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

CBSE Class 12 Mathematics: 2024 board paper, set 65/5/1

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_5_1Mathematics.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	Very Short Answer
C	26-31	Short Answer
D	32-35	Long Answer
E	36-38	Case study based

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQRelations and Functions
A function $f : R \to R$ defined as $f (x) = x^{2} - 4 x + 5$ is :
1. (A)injective but not surjective.
2. (B)surjective but not injective.
3. (C)both injective and surjective.
4. (D)neither injective nor surjective.
Try it first, then show our solutionSolution
1. $f (x) = (x - 2)^{2} + 1$ . Since $f (1) = f (3) = 2$ with $1 \neq = 3$ , $f$ is not one-one.
  ½ mark
2. Also $f (x) \geq 1$ for all $x$ , so the range is $[1, \infty) \neq = R$ ; for example, $0$ has no pre-image. Hence $f$ is not onto, and the answer is option (D).
  ½ mark
Answer: (D) neither injective nor surjective
Where the marks are
- Shows $f$ is not one-one and its range is $[1, \infty)$ , choosing (D) (1 mark)
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Q2 · 1 mark · MCQMatrices
If $A = a b 1 c 0 - 5 - 1 50$ is a skew-symmetric matrix, then the value of $2 a - (b + c)$ is :
1. (A)0
2. (B)1
3. (C) $- 10$
4. (D)10
Try it first, then show our solutionSolution
1. In a skew-symmetric matrix every diagonal entry is 0, so $a = 0$ . Also $a_{ij} = - a_{j i}$ , which gives $b = - c$ and $c = - b$ ; using $a_{13} = - 1$ , $a_{31} = 1$ confirms the pattern.
  ½ mark
2. Then $b + c = 0$ , so $2 a - (b + c) = 0 - 0 = 0$ , option (A).
  ½ mark
Answer: 0, option (A)
Where the marks are
- Uses $a = 0$ and $b = - c$ to get 0 (1 mark)
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Q3 · 1 mark · MCQDeterminants
If A is a square matrix of order 3 such that the value of $∣ adj \cdot A ∣ = 8$ , then the value of $∣ A^{T} ∣$ is :
1. (A) $2$
2. (B) $- 2$
3. (C)8
4. (D) $22$
Try it first, then show our solutionSolution
1. For a matrix of order $n$ , $∣ adj A ∣ = ∣ A ∣^{n - 1}$ . Here $n = 3$ , so $∣ A ∣^{2} = 8$ .
  ½ mark
2. This gives $∣ A ∣ = \pm 22$ . Since $∣ A^{T} ∣ = ∣ A ∣$ , the value among the options is $22$ , option (D).
  ½ mark
Answer: $22$ , option (D)
Where the marks are
- Uses $∣ adj A ∣ = ∣ A ∣^{2}$ and $∣ A^{T} ∣ = ∣ A ∣$ (1 mark)
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Q4 · 1 mark · MCQMatrices
If inverse of matrix $7 - 1 - 1 - 3 10 - 3 01$ is the matrix $111 3 λ 3 334$ , then value of $λ$ is :
1. (A) $- 4$
2. (B)1
3. (C)3
4. (D)4
Try it first, then show our solutionSolution
1. Since $A A^{- 1} = I$ , take the (2,2) entry: row 2 of $A$ is $(- 1, 1, 0)$ and column 2 of $A^{- 1}$ is $(3, λ, 3)$ .
  ½ mark
2. This gives $- 3 + λ = 1$ , so $λ = 4$ , option (D).
  ½ mark
Answer: $λ = 4$ , option (D)
Where the marks are
- Equates the (2,2) entry of $A A^{- 1}$ to 1 (1 mark)
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Q5 · 1 mark · MCQMatrices
If $[x 20] 5 - 1 x = [31] [- 2 x]$ , then value of $x$ is :
1. (A) $- 1$
2. (B)0
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. Left side: $5 x - 2$ . Right side: $- 6 + x$ .
  ½ mark
2. Setting $5 x - 2 = x - 6$ gives $4 x = - 4$ , so $x = - 1$ , option (A).
  ½ mark
Answer: $x = - 1$ , option (A)
Where the marks are
- Multiplies both sides and solves for $x$ (1 mark)
Q6 · 1 mark · MCQMatrices
Find the matrix $A^{2}$ , where $A = [a_{ij}]$ is a $2 \times 2$ matrix whose elements are given by $a_{ij} = maximum (i, j) - minimum (i, j)$ :
1. (A) $[0000]$
2. (B) $[0110]$
3. (C) $[1001]$
4. (D) $[1111]$
Try it first, then show our solutionSolution
1. $a_{11} = a_{22} = 0$ and $a_{12} = a_{21} = 1$ , so $A = [0110]$ .
  ½ mark
2. $A^{2} = [1001]$ , option (C).
  ½ mark
Answer: $I$ , option (C)
Where the marks are
- Constructs $A$ and squares it (1 mark)
Q7 · 1 mark · MCQContinuity and Differentiability
If $x e^{y} = 1$ , then the value of $\frac{d y}{d x}$ at $x = 1$ is :
1. (A) $- 1$
2. (B)1
3. (C) $- e$
4. (D) $- \frac{1}{e}$
Try it first, then show our solutionSolution
1. Differentiate $x e^{y} = 1$ : $e^{y} + x e^{y} y^{'} = 0$ , so $y^{'} = - \frac{1}{x}$ .
  ½ mark
2. At $x = 1$ , $y^{'} = - 1$ , option (A).
  ½ mark
Answer: $- 1$ , option (A)
Where the marks are
- Differentiates implicitly and substitutes $x = 1$ (1 mark)
Q8 · 1 mark · MCQContinuity and Differentiability
Derivative of $e^{s i n^{2} x}$ with respect to $cos x$ is :
1. (A) $sin x e^{s i n^{2} x}$
2. (B) $cos x e^{s i n^{2} x}$
3. (C) $- 2 cos x e^{s i n^{2} x}$
4. (D) $- 2 sin^{2} x cos x e^{s i n^{2} x}$
Try it first, then show our solutionSolution
1. $\frac{d}{d x} e^{s i n^{2} x} = 2 sin x cos x e^{s i n^{2} x}$ and $\frac{d}{d x} cos x = - sin x$ .
  ½ mark
2. The ratio is $- 2 cos x e^{s i n^{2} x}$ , option (C).
  ½ mark
Answer: $- 2 cos x e^{s i n^{2} x}$ , option (C)
Where the marks are
- Divides the two derivatives with respect to $x$ (1 mark)
Q9 · 1 mark · MCQApplication of Derivatives
The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minima at $x$ equal to :
1. (A)2
2. (B)1
3. (C)0
4. (D) $- 2$
Try it first, then show our solutionSolution
1. $f^{'} (x) = \frac{1}{2} - \frac{2}{x ^{2}} = 0$ gives $x = \pm 2$ .
  ½ mark
2. $f^{''} (x) = \frac{4}{x ^{3}}$ , so $f^{''} (2) > 0$ and $x = 2$ is a local minimum, option (A).
  ½ mark
Answer: $x = 2$ , option (A)
Where the marks are
- Finds the critical points and applies the second derivative test (1 mark)
Q10 · 1 mark · MCQApplication of Derivatives
Given a curve $y = 7 x - x^{3}$ and $x$ increases at the rate of 2 units per second. The rate at which the slope of the curve is changing, when $x = 5$ is :
1. (A) $- 60$ units/sec
2. (B)60 units/sec
3. (C) $- 70$ units/sec
4. (D) $- 140$ units/sec
Try it first, then show our solutionSolution
1. The slope is $m = 7 - 3 x^{2}$ , so $\frac{d m}{d t} = - 6 x \frac{d x}{d t}$ .
  ½ mark
2. At $x = 5$ : $\frac{d m}{d t} = - 6 (5) (2) = - 60$ units/sec, option (A).
  ½ mark
Answer: $- 60$ units/sec, option (A)
Where the marks are
- Differentiates the slope with respect to $t$ (1 mark)
Q11 · 1 mark · MCQIntegrals
$\int \frac{1}{x ( l o g x ) ^{2}} d x$ is equal to :
1. (A) $2 lo g (lo g x) + c$
2. (B) $- \frac{1}{l o g x} + c$
3. (C) $\frac{( l o g x ) ^{3}}{3} + c$
4. (D) $\frac{3}{( l o g x ) ^{3}} + c$
Try it first, then show our solutionSolution
1. Put $t = lo g x$ , so $d t = \frac{d x}{x}$ and the integral becomes $\int t^{- 2} d t$ .
  ½ mark
2. $\int t^{- 2} d t = - \frac{1}{t} + c = - \frac{1}{l o g x} + c$ , option (B).
  ½ mark
Answer: $- \frac{1}{l o g x} + c$ , option (B)
Where the marks are
- Substitutes $t = lo g x$ and integrates (1 mark)
Q12 · 1 mark · MCQIntegrals
The value of $\int_{- 1}^{1} x ∣ x ∣ d x$ is :
1. (A) $\frac{1}{6}$
2. (B) $\frac{1}{3}$
3. (C) $- \frac{1}{6}$
4. (D)0
Try it first, then show our solutionSolution
1. $f (x) = x ∣ x ∣$ satisfies $f (- x) = - f (x)$ , so it is an odd function.
  ½ mark
2. The integral of an odd function over $[- a, a]$ is 0, option (D).
  ½ mark
Answer: 0, option (D)
Where the marks are
- Identifies the integrand as odd (1 mark)
Q13 · 1 mark · MCQApplication of Integrals
Area of the region bounded by curve $y^{2} = 4 x$ and the X-axis between $x = 0$ and $x = 1$ is :
1. (A) $\frac{2}{3}$
2. (B) $\frac{8}{3}$
3. (C)3
4. (D) $\frac{4}{3}$
Try it first, then show our solutionSolution
1. The region lies above the X-axis, so take $y = 2 x$ : Area $= \int_{0}^{1} 2 x d x$ .
  ½ mark
2. $\int_{0}^{1} 2 x d x = \frac{4}{3} [x^{3/2}]_{0}^{1} = \frac{4}{3}$ , option (D).
  ½ mark
Answer: $\frac{4}{3}$ , option (D)
Where the marks are
- Integrates $2 x$ from 0 to 1 (1 mark)
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Q14 · 1 mark · MCQDifferential Equations
The order of the differential equation $\frac{d ^{4} y}{d x ^{4}} - sin (\frac{d ^{2} y}{d x ^{2}}) = 5$ is :
1. (A)4
2. (B)3
3. (C)2
4. (D)not defined
Try it first, then show our solutionSolution
1. The order is the order of the highest derivative present in the equation, which is $\frac{d ^{4} y}{d x ^{4}}$ .
  ½ mark
2. So the order is 4, option (A). (Only the degree is not defined here.)
  ½ mark
Answer: 4, option (A)
Where the marks are
- Identifies the highest order derivative (1 mark)
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Q15 · 1 mark · MCQVector Algebra
The position vectors of points P and Q are $p$ and $q$ respectively. The point R divides line segment PQ in the ratio 3 : 1 and S is the mid-point of line segment PR. The position vector of S is :
1. (A) $\frac{p + 3 q}{4}$
2. (B) $\frac{p + 3 q}{8}$
3. (C) $\frac{5 p + 3 q}{4}$
4. (D) $\frac{5 p + 3 q}{8}$
Try it first, then show our solutionSolution
1. By the section formula, $r = \frac{3 q + p}{4}$ .
  ½ mark
2. $s = \frac{p + r}{2} = \frac{5 p + 3 q}{8}$ , option (D).
  ½ mark
Answer: $\frac{5 p + 3 q}{8}$ , option (D)
Where the marks are
- Applies the section formula and then the midpoint formula (1 mark)
Q16 · 1 mark · MCQThree Dimensional Geometry
The angle which the line $\frac{x}{1} = \frac{y}{- 1} = \frac{z}{0}$ makes with the positive direction of Y-axis is :
1. (A) $\frac{5 π}{6}$
2. (B) $\frac{3 π}{4}$
3. (C) $\frac{5 π}{4}$
4. (D) $\frac{7 π}{4}$
Try it first, then show our solutionSolution
1. The direction ratios are $1, - 1, 0$ , so the direction cosines are $\frac{1}{2}, - \frac{1}{2}, 0$ .
  ½ mark
2. $cos β = - \frac{1}{2}$ , so $β = \frac{3 π}{4}$ , option (B).
  ½ mark
Answer: $\frac{3 π}{4}$ , option (B)
Where the marks are
- Finds $cos β$ from the direction cosines (1 mark)
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Q17 · 1 mark · MCQThree Dimensional Geometry
The Cartesian equation of the line passing through the point $(1, - 3, 2)$ and parallel to the line $r = (2 + λ) \hat{i} + λ \hat{j} + (2 λ - 1) \hat{k}$ is
1. (A) $\frac{x - 1}{2} = \frac{y + 3}{0} = \frac{z - 2}{- 1}$
2. (B) $\frac{x + 1}{1} = \frac{y - 3}{1} = \frac{z + 2}{2}$
3. (C) $\frac{x + 1}{2} = \frac{y - 3}{0} = \frac{z + 2}{- 1}$
4. (D) $\frac{x - 1}{1} = \frac{y + 3}{1} = \frac{z - 2}{2}$
Try it first, then show our solutionSolution
1. The given line is $r = (2 \hat{i} - \hat{k}) + λ (\hat{i} + \hat{j} + 2 \hat{k})$ , so its direction ratios are $1, 1, 2$ .
  ½ mark
2. The required line through $(1, - 3, 2)$ is $\frac{x - 1}{1} = \frac{y + 3}{1} = \frac{z - 2}{2}$ , option (D).
  ½ mark
Answer: Option (D)
Where the marks are
- Finds the direction ratios and writes the equation (1 mark)
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Q18 · 1 mark · MCQProbability
If A and B are events such that $P (A / B) = P (B / A) \neq = 0$ , then :
1. (A) $A \subset B$ , but $A \neq = B$
2. (B) $A = B$
3. (C) $A \cap B = ϕ$
4. (D) $P (A) = P (B)$
Try it first, then show our solutionSolution
1. $\frac{P ( A \cap B )}{P ( B )} = \frac{P ( A \cap B )}{P ( A )}$ with $P (A \cap B) \neq = 0$ .
  ½ mark
2. Therefore $P (A) = P (B)$ , option (D).
  ½ mark
Answer: $P (A) = P (B)$ , option (D)
Where the marks are
- Uses the definition of conditional probability (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedInverse Trigonometric Functions
Assertion (A) : Domain of $y = cos^{- 1} (x)$ is $[- 1, 1]$ .
Reason (R) : The range of the principal value branch of $y = cos^{- 1} (x)$ is $[0, π] - {\frac{π}{2}}$ .
1. (A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true and 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. The domain of $cos^{- 1} x$ is $[- 1, 1]$ , so the Assertion is true.
  ½ mark
2. The principal value range is $[0, π]$ , and it does include $\frac{π}{2}$ , so the Reason is false. The answer is option (C).
  ½ mark
Answer: (C)
Where the marks are
- A is true; R is false since the range is $[0, π]$ (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedVector Algebra
Assertion (A) : The vectors $a = 6 \hat{i} + 2 \hat{j} - 8 \hat{k}$ , $b = 10 \hat{i} - 2 \hat{j} - 6 \hat{k}$ , $c = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$ represent the sides of a right angled triangle.
Reason (R) : Three non-zero vectors of which none of two are collinear forms a triangle if their resultant is zero vector or sum of any two vectors is equal to the third.
1. (A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true and 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. $b = a + c$ , so the three vectors form a triangle. Also $a \cdot c = 24 - 8 - 16 = 0$ , so there is a right angle; the Assertion is true.
  ½ mark
2. The Reason is a true statement, but the right angle comes from the dot product being zero, not from the Reason. The answer is option (B).
  ½ mark
Answer: (B)
Where the marks are
- Verifies the triangle and the right angle, and judges the explanation (1 mark)
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Section B

Very Short Answer

Q21 · 2 marks · Very short answerInverse Trigonometric Functions
Find value of k if $sin^{- 1} [k tan (2 cos^{- 1} \frac{3}{2})] = \frac{π}{3}$ .
Try it first, then show our solutionSolution
1. Take sine of both sides: $k tan (2 cos^{- 1} \frac{3}{2}) = sin \frac{π}{3}$ .
  ½ mark
2. Since $cos^{- 1} \frac{3}{2} = \frac{π}{6}$ [principal value in $[0, π]$ ], we get $2 cos^{- 1} \frac{3}{2} = \frac{π}{3}$ . So $k tan \frac{π}{3} = \frac{3}{2}$ .
  1 mark
3. $k 3 = \frac{3}{2} \Rightarrow k = \frac{1}{2}$ .
  ½ mark
Answer: $k = \frac{1}{2}$
Where the marks are
- Removes $sin^{- 1}$ to get $k tan (\dots) = sin \frac{π}{3}$ (½ mark)
- Uses $cos^{- 1} \frac{3}{2} = \frac{π}{6}$ and writes $sin \frac{π}{3} = \frac{3}{2}$ (1 mark)
- Obtains $k = \frac{1}{2}$ (½ mark)
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Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerContinuity and Differentiability
Verify whether the function f defined by $f (x) = {x sin (\frac{1}{x}), 0, x \neq = 0 x = 0$ is continuous at $x = 0$ or not.
Try it first, then show our solutionSolution
1. For $x \neq = 0$ , $∣ sin (1/ x) ∣ \leq 1$ , so $0 \leq ∣ x sin (1/ x) ∣ \leq ∣ x ∣$ . As $x \to 0$ , $∣ x ∣ \to 0$ , hence $lim_{x \to 0} x sin \frac{1}{x} = 0$ (zero times a bounded quantity). Also $f (0) = 0$ .
  1½ marks
2. Since $lim_{x \to 0} f (x) = 0 = f (0)$ , f is continuous at $x = 0$ .
  ½ mark
Answer: f is continuous at $x = 0$ .
Where the marks are
- Shows $lim_{x \to 0} x sin \frac{1}{x} = 0$ using boundedness of sine and compares with $f (0) = 0$ (1½ marks)
- Concludes f is continuous at $x = 0$ (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerContinuity and Differentiability
Check for differentiability of the function f defined by $f (x) = ∣ x - 5∣$ , at the point $x = 5$ .
Try it first, then show our solutionSolution
1. $f (5) = 0$ . LHD $= lim_{x \to 5^{-}} \frac{∣ x - 5∣ - 0}{x - 5} = lim_{x \to 5^{-}} \frac{- ( x - 5 )}{x - 5} = - 1$ .
  1 mark
2. RHD $= lim_{x \to 5^{+}} \frac{∣ x - 5∣ - 0}{x - 5} = lim_{x \to 5^{+}} \frac{x - 5}{x - 5} = 1$ .
  ½ mark
3. Since LHD $\neq =$ RHD, f is not differentiable at $x = 5$ .
  ½ mark
Answer: LHD = −1, RHD = 1; f is not differentiable at x = 5.
Where the marks are
- Computes LHD $= - 1$ (1 mark)
- Computes RHD $= 1$ (½ mark)
- Concludes not differentiable since LHD ≠ RHD (½ mark)
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Q23 · 2 marks · Very short answerApplication of Derivatives
The area of the circle is increasing at a uniform rate of $2 cm^{2} / sec$ . How fast is the circumference of the circle increasing when the radius $r = 5$ cm ?
Try it first, then show our solutionSolution
1. Area $A = π r^{2} \Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t} = 2 \Rightarrow \frac{d r}{d t} = \frac{1}{π r}$ cm/sec.
  1 mark
2. Circumference $C = 2 π r \Rightarrow \frac{d C}{d t} = 2 π \frac{d r}{d t} = 2 π \cdot \frac{1}{π r} = \frac{2}{r}$ . At $r = 5$ cm, $\frac{d C}{d t} = \frac{2}{5}$ cm/sec.
  1 mark
Answer: $\frac{2}{5}$ cm/sec (0.4 cm/sec)
Where the marks are
- Differentiates $A = π r^{2}$ and gets $\frac{d r}{d t} = \frac{1}{π r}$ (1 mark)
- Gets $\frac{d C}{d t} = \frac{2}{r} = \frac{2}{5}$ cm/sec at $r = 5$ (1 mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerIntegrals
Find : $\int cos^{3} x e^{l o g s i n x} d x$
Try it first, then show our solutionSolution
1. Since $e^{l o g s i n x} = sin x$ , $I = \int cos^{3} x sin x d x$ . Put $cos x = t \Rightarrow - sin x d x = d t$ .
  1 mark
2. $I = - \int t^{3} d t = - \frac{t ^{4}}{4} + C$
  ½ mark
3. $= - \frac{c o s ^{4} x}{4} + C$
  ½ mark
Answer: $- \frac{c o s ^{4} x}{4} + C$
Where the marks are
- Simplifies $e^{l o g s i n x} = sin x$ and substitutes $cos x = t$ (1 mark)
- Integrates $- \int t^{3} d t$ (½ mark)
- Writes answer in terms of x (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerIntegrals
Find : $\int \frac{1}{5 + 4 x - x ^{2}} d x$
Try it first, then show our solutionSolution
1. Complete the square: $5 + 4 x - x^{2} = 9 - (x^{2} - 4 x + 4) = 3^{2} - (x - 2)^{2}$ , so $I = \int \frac{d x}{3 ^{2} - ( x - 2 ) ^{2}}$ .
  1 mark
2. Using $\int \frac{d u}{a ^{2} - u ^{2}} = \frac{1}{2 a} lo g \frac{a + u}{a - u} + C$ : $I = \frac{1}{6} lo g \frac{3 + x - 2}{3 - x + 2} + C = \frac{1}{6} lo g \frac{1 + x}{5 - x} + C$ .
  1 mark
Answer: $\frac{1}{6} lo g \frac{1 + x}{5 - x} + C$
Where the marks are
- Writes denominator as $3^{2} - (x - 2)^{2}$ (1 mark)
- Applies standard formula to get $\frac{1}{6} lo g \frac{1 + x}{5 - x} + C$ (1 mark)
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Q25 · 2 marks · Very short answerThree Dimensional Geometry
Find the vector equation of the line passing through the point $(2, 3, - 5)$ and making equal angles with the co-ordinate axes.
Try it first, then show our solutionSolution
1. Let the line make angle α with each axis, so $l = m = n = cos α$ . Using $l^{2} + m^{2} + n^{2} = 1$ : $3 l^{2} = 1 \Rightarrow l = \frac{1}{3}$ (taking the positive value; the negative value gives the same line).
  1 mark
2. Direction cosines are $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$ , so direction ratios are $1, 1, 1$ .
  ½ mark
3. Vector equation: $r = (2 \hat{i} + 3 \hat{j} - 5 \hat{k}) + λ (\hat{i} + \hat{j} + \hat{k})$ .
  ½ mark
Answer: $r = 2 \hat{i} + 3 \hat{j} - 5 \hat{k} + λ (\hat{i} + \hat{j} + \hat{k})$
Where the marks are
- Uses $l = m = n$ with $l^{2} + m^{2} + n^{2} = 1$ to get $l = \frac{1}{3}$ (1 mark)
- Direction ratios 1, 1, 1 (½ mark)
- Correct vector equation (½ mark)
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Section C

Short Answer

Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerContinuity and Differentiability
Find $\frac{d y}{d x}$ , if $(cos x)^{y} = (cos y)^{x}$ .
Try it first, then show our solutionSolution
1. Taking log on both sides: $y lo g cos x = x lo g cos y$ .
  1 mark
2. Differentiating w.r.t. x: $\frac{d y}{d x} lo g cos x + y \cdot \frac{- s i n x}{c o s x} = lo g cos y + x \cdot \frac{- s i n y}{c o s y} \frac{d y}{d x}$ , i.e. $\frac{d y}{d x} lo g cos x - y tan x = lo g cos y - x tan y \frac{d y}{d x}$ .
  1½ marks
3. $\frac{d y}{d x} (lo g cos x + x tan y) = lo g cos y + y tan x \Rightarrow \frac{d y}{d x} = \frac{l o g c o s y + y t a n x}{l o g c o s x + x t a n y}$ .
  ½ mark
Answer: $\frac{d y}{d x} = \frac{l o g c o s y + y t a n x}{l o g c o s x + x t a n y}$
Where the marks are
- Takes logarithm: $y lo g cos x = x lo g cos y$ (1 mark)
- Correct differentiation using product and chain rules (1½ marks)
- Collects terms and obtains $\frac{d y}{d x}$ (½ mark)
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OR
Q26(b) · 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 $x = sin A$ , $y = sin B$ , so $A = sin^{- 1} x$ , $B = sin^{- 1} y$ . The equation becomes $cos A + cos B = a (sin A - sin B)$ .
  ½ mark
2. $2 cos \frac{A + B}{2} cos \frac{A - B}{2} = 2 a cos \frac{A + B}{2} sin \frac{A - B}{2}$ .
  1 mark
3. $cot \frac{A - B}{2} = a \Rightarrow A - B = 2 cot^{- 1} a$ , i.e. $sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$ .
  ½ mark
4. Differentiating w.r.t. x: $\frac{1}{1 - x ^{2}} - \frac{1}{1 - y ^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}} = \frac{1 - y ^{2}}{1 - x ^{2}}$ . Hence proved.
  1 mark
Answer: $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}}$ (proved)
Where the marks are
- Substitution $x = sin A$ , $y = sin B$ (½ mark)
- Applies sum-to-product formulas (1 mark)
- Obtains $sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$ (½ mark)
- Differentiates and gets the result (1 mark)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerContinuity and Differentiability
If $x = a sin^{3} θ$ , $y = b cos^{3} θ$ , then find $\frac{d ^{2} y}{d x ^{2}}$ at $θ = \frac{π}{4}$ .
Try it first, then show our solutionSolution
1. $\frac{d x}{d θ} = 3 a sin^{2} θ cos θ$ , $\frac{d y}{d θ} = - 3 b cos^{2} θ sin θ$ .
  1 mark
2. $\frac{d y}{d x} = \frac{- 3 b c o s ^{2} θ s i n θ}{3 a s i n ^{2} θ c o s θ} = - \frac{b}{a} cot θ$ .
  ½ mark
3. $\frac{d ^{2} y}{d x ^{2}} = \frac{b}{a} csc^{2} θ \cdot \frac{d θ}{d x} = \frac{b}{a} csc^{2} θ \cdot \frac{1}{3 a s i n ^{2} θ c o s θ} = \frac{b}{3 a ^{2}} sec θ csc^{4} θ$ .
  1 mark
4. At $θ = \frac{π}{4}$ : $sec \frac{π}{4} = 2$ , $csc^{4} \frac{π}{4} = 4$ , so $\frac{d ^{2} y}{d x ^{2}} = \frac{4 2 b}{3 a ^{2}}$ .
  ½ mark
Answer: $\frac{4 2 b}{3 a ^{2}}$
Where the marks are
- Both derivatives w.r.t. θ (1 mark)
- $\frac{d y}{d x} = - \frac{b}{a} cot θ$ (½ mark)
- Second derivative $\frac{b}{3 a ^{2}} sec θ csc^{4} θ$ (1 mark)
- Value $\frac{4 2 b}{3 a ^{2}}$ at $θ = π /4$ (½ mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntegrals
Evaluate : $\int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$
Try it first, then show our solutionSolution
1. Let $I = \int_{0}^{π} \frac{e ^{c o s x}}{e ^{c o s x} + e ^{- c o s x}} d x$ …(i). Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ and $cos (π - x) = - cos x$ : $I = \int_{0}^{π} \frac{e ^{- c o s x}}{e ^{- c o s x} + e ^{c o s x}} d x$ …(ii).
  1½ marks
2. Adding (i) and (ii): $2 I = \int_{0}^{π} 1 d x = [x]_{0}^{π} = π \Rightarrow I = \frac{π}{2}$ .
  1½ marks
Answer: $\frac{π}{2}$
Where the marks are
- Applies property $x \to π - x$ to get (ii) (1½ marks)
- Adds to get $2 I = π$ , $I = \frac{π}{2}$ (1½ marks)
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OR
Q28(b) · 3 marks · Short answerIntegrals
Find : $\int \frac{2 x + 1}{( x + 1 ) ^{2} ( x - 1 )} d x$
Try it first, then show our solutionSolution
1. Let $\frac{2 x + 1}{( x + 1 ) ^{2} ( x - 1 )} = \frac{A}{x + 1} + \frac{B}{( x + 1 ) ^{2}} + \frac{C}{x - 1}$ , so $2 x + 1 = A (x + 1) (x - 1) + B (x - 1) + C (x + 1)^{2}$ . Put $x = 1$ : $3 = 4 C \Rightarrow C = \frac{3}{4}$ . Put $x = - 1$ : $- 1 = - 2 B \Rightarrow B = \frac{1}{2}$ . Compare $x^{2}$ : $A + C = 0 \Rightarrow A = - \frac{3}{4}$ . So integrand $= - \frac{3}{4} \cdot \frac{1}{x + 1} + \frac{1}{2} \cdot \frac{1}{( x + 1 ) ^{2}} + \frac{3}{4} \cdot \frac{1}{x - 1}$ .
  1½ marks
2. Integrating: $- \frac{3}{4} lo g ∣ x + 1∣ - \frac{1}{2 ( x + 1 )} + \frac{3}{4} lo g ∣ x - 1∣ + C = \frac{3}{4} lo g \frac{x - 1}{x + 1} - \frac{1}{2 ( x + 1 )} + C$ .
  1½ marks
Answer: $\frac{3}{4} lo g \frac{x - 1}{x + 1} - \frac{1}{2 ( x + 1 )} + C$
Where the marks are
- Correct partial fractions $A = - \frac{3}{4}$ , $B = \frac{1}{2}$ , $C = \frac{3}{4}$ (1½ marks)
- Correct integration of each term (1½ marks)
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Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerDifferential Equations
Find the particular solution of the differential equation $\frac{d y}{d x} - 2 x y = 3 x^{2} e^{x^{2}}$ ; $y (0) = 5$ .
Try it first, then show our solutionSolution
1. The equation is linear of the form $\frac{d y}{d x} + P y = Q$ with $P = - 2 x$ , $Q = 3 x^{2} e^{x^{2}}$ .
  ½ mark
2. I.F. $= e^{\int - 2 x d x} = e^{- x^{2}}$ .
  1 mark
3. Solution: $y e^{- x^{2}} = \int e^{- x^{2}} \cdot 3 x^{2} e^{x^{2}} d x + C = \int 3 x^{2} d x + C \Rightarrow y e^{- x^{2}} = x^{3} + C$ .
  1 mark
4. $y (0) = 5$ : $5 \cdot 1 = 0 + C \Rightarrow C = 5$ . Particular solution: $y = (x^{3} + 5) e^{x^{2}}$ .
  ½ mark
Answer: $y = (x^{3} + 5) e^{x^{2}}$
Where the marks are
- Identifies P and Q (½ mark)
- Integrating factor $e^{- x^{2}}$ (1 mark)
- General solution $y e^{- x^{2}} = x^{3} + C$ (1 mark)
- Uses $y (0) = 5$ to get $C = 5$ and particular solution (½ mark)
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OR
Q29(b) · 3 marks · Short answerDifferential Equations
Solve the following differential equation : $x^{2} d y + y (x + y) d x = 0$
Try it first, then show our solutionSolution
1. $x^{2} d y = - y (x + y) d x \Rightarrow \frac{d y}{d x} = - \frac{y}{x} - (\frac{y}{x})^{2}$ , a homogeneous equation.
  ½ mark
2. Put $y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ : $v + x \frac{d v}{d x} = - v - v^{2} \Rightarrow x \frac{d v}{d x} = - (v^{2} + 2 v)$ .
  ½ mark
3. Separating variables: $\int \frac{d v}{v ^{2} + 2 v} = - \int \frac{d x}{x}$ .
  ½ mark
4. $\int \frac{d v}{( v + 1 ) ^{2} - 1} = - \int \frac{d x}{x}$ .
  ½ mark
5. $\frac{1}{2} lo g \frac{v}{v + 2} = - lo g ∣ x ∣ + lo g ∣ C ∣ = lo g \frac{C}{x}$ .
  ½ mark
6. So $\frac{v}{v + 2} = \frac{C ^{2}}{x ^{2}}$ ; with $v = \frac{y}{x}$ : $\frac{y}{y + 2 x} = \frac{k}{x ^{2}}$ , i.e. $x^{2} y = k (y + 2 x)$ .
  ½ mark
Answer: $x^{2} y = k (y + 2 x)$ , k an arbitrary constant
Where the marks are
- Writes $\frac{d y}{d x} = - \frac{y}{x} - \frac{y ^{2}}{x ^{2}}$ (½ mark)
- Substitutes $y = v x$ (½ mark)
- Separates variables (½ mark)
- Writes $v^{2} + 2 v = (v + 1)^{2} - 1$ (½ mark)
- Integrates correctly (½ mark)
- Final solution in x and y (½ mark)
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Q30 · 3 marks · Short answerVector Algebra
Find a vector of magnitude 4 units perpendicular to each of the vectors $2 \hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - \hat{k}$ and hence verify your answer.
Try it first, then show our solutionSolution
1. Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + \hat{j} - \hat{k}$ . $c = a \times b = \hat{i} 21 \hat{j} - 1 1 \hat{k} 1 - 1 = \hat{i} (1 - 1) - \hat{j} (- 2 - 1) + \hat{k} (2 + 1) = 3 \hat{j} + 3 \hat{k}$ .
  1 mark
2. $∣ c ∣ = 32$ , so $\overset{c}{^} = \frac{j ^ + k ^}{2}$ . Required vector $d = 4 \overset{c}{^} = 22 \hat{j} + 22 \hat{k}$ (or $- 22 \hat{j} - 22 \hat{k}$ ).
  1 mark
3. Verification: $∣ d ∣ = 8 + 8 = 4$ ; $d \cdot a = 0 - 22 + 22 = 0$ ; $d \cdot b = 0 + 22 - 22 = 0$ . Hence $d ⊥ a$ and $d ⊥ b$ .
  1 mark
Answer: $\pm (22 \hat{j} + 22 \hat{k})$
Where the marks are
- Cross product $a \times b = 3 \hat{j} + 3 \hat{k}$ (1 mark)
- Scales unit vector to magnitude 4: $22 \hat{j} + 22 \hat{k}$ (1 mark)
- Verifies magnitude 4 and both dot products zero (1 mark)
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Q31 · 3 marks · Short answerProbability
The random variable X has the following probability distribution where a and b are some constants :
X 1 2 3 4 5
P(X) 0.2 a a 0.2 b
If the mean $E (X) = 3$ , then find values of a and b and hence determine $P (X \geq 3)$ .
Try it first, then show our solutionSolution
1. $E (X) = \sum x P (x) = 0.2 + 2 a + 3 a + 0.8 + 5 b = 5 a + 5 b + 1$ .
  ½ mark
2. $\sum p_{i} = 1$ : $0.2 + a + a + 0.2 + b = 1 \Rightarrow 2 a + b = 0.6$ .
  ½ mark
3. $E (X) = 3 \Rightarrow 5 a + 5 b = 2 \Rightarrow a + b = 0.4$ . Subtracting from $2 a + b = 0.6$ : $a = 0.2 = \frac{1}{5}$ , $b = 0.2 = \frac{1}{5}$ .
  1 mark
4. $P (X \geq 3) = 1 - [P (X = 1) + P (X = 2)] = 1 - (0.2 + 0.2) = 0.6 = \frac{3}{5}$ .
  1 mark
Answer: $a = \frac{1}{5}$ , $b = \frac{1}{5}$ , $P (X \geq 3) = \frac{3}{5}$
Where the marks are
- Expression for $E (X)$ (½ mark)
- Sum of probabilities gives $2 a + b = 0.6$ (½ mark)
- Solves to $a = b = \frac{1}{5}$ (1 mark)
- $P (X \geq 3) = \frac{3}{5}$ (1 mark)
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X	1	2	3	4	5
P(X)	0.2	a	a	0.2	b

Section D

Long Answer

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerMatrices
If $A = 121202 - 3 - 3 0$ , then find $A^{- 1}$ and hence solve the following system of equations :
$x + 2 y - 3 z = 1$
$2 x - 3 z = 2$
$x + 2 y = 3$
Try it first, then show our solutionSolution
1. $∣ A ∣ = 1 (0 + 6) - 2 (0 + 3) - 3 (4 - 0) = 6 - 6 - 12 = - 12 \neq = 0$ , so $A^{- 1}$ exists.
  1 mark
2. Cofactors: $A_{11} = 6, A_{12} = - 3, A_{13} = 4$ ; $A_{21} = - 6, A_{22} = 3, A_{23} = 0$ ; $A_{31} = - 6, A_{32} = - 3, A_{33} = - 4$ . So $adj A = 6 - 3 4 - 6 30 - 6 - 3 - 4$ .
  2 marks
3. $A^{- 1} = \frac{1}{∣ A ∣} adj A = - \frac{1}{12} 6 - 3 4 - 6 30 - 6 - 3 - 4$ .
  ½ mark
4. The system is $A X = B$ with $X = x y z$ , $B = 123$ . $X = A^{- 1} B = - \frac{1}{12} 6 - 12 - 18 - 3 + 6 - 9 4 + 0 - 12 = - \frac{1}{12} - 24 - 6 - 8 = 2 \frac{1}{2} \frac{2}{3}$ . Hence $x = 2$ , $y = \frac{1}{2}$ , $z = \frac{2}{3}$ .
  1½ marks
Answer: $A^{- 1} = - \frac{1}{12} 6 - 3 4 - 6 30 - 6 - 3 - 4$ ; $x = 2$ , $y = \frac{1}{2}$ , $z = \frac{2}{3}$
Where the marks are
- $∣ A ∣ = - 12$ (1 mark)
- Correct adj A (2 marks)
- $A^{- 1}$ (½ mark)
- Writes $X = A^{- 1} B$ and solves to $x = 2, y = \frac{1}{2}, z = \frac{2}{3}$ (1½ marks)
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OR
Q32(b) · 5 marks · Long answerMatrices
Find the product of the matrices $123 23 - 3 - 3 2 - 4 - 6 14 - 15 1759 13 - 8 - 1$ and hence solve the system of linear equations :
$x + 2 y - 3 z = - 4$
$2 x + 3 y + 2 z = 2$
$3 x - 3 y - 4 z = 11$
Try it first, then show our solutionSolution
1. Let $A = 123 23 - 3 - 3 2 - 4$ , $B = - 6 14 - 15 1759 13 - 8 - 1$ . Row 1: $- 6 + 28 + 45 = 67$ , $17 + 10 - 27 = 0$ , $13 - 16 + 3 = 0$ ; Row 2: $- 12 + 42 - 30 = 0$ , $34 + 15 + 18 = 67$ , $26 - 24 - 2 = 0$ ; Row 3: $- 18 - 42 + 60 = 0$ , $51 - 15 - 36 = 0$ , $39 + 24 + 4 = 67$ . So $A B = 670006700067 = 67 I$ .
  2½ marks
2. $A (\frac{1}{67} B) = I \Rightarrow A^{- 1} = \frac{1}{67} - 6 14 - 15 1759 13 - 8 - 1$ .
  1 mark
3. The system is $A X = C$ , $C = - 4 211$ . $X = A^{- 1} C = \frac{1}{67} 24 + 34 + 143 - 56 + 10 - 88 60 + 18 - 11 = \frac{1}{67} 201 - 134 67 = 3 - 2 1$ . Hence $x = 3$ , $y = - 2$ , $z = 1$ .
  1½ marks
Answer: Product $= 67 I$ ; $x = 3$ , $y = - 2$ , $z = 1$
Where the marks are
- Correct product $67 I$ (2½ marks)
- Deduces $A^{- 1} = \frac{1}{67} B$ (1 mark)
- Solves $X = A^{- 1} C$ to get $x = 3, y = - 2, z = 1$ (1½ marks)
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Q33 · 5 marks · Long answerApplication of Integrals
Find the area of the region bounded by the curve $4 x^{2} + y^{2} = 36$ using integration.
Try it first, then show our solutionSolution
1. $4 x^{2} + y^{2} = 36 \Rightarrow \frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1$ , an ellipse centred at the origin with vertices $(\pm 3, 0)$ and $(0, \pm 6)$ . Figure: ellipse elongated along the y-axis, symmetric about both axes; the first-quadrant part (from $x = 0$ to $x = 3$ ) is shaded.
  1 mark
2. By symmetry, Area $= 4 \int_{0}^{3} y d x$ with $y = 2 9 - x^{2}$ (first quadrant), so Area $= 4 \int_{0}^{3} 2 9 - x^{2} d x = 8 \int_{0}^{3} 9 - x^{2} d x$ .
  1½ marks
3. $= 8 [\frac{x}{2} 9 - x^{2} + \frac{9}{2} sin^{- 1} \frac{x}{3}]_{0}^{3} = 8 [0 + \frac{9}{2} \cdot \frac{π}{2} - 0]$
  1½ marks
4. $= 18 π$ sq. units.
  1 mark
Answer: $18 π$ sq. units
Where the marks are
- Standard form of ellipse and correct sketch (1 mark)
- Sets up Area $= 4 \int_{0}^{3} 2 9 - x^{2} d x$ (1½ marks)
- Correct integration with limits (1½ marks)
- Area $= 18 π$ sq. units (1 mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerThree Dimensional Geometry
Find the co-ordinates of the foot of the perpendicular drawn from the point $(2, 3, - 8)$ to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}$ . Also, find the perpendicular distance of the given point from the line.
Try it first, then show our solutionSolution
1. Write the line in standard form: $\frac{x - 4}{- 2} = \frac{y - 0}{6} = \frac{z - 1}{- 3}$ . It passes through $(4, 0, 1)$ and has direction ratios $- 2, 6, - 3$ .
  ½ mark
2. Let $B$ be the foot of the perpendicular from $A (2, 3, - 8)$ . A general point on the line gives $B (- 2 λ + 4, 6 λ, - 3 λ + 1)$ .
  ½ mark
3. Direction ratios of $A B$ : $(- 2 λ + 2, 6 λ - 3, - 3 λ + 9)$ .
  1 mark
4. Since $A B ⊥$ the line: $- 2 (- 2 λ + 2) + 6 (6 λ - 3) - 3 (- 3 λ + 9) = 0$ .
  This gives $4 λ - 4 + 36 λ - 18 + 9 λ - 27 = 0$ , so $49 λ = 49$ and $λ = 1$ .
  1 mark
5. Substituting $λ = 1$ , the foot of the perpendicular is $B (2, 6, - 2)$ .
  1 mark
6. Perpendicular distance:
  $A B = (2 - 2)^{2} + (6 - 3)^{2} + (- 2 + 8)^{2} = 0 + 9 + 36 = 45 = 35$ units.
  1 mark
Answer: Foot of perpendicular $(2, 6, - 2)$ ; perpendicular distance $35$ units
Where the marks are
- Writes the line in standard form with d.r.s $- 2, 6, - 3$ (½ mark)
- Takes a general point $(- 2 λ + 4, 6 λ, - 3 λ + 1)$ (½ mark)
- Writes the d.r.s of AB (1 mark)
- Uses the perpendicularity condition to get $λ = 1$ (1 mark)
- Foot of perpendicular $(2, 6, - 2)$ (1 mark)
- Distance $35$ (1 mark)
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OR
Q34(b) · 5 marks · Long answerThree Dimensional Geometry
Find the shortest distance between the lines $L_{1}$ & $L_{2}$ given below :
$L_{1}$ : The line passing through $(2, - 1, 1)$ and parallel to $\frac{x}{1} = \frac{y}{1} = \frac{z}{3}$
$L_{2}$ : $r = \hat{i} + (2 μ + 1) \hat{j} - (μ + 2) \hat{k}$ .
Try it first, then show our solutionSolution
1. $L_{1}$ passes through $(2, - 1, 1)$ and is parallel to $\hat{i} + \hat{j} + 3 \hat{k}$ :
  $L_{1} : r = (2 \hat{i} - \hat{j} + \hat{k}) + λ (\hat{i} + \hat{j} + 3 \hat{k})$ .
  1 mark
2. Rearrange $L_{2}$ by collecting the terms in $μ$ :
  $L_{2} : r = (\hat{i} + \hat{j} - 2 \hat{k}) + μ (2 \hat{j} - \hat{k})$ .
  ½ mark
3. Identify the vectors:
  $a_{1} = 2 \hat{i} - \hat{j} + \hat{k}$ , $b_{1} = \hat{i} + \hat{j} + 3 \hat{k}$
  $a_{2} = \hat{i} + \hat{j} - 2 \hat{k}$ , $b_{2} = 2 \hat{j} - \hat{k}$
  ½ mark
4. Compute the two vectors needed:
  $a_{2} - a_{1} = - \hat{i} + 2 \hat{j} - 3 \hat{k}$
  $b_{1} \times b_{2} = \hat{i} 10 \hat{j} 12 \hat{k} 3 - 1 = (- 1 - 6) \hat{i} - (- 1 - 0) \hat{j} + (2 - 0) \hat{k} = - 7 \hat{i} + \hat{j} + 2 \hat{k}$
  1½ marks
5. Find the shortest distance:
  $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 7 + 2 - 6 = 3$
  $∣ b_{1} \times b_{2} ∣ = 49 + 1 + 4 = 54 = 36$
  $S D = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣} = \frac{3}{3 6} = \frac{1}{6}$ units.
  1½ marks
Answer: Shortest distance $= \frac{1}{6}$ units
Where the marks are
- Vector equation of $L_{1}$ (1 mark)
- Rewrites $L_{2}$ as $(\hat{i} + \hat{j} - 2 \hat{k}) + μ (2 \hat{j} - \hat{k})$ (½ mark)
- Identifies $a_{1}, b_{1}, a_{2}, b_{2}$ (½ mark)
- Finds $a_{2} - a_{1}$ (½ mark) and $b_{1} \times b_{2} = - 7 \hat{i} + \hat{j} + 2 \hat{k}$ (1 mark)
- Applies the SD formula to get $1/ 6$ (1½ marks)
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Q35 · 5 marks · Long answerLinear Programming
Solve the following L.P.P. graphically :
Maximise $Z = 60 x + 40 y$
Subject to $x + 2 y \leq 12$ , $2 x + y \leq 12$ , $4 x + 5 y \geq 20$ , $x, y \geq 0$
Try it first, then show our solutionSolution
1. Graph. Draw the boundary lines:
  $x + 2 y = 12$ through $(12, 0)$ and $(0, 6)$
  $2 x + y = 12$ through $(6, 0)$ and $(0, 12)$
  $4 x + 5 y = 20$ through $(5, 0)$ and $(0, 4)$
  Shade the region that satisfies all the constraints:
  below $x + 2 y = 12$ and below $2 x + y = 12$
  above $4 x + 5 y = 20$
  in the first quadrant
  The two lines $x + 2 y = 12$ and $2 x + y = 12$ meet at $C (4, 4)$ . The feasible region is bounded, with corner points $A (0, 4)$ , $B (0, 6)$ , $C (4, 4)$ , $D (6, 0)$ and $E (5, 0)$ .
  3½ marks
2. Evaluate $Z = 60 x + 40 y$ at each corner point:
  Corner point $Z = 60 x + 40 y$
  $A (0, 4)$ 160
  $B (0, 6)$ 240
  $C (4, 4)$ 400
  $D (6, 0)$ 360
  $E (5, 0)$ 300
  1 mark
3. The largest value in the table is at $C$ . Hence the maximum value of $Z$ is $400$ , at $x = 4, y = 4$ .
  ½ mark
Answer: Max $Z = 400$ at $x = 4, y = 4$
Where the marks are
- Correct graph with the feasible region shaded and corner points $(0, 4), (0, 6), (4, 4), (6, 0), (5, 0)$ (3½ marks)
- Table of Z values at the corner points (1 mark)
- Conclusion: max Z = 400 at (4,4) (½ mark)
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Corner point	$Z = 60 x + 40 y$
$A (0, 4)$	160
$B (0, 6)$	240
$C (4, 4)$	400
$D (6, 0)$	360
$E (5, 0)$	300

Section E

Case study based

Answer one of these 2 alternatives.
Q36(a) · 4 marks · Case-basedCompetency-basedRelations and Functions
Students of a school are taken to a railway museum to learn about railways heritage and its history. An exhibit in the museum depicted many rail lines on the track near the railway station. Let L be the set of all rail lines on the railway track and R be the relation on L defined by $R = {(l_{1}, l_{2}) : l_{1} is parallel to l_{2}}$
On the basis of the above information, answer the following questions :
1. (i)
  Find whether the relation R is symmetric or not.
  1 mark
2. (ii)
  Find whether the relation R is transitive or not.
  1 mark
3. (iii)
  If one of the rail lines on the railway track is represented by the equation $y = 3 x + 2$ , then find the set of rail lines in R related to it.
  2 marks
Figure in the paper: Photograph of several parallel rail lines on a track near a railway station, with a train engine and hills in the background. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Let $(l_{1}, l_{2}) \in R$ . Then $l_{1} ∥ l_{2}$ , so $l_{2} ∥ l_{1}$ , and hence $(l_{2}, l_{1}) \in R$ . Therefore R is symmetric.
  1 mark
2. (ii) Let $(l_{1}, l_{2}) \in R$ and $(l_{2}, l_{3}) \in R$ . Then $l_{1} ∥ l_{2}$ and $l_{2} ∥ l_{3}$ , so $l_{1} ∥ l_{3}$ [lines parallel to the same line are parallel to each other]. Hence $(l_{1}, l_{3}) \in R$ , and R is transitive.
  1 mark
3. (iii) The lines related to $y = 3 x + 2$ are exactly the lines parallel to it, i.e. the lines with the same slope $3$ .
  The required set is ${l : l is a line y = 3 x + c, c \in R}$ .
  2 marks
Answer: (i) Symmetric; (ii) Transitive; (iii) ${l : y = 3 x + c, c \in R}$
Where the marks are
- Shows $l_{1} ∥ l_{2} \Rightarrow l_{2} ∥ l_{1}$ , so R is symmetric (1 mark)
- Shows $l_{1} ∥ l_{2}, l_{2} ∥ l_{3} \Rightarrow l_{1} ∥ l_{3}$ , so R is transitive (1 mark)
- Set of lines $y = 3 x + c, c \in R$ (2 marks)
Open Q36(a) on its own page
OR
Q36(b) · 4 marks · Case-basedCompetency-basedRelations and Functions
Let S be the relation defined by $S = {(l_{1}, l_{2}) : l_{1} is perpendicular to l_{2}}$ check whether the relation S is symmetric and transitive.
Try it first, then show our solutionSolution
1. Symmetric: Let $(l_{1}, l_{2}) \in S$ . Then $l_{1} ⊥ l_{2}$ , which means $l_{2} ⊥ l_{1}$ , so $(l_{2}, l_{1}) \in S$ . Hence S is symmetric.
  2 marks
2. Transitive: Let $(l_{1}, l_{2}) \in S$ and $(l_{2}, l_{3}) \in S$ , so $l_{1} ⊥ l_{2}$ and $l_{2} ⊥ l_{3}$ . Then $l_{1} ∥ l_{3}$ [lines perpendicular to the same line are parallel], so $l_{1}$ is not perpendicular to $l_{3}$ and $(l_{1}, l_{3}) \in / S$ .
  Example: take $l_{1} : y = x$ , $l_{2} : y = - x$ and $l_{3} : y = x + 1$ . Here $l_{1} ⊥ l_{2}$ and $l_{2} ⊥ l_{3}$ , but $l_{1} ∥ l_{3}$ .
  Hence S is not transitive.
  2 marks
Answer: S is symmetric but not transitive
Where the marks are
- Proves S is symmetric (2 marks)
- Shows $l_{1} ⊥ l_{2}, l_{2} ⊥ l_{3} \Rightarrow l_{1} ∥ l_{3}$ , so S is not transitive (2 marks)
Open Q36(b) on its own page
Q37 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
A rectangular visiting card is to contain 24 sq.cm. of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins on the left and right are to be $1 \frac{1}{2}$ cm as shown below :
On the basis of the above information, answer the following questions :
1. (i)
  Write the expression for the area of the visiting card in terms of $x$ .
  2 marks
2. (ii)
  Obtain the dimensions of the card of minimum area.
  2 marks
Figure in the paper: Two rectangles. Left: visiting card with handwritten text inside a dashed rectangle, margins 1 cm at top and bottom, 1½ cm at left and right. Right: same card with the shaded inner rectangle labelled 'Printed matter', width x (top) and height y (right), with the same margins marked. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The printed area is $x y = 24$ , so $y = \frac{24}{x}$ .
  The card measures $(x + 3)$ cm by $(y + 2)$ cm, so its area is
  $A (x) = (x + 3) (y + 2) = x y + 2 x + 3 y + 6 = 24 + 2 x + \frac{72}{x} + 6$
  $\Rightarrow A (x) = 2 x + \frac{72}{x} + 30$ .
  2 marks
2. (ii) Differentiate: $A^{'} (x) = 2 - \frac{72}{x ^{2}}$ and $A^{''} (x) = \frac{144}{x ^{3}}$ .
  1 mark
3. Set $A^{'} (x) = 0$ : $x^{2} = 36$ , so $x = 6$ (since $x > 0$ ). This is the critical point.
  ½ mark
4. Check: $A^{''} (6) = \frac{144}{216} > 0$ , so the area is minimum at $x = 6$ , which gives $y = \frac{24}{6} = 4$ .
  Dimensions of the card: length $= 6 + 3 = 9$ cm and breadth $= 4 + 2 = 6$ cm (minimum area $54$ cm²).
  ½ mark
Answer: (i) $A (x) = 2 x + \frac{72}{x} + 30$ ; (ii) card is 9 cm × 6 cm
Where the marks are
- Uses $x y = 24$ and forms $A (x) = 2 x + \frac{72}{x} + 30$ (2 marks)
- Finds $A^{'} (x)$ and $A^{''} (x)$ (1 mark)
- Critical point $x = 6$ (½ mark)
- $A^{''} (6) > 0$ , dimensions 9 cm × 6 cm (½ mark)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
A departmental store sends bills to charge its customers once a month. Past experience shows that 70% of its customers pay their first month bill in time. The store also found that the customer who pays the bill in time has the probability of 0.8 of paying in time next month and the customer who doesn't pay in time has the probability of 0.4 of paying in time the next month.
Based on the above information, answer the following questions :
1. (i)
  Let $E_{1}$ and $E_{2}$ respectively denote the event of customer paying or not paying the first month bill in time. Find $P (E_{1})$ , $P (E_{2})$ .
  1 mark
2. (ii)
  Let A denotes the event of customer paying second month's bill in time, then find $P (A ∣ E_{1})$ and $P (A ∣ E_{2})$ .
  1 mark
3. (iii)
  Find the probability of customer paying second month's bill in time.
  2 marks
4. OR
  (iii)
  Find the probability of customer paying first month's bill in time if it is found that customer has paid the second month's bill in time.
  2 marks
Try it first, then show our solutionSolution
1. (i) 70% of customers pay the first bill in time, so $P (E_{1}) = 0.7$ and $P (E_{2}) = 1 - 0.7 = 0.3$ .
  1 mark
2. (ii) From the given data, $P (A ∣ E_{1}) = 0.8$ and $P (A ∣ E_{2}) = 0.4$ .
  1 mark
3. (iii) By the theorem of total probability:
  $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) = 0.7 \times 0.8 + 0.3 \times 0.4 = 0.56 + 0.12 = 0.68 = \frac{17}{25}$ .
  2 marks
4. (iii) OR: By Bayes' theorem:
  $P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} )} = \frac{0.56}{0.68} = \frac{14}{17}$ .
Answer: (i) $P (E_{1}) = 0.7, P (E_{2}) = 0.3$ ; (ii) $P (A ∣ E_{1}) = 0.8, P (A ∣ E_{2}) = 0.4$ ; (iii) $0.68$ ; (iii) OR $\frac{14}{17}$
Where the marks are
- $P (E_{1}) = 0.7, P (E_{2}) = 0.3$ (1 mark)
- $P (A ∣ E_{1}) = 0.8, P (A ∣ E_{2}) = 0.4$ (1 mark)
- Total probability gives $P (A) = 0.68$ (2 marks)
- OR: Bayes' theorem gives $P (E_{1} ∣ A) = 14/17$ (2 marks)
Open Q38 on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.