MathsCBSE Class 12 · Board exam 2025 (session 2024-25)

CBSE Class 12 Mathematics: 2025 board paper, set 65/1/1

The question paper set in the CBSE Class 12 board exam of 2025, 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/2025/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 of Section B, 3 of C, 2 of D and 2 of E; calculator not allowed.

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQMatrices
If $A = - 1 00 010001$ , then $A^{- 1}$ is
1. (A) $- 1 00 0 - 1 0 00 - 1$
2. (B) $100 0 - 1 0 00 - 1$
3. (C) $- 1 00 0 - 1 0 001$
4. (D) $- 1 00 010001$
Try it first, then show our solutionSolution
1. $A$ is diagonal, so $A^{- 1}$ is diagonal with reciprocals of the diagonal entries: $\frac{1}{- 1} = - 1, 1, 1$ . Hence $A^{- 1} = A$ , which is option (D).
  1 mark
Answer: (D)
Where the marks are
- Inverse of a diagonal matrix taken as reciprocals of the diagonal entries, giving $A^{- 1} = A$ (1 mark)
Q2 · 1 mark · MCQVector Algebra
If vector $a = 3 \hat{i} + 2 \hat{j} - \hat{k}$ and vector $b = \hat{i} - \hat{j} + \hat{k}$ , then which of the following is correct ?
1. (A) $a ∥ b$
2. (B) $a ⊥ b$
3. (C) $∣ b ∣ > ∣ a ∣$
4. (D) $∣ a ∣ = ∣ b ∣$
Try it first, then show our solutionSolution
1. $a \cdot b = 3 (1) + 2 (- 1) + (- 1) (1) = 0$ , so $a ⊥ b$ . Also $∣ a ∣ = 14$ and $∣ b ∣ = 3$ , which rules out (C) and (D). Hence option (B).
  1 mark
Answer: (B)
Where the marks are
- Shows the dot product is 0, so the vectors are perpendicular (1 mark)
Q3 · 1 mark · MCQIntegrals
$\int_{- 1}^{1} \frac{∣ x ∣}{x} d x, x \neq = 0$ is equal to
1. (A) $- 1$
2. (B)0
3. (C)1
4. (D)2
Try it first, then show our solutionSolution
1. $\frac{∣ x ∣}{x} = - 1$ for $x < 0$ and $1$ for $x > 0$ , so the integral is $\int_{- 1}^{0} (- 1) d x + \int_{0}^{1} 1 d x = - 1 + 1 = 0$ . Hence option (B).
  1 mark
Answer: 0, option (B)
Where the marks are
- Splits the integral at 0 and gets $- 1 + 1 = 0$ (1 mark)
Q4 · 1 mark · MCQDifferential Equations
Which of the following is not a homogeneous function of $x$ and y ?
1. (A) $y^{2} - x y$
2. (B) $x - 3 y$
3. (C) $sin^{2} \frac{y}{x} + \frac{y}{x}$
4. (D) $tan x - sec y$
Try it first, then show our solutionSolution
1. Replacing $x, y$ by $λ x, λ y$ : (A), (B) and (C) become $λ^{n} f (x, y)$ (degrees 2, 1, 0). But $tan λ x - sec λ y$ cannot be written as $λ^{n} (tan x - sec y)$ , so (D) is not homogeneous.
  1 mark
Answer: (D)
Where the marks are
- Applies the test $f (λ x, λ y) = λ^{n} f (x, y)$ and identifies (D) (1 mark)
Q5 · 1 mark · MCQContinuity and Differentiability
If $f (x) = ∣ x ∣ + ∣ x - 1∣$ , then which of the following is correct ?
1. (A) $f (x)$ is both continuous and differentiable, at $x = 0$ and $x = 1$ .
2. (B) $f (x)$ is differentiable but not continuous, at $x = 0$ and $x = 1$ .
3. (C) $f (x)$ is continuous but not differentiable, at $x = 0$ and $x = 1$ .
4. (D) $f (x)$ is neither continuous nor differentiable, at $x = 0$ and $x = 1$ .
Try it first, then show our solutionSolution
1. The modulus functions are continuous everywhere, so $f$ is continuous at both points.
2. At $x = 0$ : LHD $= - 2$ , RHD $= 0$ . At $x = 1$ : LHD $= 0$ , RHD $= 2$ . The one-sided derivatives differ, so $f$ is not differentiable at either point. Hence option (C).
  1 mark
Answer: (C)
Where the marks are
- Notes continuity and shows the left and right derivatives differ at 0 and 1 (1 mark)
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Q6 · 1 mark · MCQDeterminants
If A is a square matrix of order 2 such that det (A) = 4, then det (4 adj A) is equal to :
1. (A)16
2. (B)64
3. (C)256
4. (D)512
Try it first, then show our solutionSolution
1. For order $n = 2$ : $∣ adj A ∣ = ∣ A ∣^{n - 1} = 4$ .
2. $∣4 adj A ∣ = 4^{2} \cdot 4 = 64$ . Hence option (B).
  1 mark
Answer: 64, option (B)
Where the marks are
- Uses $∣ k B ∣ = k^{n} ∣ B ∣$ and $∣ adj A ∣ = ∣ A ∣^{n - 1}$ to get 64 (1 mark)
Q7 · 1 mark · MCQProbability
If E and F are two independent events such that $P (E) = \frac{2}{3}$ , $P (F) = \frac{3}{7}$ , then $P (E / \overline{F})$ is equal to :
1. (A) $\frac{1}{6}$
2. (B) $\frac{1}{2}$
3. (C) $\frac{2}{3}$
4. (D) $\frac{7}{9}$
Try it first, then show our solutionSolution
1. If $E$ and $F$ are independent, then $E$ and $\overline{F}$ are also independent, so $P (E / \overline{F}) = P (E) = \frac{2}{3}$ . Hence option (C).
  1 mark
Answer: $\frac{2}{3}$ , option (C)
Where the marks are
- Uses independence of $E$ and $\overline{F}$ to get $\frac{2}{3}$ (1 mark)
Q8 · 1 mark · MCQApplication of Derivatives
The absolute maximum value of function $f (x) = x^{3} - 3 x + 2$ in $[0, 2]$ is :
1. (A)0
2. (B)2
3. (C)4
4. (D)5
Try it first, then show our solutionSolution
1. $f^{'} (x) = 3 x^{2} - 3 = 0 \Rightarrow x = 1$ in $[0, 2]$ .
2. $f (0) = 2$ , $f (1) = 0$ , $f (2) = 4$ . The absolute maximum is 4, option (C).
  1 mark
Answer: 4, option (C)
Where the marks are
- Finds the critical point and compares values at the endpoints to get 4 (1 mark)
Q9 · 1 mark · MCQMatrices
Let $A = 10 - 3 - 2 42 - 1 - 1 1$ , $B = - 2 - 5 - 7$ , $C = [987]$ , which of the following is defined ?
1. (A)Only AB
2. (B)Only AC
3. (C)Only BA
4. (D)All AB, AC and BA
Try it first, then show our solutionSolution
1. Orders: $A$ is $3 \times 3$ , $B$ is $3 \times 1$ , $C$ is $1 \times 3$ . $A B$ is defined ( $3 = 3$ ). $A C$ is not ( $3 \neq = 1$ ). $B A$ is not ( $1 \neq = 3$ ). Hence only $A B$ , option (A).
  1 mark
Answer: (A)
Where the marks are
- Checks the order compatibility of each product (1 mark)
Q10 · 1 mark · MCQIntegrals
If $\int \frac{2 ^{\frac{1}{x}}}{x ^{2}} d x = k \cdot 2^{\frac{1}{x}} + C$ , then k is equal to
1. (A) $\frac{- 1}{l o g 2}$
2. (B) $- lo g 2$
3. (C) $- 1$
4. (D) $\frac{1}{2}$
Try it first, then show our solutionSolution
1. Put $t = \frac{1}{x}$ , so $d t = - \frac{1}{x ^{2}} d x$ .
2. The integral becomes $- \int 2^{t} d t = - \frac{2 ^{t}}{l o g 2} + C = - \frac{1}{l o g 2} 2^{1/ x} + C$ , so $k = - \frac{1}{l o g 2}$ , option (A).
  1 mark
Answer: $k = - \frac{1}{l o g 2}$ , option (A)
Where the marks are
- Substitutes $t = 1/ x$ and integrates $2^{t}$ to get $k$ (1 mark)
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Q11 · 1 mark · MCQVector Algebra
If $a + b + c = 0$ , $∣ a ∣ = 37$ , $∣ b ∣ = 3$ and $∣ c ∣ = 4$ , then angle between $b$ and $c$ is
1. (A) $\frac{π}{6}$
2. (B) $\frac{π}{4}$
3. (C) $\frac{π}{3}$
4. (D) $\frac{π}{2}$
Try it first, then show our solutionSolution
1. $b + c = - a$ , so $∣ a ∣^{2} = ∣ b ∣^{2} + ∣ c ∣^{2} + 2∣ b ∣∣ c ∣ cos θ$ .
2. $37 = 9 + 16 + 24 cos θ \Rightarrow cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$ , option (C).
  1 mark
Answer: $\frac{π}{3}$ , option (C)
Where the marks are
- Squares $b + c = - a$ and gets $cos θ = \frac{1}{2}$ (1 mark)
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Q12 · 1 mark · MCQDifferential Equations
The integrating factor of differential equation $(x + 2 y^{3}) \frac{d y}{d x} = 2 y$ is
1. (A) $e^{\frac{y ^{2}}{2}}$
2. (B) $\frac{1}{y}$
3. (C) $\frac{1}{y ^{2}}$
4. (D) $e^{- \frac{1}{y ^{2}}}$
Try it first, then show our solutionSolution
1. Treat $x$ as the dependent variable: $\frac{d x}{d y} = \frac{x + 2 y ^{3}}{2 y} \Rightarrow \frac{d x}{d y} - \frac{1}{2 y} x = y^{2}$ .
2. I.F. $= e^{- \int \frac{1}{2 y} d y} = e^{- \frac{1}{2} l o g y} = \frac{1}{y}$ , option (B).
  1 mark
Answer: $\frac{1}{y}$ , option (B)
Where the marks are
- Writes the equation in linear form in $x$ and finds the I.F. (1 mark)
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Q13 · 1 mark · MCQMatrices
If $A = 700070 x 0 y$ is a scalar matrix, then $y^{x}$ is equal to
1. (A)0
2. (B)1
3. (C)7
4. (D) $\pm 7$
Try it first, then show our solutionSolution
1. In a scalar matrix, off-diagonal entries are 0 and diagonal entries are equal, so $x = 0$ and $y = 7$ .
2. $y^{x} = 7^{0} = 1$ , option (B).
  1 mark
Answer: 1, option (B)
Where the marks are
- Gets $x = 0$ , $y = 7$ and hence $y^{x} = 1$ (1 mark)
Q14 · 1 mark · MCQLinear Programming
The corner points of the feasible region in graphical representation of a L.P.P. are (2, 72), (15, 20) and (40, 15). If $Z = 18 x + 9 y$ be the objective function, then
1. (A)Z is maximum at (2, 72), minimum at (15, 20)
2. (B)Z is maximum at (15, 20) minimum at (40, 15)
3. (C)Z is maximum at (40, 15), minimum at (15, 20)
4. (D)Z is maximum at (40, 15), minimum at (2, 72)
Try it first, then show our solutionSolution
1. $Z (2, 72) = 36 + 648 = 684$ , $Z (15, 20) = 270 + 180 = 450$ , $Z (40, 15) = 720 + 135 = 855$ .
2. Z is maximum at $(40, 15)$ and minimum at $(15, 20)$ , option (C).
  1 mark
Answer: (C)
Where the marks are
- Evaluates Z at all three corners and compares (1 mark)
Q15 · 1 mark · MCQMatrices
If A and B are invertible matrices, then which of the following is not correct ?
1. (A) $(A + B)^{- 1} = B^{- 1} + A^{- 1}$
2. (B) $(A B)^{- 1} = B^{- 1} A^{- 1}$
3. (C) $adj (A) = ∣ A ∣ A^{- 1}$
4. (D) $∣ A ∣^{- 1} = ∣ A^{- 1} ∣$
Try it first, then show our solutionSolution
1. (B), (C) and (D) are standard results. $(A + B)^{- 1} = A^{- 1} + B^{- 1}$ is false in general (e.g. $A = B = I$ gives $\frac{1}{2} I$ on the left but $2 I$ on the right). Hence option (A).
  1 mark
Answer: (A)
Where the marks are
- Identifies (A) as false, e.g. by a counterexample (1 mark)
Q16 · 1 mark · MCQLinear Programming
If the feasible region of a linear programming problem with objective function $Z = a x + b y$ , is bounded, then which of the following is correct ?
1. (A)It will only have a maximum value.
2. (B)It will only have a minimum value.
3. (C)It will have both maximum and minimum values.
4. (D)It will have neither maximum nor minimum value.
Try it first, then show our solutionSolution
1. On a bounded feasible region, the objective function attains both a maximum and a minimum, each at a corner point. Hence option (C).
  1 mark
Answer: (C)
Where the marks are
- States the corner point theorem for a bounded region (1 mark)
Q17 · 1 mark · MCQApplication of Integrals
The area of the shaded region bounded by the curves $y^{2} = x$ , $x = 4$ and the $x$ -axis is given by
1. (A) $\int_{0}^{4} x d x$
2. (B) $\int_{0}^{2} y^{2} d y$
3. (C) $2 \int_{0}^{4} x d x$
4. (D) $\int_{0}^{4} x d x$
Figure in the paper: Graph of parabola $y^{2} = x$ opening to the right, with vertical line $x = 4$ ; the region above the x-axis between the curve and x-axis from x=0 to x=4 is shaded. Axes marked 1 to 6 on x, -1, 1, 2 on y. See it in the official paper.
Try it first, then show our solutionSolution
1. The shaded region lies above the x-axis under $y = x$ from $x = 0$ to $x = 4$ , so its area is $\int_{0}^{4} x d x$ , option (D).
  1 mark
Answer: (D)
Where the marks are
- Takes the upper branch $y = x$ with limits 0 to 4 (1 mark)
Q18 · 1 mark · MCQInverse Trigonometric Functions
The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse ?
1. (A)Graph of an increasing odd curve through origin on $[- π /2, π /2]$ with y from -1 to 1, steep at both ends (like sin inverse)
2. (B)Graph of an increasing curve through origin on $[- π /2, π /2]$ with y from -1 to 1, steep near origin and flattening at ends
3. (C)Graph decreasing from point at x=-1, y=π to point at x=1, y=0, passing y=π/2 at x=0
4. (D)Graph decreasing from y=π (dashed line) to y=0 over x axis, passing y=π/2 at x=0
Figure in the paper: Given graph: a cosine-like curve with maximum at x=0, zeros at -π/2 and π/2, minimum at x=π, on axes with Y marked. Options are graphs as described. See it in the official paper.
Try it first, then show our solutionSolution
1. The given graph is $y = cos x$ , whose principal branch is $[0, π]$ . Its inverse $cos^{- 1} x$ has domain $[- 1, 1]$ , range $[0, π]$ , and decreases from $(- 1, π)$ through $(0, \frac{π}{2})$ to $(1, 0)$ . This matches graph (C).
  1 mark
Answer: (C)
Where the marks are
- Identifies the function as cosine and matches the graph of $cos^{- 1} x$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedRelations and Functions
Assertion (A) : Let Z be the set of integers. A function $f : Z \to Z$ defined as $f (x) = 3 x - 5$ , $\forall x \in Z$ is a bijective.
Reason (R) : A function is a bijective if it is both surjective and injective.
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, 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. $f$ is one-one, but it is not onto: for $f (x) = 0$ we need $x = \frac{5}{3} \in / Z$ . So the Assertion is false.
2. The Reason is the correct definition of a bijection, so it is true. Hence option (D).
  1 mark
Answer: (D)
Where the marks are
- Shows $f$ is not onto (e.g. 0 has no preimage) and that R is true (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedContinuity and Differentiability
Assertion (A) : $f (x) = {3 x - 8, 2 k, x \leq 5 x > 5$ is continuous at $x = 5$ for $k = \frac{5}{2}$ .
Reason (R) : For a function f to be continuous at $x = a$ , $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$ .
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, 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. $f (5) = 15 - 8 = 7$ and the RHL is $2 k$ . Continuity needs $2 k = 7$ , i.e. $k = \frac{7}{2}$ , not $\frac{5}{2}$ . So the Assertion is false.
2. The Reason is the correct condition for continuity, so it is true. Hence option (D).
  1 mark
Answer: (D)
Where the marks are
- Finds $k = \frac{7}{2}$ , so A is false, and recognises R is true (1 mark)
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Section B

VSA

Answer one of these 2 alternatives.
Q21(a) · 2 marks · Very short answerContinuity and Differentiability
Differentiate $2^{c o s^{2} x}$ w.r.t $cos^{2} x$ .
Try it first, then show our solutionSolution
1. Let $u = 2^{c o s^{2} x}$ . By the chain rule, $\frac{d u}{d x} = 2^{c o s^{2} x} lo g 2 \cdot (- 2 cos x sin x)$ .
  1 mark
2. Let $v = cos^{2} x$ . Then $\frac{d v}{d x} = - 2 cos x sin x$ .
  ½ mark
3. Divide the two derivatives: $\frac{d u}{d v} = \frac{d u / d x}{d v / d x} = \frac{2 ^{c o s^{2} x} lo g 2 ( - 2 cos x sin x )}{- 2 cos x sin x} = 2^{c o s^{2} x} lo g 2$ .
  ½ mark
Answer: $\frac{d}{d ( cos ^{2} x )} (2^{c o s^{2} x}) = 2^{c o s^{2} x} lo g 2$
Where the marks are
- Correct derivative of $2^{c o s^{2} x}$ with respect to $x$ (1 mark)
- Correct derivative of $cos^{2} x$ with respect to $x$ (½ mark)
- Divides the two derivatives to get $2^{c o s^{2} x} lo g 2$ (½ mark)
Open Q21(a) on its own page
OR
Q21(b) · 2 marks · Very short answerContinuity and Differentiability
If $tan^{- 1} (x^{2} + y^{2}) = a^{2}$ , then find $\frac{d y}{d x}$ .
Try it first, then show our solutionSolution
1. Take tan of both sides: $tan^{- 1} (x^{2} + y^{2}) = a^{2} \Rightarrow x^{2} + y^{2} = tan a^{2}$ , which is a constant.
  ½ mark
2. Differentiate both sides with respect to $x$ : $2 x + 2 y \frac{d y}{d x} = 0$ .
  1 mark
3. Solve for the derivative: $\frac{d y}{d x} = - \frac{x}{y}$ .
  ½ mark
Answer: $\frac{d y}{d x} = - \frac{x}{y}$
Where the marks are
- Rewrites the equation as $x^{2} + y^{2} = tan a^{2}$ (½ mark)
- Differentiates to get $2 x + 2 y y^{'} = 0$ (1 mark)
- Obtains $\frac{d y}{d x} = - \frac{x}{y}$ (½ mark)
Open Q21(b) on its own page
Q22 · 2 marks · Very short answerInverse Trigonometric Functions
Evaluate : $tan^{- 1} [2 sin (2 cos^{- 1} \frac{3}{2})]$
Try it first, then show our solutionSolution
1. Since $cos \frac{π}{6} = \frac{3}{2}$ and $\frac{π}{6} \in [0, π]$ , we have $cos^{- 1} \frac{3}{2} = \frac{π}{6}$ . So the expression becomes $tan^{- 1} [2 sin (2 \times \frac{π}{6})] = tan^{- 1} [2 sin \frac{π}{3}]$ .
  1 mark
2. Now $2 sin \frac{π}{3} = 2 \times \frac{3}{2} = 3$ . Since $\frac{π}{3} \in (- \frac{π}{2}, \frac{π}{2})$ , $tan^{- 1} 3 = \frac{π}{3}$ .
  1 mark
Answer: $\frac{π}{3}$
Where the marks are
- Uses the principal value $cos^{- 1} \frac{3}{2} = \frac{π}{6}$ and simplifies to $tan^{- 1} (2 sin \frac{π}{3})$ (1 mark)
- Simplifies to $tan^{- 1} 3 = \frac{π}{3}$ (1 mark)
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Q23 · 2 marks · Very short answerVector Algebra
The diagonals of a parallelogram are given by $a = 2 \hat{i} - \hat{j} + \hat{k}$ and $b = \hat{i} + 3 \hat{j} - \hat{k}$ . Find the area of the parallelogram.
Try it first, then show our solutionSolution
1. Find the cross product of the diagonals: $a \times b = \hat{i} 21 \hat{j} - 1 3 \hat{k} 1 - 1 = \hat{i} (1 - 3) - \hat{j} (- 2 - 1) + \hat{k} (6 + 1) = - 2 \hat{i} + 3 \hat{j} + 7 \hat{k}$ .
  1 mark
2. When $a$ and $b$ are the diagonals, area of parallelogram $= \frac{1}{2} ∣ a \times b ∣ = \frac{1}{2} (- 2)^{2} + 3^{2} + 7^{2} = \frac{1}{2} 62 = \frac{62}{2}$ sq units.
  1 mark
Answer: Area $= \frac{62}{2}$ sq units
Where the marks are
- Correct $a \times b = - 2 \hat{i} + 3 \hat{j} + 7 \hat{k}$ (1 mark)
- Uses area $= \frac{1}{2} ∣ a \times b ∣$ to get $\frac{62}{2}$ (1 mark)
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Q24 · 2 marks · Very short answerApplication of Derivatives
Find the intervals in which function $f (x) = 5 x^{\frac{3}{2}} - 3 x^{\frac{5}{2}}$ is (i) increasing (ii) decreasing.
Try it first, then show our solutionSolution
1. The function is defined for $x \geq 0$ . Differentiate: $f^{'} (x) = \frac{15}{2} x^{1/2} - \frac{15}{2} x^{3/2} = \frac{15}{2} x (1 - x)$ . Setting $f^{'} (x) = 0$ gives $x = 0, 1$ .
  1 mark
2. (i) For $x \in [0, 1]$ , $x \geq 0$ and $1 - x \geq 0$ , so $f^{'} (x) \geq 0$ . Hence $f$ is increasing on $[0, 1]$ .
  ½ mark
3. (ii) For $x \in [1, \infty)$ , $1 - x \leq 0$ , so $f^{'} (x) \leq 0$ . Hence $f$ is decreasing on $[1, \infty)$ .
  ½ mark
Answer: (i) Increasing on $[0, 1]$ ; (ii) decreasing on $[1, \infty)$
Where the marks are
- Finds $f^{'} (x) = \frac{15}{2} x (1 - x)$ and critical points $x = 0, 1$ (1 mark)
- States increasing on $[0, 1]$ ; open or half-open forms also accepted (½ mark)
- States decreasing on $[1, \infty)$ ; $(1, \infty)$ also accepted (½ mark)
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Answer one of these 2 alternatives.
Q25(a) · 2 marks · Very short answerCompetency-basedVector Algebra
Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors $a = 3 \hat{i} + \hat{j} + 2 \hat{k}$ and $b = 2 \hat{i} - 2 \hat{j} + 4 \hat{k}$ . Determine the angle formed between the kite strings. Assume there is no slack in the strings.
Try it first, then show our solutionSolution
1. Let $θ$ be the angle between the strings, so $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ . Here $a \cdot b = 6 - 2 + 8 = 12$ , $∣ a ∣ = 9 + 1 + 4 = 14$ and $∣ b ∣ = 4 + 4 + 16 = 24$ . So $cos θ = \frac{12}{336} = \frac{12}{4 21} = \frac{3}{21}$ .
  1½ marks
2. Therefore $θ = cos^{- 1} (\frac{3}{21})$ .
  ½ mark
Answer: $θ = cos^{- 1} (\frac{3}{21})$ , which equals $cos^{- 1} (\frac{12}{336})$
Where the marks are
- Applies the dot-product formula and gets $cos θ = \frac{12}{336} = \frac{3}{21}$ (1½ marks)
- States $θ = cos^{- 1} \frac{3}{21}$ (½ mark)
Open Q25(a) on its own page
OR
Q25(b) · 2 marks · Very short answerVector Algebra
Find a vector of magnitude 21 units in the direction opposite to that of $A B$ where A and B are the points A(2, 1, 3) and B(8, $- 1$ , 0) respectively.
Try it first, then show our solutionSolution
1. The direction opposite to $A B$ is that of $B A$ : $B A = (2 - 8) \hat{i} + (1 + 1) \hat{j} + (3 - 0) \hat{k} = - 6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ .
  1 mark
2. Its magnitude is $∣ B A ∣ = 36 + 4 + 9 = 7$ . The required vector is $21 \times \frac{- 6 i ^ + 2 j ^ + 3 k ^}{7}$ .
  ½ mark
3. This simplifies to $3 (- 6 \hat{i} + 2 \hat{j} + 3 \hat{k}) = - 18 \hat{i} + 6 \hat{j} + 9 \hat{k}$ .
  ½ mark
Answer: $- 18 \hat{i} + 6 \hat{j} + 9 \hat{k}$
Where the marks are
- Finds $B A = - 6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ (1 mark)
- Multiplies the unit vector by 21 (½ mark)
- Gets $- 18 \hat{i} + 6 \hat{j} + 9 \hat{k}$ (½ mark)
Open Q25(b) on its own page

Section C

Q26 · 3 marks · Short answerApplication of Derivatives
The side of an equilateral triangle is increasing at the rate of 3 cm/s. At what rate its area increasing when the side of the triangle is 15 cm ?
Try it first, then show our solutionSolution
1. Let $a$ be the side of the triangle. Then $\frac{d a}{d t} = 3$ cm/s.
  ½ mark
2. Area of an equilateral triangle: $A = \frac{3}{4} a^{2}$ . Differentiating with respect to $t$ : $\frac{d A}{d t} = \frac{3}{2} a \frac{d a}{d t}$ .
  1½ marks
3. At $a = 15$ cm: $\frac{d A}{d t} = \frac{3 \times 15}{2} \times 3 = \frac{45 3}{2}$ cm²/s.
  1 mark
Answer: The area is increasing at $\frac{45 3}{2}$ cm²/s
Where the marks are
- Writes $\frac{d a}{d t} = 3$ cm/s (½ mark)
- Uses $A = \frac{3}{4} a^{2}$ and gets $\frac{d A}{d t} = \frac{3}{2} a \frac{d a}{d t}$ (1½ marks)
- Substitutes to get $\frac{45 3}{2}$ cm²/s (1 mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerLinear Programming
Solve the following linear programming problem graphically :
Maximise $Z = x + 2 y$
Subject to the constraints:
$x - y \geq 0$
$x - 2 y \geq - 2$
$x \geq 0, y \geq 0$
Try it first, then show our solutionSolution
1. Graph. Draw the lines $x - y = 0$ (through $(0, 0)$ and $(4, 4)$ ) and $x - 2 y = - 2$ (through $(- 2, 0)$ and $(0, 1)$ ). The region $x - y \geq 0$ lies on and below $y = x$ . The region $x - 2 y \geq - 2$ lies on and below $x - 2 y = - 2$ (the origin satisfies it). Together with $x \geq 0, y \geq 0$ , the feasible region lies above the $x$ -axis, under $y = x$ from $O$ to $A$ , and under $x - 2 y = - 2$ beyond $A$ . It is unbounded. The lines $y = x$ and $x - 2 y = - 2$ meet where $x - 2 x = - 2$ , i.e. at $A (2, 2)$ . So the corner points are $O (0, 0)$ and $A (2, 2)$ .
  1½ marks
2. Values of Z at the corner points
  Corner point $Z = x + 2 y$
  $O (0, 0)$ 0
  $A (2, 2)$ 6
  1 mark
3. The largest value at a corner is 6, but the region is unbounded. Draw the open half-plane $x + 2 y > 6$ (boundary line through $(6, 0)$ and $(0, 3)$ ). It has points in common with the feasible region, e.g. $(10, 0)$ gives $Z = 10 > 6$ . Hence $Z$ has no maximum value.
  ½ mark
Answer: Corner points O(0,0) and A(2,2) give Z = 0 and 6. The feasible region is unbounded and $x + 2 y > 6$ meets it, so Z has no maximum value.
Where the marks are
- Correct graph of the constraints with the unbounded feasible region shaded and corner point A(2,2) (1½ marks)
- Correct table of Z at the corner points O and A (1 mark)
- Checks the half-plane $x + 2 y > 6$ and concludes there is no maximum (½ mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerIntegrals
Find : $\int \frac{x + s i n x}{1 + c o s x} d x$
Try it first, then show our solutionSolution
1. Use $sin x = 2 sin \frac{x}{2} cos \frac{x}{2}$ and $1 + cos x = 2 cos^{2} \frac{x}{2}$ : $I = \int \frac{x + 2 sin \frac{x}{2} cos \frac{x}{2}}{2 cos ^{2} \frac{x}{2}} d x$ .
  1 mark
2. Split the integral: $I = \int x \cdot \frac{1}{2} sec^{2} \frac{x}{2} d x + \int tan \frac{x}{2} d x$ .
  ½ mark
3. Integrate the first part by parts, taking $x$ as the first function. Since $\int \frac{1}{2} sec^{2} \frac{x}{2} d x = tan \frac{x}{2}$ : $I = x tan \frac{x}{2} - \int tan \frac{x}{2} d x + \int tan \frac{x}{2} d x$ .
  1 mark
4. The two remaining integrals cancel, so $I = x tan \frac{x}{2} + C$ .
  ½ mark
Answer: $x tan \frac{x}{2} + C$
Where the marks are
- Uses half-angle identities for numerator and denominator (1 mark)
- Splits into $\int \frac{x}{2} sec^{2} \frac{x}{2} d x + \int tan \frac{x}{2} d x$ (½ mark)
- Applies integration by parts correctly (1 mark)
- Gets $x tan \frac{x}{2} + C$ (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerIntegrals
Evaluate : $\int_{0}^{\frac{π}{4}} \frac{d x}{c o s ^{3} x 2 s i n 2 x}$
Try it first, then show our solutionSolution
1. Since $2 sin 2 x = 4 sin x cos x = 2 cos x tan x$ , we get $I = \frac{1}{2} \int_{0}^{π /4} \frac{d x}{cos ^{4} x tan x} = \frac{1}{2} \int_{0}^{π /4} \frac{( 1 + tan ^{2} x ) sec ^{2} x}{tan x} d x$ .
  ½ mark
2. Put $tan x = t \Rightarrow sec^{2} x d x = d t$ . When $x = 0$ , $t = 0$ ; when $x = \frac{π}{4}$ , $t = 1$ .
  ½ mark
3. Then $I = \frac{1}{2} \int_{0}^{1} \frac{1 + t ^{2}}{t} d t = \frac{1}{2} \int_{0}^{1} (t^{- 1/2} + t^{3/2}) d t$ .
  ½ mark
4. Integrate: $I = \frac{1}{2} [2 t + \frac{2}{5} t^{5/2}]_{0}^{1}$ .
  1 mark
5. Evaluate: $I = \frac{1}{2} (2 + \frac{2}{5}) = \frac{1}{2} \times \frac{12}{5} = \frac{6}{5}$ .
  ½ mark
Answer: $\frac{6}{5}$
Where the marks are
- Rewrites the integrand as $\frac{1}{2} sec^{4} x / tan x$ (½ mark)
- Substitutes $tan x = t$ (½ mark)
- Changes limits and splits into $t^{- 1/2} + t^{3/2}$ (½ mark)
- Integrates correctly (1 mark)
- Gets $\frac{6}{5}$ (½ mark)
Open Q28(b) on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerThree Dimensional Geometry
Verify that lines given by $r = (1 - λ) \hat{i} + (λ - 2) \hat{j} + (3 - 2 λ) \hat{k}$ and $r = (μ + 1) \hat{i} + (2 μ - 1) \hat{j} - (2 μ + 1) \hat{k}$ are skew lines. Hence, find shortest distance between the lines.
Try it first, then show our solutionSolution
1. Rewrite the lines in vector form: $r = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (- \hat{i} + \hat{j} - 2 \hat{k})$ and $r = (\hat{i} - \hat{j} - \hat{k}) + μ (\hat{i} + 2 \hat{j} - 2 \hat{k})$ . So $a_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ , $b_{1} = - \hat{i} + \hat{j} - 2 \hat{k}$ , $a_{2} = \hat{i} - \hat{j} - \hat{k}$ , $b_{2} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ . The direction ratios $- 1, 1, - 2$ and $1, 2, - 2$ are not proportional, so the lines are not parallel.
  ½ mark
2. $a_{2} - a_{1} = \hat{j} - 4 \hat{k}$ .
  ½ mark
3. $b_{1} \times b_{2} = \hat{i} - 1 1 \hat{j} 12 \hat{k} - 2 - 2 = \hat{i} (- 2 + 4) - \hat{j} (2 + 2) + \hat{k} (- 2 - 1) = 2 \hat{i} - 4 \hat{j} - 3 \hat{k}$ .
  ½ mark
4. Compute $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 0 - 4 + 12 = 8 \neq = 0$ . So the lines do not intersect. Being neither parallel nor intersecting, the lines are skew.
  ½ mark
5. Shortest distance $= \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣} = \frac{8}{4 + 16 + 9} = \frac{8}{29}$ units.
  1 mark
Answer: The lines are skew; shortest distance $= \frac{8}{29}$ units
Where the marks are
- Rewrites the lines as $a + λ b$ and notes they are not parallel (½ mark)
- Finds $a_{2} - a_{1}$ (½ mark)
- Finds $b_{1} \times b_{2} = 2 \hat{i} - 4 \hat{j} - 3 \hat{k}$ (½ mark)
- Shows the scalar triple product is $8 \neq = 0$ , so the lines are skew (½ mark)
- Gets shortest distance $\frac{8}{29}$ (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerCompetency-basedVector Algebra
During a cricket match, the position of the bowler, the wicket keeper and the leg slip fielder are in a line given by $B = 2 \hat{i} + 8 \hat{j}$ , $W = 6 \hat{i} + 12 \hat{j}$ and $F = 12 \hat{i} + 18 \hat{j}$ respectively. Calculate the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder.
Try it first, then show our solutionSolution
1. Let W divide BF in the ratio $k : 1$ . By the section formula, $W = \frac{k F + 1 \cdot B}{k + 1}$ .
  1 mark
2. So $6 \hat{i} + 12 \hat{j} = (\frac{12 k + 2}{k + 1}) \hat{i} + (\frac{18 k + 8}{k + 1}) \hat{j}$ .
  1 mark
3. Compare the $\hat{i}$ components: $12 k + 2 = 6 k + 6 \Rightarrow k = \frac{2}{3}$ . The $\hat{j}$ components give $18 k + 8 = 12 k + 12$ , which also gives $k = \frac{2}{3}$ , so the points are collinear. Hence the wicket keeper divides BF in the ratio $2 : 3$ (internally).
  1 mark
Answer: $2 : 3$
Where the marks are
- Applies the section formula with ratio $k : 1$ (1 mark)
- Writes the components in terms of $k$ (1 mark)
- Solves $k = \frac{2}{3}$ and states the ratio $2 : 3$ (1 mark)
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Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerCompetency-basedProbability
The probability distribution for the number of students being absent in a class on a Saturday is as follows :
X 0 2 4 5
P(X) p 2p 3p p
Where X is the number of students absent.
1. (i)
  Calculate p.
  1 mark
2. (ii)
  Calculate the mean of the number of absent students on Saturday.
  2 marks
Try it first, then show our solutionSolution
1. (i) Since $\sum P (X) = 1$ : $p + 2 p + 3 p + p = 1$ .
  ½ mark
2. (i) So $7 p = 1 \Rightarrow p = \frac{1}{7}$ .
  ½ mark
3. (ii) Mean $= \sum X P (X) = 0 (p) + 2 (2 p) + 4 (3 p) + 5 (p)$ .
  1 mark
4. (ii) Mean $= 21 p = 21 \times \frac{1}{7} = 3$ .
  1 mark
Answer: (i) $p = \frac{1}{7}$ ; (ii) Mean = 3
Where the marks are
- Uses $\sum P (X) = 1$ (½ mark)
- Gets $p = \frac{1}{7}$ (½ mark)
- Writes the mean as $\sum X P (X)$ with the values substituted (1 mark)
- Gets mean $= 3$ (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerCompetency-basedProbability
For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.
Try it first, then show our solutionSolution
1. Define the events. $E_{1}$ : the applicant is male. $E_{2}$ : the applicant is female. $A$ : the chosen candidate gets a distinction in the written test.
  ½ mark
2. $P (E_{1}) = \frac{1}{3}$ , $P (E_{2}) = \frac{2}{3}$ , $P (A ∣ E_{1}) = 0.4$ , $P (A ∣ E_{2}) = 0.35$ .
  1 mark
3. By the theorem of total probability: $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) = \frac{1}{3} \times 0.4 + \frac{2}{3} \times 0.35$ .
  1 mark
4. $P (A) = \frac{0.4 + 0.7}{3} = \frac{1.1}{3} = \frac{11}{30}$ .
  ½ mark
Answer: $P (distinction) = \frac{11}{30}$
Where the marks are
- Defines the events $E_{1}$ , $E_{2}$ and $A$ (½ mark)
- Writes the correct probabilities $\frac{1}{3}, \frac{2}{3}, 0.4, 0.35$ (1 mark)
- Applies the total probability theorem (1 mark)
- Gets $\frac{11}{30}$ (½ mark)
Open Q30(b) on its own page
Q31 · 3 marks · Short answerApplication of Integrals
Sketch the graph of $y = ∣ x + 3∣$ and find the area of the region enclosed by the curve, $x$ -axis, between $x = - 6$ and $x = 0$ , using integration.
Try it first, then show our solutionSolution
1. Graph. $y = ∣ x + 3∣$ is V-shaped with its vertex at $(- 3, 0)$ . For $x \geq - 3$ it is the line $y = x + 3$ , through $(0, 3)$ . For $x < - 3$ it is the line $y = - (x + 3)$ , through $(- 6, 3)$ . The region between $x = - 6$ and $x = 0$ above the $x$ -axis consists of two congruent triangles, one on each side of $x = - 3$ , each of height 3. Shade both.
  1 mark
2. Required area $= \int_{- 6}^{0} y d x = \int_{- 6}^{0} ∣ x + 3∣ d x$ .
  ½ mark
3. The graph is symmetric about $x = - 3$ , so the area $= 2 \int_{- 3}^{0} (x + 3) d x$ .
  ½ mark
4. Integrate: $= 2 [\frac{( x + 3 ) ^{2}}{2}]_{- 3}^{0}$ .
  ½ mark
5. Evaluate: $= 2 (\frac{9}{2} - 0) = 9$ sq units.
  ½ mark
Answer: Area = 9 sq units
Where the marks are
- Correct sketch of $y = ∣ x + 3∣$ with the region shaded (1 mark)
- Sets up the area as $\int_{- 6}^{0} y d x$ (½ mark)
- Uses symmetry to write $2 \int_{- 3}^{0} (x + 3) d x$ , or splits at $x = - 3$ (½ mark)
- Integrates correctly (½ mark)
- Gets area 9 sq units (½ mark)
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Corner point	$Z = x + 2 y$
$O (0, 0)$	0
$A (2, 2)$	6

X	0	2	4	5
P(X)	p	2p	3p	p

Section D

Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerContinuity and Differentiability
If $1 - x^{2} + 1 - y^{2} = a (x - y)$ , then prove that $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}}$ .
Try it first, then show our solutionSolution
1. Put $x = sin A$ and $y = sin B$ , so $A = sin^{- 1} x$ and $B = sin^{- 1} y$ . Then $1 - x^{2} = cos A$ and $1 - y^{2} = cos B$ .
  1 mark
2. The equation becomes $cos A + cos B = a (sin A - sin B)$ . Using sum-to-product formulas: $2 cos \frac{A + B}{2} cos \frac{A - B}{2} = 2 a cos \frac{A + B}{2} sin \frac{A - B}{2}$ .
  1 mark
3. Cancel $2 cos \frac{A + B}{2}$ : $cot \frac{A - B}{2} = a \Rightarrow A - B = 2 cot^{- 1} a$ .
  1 mark
4. So $sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$ , which is a constant.
  ½ mark
5. Differentiate both sides with respect to $x$ : $\frac{1}{1 - x ^{2}} - \frac{1}{1 - y ^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}} = \frac{1 - y ^{2}}{1 - x ^{2}}$ . Hence proved.
  1½ marks
Answer: Proved: $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}}$
Where the marks are
- Substitutes $x = sin A$ , $y = sin B$ (1 mark)
- Applies sum-to-product formulas correctly (1 mark)
- Gets $cot \frac{A - B}{2} = a$ , i.e. $A - B = 2 cot^{- 1} a$ (1 mark)
- Rewrites as $sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$ (½ mark)
- Differentiates to reach the required result (1½ marks)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerContinuity and Differentiability
If $x = a (cos θ + lo g tan \frac{θ}{2})$ and $y = sin θ$ , then find $\frac{d ^{2} y}{d x ^{2}}$ at $θ = \frac{π}{4}$ .
Try it first, then show our solutionSolution
1. Differentiate $x$ : $\frac{d x}{d θ} = a (- sin θ + \frac{1}{tan \frac{θ}{2}} \cdot sec^{2} \frac{θ}{2} \cdot \frac{1}{2})$ .
  ½ mark
2. Since $\frac{sec ^{2} \frac{θ}{2}}{2 tan \frac{θ}{2}} = \frac{1}{2 sin \frac{θ}{2} cos \frac{θ}{2}} = \frac{1}{sin θ}$ , we get $\frac{d x}{d θ} = a (- sin θ + \frac{1}{sin θ}) = a (\frac{1 - sin ^{2} θ}{sin θ})$ .
  ½ mark
3. So $\frac{d x}{d θ} = a \frac{cos ^{2} θ}{sin θ} = a cot θ cos θ$ .
  ½ mark
4. Also, $y = sin θ \Rightarrow \frac{d y}{d θ} = cos θ$ .
  ½ mark
5. $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{cos θ sin θ}{a cos ^{2} θ} = \frac{tan θ}{a}$ .
  1 mark
6. Differentiate again with respect to $x$ : $\frac{d ^{2} y}{d x ^{2}} = \frac{sec ^{2} θ}{a} \cdot \frac{d θ}{d x} = \frac{sec ^{2} θ}{a} \cdot \frac{sin θ}{a cos ^{2} θ} = \frac{sec ^{3} θ tan θ}{a ^{2}}$ .
  1 mark
7. At $θ = \frac{π}{4}$ : $sec \frac{π}{4} = 2$ and $tan \frac{π}{4} = 1$ . So $\frac{d ^{2} y}{d x ^{2}} = \frac{( 2 ) ^{3} \cdot 1}{a ^{2}} = \frac{2 2}{a ^{2}}$ .
  1 mark
Answer: $\frac{d ^{2} y}{d x ^{2}}_{θ = π /4} = \frac{2 2}{a ^{2}}$
Where the marks are
- Differentiates $x$ with respect to $θ$ using the chain rule (½ mark)
- Simplifies using $2 sin \frac{θ}{2} cos \frac{θ}{2} = sin θ$ (½ mark)
- Gets $\frac{d x}{d θ} = a cot θ cos θ$ (½ mark)
- Finds $\frac{d y}{d θ} = cos θ$ (½ mark)
- Gets $\frac{d y}{d x} = \frac{t a n θ}{a}$ (1 mark)
- Gets $\frac{d ^{2} y}{d x ^{2}} = \frac{s e c ^{3} θ t a n θ}{a ^{2}}$ (1 mark)
- Evaluates $\frac{2 2}{a ^{2}}$ at $θ = \frac{π}{4}$ (1 mark)
Open Q32(b) on its own page
Q33 · 5 marks · Long answerApplication of Derivatives
Find the absolute maximum and absolute minimum of function $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1$ on $[1, 5]$ .
Try it first, then show our solutionSolution
1. Differentiate: $f^{'} (x) = 6 x^{2} - 30 x + 36 = 6 (x^{2} - 5 x + 6) = 6 (x - 2) (x - 3)$ .
  1 mark
2. $f^{'} (x) = 0 \Rightarrow x = 2, 3$ . Both points lie in $[1, 5]$ .
  1 mark
3. Evaluate $f$ at the critical points and the end points:
  $f (1) = 2 - 15 + 36 + 1 = 24$
  $f (2) = 16 - 60 + 72 + 1 = 29$
  $f (3) = 54 - 135 + 108 + 1 = 28$
  $f (5) = 250 - 375 + 180 + 1 = 56$
  2 marks
4. Hence the absolute maximum value is 56 (at $x = 5$ ) and the absolute minimum value is 24 (at $x = 1$ ).
  1 mark
Answer: Absolute maximum = 56 at x = 5; absolute minimum = 24 at x = 1
Where the marks are
- Finds $f^{'} (x) = 6 (x - 2) (x - 3)$ (1 mark)
- Gets critical points $x = 2, 3$ in $[1, 5]$ (1 mark)
- Evaluates $f (1) = 24$ , $f (2) = 29$ , $f (3) = 28$ , $f (5) = 56$ (2 marks)
- States absolute maximum 56 and absolute minimum 24 (1 mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerThree Dimensional Geometry
Find the image A' of the point A(1, 6, 3) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Also, find the equation of the line joining A and A'.
Try it first, then show our solutionSolution
1. Write the line as $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = λ$ . A general point on it is $M (λ, 2 λ + 1, 3 λ + 2)$ . Take $M$ as the foot of the perpendicular from $A (1, 6, 3)$ .
  1 mark
2. Direction ratios of $A M$ are $(λ - 1, 2 λ - 5, 3 λ - 1)$ . Since $A M ⊥$ the line, whose d.r.'s are $(1, 2, 3)$ :
  $1 (λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0$
  1 mark
3. $14 λ - 14 = 0 \Rightarrow λ = 1$ , so the foot of the perpendicular is $M (1, 3, 5)$ .
  ½ mark
4. Let the image be $A^{'} (α, β, γ)$ . $M$ is the mid-point of $A A^{'}$ :
  $(\frac{1 + α}{2}, \frac{6 + β}{2}, \frac{3 + γ}{2}) = (1, 3, 5)$
  ½ mark
5. $α = 1, β = 0, γ = 7$ , so the image is $A^{'} (1, 0, 7)$ .
  1 mark
6. D.r.'s of $A A^{'}$ : $(1 - 1, 0 - 6, 7 - 3) = (0, - 6, 4)$ , i.e. proportional to $(0, - 3, 2)$ . The line through $A (1, 6, 3)$ is
  $\frac{x - 1}{0} = \frac{y - 6}{- 3} = \frac{z - 3}{2}$
  1 mark
Answer: $A^{'} (1, 0, 7)$ ; line $A A^{'}$ : $\frac{x - 1}{0} = \frac{y - 6}{- 3} = \frac{z - 3}{2}$
Where the marks are
- Writes the general point $M (λ, 2 λ + 1, 3 λ + 2)$ on the line (1 mark)
- Uses the perpendicularity condition $1 (λ - 1) + 2 (2 λ - 5) + 3 (3 λ - 1) = 0$ (1 mark)
- Gets $λ = 1$ and foot $M (1, 3, 5)$ (½ mark)
- Uses the mid-point condition for $A A^{'}$ (½ mark)
- Finds the image $A^{'} (1, 0, 7)$ (1 mark)
- Writes the equation of $A A^{'}$ (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerThree Dimensional Geometry
Find a point P on the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}$ such that its distance from point Q(2, 4, $- 1$ ) is 7 units. Also, find the equation of line joining P and Q.
Try it first, then show our solutionSolution
1. Write the line as $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} = λ$ . Any point on it is $P (λ - 5, 4 λ - 3, - 9 λ + 6)$ , and $Q$ is $(2, 4, - 1)$ .
  1 mark
2. Since $P Q = 7$ :
  $(λ - 7)^{2} + (4 λ - 7)^{2} + (- 9 λ + 7)^{2} = 7$
  1 mark
3. Squaring gives $98 λ^{2} - 196 λ + 147 = 49$ .
  So $98 (λ^{2} - 2 λ + 1) = 0 \Rightarrow (λ - 1)^{2} = 0 \Rightarrow λ = 1$ .
  1 mark
4. Hence the required point is $P (- 4, 1, - 3)$ .
  1 mark
5. D.r.'s of $P Q$ : $(2 - (- 4), 4 - 1, - 1 - (- 3)) = (6, 3, 2)$ . The line through $P$ is
  $\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}$
  (equivalently, through $Q$ : $\frac{x - 2}{6} = \frac{y - 4}{3} = \frac{z + 1}{2}$ ).
  1 mark
Answer: $P (- 4, 1, - 3)$ ; line $P Q$ : $\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}$
Where the marks are
- Writes the general point $P (λ - 5, 4 λ - 3, - 9 λ + 6)$ (1 mark)
- Uses the distance formula with $P Q = 7$ (1 mark)
- Simplifies to $98 (λ - 1)^{2} = 0$ , giving $λ = 1$ (1 mark)
- Finds $P (- 4, 1, - 3)$ (1 mark)
- Writes the equation of line $P Q$ (1 mark)
Open Q34(b) on its own page
Q35 · 5 marks · Long answerCompetency-basedMatrices
A school wants to allocate students into three clubs : Sports, Music and Drama, under following conditions :
- The number of students in Sports club should be equal to the sum of the number of students in Music and Drama club.
- The number of students in Music club should be 20 more than half the number of students in Sports club.
- The total number of students to be allocated in all three clubs are 180.
Find the number of students allocated to different clubs, using matrix method.
Try it first, then show our solutionSolution
1. Let $x, y, z$ be the number of students in the Sports, Music and Drama clubs respectively. The conditions give:
  $x = y + z, y = \frac{x}{2} + 20, x + y + z = 180$
  That is,
  $x - y - z = 0, x - 2 y = - 40, x + y + z = 180$
  1½ marks
2. Write the system as $A X = B$ , where
  $A = 111 - 1 - 2 1 - 1 01, X = x y z, B = 0 - 40 180$
  ½ mark
3. $∣ A ∣ = 1 (- 2 - 0) + 1 (1 - 0) - 1 (1 + 2) = - 4 \neq = 0$ , so $A^{- 1}$ exists.
  ½ mark
4. The cofactors are:
  $C_{11} = - 2, C_{12} = - 1, C_{13} = 3$
  $C_{21} = 0, C_{22} = 2, C_{23} = - 2$
  $C_{31} = - 2, C_{32} = - 1, C_{33} = - 1$
  Therefore
  $adj A = - 2 - 1 3 02 - 2 - 2 - 1 - 1$
  1 mark
5. $A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{4} 21 - 3 0 - 2 2 211$
  ½ mark
6. $X = A^{- 1} B = \frac{1}{4} 21 - 3 0 - 2 2 211 0 - 40 180 = \frac{1}{4} 360260100 = 906525$
  So $x = 90, y = 65, z = 25$ . The Sports, Music and Drama clubs get 90, 65 and 25 students respectively.
  1 mark
Answer: Sports = 90, Music = 65, Drama = 25
Where the marks are
- Forms the three linear equations correctly (1½ marks)
- Writes the system in the form $A X = B$ (½ mark)
- Finds $∣ A ∣ = - 4 \neq = 0$ (½ mark)
- Finds adj $A$ correctly (1 mark)
- Writes $A^{- 1}$ (½ mark)
- Computes $X = A^{- 1} B$ , giving 90, 65, 25 (1 mark)
Open Q35 on its own page

Section E

case study based

Q36 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.
Let the length of the side perpendicular to the partition be $x$ metres and with parallel to the partition be y metres.
Based on this information, answer the following questions :
1. (i)
  Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of $x$ and y.
  1 mark
2. (ii)
  Write the area of the solar panel as a function of $x$ .
  1 mark
3. (iii)(a)
  Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.
  2 marks
4. OR
  (iii)(b)
  Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.
  2 marks
Figure in the paper: Photograph of two rectangular solar panels installed side by side on a roof. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The two sides perpendicular to the partition use $2 x$ m of material. The two sides parallel to the partition, plus the partition itself, use $3 y$ m. Hence
  $2 x + 3 y = 300$
  1 mark
2. (ii) From (i), $y = \frac{300 - 2 x}{3}$ . So
  $A (x) = x y = \frac{x}{3} (300 - 2 x) = \frac{1}{3} (300 x - 2 x^{2})$
  1 mark
3. (iii)(a) $\frac{d A}{d x} = \frac{1}{3} (300 - 4 x)$
  ½ mark
4. (iii)(a) For critical points, $\frac{d A}{d x} = 0 \Rightarrow x = 75$ .
  ½ mark
5. (iii)(a) $\frac{d ^{2} A}{d x ^{2}} = - \frac{4}{3} < 0$ , so $A$ is maximum at $x = 75$ .
  ½ mark
6. (iii)(a) Maximum area $= \frac{75}{3} (300 - 150) = 25 \times 150 = 3750 m^{2}$ (with $y = 50$ m).
  ½ mark
7. (iii)(b) $A = \frac{1}{3} (300 x - 2 x^{2})$ , so $\frac{d A}{d x} = \frac{1}{3} (300 - 4 x)$ . Setting $\frac{d A}{d x} = 0$ gives the critical point $x = 75$ .
  1 mark
8. (iii)(b) For $x < 75$ , $\frac{d A}{d x} > 0$ , and for $x > 75$ , $\frac{d A}{d x} < 0$ . Since $\frac{d A}{d x}$ changes sign from positive to negative at $x = 75$ , the area is maximum there [first derivative test].
  ½ mark
9. (iii)(b) Maximum area $= \frac{75}{3} (300 - 150) = 3750 m^{2}$ .
  ½ mark
Answer: (i) $2 x + 3 y = 300$ ; (ii) $A = \frac{x}{3} (300 - 2 x)$ ; (iii) critical point $x = 75$ , maximum area $3750 m^{2}$
Where the marks are
- (i) Writes $2 x + 3 y = 300$ (1 mark)
- (ii) Writes $A = \frac{x}{3} (300 - 2 x)$ (1 mark)
- (iii)(a) Finds $A^{'} (x)$ (½ mark), $x = 75$ (½ mark), $A^{''} < 0$ (½ mark), maximum area 3750 m² (½ mark)
- (iii)(b) Finds $A^{'} (x)$ and $x = 75$ (½ + ½ mark), uses the sign change of $A^{'}$ (½ mark), maximum area 3750 m² (½ mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedRelations and Functions
A class-room teacher is keen to assess the learning of her students the concept of "relations" taught to them. She writes the following five relations each defined on the set $A = {1, 2, 3}$ :
$R_{1} = {(2, 3), (3, 2)}$
$R_{2} = {(1, 2), (1, 3), (3, 2)}$
$R_{3} = {(1, 2), (2, 1), (1, 1)}$
$R_{4} = {(1, 1), (1, 2), (3, 3), (2, 2)}$
$R_{5} = {(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)}$
The students are asked to answer the following questions about the above relations :
1. (i)
  Identify the relation which is reflexive, transitive but not symmetric.
  1 mark
2. (ii)
  Identify the relation which is reflexive and symmetric but not transitive.
  1 mark
3. (iii)(a)
  Identify the relations which are symmetric but neither reflexive nor transitive.
  2 marks
4. OR
  (iii)(b)
  What pairs should be added to the relation $R_{2}$ to make it an equivalence relation ?
  2 marks
Try it first, then show our solutionSolution
1. (i) $R_{4}$ contains $(1, 1), (2, 2), (3, 3)$ , so it is reflexive. Every chain stays inside $R_{4}$ (e.g. $(1, 1), (1, 2) \to (1, 2)$ and $(1, 2), (2, 2) \to (1, 2)$ ), so it is transitive. But $(1, 2) \in R_{4}$ while $(2, 1) \in / R_{4}$ , so it is not symmetric. Answer: $R_{4}$ .
  1 mark
2. (ii) $R_{5}$ contains all of $(1, 1), (2, 2), (3, 3)$ , so it is reflexive. Every pair in it has its reverse, so it is symmetric. But $(1, 2), (2, 3) \in R_{5}$ while $(1, 3) \in / R_{5}$ , so it is not transitive. Answer: $R_{5}$ .
  1 mark
3. (iii)(a) $R_{1}$ is symmetric. It is not reflexive since $(1, 1) \in / R_{1}$ . It is not transitive, because $(2, 3), (3, 2) \in R_{1}$ but $(2, 2) \in / R_{1}$ .
  1 mark
4. (iii)(a) $R_{3}$ is symmetric. It is not reflexive since $(2, 2) \in / R_{3}$ . It is not transitive, because $(2, 1), (1, 2) \in R_{3}$ but $(2, 2) \in / R_{3}$ . So the required relations are $R_{1}$ and $R_{3}$ .
  1 mark
5. (iii)(b) For reflexivity, add $(1, 1), (2, 2), (3, 3)$ . For symmetry, add the reverses $(2, 1), (3, 1), (2, 3)$ .
  1 mark
6. (iii)(b) The relation then becomes $A \times A$ , which is transitive and hence an equivalence relation. The pairs to add are $(1, 1), (2, 2), (3, 3), (2, 1), (3, 1), (2, 3)$ .
  1 mark
Answer: (i) $R_{4}$ ; (ii) $R_{5}$ ; (iii)(a) $R_{1}$ and $R_{3}$ ; (iii)(b) add $(1, 1), (2, 2), (3, 3), (2, 1), (3, 1), (2, 3)$
Where the marks are
- (i) Identifies $R_{4}$ (1 mark)
- (ii) Identifies $R_{5}$ (1 mark)
- (iii)(a) Identifies $R_{1}$ (1 mark) and $R_{3}$ (1 mark)
- (iii)(b) Lists all six pairs to be added (2 marks)
Open Q37 on its own page
Q38 · 4 marks · Case-basedCompetency-basedProbability
A bank offers loan to its customers on different types of interest namely, fixed rate, floating rate and variable rate. From the past data with the bank, it is known that a customer avails loan on fixed rate, floating rate or variable rate with probabilities 10%, 20% and 70% respectively. A customer after availing loan can pay the loan or default on loan repayment. The bank data suggests that the probability that a person defaults on loan after availing it at fixed rate, floating rate and variable rate is 5%, 3% and 1% respectively.
Based on the above information, answer the following :
1. (i)
  What is the probability that a customer after availing the loan will default on the loan repayment ?
  2 marks
2. (ii)
  A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest ?
  2 marks
Figure in the paper: Illustration of stacks of coins decreasing in height with an arrow labelled 'INTEREST RATES' and a % sign. See it in the official paper.
Try it first, then show our solutionSolution
1. Define the events:
  $E_{1}$ : the loan is at fixed rate
  $E_{2}$ : the loan is at floating rate
  $E_{3}$ : the loan is at variable rate
  $A$ : the customer defaults
  Then $P (E_{1}) = \frac{1}{10}, P (E_{2}) = \frac{2}{10}, P (E_{3}) = \frac{7}{10}$ and $P (A ∣ E_{1}) = \frac{5}{100}, P (A ∣ E_{2}) = \frac{3}{100}, P (A ∣ E_{3}) = \frac{1}{100}$ .
2. (i) By the theorem of total probability:
  $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) + P (E_{3}) P (A ∣ E_{3})$
  $= \frac{1}{10} \cdot \frac{5}{100} + \frac{2}{10} \cdot \frac{3}{100} + \frac{7}{10} \cdot \frac{1}{100}$
  1 mark
3. (i) $P (A) = \frac{5 + 6 + 7}{1000} = \frac{18}{1000} = \frac{9}{500}$
  1 mark
4. (ii) By Bayes' theorem:
  $P (E_{3} ∣ A) = \frac{P ( E _{3} ) P ( A ∣ E _{3} )}{P ( A )} = \frac{\frac{7}{10} \cdot \frac{1}{100}}{\frac{18}{1000}}$
  1 mark
5. (ii) $P (E_{3} ∣ A) = \frac{7}{18}$
  1 mark
Answer: (i) $\frac{9}{500}$ (= 0.018); (ii) $\frac{7}{18}$
Where the marks are
- (i) Applies total probability with the correct values (1 mark)
- (i) Obtains $\frac{18}{1000} = \frac{9}{500}$ (1 mark)
- (ii) Applies Bayes' theorem correctly (1 mark)
- (ii) Obtains $\frac{7}{18}$ (1 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.