} \frac{d y}{d x} - x y = x^{2} cos^{2} (\frac{y}{2 x})$ , given that when $x = 1$ , $y = \frac{π}{2}$ .
Try it first, then show our solutionSolution
1. Dividing by $x^{2}$ : $\frac{d y}{d x} = \frac{y}{x} + cos^{2} (\frac{y}{2 x})$ , a homogeneous equation.
  ½ mark
2. Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
  ½ mark
3. $v + x \frac{d v}{d x} = v + cos^{2} \frac{v}{2} \Rightarrow \int sec^{2} \frac{v}{2} d v = \int \frac{d x}{x}$
  ½ mark
4. $2 tan \frac{v}{2} = lo g ∣ x ∣ + C \Rightarrow 2 tan (\frac{y}{2 x}) = lo g ∣ x ∣ + C$
  1 mark
5. At $x = 1, y = \frac{π}{2}$ : $2 tan \frac{π}{4} = lo g 1 + C \Rightarrow C = 2$ . Particular solution: $2 tan (\frac{y}{2 x}) = lo g ∣ x ∣ + 2$
  ½ mark
Answer: $2 tan (\frac{y}{2 x}) = lo g ∣ x ∣ + 2$
Where the marks are
- Writes in form dy/dx = y/x + cos²(y/2x) (0.5 mark)
- Substitutes y = vx (0.5 mark)
- Separates variables (0.5 mark)
- Integrates to get general solution (1 mark)
- Finds C = 2 and particular solution (0.5 mark)
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Q30 · 3 marks · Short answerLinear Programming
Solve the following linear programming problem graphically :
Maximise $z = 500 x + 300 y$ ,
subject to constraints
$x + 2 y \leq 12$
$2 x + y \leq 12$
$4 x + 5 y \geq 20$
$x \geq 0, y \geq 0$
Try it first, then show our solutionSolution
1. Graph: draw $x + 2 y = 12$ through $(12, 0), (0, 6)$ ; $2 x + y = 12$ through $(6, 0), (0, 12)$ ; $4 x + 5 y = 20$ through $(5, 0), (0, 4)$ . The feasible region (in the first quadrant, below both $x + 2 y = 12$ and $2 x + y = 12$ , above $4 x + 5 y = 20$ ) is the bounded pentagon with corners $A (0, 4)$ , $B (0, 6)$ , $C (4, 4)$ [intersection of $x + 2 y = 12$ and $2 x + y = 12$ ], $D (6, 0)$ , $E (5, 0)$ .
  1½ marks
2. Values of $z = 500 x + 300 y$ at corners:
  Corner point $z$
  A(0, 4) 1200
  B(0, 6) 1800
  C(4, 4) 3200
  D(6, 0) 3000
  E(5, 0) 2500
  1 mark
3. Maximum $z = 3200$ at $x = 4, y = 4$ .
  ½ mark
Answer: Max z = 3200 at x = 4, y = 4
Where the marks are
- Correct graph with shaded feasible region and corner points (1.5 marks)
- Correct corner-point table of z values (1 mark)
- States maximum 3200 at (4, 4) (0.5 mark)
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Q31 · 3 marks · Short answerProbability
E and F are two independent events such that $P (\overline{E}) = 0.6$ and $P (E \cup F) = 0.6$ . Find $P (F)$ and $P (\overline{E} \cup \overline{F})$ .
Try it first, then show our solutionSolution
1. $P (\overline{E}) = 0.6 \Rightarrow P (E) = 1 - 0.6 = 0.4$
  ½ mark
2. $P (E \cup F) = P (E) + P (F) - P (E \cap F)$ , and $P (E \cap F) = P (E) P (F)$ [E, F independent]
  ½ mark
3. $0.6 = 0.4 + P (F) - 0.4 P (F) \Rightarrow 0.6 P (F) = 0.2 \Rightarrow P (F) = \frac{1}{3}$
  1 mark
4. $P (\overline{E} \cup \overline{F}) = P (\overline{E \cap F}) = 1 - P (E \cap F)$ [De Morgan's law]
  ½ mark
5. $= 1 - 0.4 \times \frac{1}{3} = 1 - \frac{2}{15} = \frac{13}{15}$
  ½ mark
Answer: P(F) = 1/3; $P (\overline{E} \cup \overline{F}) = 13/15$
Where the marks are
- Finds P(E) = 0.4 (0.5 mark)
- Uses addition law with independence (0.5 mark)
- Solves P(F) = 1/3 (1 mark)
- Uses De Morgan: $1 - P (E \cap F)$ (0.5 mark)
- Gets 13/15 (0.5 mark)
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Corner point	$z$
A(0, 4)	1200
B(0, 6)	1800
C(4, 4)	3200
D(6, 0)	3000
E(5, 0)	2500

Section D

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerMatrices
If $A = 120 - 2 - 1 - 2 0 - 1 1$ , find $A^{- 1}$ and use it to solve the following system of equations :
$x - 2 y = 10$ , $2 x - y - z = 8$ , $- 2 y + z = 7$
Try it first, then show our solutionSolution
1. $∣ A ∣ = 1 [(- 1) (1) - (- 1) (- 2)] - (- 2) [2 \cdot 1 - (- 1) \cdot 0] + 0 = - 3 + 4 = 1 \neq = 0$ , so $A^{- 1}$ exists.
  1 mark
2. Cofactors: $C_{11} = - 3, C_{12} = - 2, C_{13} = - 4; C_{21} = 2, C_{22} = 1, C_{23} = 2; C_{31} = 2, C_{32} = 1, C_{33} = 3$ . So $adj A = - 3 - 2 - 4 212213$
  2 marks
3. $A^{- 1} = \frac{1}{∣ A ∣} adj A = - 3 - 2 - 4 212213$
  ½ mark
4. The system is $A X = B$ with $X = x y z$ , $B = 1087$ . So $X = A^{- 1} B = - 30 + 16 + 14 - 20 + 8 + 7 - 40 + 16 + 21 = 0 - 5 - 3$ , i.e. $x = 0, y = - 5, z = - 3$ .
  1½ marks
Answer: $A^{- 1} = - 3 - 2 - 4 212213$ ; x = 0, y = −5, z = −3
Where the marks are
- |A| = 1 ≠ 0 (1 mark)
- Correct adj A (2 marks)
- Writes A⁻¹ (0.5 mark)
- Writes AX = B and solves X = A⁻¹B to get x = 0, y = −5, z = −3 (1.5 marks)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerMatrices
If $A = - 1 13 a 21 2 x 1$ and $A^{- 1} = 1 - 8 b - 1 7 y 1 - 5 3$ , find the value of $(a + x) - (b + y)$ .
Try it first, then show our solutionSolution
1. Since $A^{- 1}$ is the inverse, $A A^{- 1} = I$ : $- 1 13 a 21 2 x 1 1 - 8 b - 1 7 y 1 - 5 3 = 100010001$
  1 mark
2. Multiplying: $- 1 - 8 a + 2 b - 15 + b x b - 5 1 + 7 a + 2 y 13 + x y 4 + y 5 - 5 a 3 x - 9 1 = 100010001$
  1½ marks
3. Equating: $b - 5 = 0 \Rightarrow b = 5$ ; $5 - 5 a = 0 \Rightarrow a = 1$
  1 mark
4. $4 + y = 0 \Rightarrow y = - 4$ ; $3 x - 9 = 0 \Rightarrow x = 3$ (these also satisfy the other entries, e.g. $- 1 - 8 + 10 = 1$ , $13 - 12 = 1$ ).
  1 mark
5. $(a + x) - (b + y) = (1 + 3) - (5 - 4) = 4 - 1 = 3$
  ½ mark
Answer: 3
Where the marks are
- Uses $A A^{- 1} = I$ (1 mark)
- Correct product matrix (1.5 marks)
- Finds b = 5, a = 1 (1 mark)
- Finds y = −4, x = 3 (1 mark)
- Final value 3 (0.5 mark)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerIntegrals
Evaluate : $\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x$
Try it first, then show our solutionSolution
1. Put $sin x - cos x = t$ , so $(cos x + sin x) d x = d t$ . Limits: $x = 0 \Rightarrow t = - 1$ ; $x = \frac{π}{4} \Rightarrow t = 0$ .
  1 mark
2. Squaring: $sin^{2} x + cos^{2} x - 2 sin x cos x = t^{2} \Rightarrow sin 2 x = 1 - t^{2}$ , so $9 + 16 sin 2 x = 25 - 16 t^{2}$ .
  ½ mark
3. $I = \int_{- 1}^{0} \frac{d t}{25 - 16 t ^{2}} = \frac{1}{16} \int_{- 1}^{0} \frac{d t}{( \frac{5}{4} ) ^{2} - t ^{2}}$
  1 mark
4. Using $\int \frac{d t}{a ^{2} - t ^{2}} = \frac{1}{2 a} lo g \frac{a + t}{a - t}$ : $I = \frac{1}{16} \cdot \frac{1}{2 \cdot \frac{5}{4}} [lo g \frac{\frac{5}{4} + t}{\frac{5}{4} - t}]_{- 1}^{0} = \frac{1}{40} [lo g \frac{5 + 4 t}{5 - 4 t}]_{- 1}^{0}$
  1½ marks
5. $= \frac{1}{40} [lo g 1 - lo g \frac{1}{9}] = \frac{1}{40} lo g 9 = \frac{1}{20} lo g 3$
  1 mark
Answer: $\frac{1}{40} lo g 9 = \frac{1}{20} lo g 3$
Where the marks are
- Substitution $sin x - cos x = t$ with $d t$ and new limits (1 mark)
- Expresses $sin 2 x = 1 - t^{2}$ (0.5 mark)
- Reduces to $\frac{1}{16} \int_{- 1}^{0} \frac{d t}{( 5/4 ) ^{2} - t ^{2}}$ (1 mark)
- Correct standard integration (1.5 marks)
- Evaluates to $\frac{1}{20} lo g 3$ (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerIntegrals
Evaluate : $\int_{0}^{\frac{π}{2}} sin 2 x tan^{- 1} (sin x) d x$
Try it first, then show our solutionSolution
1. Let $I = \int_{0}^{π /2} sin 2 x tan^{- 1} (sin x) d x = \int_{0}^{π /2} 2 sin x cos x tan^{- 1} (sin x) d x$ .
  Put $sin x = t \Rightarrow cos x d x = d t$ . When $x = 0$ , $t = 0$ ; when $x = \frac{π}{2}$ , $t = 1$ .
  1 mark
2. $I = 2 \int_{0}^{1} t tan^{- 1} t d t$
  1 mark
3. Integrate by parts, taking $tan^{- 1} t$ as the first function:
  $I = 2 [\frac{t ^{2}}{2} tan^{- 1} t - \frac{1}{2} \int \frac{t ^{2}}{1 + t ^{2}} d t]_{0}^{1}$
  1½ marks
4. Since $\frac{t ^{2}}{1 + t ^{2}} = 1 - \frac{1}{1 + t ^{2}}$ :
  $I = 2 [\frac{t ^{2}}{2} tan^{- 1} t - \frac{t}{2} + \frac{1}{2} tan^{- 1} t]_{0}^{1}$
  1 mark
5. $I = 2 (\frac{π}{8} - \frac{1}{2} + \frac{π}{8}) = 2 (\frac{π}{4} - \frac{1}{2}) = \frac{π}{2} - 1$
  ½ mark
Answer: $\frac{π}{2} - 1$
Where the marks are
- Writes $sin 2 x = 2 sin x cos x$ , substitutes $sin x = t$ and changes the limits to 0 and 1 (1 mark)
- Reduces the integral to $2 \int_{0}^{1} t tan^{- 1} t d t$ (1 mark)
- Applies integration by parts correctly (1½ marks)
- Integrates $\frac{t ^{2}}{1 + t ^{2}}$ as $t - tan^{- 1} t$ (1 mark)
- Substitutes the limits to get $\frac{π}{2} - 1$ (½ mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerApplication of Integrals
Using integration, find the area of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1$ , included between the lines $x = - 2$ and $x = 2$ .
Try it first, then show our solutionSolution
1. Figure: draw the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1$ centred at the origin, with vertices $(\pm 4, 0)$ and $(0, \pm 2)$ . Draw the vertical lines $x = - 2$ and $x = 2$ and shade the part of the ellipse between them.
  1 mark
2. From the equation, $y = \frac{1}{2} 16 - x^{2}$ for the upper half. The region is symmetric about both axes, so
  Area $= 4 \int_{0}^{2} y d x = 4 \cdot \frac{1}{2} \int_{0}^{2} 4^{2} - x^{2} d x$
  1 mark
3. Using $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ :
  Area $= 2 [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{0}^{2}$
  2 marks
4. $= 2 [12 + 8 \cdot \frac{π}{6}] = 2 [23 + \frac{4 π}{3}] = 43 + \frac{8 π}{3}$ sq units
  1 mark
Answer: $43 + \frac{8 π}{3}$ square units
Where the marks are
- Correct figure of the ellipse with the region between $x = - 2$ and $x = 2$ shaded (1 mark)
- Sets up the area as $4 \int_{0}^{2} \frac{1}{2} 16 - x^{2} d x$ using symmetry (1 mark)
- Integrates correctly using the standard formula for $a^{2} - x^{2}$ (2 marks)
- Substitutes the limits to get $43 + \frac{8 π}{3}$ (1 mark)
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Q35 · 5 marks · Long answerThree Dimensional Geometry
The image of point $P (x, y, z)$ with respect to line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ is $P^{'} (1, 0, 7)$ . Find the coordinates of point P.
Try it first, then show our solutionSolution
1. Any point on the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = λ$ is $(λ, 2 λ + 1, 3 λ + 2)$ . Let $M (λ, 2 λ + 1, 3 λ + 2)$ be the foot of the perpendicular from $P^{'}$ (and hence from $P$ ) to the line.
  1 mark
2. Direction ratios of $P^{'} M$ are $(λ - 1, 2 λ + 1, 3 λ - 5)$ .
  1 mark
3. $P^{'} M ⊥$ line, whose direction ratios are $1, 2, 3$ :
  $1 (λ - 1) + 2 (2 λ + 1) + 3 (3 λ - 5) = 0 \Rightarrow 14 λ - 14 = 0 \Rightarrow λ = 1$
  1 mark
4. So the foot of the perpendicular is $M (1, 3, 5)$ .
  1 mark
5. $M$ is the mid-point of $P P^{'}$ [the line is the perpendicular bisector of $P P^{'}$ ]:
  $\frac{x + 1}{2} = 1, \frac{y + 0}{2} = 3, \frac{z + 7}{2} = 5 \Rightarrow x = 1, y = 6, z = 3$
  Hence $P (1, 6, 3)$ .
  1 mark
Answer: $P (1, 6, 3)$
Where the marks are
- Takes a general point $M (λ, 2 λ + 1, 3 λ + 2)$ on the line as the foot of the perpendicular (1 mark)
- Writes the direction ratios of $P^{'} M$ (1 mark)
- Uses perpendicularity to get $λ = 1$ (1 mark)
- Finds the foot $M (1, 3, 5)$ (1 mark)
- Uses the mid-point formula to get $P (1, 6, 3)$ (1 mark)
Open Q35 on its own page

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
The traffic police has installed Over Speed Violation Detection (OSVD) system at various locations in a city. These cameras can capture a speeding vehicle from a distance of 300 m and even function in the dark.
A camera is installed on a pole at the height of 5 m. It detects a car travelling away from the pole at the speed of 20 m/s. At any point, x m away from the base of the pole, the angle of elevation of the speed camera from the car C is $θ$ .
On the basis of the above information, answer the following questions :
1. (i)
  Express $θ$ in terms of height of the camera installed on the pole and x.
  1 mark
2. (ii)
  Find $\frac{d θ}{d x}$ .
  1 mark
3. (iii)
  (a) Find the rate of change of angle of elevation with respect to time at an instant when the car is 50 m away from the pole.
  2 marks
4. OR
  (iii)
  (b) If the rate of change of angle of elevation with respect to time of another car at a distance of 50 m from the base of the pole is $\frac{3}{101}$ rad/s, then find the speed of the car.
  2 marks
Figure in the paper: Illustration of speed cameras on poles: average speed detection with Point A and Point B cameras (Speed = Distance / (Time B - Time A)), and RADAR speed detection where radar measures change in frequency of returned radio waves (Doppler effect), with cars passing below. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In the right triangle formed by the pole (height 5 m) and the car at distance $x$ m from its base, $tan θ = \frac{5}{x}$ , so $θ = tan^{- 1} (\frac{5}{x})$ .
  1 mark
2. (ii) $\frac{d θ}{d x} = \frac{1}{1 + \frac{25}{x ^{2}}} \cdot (- \frac{5}{x ^{2}}) = \frac{- 5}{25 + x ^{2}}$
  1 mark
3. (iii)(a) The car moves away, so $\frac{d x}{d t} = 20$ m/s.
  $\frac{d θ}{d t} = \frac{d θ}{d x} \cdot \frac{d x}{d t} = \frac{- 5}{25 + x ^{2}} \times 20$ . At $x = 50$ : $\frac{d θ}{d t} = \frac{- 100}{25 + 2500}$
  1½ marks
4. (iii)(a) $= \frac{- 100}{2525} = - \frac{4}{101}$ rad/s. The angle of elevation decreases at $\frac{4}{101}$ rad/s.
  ½ mark
5. (iii)(b) $\frac{d θ}{d t} = \frac{d θ}{d x} \cdot \frac{d x}{d t}$ . At $x = 50$ : $\frac{3}{101} = \frac{- 5}{2525} \cdot \frac{d x}{d t}$
  1½ marks
6. (iii)(b) $\frac{d x}{d t} = \frac{3}{101} \times \frac{2525}{- 5} = - 15$ m/s. The negative sign means the car is moving towards the pole. Speed of the car = 15 m/s.
  ½ mark
Answer: (i) $θ = tan^{- 1} (5/ x)$ ; (ii) $\frac{d θ}{d x} = \frac{- 5}{25 + x ^{2}}$ ; (iii)(a) $- \frac{4}{101}$ rad/s; (iii)(b) 15 m/s
Where the marks are
- (i) Uses $tan θ = 5/ x$ to write $θ = tan^{- 1} (5/ x)$ (1 mark)
- (ii) Differentiates correctly to get $\frac{- 5}{25 + x ^{2}}$ (1 mark)
- (iii)(a) Applies the chain rule with $d x / d t = 20$ at $x = 50$ (1½ marks)
- (iii)(a) Gets $- \frac{4}{101}$ rad/s (½ mark)
- (iii)(b) Sets up $\frac{3}{101} = \frac{- 5}{2525} \cdot \frac{d x}{d t}$ (1½ marks)
- (iii)(b) Gets $d x / d t = - 15$ , so speed = 15 m/s (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedProbability
According to recent research, air turbulence has increased in various regions around the world due to climate change. Turbulence makes flights bumpy and often delays the flights.
Assume that, an airplane observes severe turbulence, moderate turbulence or light turbulence with equal probabilities. Further, the chance of an airplane reaching late to the destination are 55%, 37% and 17% due to severe, moderate and light turbulence respectively.
On the basis of the above information, answer the following questions :
1. (i)
  Find the probability that an airplane reached its destination late.
  2 marks
2. (ii)
  If the airplane reached its destination late, find the probability that it was due to moderate turbulence.
  2 marks
Figure in the paper: Picture titled 'Turbulence intensity' showing three airplanes: Light (±1 meter), Moderate (±5 meters), Severe (±30 meters) vertical displacement with arrows. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let $E_{1}, E_{2}, E_{3}$ be the events of severe, moderate and light turbulence, and $A$ the event that the airplane reaches late.
  $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$ ; $P (A ∣ E_{1}) = \frac{55}{100}, P (A ∣ E_{2}) = \frac{37}{100}, P (A ∣ E_{3}) = \frac{17}{100}$
  ½ mark
2. (i) By the theorem of total probability:
  $P (A) = \frac{1}{3} \cdot \frac{55}{100} + \frac{1}{3} \cdot \frac{37}{100} + \frac{1}{3} \cdot \frac{17}{100}$
  1 mark
3. (i) $= \frac{1}{3} \cdot \frac{109}{100} = \frac{109}{300}$
  ½ mark
4. (ii) By Bayes' theorem:
  $P (E_{2} ∣ A) = \frac{P ( E _{2} ) P ( A ∣ E _{2} )}{P ( A )} = \frac{\frac{1}{3} \cdot \frac{37}{100}}{\frac{109}{300}}$
  1½ marks
5. (ii) $= \frac{37}{109}$
  ½ mark
Answer: (i) $\frac{109}{300}$ ; (ii) $\frac{37}{109}$
Where the marks are
- (i) Defines the events and writes their probabilities (½ mark)
- (i) Applies total probability correctly (1 mark)
- (i) Gets $P (A) = \frac{109}{300}$ (½ mark)
- (ii) Applies Bayes' theorem correctly (1½ marks)
- (ii) Gets $\frac{37}{109}$ (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedInverse Trigonometric Functions
If a function $f : X \to Y$ defined as $f (x) = y$ is one-one and onto, then we can define a unique function $g : Y \to X$ such that $g (y) = x$ , where $x \in X$ and $y = f (x), y \in Y$ . Function g is called the inverse of function f.
The domain of sine function is R and function sine : $R \to R$ is neither one-one nor onto. The following graph shows the sine function.
Let sine function be defined from set A to $[- 1, 1]$ such that inverse of sine function exists, i.e., $sin^{- 1} x$ is defined from $[- 1, 1]$ to A.
On the basis of the above information, answer the following questions :
1. (i)
  If A is the interval other than principal value branch, give an example of one such interval.
  1 mark
2. (ii)
  If $sin^{- 1} (x)$ is defined from $[- 1, 1]$ to its principal value branch, find the value of $sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1)$ .
  1 mark
3. (iii)
  (a) Draw the graph of $sin^{- 1} x$ from $[- 1, 1]$ to its principal value branch.
  2 marks
4. OR
  (iii)
  (b) Find the domain and range of $f (x) = 2 sin^{- 1} (1 - x)$ .
  2 marks
Figure in the paper: Graph of y = sin x on X'X and Y'Y axes, a wave passing through the origin, with marked x-values at -5π/2, -2π, -π, -π/2, π/2, π, 3π/2, 2π, 5π/2 and y-values 1 and -1. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) On $[\frac{π}{2}, \frac{3 π}{2}]$ , sine is one-one and takes every value in $[- 1, 1]$ (it falls from 1 to −1). So $A = [\frac{π}{2}, \frac{3 π}{2}]$ works. Any such interval, e.g. $[- \frac{3 π}{2}, - \frac{π}{2}]$ , is also correct.
  1 mark
2. (ii) With principal values in $[- \frac{π}{2}, \frac{π}{2}]$ :
  $sin^{- 1} (- \frac{1}{2}) - sin^{- 1} (1) = - \frac{π}{6} - \frac{π}{2} = - \frac{2 π}{3}$
  1 mark
3. (iii)(a) **Graph of $y = sin^{- 1} x$ , domain $[- 1, 1]$ , range $[- \frac{π}{2}, \frac{π}{2}]$ :**
  Plot the points $(- 1, - \frac{π}{2})$ , $(- \frac{1}{2}, - \frac{π}{6})$ , $(0, 0)$ , $(\frac{1}{2}, \frac{π}{6})$ , $(1, \frac{π}{2})$ .
  Join them with a smooth increasing curve through the origin. It is symmetric about the origin and becomes vertical at the end-points $x = \pm 1$ .
  This curve is the mirror image of $y = sin x$ , $x \in [- \frac{π}{2}, \frac{π}{2}]$ , in the line $y = x$ .
  Mark $\pm \frac{π}{2}$ on the y-axis and $\pm 1$ on the x-axis.
  2 marks
4. (iii)(b) Domain: $sin^{- 1} u$ is defined only for $- 1 \leq u \leq 1$ , so $- 1 \leq 1 - x \leq 1 \Rightarrow - 2 \leq - x \leq 0 \Rightarrow 0 \leq x \leq 2$ .
  ½ mark
5. (iii)(b) Domain $= [0, 2]$ .
  ½ mark
6. (iii)(b) Range: $- \frac{π}{2} \leq sin^{- 1} (1 - x) \leq \frac{π}{2}$ ; multiplying by 2, $- π \leq 2 sin^{- 1} (1 - x) \leq π$ .
  ½ mark
7. (iii)(b) Range $= [- π, π]$ .
  ½ mark
Answer: (i) $[\frac{π}{2}, \frac{3 π}{2}]$ (or any similar interval); (ii) $- \frac{2 π}{3}$ ; (iii)(a) increasing curve from $(- 1, - π /2)$ through $(0, 0)$ to $(1, π /2)$ ; (iii)(b) domain $[0, 2]$ , range $[- π, π]$
Where the marks are
- (i) Gives a valid interval other than the principal branch, e.g. $[π /2, 3 π /2]$ (1 mark)
- (ii) Uses the principal values $- π /6$ and $π /2$ to get $- 2 π /3$ (1 mark)
- (iii)(a) Correct graph of $sin^{- 1} x$ on $[- 1, 1]$ with range $[- π /2, π /2]$ marked (2 marks)
- (iii)(b) Writes the condition $- 1 \leq 1 - x \leq 1$ (½ mark)
- (iii)(b) Gets domain $[0, 2]$ (½ mark)
- (iii)(b) Uses the bounds $- π /2 \leq sin^{- 1} (1 - x) \leq π /2$ (½ mark)
- (iii)(b) Gets range $[- π, π]$ (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.