MathsCBSE Class 12 · Board exam 2026 (session 2025-26)

CBSE Class 12 Mathematics: 2026 board paper, set 65/5/1

The question paper set in the CBSE Class 12 board exam of 2026, 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/2026/XII/Mathematics.zip (a zip; this paper is the file 65-5-1 R.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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

How the paper is set

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

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQMatrices
If matrix $A = [- p r q p]$ is such that $A^{2} = I$ , then :
1. (A) $1 + p^{2} + q r = 0$
2. (B) $1 - p^{2} - q r = 0$
3. (C) $1 - p^{2} + q r = 0$
4. (D) $1 + p^{2} - q r = 0$
Try it first, then show our solutionSolution
1. $A^{2} = [p^{2} + q r 0 0 q r + p^{2}] = I \Rightarrow p^{2} + q r = 1$ , i.e. $1 - p^{2} - q r = 0$ . Option (B).
  1 mark
Answer: $1 - p^{2} - q r = 0$ , option (B)
Where the marks are
- Squares A and equates diagonal entry to 1 (1 mark)
Q2 · 1 mark · MCQMatrices
If A is a square matrix such that $A^{2} = A$ , then $(A - I)^{3} - A$ is equal to :
1. (A)I
2. (B) $- I$
3. (C)A
4. (D) $A^{2}$
Try it first, then show our solutionSolution
1. $(A - I)^{3} = A^{3} - 3 A^{2} + 3 A - I$ . Since $A^{2} = A$ , $A^{3} = A$ , so $(A - I)^{3} = A - 3 A + 3 A - I = A - I$ .
  ½ mark
2. $(A - I)^{3} - A = - I$ . Option (B).
  ½ mark
Answer: $- I$ , option (B)
Where the marks are
- Expands and uses $A^{n} = A$ (½ mark)
- Gets $- I$ (½ mark)
Q3 · 1 mark · MCQInverse Trigonometric Functions
For the inverse trigonometric functions, which of the following Principal Value Branch is not correctly defined ?
1. (A) $tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2})$
2. (B) $sec^{- 1} : R - (- 1, 1) \to [0, π] - {\frac{π}{2}}$
3. (C) $cot^{- 1} : R \to (0, π)$
4. (D) $cosec^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}]$
Try it first, then show our solutionSolution
1. cosec x is never defined as 0 → value, so $cosec^{- 1}$ range must exclude 0: $[- \frac{π}{2}, \frac{π}{2}] - {0}$ . So (D) is wrongly defined.
  1 mark
Answer: Option (D)
Where the marks are
- Notes 0 must be excluded from range of $cosec^{- 1}$ (1 mark)
Q4 · 1 mark · MCQMatrices
Let $A = [01 - 3 0 42]$ and $B = [- 3 2 0410]$ . If $A + B + C = O$ , then matrix C is :
1. (A) $[- 3 3 - 3 4 52]$
2. (B) $[3 - 3 3 - 4 5 - 2]$
3. (C) $[3 - 3 3 - 4 - 5 - 2]$
4. (D) $[- 3 3 - 3 4 - 5 2]$
Try it first, then show our solutionSolution
1. $A + B = [- 3 3 - 3 4 52]$ , so $C = - (A + B) = [3 - 3 3 - 4 - 5 - 2]$ . Option (C).
  1 mark
Answer: Option (C)
Where the marks are
- C = −(A+B) computed (1 mark)
Q5 · 1 mark · MCQMatrices
If A is a non-singular matrix, then which of the following is not true ?
1. (A)adj A is singular
2. (B) $(adj A)^{- 1} = (adj A^{- 1})$
3. (C) $∣ A ∣ \neq = 0$
4. (D) $A^{- 1}$ exists
Try it first, then show our solutionSolution
1. $∣ adj A ∣ = ∣ A ∣^{n - 1} \neq = 0$ , so adj A is non-singular; statement (A) is false. Others are true.
  1 mark
Answer: Option (A)
Where the marks are
- Uses $∣ adj A ∣ = ∣ A ∣^{n - 1} \neq = 0$ (1 mark)
Q6 · 1 mark · MCQContinuity and Differentiability
If $f (x) = {\frac{x ^{2} - 4 x - 5}{x + 1}, k, x \neq = - 1 x = - 1$ is continuous at $x = - 1$ , then the value of k is :
1. (A)Any real value
2. (B)6
3. (C) $- 1$
4. (D) $- 6$
Try it first, then show our solutionSolution
1. $\frac{( x - 5 ) ( x + 1 )}{x + 1} = x - 5 \to - 6$ as $x \to - 1$ ; continuity gives $k = - 6$ . Option (D).
  1 mark
Answer: k = −6, option (D)
Where the marks are
- Factorises and finds limit −6 (1 mark)
Q7 · 1 mark · MCQDeterminants
If the area of $Δ$ ABC with vertices A(3, 1), B( $- 2$ , 1) and C(0, k) is 5 sq. units, then values of k are :
1. (A)3, 1
2. (B) $- 1$ , 3
3. (C) $- 1$ , 2
4. (D)0, 2
Try it first, then show our solutionSolution
1. Base AB = 5 (on y = 1), height $∣ k - 1∣$ . $\frac{1}{2} \cdot 5∣ k - 1∣ = 5 \Rightarrow ∣ k - 1∣ = 2 \Rightarrow k = 3, - 1$ . Option (B).
  1 mark
Answer: k = −1, 3, option (B)
Where the marks are
- Area equation gives $∣ k - 1∣ = 2$ (1 mark)
Q8 · 1 mark · MCQContinuity and Differentiability
Derivative of $cos^{- 1} (\frac{s i n x + c o s x}{2})$ , $- \frac{π}{4} < x < \frac{π}{4}$ with respect to x is :
1. (A) $- 1$
2. (B)1
3. (C) $\frac{π}{4}$
4. (D) $- \frac{π}{4}$
Try it first, then show our solutionSolution
1. $\frac{s i n x + c o s x}{2} = cos (x - \frac{π}{4})$ , and $x - \frac{π}{4} \in (- \frac{π}{2}, 0)$ , so $cos^{- 1} = \frac{π}{4} - x$ .
  ½ mark
2. Derivative $= - 1$ . Option (A).
  ½ mark
Answer: −1, option (A)
Where the marks are
- Simplifies to $\frac{π}{4} - x$ (½ mark)
- Derivative −1 (½ mark)
Q9 · 1 mark · MCQApplication of Derivatives
Absolute minimum value of $f (x) = (x - 2)^{2} + 5$ in the interval $[- 3, 2]$ is :
1. (A) $- 3$
2. (B)2
3. (C)5
4. (D)30
Try it first, then show our solutionSolution
1. $(x - 2)^{2} \geq 0$ with equality at x = 2 ∈ [−3, 2], so minimum = 5 (f(−3)=30 is max). Option (C).
  1 mark
Answer: 5, option (C)
Where the marks are
- Minimum at x = 2 equals 5 (1 mark)
Q10 · 1 mark · MCQIntegrals
$\int \frac{1}{1 + c o s 2 x} d x$ is equal to :
1. (A) $lo g cos x + C$
2. (B) $\frac{1}{2} lo g ∣ sec x + tan x ∣ + C$
3. (C) $\frac{1}{2} lo g ∣ sec x - tan x ∣ + C$
4. (D) $lo g sin 2 x + C$
Try it first, then show our solutionSolution
1. $1 + cos 2 x = 2 cos^{2} x$ , so integrand $= \frac{1}{2} sec x$ ; integral $= \frac{1}{2} lo g ∣ sec x + tan x ∣ + C$ . Option (B).
  1 mark
Answer: Option (B)
Where the marks are
- Uses $1 + cos 2 x = 2 cos^{2} x$ and integrates sec x (1 mark)
Q11 · 1 mark · MCQIntegrals
The value of $\int_{- 5}^{- 1} \frac{1}{x} d x$ is equal to :
1. (A) $- lo g 5$
2. (B) $x^{6}$
3. (C) $lo g (- 5)$
4. (D) $x^{- 6}$
Try it first, then show our solutionSolution
1. $[lo g ∣ x ∣]_{- 5}^{- 1} = lo g 1 - lo g 5 = - lo g 5$ . Option (A).
  1 mark
Answer: $- lo g 5$ , option (A)
Where the marks are
- Uses $lo g ∣ x ∣$ correctly (1 mark)
Q12 · 1 mark · MCQCompetency-basedApplication of Integrals
An ant is observed crawling on a sheet of paper along a straight line given by equation $y = 2 x - 4$ . Area of the surface covered by the ant bounded by y-axis, x-axis and $x = 1$ is :
1. (A)1 sq. unit
2. (B)3 sq. units
3. (C)2 sq. units
4. (D)4 sq. units
Try it first, then show our solutionSolution
1. On [0,1], y = 2x − 4 < 0, so area $= \int_{0}^{1} (4 - 2 x) d x = [4 x - x^{2}]_{0}^{1} = 3$ . Option (B).
  1 mark
Answer: 3 sq. units, option (B)
Where the marks are
- Integrates |2x−4| from 0 to 1 = 3 (1 mark)
Q13 · 1 mark · MCQDifferential Equations
The order and degree of the differential equation $1 + (\frac{d ^{3} y}{d x ^{3}})^{3} = λ \frac{d ^{2} y}{d x ^{2}}$ is :
1. (A)Order = 3, Degree = 3
2. (B)Order = 2, Degree = 2
3. (C)Order = 3, Degree = 1
4. (D)Order = 2, Degree = 1
Try it first, then show our solutionSolution
1. Highest derivative $\frac{d ^{3} y}{d x ^{3}}$ (order 3), its power is 3 (degree 3). Option (A).
  1 mark
Answer: Order 3, degree 3, option (A)
Where the marks are
- Identifies order 3 and degree 3 (1 mark)
Q14 · 1 mark · MCQDifferential Equations
The general solution for the differential equation $\frac{d y}{d x} = e^{3 x - y}$ is :
1. (A) $3 e^{y} = e^{3 x} + C$
2. (B) $lo g (3 x - y) = C$
3. (C) $e^{3 x - y} = C$
4. (D) $- e^{y} + 3 e^{3 x} = C$
Try it first, then show our solutionSolution
1. $e^{y} d y = e^{3 x} d x \Rightarrow e^{y} = \frac{e ^{3 x}}{3} + C_{1} \Rightarrow 3 e^{y} = e^{3 x} + C$ . Option (A).
  1 mark
Answer: $3 e^{y} = e^{3 x} + C$ , option (A)
Where the marks are
- Separates variables and integrates (1 mark)
Q15 · 1 mark · MCQLinear Programming
The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40) (60, 20) and (60, 0). If the objective function of an LPP is $Z = 4 x + 3 y$ , then the maximum value is :
1. (A)200
2. (B)300
3. (C)240
4. (D)120
Try it first, then show our solutionSolution
1. Z: (0,0)=0, (0,40)=120, (20,40)=200, (60,20)=300, (60,0)=240. Max = 300. Option (B).
  1 mark
Answer: 300, option (B)
Where the marks are
- Evaluates Z at corners, max 300 (1 mark)
Q16 · 1 mark · MCQVector Algebra
If position vector $p$ of a point (24, n) is such that $∣ p ∣ = 25$ , then the value of n is :
1. (A) $\pm 49$
2. (B) $\pm 5$
3. (C) $\pm 1$
4. (D) $\pm 7$
Try it first, then show our solutionSolution
1. $2 4^{2} + n^{2} = 625 \Rightarrow n^{2} = 49 \Rightarrow n = \pm 7$ . Option (D).
  1 mark
Answer: ±7, option (D)
Where the marks are
- Uses magnitude formula (1 mark)
Q17 · 1 mark · MCQCompetency-basedVector Algebra
If vectors $a = 3 \hat{i} + 2 \hat{j} + λ \hat{k}$ and $b = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ , represent the two strips of the Red Cross sign placed outside a doctor's clinic, then the value of $λ$ is :
1. (A)1
2. (B) $\frac{5}{2}$
3. (C) $\frac{2}{5}$
4. (D)0
Try it first, then show our solutionSolution
1. Red Cross strips are perpendicular: $a \cdot b = 6 - 8 + 5 λ = 0 \Rightarrow λ = \frac{2}{5}$ . Option (C).
  1 mark
Answer: λ = 2/5, option (C)
Where the marks are
- Sets dot product zero (1 mark)
Q18 · 1 mark · MCQProbability
If $3 P (A) = P (B) = \frac{3}{5}$ and $P (A ∣ B) = \frac{2}{3}$ , then $P (A \cup B)$ is :
1. (A) $\frac{3}{5}$
2. (B) $\frac{1}{5}$
3. (C) $\frac{2}{15}$
4. (D) $\frac{2}{5}$
Try it first, then show our solutionSolution
1. P(A)=1/5, P(B)=3/5, P(A∩B)=(2/3)(3/5)=2/5.
  ½ mark
2. P(A∪B)=1/5+3/5−2/5=2/5. Option (D).
  ½ mark
Answer: 2/5, option (D)
Where the marks are
- Finds P(A∩B) (½ mark)
- Addition rule gives 2/5 (½ mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedRelations and Functions
Assertion (A) : A relation R on the set {1, 2, 3} defined as R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} is an equivalence relation.
Reason (R) : A relation that is reflexive, symmetric and transitive is an equivalence relation.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. R contains (1,1),(2,2),(3,3): reflexive; (1,2),(2,1) both present: symmetric; all compositions stay in R: transitive. So A true.
  ½ mark
2. R is the definition of equivalence relation and explains A. Option (A).
  ½ mark
Answer: Option (A)
Where the marks are
- Checks R reflexive, symmetric, transitive (½ mark)
- Recognises R explains A (½ mark)
Open Q19 on its own page
Q20 · 1 mark · Assertion-reasonCompetency-basedLinear Programming
Assertion (A) : Consider a Linear Programming Problem with minimise $Z = x + 2 y$ subject to constraints $2 x + y \geq 3$ , $x + 2 y \geq 6$ , $x, y \geq 0$ which gives minimum Z at infinitely many points. The corner points of feasible region are (0, 3) and (6, 0).
Reason (R) : If two corner points produce the same minimum value of the objective function, then every point on the line segment joining the points will give the same minimum value.
1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3. (C)Assertion (A) is true, but Reason (R) is false.
4. (D)Assertion (A) is false, but Reason (R) is true.
Try it first, then show our solutionSolution
1. Unbounded region with corners (0,3) and (6,0): Z=6 at both. Half-plane x+2y<6 has no common point with region, so min 6 holds on whole segment — infinitely many points. A true.
  ½ mark
2. R is the standard result that explains A. Option (A).
  ½ mark
Answer: Option (A)
Where the marks are
- Z = 6 at both corners, min verified (½ mark)
- R explains A (½ mark)
Open Q20 on its own page

Section B

VSA

Q21 · 2 marks · Very short answerInverse Trigonometric Functions
Evaluate $sin [tan^{- 1} tan (\frac{3 π}{4})]$ .
Try it first, then show our solutionSolution
1. $tan \frac{3 π}{4} = tan (π - \frac{π}{4}) = - 1$ , so the expression is $sin [tan^{- 1} (- 1)]$ .
  1 mark
2. The principal value branch of $tan^{- 1}$ is $(- \frac{π}{2}, \frac{π}{2})$ , so $tan^{- 1} (- 1) = - \frac{π}{4}$ . The expression becomes $sin (- \frac{π}{4})$ .
  ½ mark
3. $sin (- \frac{π}{4}) = - \frac{1}{2}$
  ½ mark
Answer: $- \frac{1}{2}$
Where the marks are
- Writes $tan \frac{3 π}{4} = - 1$ , i.e. $tan (π - \frac{π}{4})$ (1 mark)
- Uses the principal branch to get $tan^{- 1} (- 1) = - \frac{π}{4}$ (½ mark)
- Final value $- \frac{1}{2}$ (½ mark)
Open Q21 on its own page
Answer one of these 2 alternatives.
Q22(a) · 2 marks · Very short answerContinuity and Differentiability
Differentiate $x^{x}$ with respect to $x lo g x$ .
Try it first, then show our solutionSolution
1. Let $u = x^{x}$ and $v = x lo g x$ . We need $\frac{d u}{d v}$ .
  ½ mark
2. $lo g u = x lo g x \Rightarrow \frac{1}{u} \frac{d u}{d x} = 1 + lo g x \Rightarrow \frac{d u}{d x} = x^{x} (1 + lo g x)$
  ½ mark
3. $\frac{d v}{d x} = x \cdot \frac{1}{x} + lo g x = 1 + lo g x$
  ½ mark
4. $\frac{d u}{d v} = \frac{d u / d x}{d v / d x} = \frac{x ^{x} ( 1 + l o g x )}{1 + l o g x} = x^{x}$
  ½ mark
Answer: $\frac{d ( x ^{x} )}{d ( x l o g x )} = x^{x}$
Where the marks are
- Takes $u = x^{x}$ , $v = x lo g x$ (½ mark)
- $\frac{d u}{d x} = x^{x} (1 + lo g x)$ (½ mark)
- $\frac{d v}{d x} = 1 + lo g x$ (½ mark)
- $\frac{d u}{d v} = x^{x}$ (½ mark)
Open Q22(a) on its own page
OR
Q22(b) · 2 marks · Very short answerContinuity and Differentiability
If $y = P cos ux + Q sin ux$ , show that $\frac{d ^{2} y}{d x ^{2}} + u^{2} y = 0$ .
Try it first, then show our solutionSolution
1. $\frac{d y}{d x} = - P u sin ux + Q u cos ux$
  ½ mark
2. $\frac{d ^{2} y}{d x ^{2}} = - P u^{2} cos ux - Q u^{2} sin ux$
  ½ mark
3. $\frac{d ^{2} y}{d x ^{2}} = - u^{2} (P cos ux + Q sin ux) = - u^{2} y$
  ½ mark
4. Therefore $\frac{d ^{2} y}{d x ^{2}} + u^{2} y = 0$ . Hence proved.
  ½ mark
Answer: $\frac{d ^{2} y}{d x ^{2}} = - u^{2} y$ , so $\frac{d ^{2} y}{d x ^{2}} + u^{2} y = 0$ (proved)
Where the marks are
- Correct first derivative (½ mark)
- Correct second derivative (½ mark)
- Factorises as $- u^{2} (P cos ux + Q sin ux)$ (½ mark)
- Concludes the required result (½ mark)
Open Q22(b) on its own page
Q23 · 2 marks · Very short answerApplication of Derivatives
Determine the values of x for which $f (x) = \frac{x - 3}{x + 1}$ , $x \neq = - 1$ is an increasing function.
Try it first, then show our solutionSolution
1. By the quotient rule, $f^{'} (x) = \frac{( x + 1 ) \cdot 1 - ( x - 3 ) \cdot 1}{( x + 1 ) ^{2}}$
  1 mark
2. $f^{'} (x) = \frac{4}{( x + 1 ) ^{2}} > 0$ for every $x \neq = - 1$ , since $(x + 1)^{2} > 0$ .
  ½ mark
3. Hence $f$ is increasing for all $x \in R - {- 1}$ , i.e. on $(- \infty, - 1)$ and on $(- 1, \infty)$ .
  ½ mark
Answer: $f$ is increasing for all $x \in R - {- 1}$
Where the marks are
- Correct quotient-rule derivative (1 mark)
- Simplifies to $\frac{4}{( x + 1 ) ^{2}} > 0$ (½ mark)
- Concludes increasing on $R - {- 1}$ (½ mark)
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Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerCompetency-basedVector Algebra
Three honey bees were found flying along the vectors $a = 2 \hat{i} - 3 \hat{j} + \hat{k}$ , $b = 4 \hat{j} - 2 \hat{k}$ and $c = 3 \hat{i} + 2 \hat{k}$ respectively. Find the value of $λ$ such that the path for $a + λ b$ is perpendicular to $c$ .
Try it first, then show our solutionSolution
1. $a + λ b ⊥ c \Rightarrow (a + λ b) \cdot c = 0$ , i.e. $a \cdot c + λ b \cdot c = 0$
  ½ mark
2. $a \cdot c = 2 (3) + (- 3) (0) + 1 (2) = 8$ and $b \cdot c = 0 (3) + 4 (0) + (- 2) (2) = - 4$ , so $8 - 4 λ = 0$
  1 mark
3. $λ = 2$
  ½ mark
Answer: $λ = 2$
Where the marks are
- Uses the perpendicularity condition $(a + λ b) \cdot c = 0$ (½ mark)
- Gets $8 - 4 λ = 0$ (1 mark)
- $λ = 2$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerVector Algebra
If A, B and C be three non-collinear points such that $A B = \hat{i} + 2 \hat{j} - \hat{k}$ and $A C = 2 \hat{i} - 3 \hat{j}$ , then find the area of $Δ$ ABC.
Try it first, then show our solutionSolution
1. Area of $Δ A B C = \frac{1}{2} ∣ A B \times A C ∣$ .
  $A B \times A C = \hat{i} 12 \hat{j} 2 - 3 \hat{k} - 1 0 = \hat{i} (0 - 3) - \hat{j} (0 + 2) + \hat{k} (- 3 - 4) = - 3 \hat{i} - 2 \hat{j} - 7 \hat{k}$
  1 mark
2. $∣ A B \times A C ∣ = 9 + 4 + 49 = 62$ , so the area is $\frac{62}{2}$ sq. units.
  1 mark
Answer: $\frac{62}{2}$ sq. units
Where the marks are
- Uses the area formula and finds $A B \times A C = - 3 \hat{i} - 2 \hat{j} - 7 \hat{k}$ (1 mark)
- Area $= \frac{62}{2}$ sq. units (1 mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerThree Dimensional Geometry
Find the angle between the following pair of lines :
$\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{1 - z}{- 6}$ and
$\frac{x - 7}{1} = \frac{y}{2} = \frac{6 - z}{- 2}$
Try it first, then show our solutionSolution
1. Write the lines in standard form: $\frac{x - 2}{3} = \frac{y + 5}{2} = \frac{z - 1}{6}$ and $\frac{x - 7}{1} = \frac{y}{2} = \frac{z - 6}{2}$ . Their direction ratios are $⟨ 3, 2, 6 ⟩$ and $⟨ 1, 2, 2 ⟩$ .
  ½ mark
2. $cos θ = \frac{∣3 ( 1 ) + 2 ( 2 ) + 6 ( 2 ) ∣}{9 + 4 + 36 1 + 4 + 4} = \frac{19}{7 \times 3} = \frac{19}{21}$
  1 mark
3. $θ = cos^{- 1} (\frac{19}{21})$
  ½ mark
Answer: $θ = cos^{- 1} (\frac{19}{21})$
Where the marks are
- Correct direction ratios $⟨ 3, 2, 6 ⟩$ , $⟨ 1, 2, 2 ⟩$ (½ mark)
- Correct substitution in the cosine formula (1 mark)
- $θ = cos^{- 1} (19/21)$ (½ mark)
Open Q25 on its own page

Section C

Q26 · 3 marks · Short answerCompetency-basedDifferential Equations
A spherical balloon loses its volume due to escape of air from it in such a way that decrease of volume at any instant is proportional to its surface area. Show that the radius is decreasing at a constant rate.
Try it first, then show our solutionSolution
1. Let $r$ be the radius and $V = \frac{4}{3} π r^{3}$ the volume at time $t$ . The volume decreases at a rate proportional to the surface area, so $\frac{d V}{d t} = - k (4 π r^{2})$ , where $k > 0$ is a constant.
  1 mark
2. Differentiating $V = \frac{4}{3} π r^{3}$ with respect to $t$ : $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ . Hence $4 π r^{2} \frac{d r}{d t} = - 4 k π r^{2}$ .
  1 mark
3. Therefore $\frac{d r}{d t} = - k$ , which is a negative constant. So the radius decreases at a constant rate.
  1 mark
Answer: $\frac{d r}{d t} = - k$ (a constant), so the radius decreases at a constant rate
Where the marks are
- Forms $\frac{d V}{d t} = - k (4 π r^{2})$ with $k > 0$ (1 mark)
- Uses $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ and equates the two (1 mark)
- Gets $\frac{d r}{d t} = - k$ and concludes (1 mark)
Open Q26 on its own page
Answer one of these 2 alternatives.
Q27(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. Split the integral: $I = \int \frac{x}{1 - c o s x} d x - \int \frac{s i n x}{1 - c o s x} d x$
  ½ mark
2. Use $1 - cos x = 2 sin^{2} \frac{x}{2}$ and $sin x = 2 sin \frac{x}{2} cos \frac{x}{2}$ :
  $I = \frac{1}{2} \int x cosec^{2} \frac{x}{2} d x - \int cot \frac{x}{2} d x$
  1 mark
3. Integrate the first term by parts, taking $x$ as the first function and $\int cosec^{2} \frac{x}{2} d x = - 2 cot \frac{x}{2}$ :
  $I = \frac{1}{2} [- 2 x cot \frac{x}{2} - \int (- 2 cot \frac{x}{2}) d x] - \int cot \frac{x}{2} d x = - x cot \frac{x}{2} + \int cot \frac{x}{2} d x - \int cot \frac{x}{2} d x$
  1 mark
4. The two remaining integrals cancel, so $I = - x cot \frac{x}{2} + C$ .
  ½ mark
Answer: $- x cot \frac{x}{2} + C$
Where the marks are
- Splits the integrand into two integrals (½ mark)
- Converts using half-angle formulae (1 mark)
- Applies integration by parts correctly (1 mark)
- Final answer $- x cot \frac{x}{2} + C$ (½ mark)
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Q27(b) · 3 marks · Short answerIntegrals
Evaluate :
$\int_{0}^{2} \frac{1}{x ^{2} + 2 x + 3} d x$
Try it first, then show our solutionSolution
1. Complete the square: $x^{2} + 2 x + 3 = (x + 1)^{2} + (2)^{2}$ , so $I = \int_{0}^{2} \frac{d x}{( x + 1 ) ^{2} + ( 2 ) ^{2}}$
  1 mark
2. Using $\int \frac{d x}{t ^{2} + a ^{2}} = lo g ∣ t + t^{2} + a^{2} ∣$ :
  $I = [lo g (x + 1 + x^{2} + 2 x + 3)]_{0}^{2}$
  1 mark
3. $I = lo g (3 + 11) - lo g (1 + 3) = lo g (\frac{3 + 11}{1 + 3})$
  1 mark
Answer: $lo g (\frac{3 + 11}{1 + 3})$
Where the marks are
- Completes the square as $(x + 1)^{2} + 2$ (1 mark)
- Correct antiderivative $lo g (x + 1 + x^{2} + 2 x + 3)$ (1 mark)
- Correct evaluation at the limits (1 mark)
Open Q27(b) on its own page
Q28 · 3 marks · Short answerDifferential Equations
Solve the differential equation $(x + 2 y^{3}) d y = y d x$ .
Try it first, then show our solutionSolution
1. Rewrite as a linear equation in $x$ : $\frac{d x}{d y} = \frac{x + 2 y ^{3}}{y}$ , i.e. $\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ . Here $P = - \frac{1}{y}$ and $Q = 2 y^{2}$ .
  ½ mark
2. I.F. $= e^{\int - \frac{1}{y} d y} = e^{- l o g y} = \frac{1}{y}$ (taking $y > 0$ )
  1 mark
3. The solution is $x \cdot \frac{1}{y} = \int 2 y^{2} \cdot \frac{1}{y} d y + C = \int 2 y d y + C$
  1 mark
4. $\frac{x}{y} = y^{2} + C$ , i.e. $x = y^{3} + C y$
  ½ mark
Answer: $x = y^{3} + C y$ (or $\frac{x}{y} = y^{2} + C$ )
Where the marks are
- Writes the equation as $\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ (½ mark)
- Integrating factor $\frac{1}{y}$ (1 mark)
- Writes the solution form $x \cdot \frac{1}{y} = \int 2 y d y + C$ (1 mark)
- Final solution $x = y^{3} + C y$ (½ mark)
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Q29 · 3 marks · Short answerLinear Programming
Solve the following Linear Programming Problem graphically :
Maximize $Z = \frac{2 x}{5} + \frac{3 y}{10}$
subject to constraints
$2 x + y \leq 1000$
$x + y \leq 800$
$x, y \geq 0$ .
Try it first, then show our solutionSolution
1. Draw the lines $2 x + y = 1000$ (through $(500, 0)$ and $(0, 1000)$ ) and $x + y = 800$ (through $(800, 0)$ and $(0, 800)$ ). The lines meet where $x = 200$ , $y = 600$ .
  The feasible region is bounded and lies in the first quadrant, below both lines (towards the origin). It is shaded with vertices $O (0, 0)$ , $B (500, 0)$ , $C (200, 600)$ and $D (0, 800)$ .
  1½ marks
2. Evaluate $Z = \frac{2 x}{5} + \frac{3 y}{10}$ at the corner points:
  Corner point $Z$
  $(0, 0)$ 0
  $(500, 0)$ 200
  $(200, 600)$ 80 + 180 = 260
  $(0, 800)$ 240
  1 mark
3. Maximum $Z = 260$ at $x = 200$ , $y = 600$ .
  ½ mark
Answer: $Z_{m a x} = 260$ at $(200, 600)$
Where the marks are
- Correct lines and shaded feasible region with its corner points (1½ marks)
- Correct values of Z at all corner points (1 mark)
- Maximum 260 at (200, 600) (½ mark)
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Answer one of these 2 alternatives.
Q30(a) · 3 marks · Short answerCompetency-basedVector Algebra
Let three toys A, B and C be placed in the same straight line. If the position vectors of A, B and C are $55 \hat{i} - 2 \hat{j}$ , $5 \hat{i} + 8 \hat{j}$ and $a \hat{i} - 52 \hat{j}$ respectively, find the value of 'a'.
Try it first, then show our solutionSolution
1. $A B = (5 - 55) \hat{i} + (8 + 2) \hat{j} = - 50 \hat{i} + 10 \hat{j}$
  $B C = (a - 5) \hat{i} + (- 52 - 8) \hat{j} = (a - 5) \hat{i} - 60 \hat{j}$
  1 mark
2. A, B and C lie on one line, so $A B$ and $B C$ are collinear and their components are proportional: $\frac{- 50}{a - 5} = \frac{10}{- 60}$
  1 mark
3. $10 (a - 5) = 3000 \Rightarrow a - 5 = 300 \Rightarrow a = 305$
  1 mark
Answer: $a = 305$
Where the marks are
- Correct $A B$ and $B C$ (1 mark)
- Uses the proportionality condition for collinear vectors (1 mark)
- $a = 305$ (1 mark)
Open Q30(a) on its own page
OR
Q30(b) · 3 marks · Short answerVector Algebra
If $a$ , $b$ and $c$ are unit vectors, then prove that $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} \leq 9$ .
Try it first, then show our solutionSolution
1. Since the vectors are unit vectors, $∣ a ∣ = ∣ b ∣ = ∣ c ∣ = 1$ .
  ½ mark
2. Expand each square using $∣ p - q ∣^{2} = ∣ p ∣^{2} + ∣ q ∣^{2} - 2 p \cdot q$ :
  $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 2∣ a ∣^{2} + 2∣ b ∣^{2} + 2∣ c ∣^{2} - 2 (a \cdot b + b \cdot c + c \cdot a)$
  1½ marks
3. Rewrite this as $3 (∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2}) - (∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 a \cdot b + 2 b \cdot c + 2 c \cdot a) = 9 - ∣ a + b + c ∣^{2}$
  ½ mark
4. Since $∣ a + b + c ∣^{2} \geq 0$ , it follows that $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} \leq 9$ . Hence proved.
  ½ mark
Answer: LHS $= 9 - ∣ a + b + c ∣^{2} \leq 9$ (proved)
Where the marks are
- States the magnitudes are 1 (½ mark)
- Correct expansion of the three squares (1½ marks)
- Expresses the sum as $9 - ∣ a + b + c ∣^{2}$ (½ mark)
- Concludes the sum is at most 9 (½ mark)
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Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerProbability
A die is rolled. Consider events :
A = {1, 2, 5}, B = {3, 5}, C = {2, 3, 4, 5}
and hence find :
1. (i)
  $P (A ∣ C)$ and $P (C ∣ A)$
  0 marks
2. (ii)
  $P (A \cap B ∣ C)$ and $P (A \cup B ∣ C)$
  0 marks
Try it first, then show our solutionSolution
1. (i) $A \cap C = {2, 5}$ and $n (C) = 4$ , so $P (A ∣ C) = \frac{n ( A \cap C )}{n ( C )} = \frac{2}{4} = \frac{1}{2}$
  1 mark
2. (i) $n (A) = 3$ , so $P (C ∣ A) = \frac{n ( C \cap A )}{n ( A )} = \frac{2}{3}$
  ½ mark
3. (ii) $A \cap B = {5}$ and $(A \cap B) \cap C = {5}$ , so $P (A \cap B ∣ C) = \frac{1}{4}$
  1 mark
4. (ii) $A \cup B = {1, 2, 3, 5}$ and $(A \cup B) \cap C = {2, 3, 5}$ , so $P (A \cup B ∣ C) = \frac{3}{4}$
  ½ mark
Answer: (i) $P (A ∣ C) = \frac{1}{2}$ , $P (C ∣ A) = \frac{2}{3}$ ; (ii) $P (A \cap B ∣ C) = \frac{1}{4}$ , $P (A \cup B ∣ C) = \frac{3}{4}$
Where the marks are
- $P (A ∣ C) = \frac{1}{2}$ (1 mark)
- $P (C ∣ A) = \frac{2}{3}$ (½ mark)
- $P (A \cap B ∣ C) = \frac{1}{4}$ (1 mark)
- $P (A \cup B ∣ C) = \frac{3}{4}$ (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerProbability
A box contains 6 cards numbered 1 to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as 10, and B, the event of a number other than 4 on the first card selected.
Find P(A and B) and find whether the events A and B are independent events or not.
Try it first, then show our solutionSolution
1. The sample space has $6 \times 6 = 36$ outcomes. $A = {(4, 6), (5, 5), (6, 4)}$ , so $n (A) = 3$ .
  ½ mark
2. B: the first card is any of 1, 2, 3, 5, 6 and the second is any of 6 numbers, so $n (B) = 5 \times 6 = 30$ .
  ½ mark
3. $A \cap B = {(5, 5), (6, 4)}$ , so $P (A \cap B) = \frac{2}{36} = \frac{1}{18}$
  ½ mark
4. $P (A) = \frac{3}{36} = \frac{1}{12}$
  ½ mark
5. $P (B) = \frac{30}{36} = \frac{5}{6}$ , so $P (A) P (B) = \frac{1}{12} \times \frac{5}{6} = \frac{5}{72} \neq = \frac{1}{18} = P (A \cap B)$
  ½ mark
6. Hence A and B are not independent events.
  ½ mark
Answer: $P (A and B) = \frac{1}{18}$ ; A and B are not independent
Where the marks are
- Lists A, $n (A) = 3$ (½ mark)
- $n (B) = 30$ (½ mark)
- $n (A \cap B) = 2$ , $P (A \cap B) = \frac{2}{36}$ (½ mark)
- $P (A) = \frac{1}{12}$ (½ mark)
- $P (B) = \frac{5}{6}$ and $P (A) P (B) = \frac{5}{72} \neq = P (A \cap B)$ (½ mark)
- Concludes not independent (½ mark)
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Corner point	$Z$
$(0, 0)$	0
$(500, 0)$	200
$(200, 600)$	80 + 180 = 260
$(0, 800)$	240

Section D

Q32 · 5 marks · Long answerCompetency-basedMatrices
A man goes to buy fruits from the market. The shopkeeper informs him that 4 apples, 3 oranges and 2 bananas cost ₹ 60; 2 apples, 4 oranges and 6 bananas cost ₹ 90; whereas 6 apples, 2 oranges and 3 bananas cost ₹ 70. Using matrix method, find the cost of one fruit of each kind.
Try it first, then show our solutionSolution
1. Let the costs of one apple, one orange and one banana be ₹ $x$ , ₹ $y$ and ₹ $z$ . Then:
  $4 x + 3 y + 2 z = 60$
  $2 x + 4 y + 6 z = 90$
  $6 x + 2 y + 3 z = 70$
  1 mark
2. In matrix form $A X = B$ with $A = 426342263$ , $X = x y z$ , $B = 609070$ .
  $∣ A ∣ = 4 (12 - 12) - 3 (6 - 36) + 2 (4 - 24) = 0 + 90 - 40 = 50 \neq = 0$ , so $A^{- 1}$ exists and $X = A^{- 1} B$ .
  1 mark
3. Cofactors:
  $A_{11} = 0$ , $A_{12} = 30$ , $A_{13} = - 20$
  $A_{21} = - 5$ , $A_{22} = 0$ , $A_{23} = 10$
  $A_{31} = 10$ , $A_{32} = - 20$ , $A_{33} = 10$
  So $adj A = 030 - 20 - 5 010 10 - 20 10$
  1½ marks
4. $A^{- 1} = \frac{1}{50} 030 - 20 - 5 010 10 - 20 10$
  ½ mark
5. $X = \frac{1}{50} 0 - 450 + 700 1800 + 0 - 1400 - 1200 + 900 + 700 = \frac{1}{50} 250400400 = 588$
  So one apple costs ₹5, one orange costs ₹8 and one banana costs ₹8.
  1 mark
Answer: Apple ₹5, orange ₹8, banana ₹8
Where the marks are
- Forms the three linear equations (1 mark)
- $∣ A ∣ = 50 \neq = 0$ , so $A^{- 1}$ exists and $X = A^{- 1} B$ (1 mark)
- Correct adj A (1½ marks)
- Correct $A^{- 1}$ (½ mark)
- Computes $X$ : $x = 5$ , $y = 8$ , $z = 8$ (1 mark)
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Q33(a) · 5 marks · Long answerContinuity and Differentiability
If $y x^{2} + 1 = lo g (x^{2} + 1 - x)$ , show that
$(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ .
Try it first, then show our solutionSolution
1. Differentiate both sides with respect to $x$ (product rule on the left, chain rule on the right):
  $x^{2} + 1 \frac{d y}{d x} + y \cdot \frac{x}{x ^{2} + 1} = \frac{1}{x ^{2} + 1 - x} (\frac{x}{x ^{2} + 1} - 1)$
  2 marks
2. Simplify the right side: $\frac{1}{x ^{2} + 1 - x} \cdot \frac{x - x ^{2} + 1}{x ^{2} + 1}$
  1 mark
3. $= - \frac{1}{x ^{2} + 1}$ , so $x^{2} + 1 \frac{d y}{d x} + \frac{x y}{x ^{2} + 1} = - \frac{1}{x ^{2} + 1}$
  1 mark
4. Multiply by $x^{2} + 1$ : $(x^{2} + 1) \frac{d y}{d x} + x y = - 1$ , i.e. $(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ . Hence proved.
  1 mark
Answer: $(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ (proved)
Where the marks are
- Correct differentiation of both sides (2 marks)
- Combines the bracket as $\frac{x - x ^{2} + 1}{x ^{2} + 1}$ (1 mark)
- Simplifies the RHS to $- \frac{1}{x ^{2} + 1}$ (1 mark)
- Multiplies through by $x^{2} + 1$ to get the result (1 mark)
Open Q33(a) on its own page
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Q33(b) · 5 marks · Long answerContinuity and Differentiability
Find the differential of $x^{c o t x} + \frac{2 x ^{2} - 3}{2 x ^{2} - x + 2}$ with respect to x.
Try it first, then show our solutionSolution
1. Let $u = x^{c o t x}$ and $v = \frac{2 x ^{2} - 3}{2 x ^{2} - x + 2}$ , so the required derivative is $\frac{d u}{d x} + \frac{d v}{d x}$ .
  ½ mark
2. $lo g u = cot x \cdot lo g x$
  ½ mark
3. $\frac{1}{u} \frac{d u}{d x} = \frac{c o t x}{x} + lo g x (- cosec^{2} x)$
  1 mark
4. $\frac{d u}{d x} = x^{c o t x} (\frac{c o t x}{x} - cosec^{2} x lo g x)$
  ½ mark
5. By the quotient rule, $\frac{d v}{d x} = \frac{( 2 x ^{2} - x + 2 ) ( 4 x ) - ( 2 x ^{2} - 3 ) ( 4 x - 1 )}{( 2 x ^{2} - x + 2 ) ^{2}}$
  1 mark
6. Numerator $= (8 x^{3} - 4 x^{2} + 8 x) - (8 x^{3} - 2 x^{2} - 12 x + 3) = - 2 x^{2} + 20 x - 3$ , so $\frac{d v}{d x} = \frac{- 2 x ^{2} + 20 x - 3}{( 2 x ^{2} - x + 2 ) ^{2}}$
  1 mark
7. Required derivative $= x^{c o t x} (\frac{c o t x}{x} - cosec^{2} x lo g x) + \frac{- 2 x ^{2} + 20 x - 3}{( 2 x ^{2} - x + 2 ) ^{2}}$
  ½ mark
Answer: $x^{c o t x} (\frac{c o t x}{x} - cosec^{2} x lo g x) + \frac{- 2 x ^{2} + 20 x - 3}{( 2 x ^{2} - x + 2 ) ^{2}}$
Where the marks are
- Splits into $u$ and $v$ (½ mark)
- Takes log: $lo g u = cot x lo g x$ (½ mark)
- Differentiates $lo g u$ correctly (1 mark)
- Correct $\frac{d u}{d x}$ (½ mark)
- Applies the quotient rule to $v$ (1 mark)
- Simplifies the numerator to $- 2 x^{2} + 20 x - 3$ (1 mark)
- Adds the two results (½ mark)
Open Q33(b) on its own page
Q34 · 5 marks · Long answerApplication of Integrals
Sketch the curve ${(x, y) : 100 x^{2} + 25 y^{2} = 2500}$ and find the area of the region enclosed by it, using integration.
Try it first, then show our solutionSolution
1. Divide by 2500: $\frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ . This is an ellipse centred at the origin with major axis along the y-axis. It meets the x-axis at $(\pm 5, 0)$ and the y-axis at $(0, \pm 10)$ .
  Sketch: draw a vertical ellipse through $(5, 0), (- 5, 0), (0, 10), (0, - 10)$ and shade the region inside it.
  1½ marks
2. In the first quadrant $y = \frac{10}{5} 25 - x^{2} = 2 25 - x^{2}$ . The ellipse is symmetric about both axes, so
  Required area $= 4 \int_{0}^{5} 2 25 - x^{2} d x = 8 \int_{0}^{5} 25 - x^{2} d x$ .
  1½ marks
3. Using $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ :
  $= 8 [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{0}^{5}$
  1 mark
4. $= 8 [(0 + \frac{25}{2} \cdot \frac{π}{2}) - (0 + 0)] = 8 \cdot \frac{25 π}{4} = 50 π$ sq. units.
  1 mark
Answer: Area $= 50 π$ sq. units
Where the marks are
- Writes the ellipse in standard form $\frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ and draws a correct sketch (1½ marks)
- Sets up the area as $4 \int_{0}^{5} 2 25 - x^{2} d x$ (1½ marks)
- Integrates correctly using the standard formula (1 mark)
- Obtains the area $50 π$ sq. units (1 mark)
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Q35(a) · 5 marks · Long answerThree Dimensional Geometry
Find the foot of the perpendicular from the point (0, 2, 3) on the line $\frac{- x - 3}{- 5} = \frac{1 - y}{- 2} = \frac{3 z + 12}{9}$ and hence find the length of the perpendicular.
Try it first, then show our solutionSolution
1. Write the line in standard form: $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = λ$ .
  1 mark
2. Any point on the line is $Q (5 λ - 3, 2 λ + 1, 3 λ - 4)$ . The direction ratios of the line are $⟨ 5, 2, 3 ⟩$ .
  ½ mark
3. With $P (0, 2, 3)$ , the direction ratios of $P Q$ are $⟨ 5 λ - 3, 2 λ - 1, 3 λ - 7 ⟩$ .
  ½ mark
4. $P Q ⊥$ the line, so $5 (5 λ - 3) + 2 (2 λ - 1) + 3 (3 λ - 7) = 0$ .
  This gives $38 λ - 38 = 0$ , so $λ = 1$ .
  1 mark
5. Foot of the perpendicular: $Q = (5 - 3, 2 + 1, 3 - 4) = (2, 3, - 1)$ .
  1 mark
6. Length $P Q = (2 - 0)^{2} + (3 - 2)^{2} + (- 1 - 3)^{2} = 4 + 1 + 16 = 21$ units.
  1 mark
Answer: Foot of perpendicular $(2, 3, - 1)$ ; length $21$ units
Where the marks are
- Converts the line to standard form $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ (1 mark)
- Writes the general point $(5 λ - 3, 2 λ + 1, 3 λ - 4)$ (½ mark)
- Writes the direction ratios of $P Q$ (½ mark)
- Uses the perpendicularity condition to get $λ = 1$ (1 mark)
- Finds the foot $(2, 3, - 1)$ (1 mark)
- Finds the length $21$ (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerThree Dimensional Geometry
Find the value of p if the shortest distance between the lines
$r = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and
$r = (p \hat{i} - \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$
is $\frac{3}{2}$ units.
Try it first, then show our solutionSolution
1. Here $a_{1} = \hat{i} + 2 \hat{j} + \hat{k}$ , $b_{1} = \hat{i} - \hat{j} + \hat{k}$ , $a_{2} = p \hat{i} - \hat{j} - \hat{k}$ and $b_{2} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ .
  So $a_{2} - a_{1} = (p - 1) \hat{i} - 3 \hat{j} - 2 \hat{k}$ .
  1 mark
2. $b_{1} \times b_{2} = \hat{i} 12 \hat{j} - 1 1 \hat{k} 12 = (- 2 - 1) \hat{i} - (2 - 2) \hat{j} + (1 + 2) \hat{k} = - 3 \hat{i} + 3 \hat{k}$ .
  Hence $∣ b_{1} \times b_{2} ∣ = 9 + 9 = 32$ .
  1½ marks
3. Shortest distance $d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣} = \frac{- 3 ( p - 1 ) + 0 - 6}{3 2} = \frac{- 3 p - 3}{3 2}$ .
  Setting this equal to the given distance: $\frac{- 3 p - 3}{3 2} = \pm \frac{3}{2}$ .
  1½ marks
4. This gives $- 3 p - 3 = \pm 9$ , i.e. $∣ p + 1∣ = 3$ .
  Therefore $p = 2$ or $p = - 4$ .
  1 mark
Answer: $p = 2$ or $p = - 4$
Where the marks are
- Identifies the vectors and finds $a_{2} - a_{1} = (p - 1) \hat{i} - 3 \hat{j} - 2 \hat{k}$ (1 mark)
- Computes $b_{1} \times b_{2} = - 3 \hat{i} + 3 \hat{k}$ with modulus $32$ (1½ marks)
- Applies the shortest distance formula and equates it to $\frac{3}{2}$ (1½ marks)
- Solves to get $p = 2, - 4$ (1 mark)
Open Q35(b) on its own page

Section E

Case study based

Q36 · 4 marks · Case-basedCompetency-basedRelations and Functions
A school wants the students of class XII to do a project on 'Sustainability' keeping the world environment in mind. They select the student participants on the basis of an essay writing competition.
7 students out of 80 are selected for the project and are categorized into two sets such that :
Girl students belong to Set A = {G $_{1}$ , G $_{2}$ , G $_{3}$ , G $_{4}$ },
Boy students belong to Set B = {B $_{1}$ , B $_{2}$ , B $_{3}$ }.
Based on the above information, answer the following questions :
1. (i)
  How many relations are possible from Set A $\to$ Set B ?
  1 mark
2. (ii)
  Let R be a relation from A $\to$ B such that R = {(G $_{1}$ , B $_{1}$ ), (G $_{2}$ , B $_{2}$ ), (G $_{3}$ , B $_{2}$ ), (G $_{4}$ , B $_{3}$ ), (G $_{1}$ , B $_{2}$ )}. Is R an injective function ? Justify your answer.
  1 mark
3. (iii)(a)
  Let the relation R from A $\to$ A be such that R = {(x, y), x, y $\in$ A, x and y are students from the same colony in the city}. Verify if R is an equivalence relation.
  2 marks
4. OR
  (iii)(b)
  Verify if any function f : B $\to$ A is bijective. Give reason to support your answer.
  2 marks
Try it first, then show our solutionSolution
1. (i) $n (A \times B) = 4 \times 3 = 12$ . Every subset of $A \times B$ is a relation from A to B, so the number of relations is $2^{12} = 4096$ .
  1 mark
2. (ii) No, R is not an injective function.
  ½ mark
3. (ii) R is not even a function, because $G_{1}$ is related to two images, $B_{1}$ and $B_{2}$ .
  ½ mark
4. (iii)(a) Reflexive: every student x lives in the same colony as himself or herself, so $(x, x) \in R$ for all $x \in A$ .
  ½ mark
5. (iii)(a) Symmetric: if $(x, y) \in R$ , then x and y live in the same colony. So y and x live in the same colony, and $(y, x) \in R$ .
  ½ mark
6. (iii)(a) Transitive: if $(x, y) \in R$ and $(y, z) \in R$ , then x, y live in the same colony and y, z live in the same colony. So x and z live in the same colony, and $(x, z) \in R$ .
  ½ mark
7. (iii)(a) R is reflexive, symmetric and transitive, hence R is an equivalence relation.
  ½ mark
8. (iii)(b) No function $f : B \to A$ is bijective.
  1 mark
9. (iii)(b) Reason: $n (B) = 3 < n (A) = 4$ . The range of f has at most 3 elements, so at least one girl in A has no pre-image. Thus f cannot be onto, and so it cannot be bijective.
  1 mark
Answer: (i) $2^{12} = 4096$ ; (ii) No, R is not a function ( $G_{1}$ has two images), so not injective; (iii)(a) R is an equivalence relation; (iii)(b) No $f : B \to A$ is bijective, since $n (B) < n (A)$ so f cannot be onto
Where the marks are
- (i) Number of relations $2^{12} = 4096$ (1 mark)
- (ii) States R is not injective (½ mark)
- (ii) Justifies that R is not a function since $G_{1}$ has two images (½ mark)
- (iii)(a) Proves R is reflexive (½ mark)
- (iii)(a) Proves R is symmetric (½ mark)
- (iii)(a) Proves R is transitive (½ mark)
- (iii)(a) Concludes R is an equivalence relation (½ mark)
- (iii)(b) States no function from B to A is bijective (1 mark)
- (iii)(b) Reason: $n (B) < n (A)$ , so f cannot be onto (1 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedProbability
There are three types of vaccines A $_{1}$ , A $_{2}$ , A $_{3}$ , available in the market to protect the population of the country from spread of certain infection. According to a survey conducted, it was found that 25% of the population was given Vaccine A $_{1}$ , 35% of the population was given Vaccine A $_{2}$ and 40% of the population was given Vaccine A $_{3}$ . The survey also stated that the probabilities that Vaccines A $_{1}$ , A $_{2}$ and A $_{3}$ would protect against the infection were 60%, 55% and 50% respectively.
Based on the above information, answer the following questions :
Find the probability that :
1. (i)
  The person taking vaccine A $_{2}$ will get infected.
  1 mark
2. (ii)
  If a person is chosen randomly, he/she will be protected from the infection.
  1 mark
3. (iii)(a)
  The person was given Vaccine A $_{1}$ , given that the randomly chosen person is infected.
  2 marks
4. OR
  (iii)(b)
  The person was given Vaccine A $_{3}$ , given that the randomly chosen person is not infected.
  2 marks
Try it first, then show our solutionSolution
1. Let $E_{1}, E_{2}, E_{3}$ be the events that a person was given $A_{1}, A_{2}, A_{3}$ . Let I be the event that the person is infected.
  Then $P (E_{1}) = 0.25$ , $P (E_{2}) = 0.35$ , $P (E_{3}) = 0.40$ and $P (I^{'} ∣ E_{1}) = 0.60$ , $P (I^{'} ∣ E_{2}) = 0.55$ , $P (I^{'} ∣ E_{3}) = 0.50$ .
  Hence $P (I ∣ E_{1}) = 0.40$ , $P (I ∣ E_{2}) = 0.45$ , $P (I ∣ E_{3}) = 0.50$ .
2. (i) $P (infected ∣ A_{2}) = 1 - 0.55 = 0.45$ (i.e. 45%).
  1 mark
3. (ii) By the total probability theorem:
  $P (I^{'}) = 0.25 (0.60) + 0.35 (0.55) + 0.40 (0.50)$
  ½ mark
4. (ii) $= 0.15 + 0.1925 + 0.20 = 0.5425$ .
  ½ mark
5. (iii)(a) $P (I) = 0.25 (0.40) + 0.35 (0.45) + 0.40 (0.50) = 0.1 + 0.1575 + 0.2 = 0.4575$ .
  By Bayes' theorem: $P (E_{1} ∣ I) = \frac{P ( E _{1} ) P ( I ∣ E _{1} )}{P ( I )} = \frac{0.25 \times 0.40}{0.4575}$ .
  1½ marks
6. (iii)(a) $= \frac{0.1}{0.4575} = \frac{1000}{4575} = \frac{40}{183}$ .
  ½ mark
7. (iii)(b) By Bayes' theorem: $P (E_{3} ∣ I^{'}) = \frac{P ( E _{3} ) P ( I ^{'} ∣ E _{3} )}{P ( I ^{'} )} = \frac{0.40 \times 0.50}{0.25 ( 0.60 ) + 0.35 ( 0.55 ) + 0.40 ( 0.50 )}$ .
  1½ marks
8. (iii)(b) $= \frac{0.2}{0.5425} = \frac{2000}{5425} = \frac{80}{217}$ .
  ½ mark
Answer: (i) 0.45; (ii) 0.5425; (iii)(a) $\frac{40}{183}$ ; (iii)(b) $\frac{80}{217}$
Where the marks are
- (i) Probability of infection with $A_{2}$ is 0.45 (1 mark)
- (ii) Writes the total probability expression for protection (½ mark)
- (ii) Evaluates it as 0.5425 (½ mark)
- (iii)(a) Correct Bayes' theorem expression for $P (A_{1} ∣ infected)$ (1½ marks)
- (iii)(a) Simplifies to $\frac{40}{183}$ (½ mark)
- (iii)(b) Correct Bayes' theorem expression for $P (A_{3} ∣ not infected)$ (1½ marks)
- (iii)(b) Simplifies to $\frac{80}{217}$ (½ mark)
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Q38 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they fix the surface area of the tumblers produced.
Based on the above information, answer the following questions :
If for a tumbler, V is its volume, h the height and r the radius of the circular base, then :
1. (i)
  Differentiate its volume with respect to radius of the base, where the surface area is constant.
  2 marks
2. (ii)
  If the company wants to maximize the volume of each tumbler, then establish a relation between its height and the radius of the base.
  2 marks
Figure in the paper: A right circular cylinder labelled 'Tumbler', open at the top, with height h marked by a double-headed arrow on the right side and radius r marked on the circular base with centre O. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The tumbler is open at the top, so the surface area is curved surface plus base:
  $S = 2 π r h + π r^{2}$ (constant). Hence $h = \frac{S - π r ^{2}}{2 π r}$ .
  ½ mark
2. (i) $V = π r^{2} h = π r^{2} \cdot \frac{S - π r ^{2}}{2 π r} = \frac{S r - π r ^{3}}{2}$ .
  1 mark
3. (i) $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ .
  ½ mark
4. (ii) For a stationary point, $\frac{d V}{d r} = 0$ , which gives $S = 3 π r^{2}$ .
  1 mark
5. (ii) $\frac{d ^{2} V}{d r ^{2}} = \frac{- 6 π r}{2} = - 3 π r < 0$ (since $r > 0$ ), so V is maximum when $S = 3 π r^{2}$ .
  ½ mark
6. (ii) Then $h = \frac{3 π r ^{2} - π r ^{2}}{2 π r} = \frac{2 π r ^{2}}{2 π r} = r$ .
  So for maximum volume, the height equals the radius of the base: $h = r$ .
  ½ mark
Answer: (i) $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ ; (ii) $h = r$
Where the marks are
- (i) Writes $S = 2 π r h + π r^{2}$ for the open cylinder (½ mark)
- (i) Expresses V in terms of r: $V = \frac{S r - π r ^{3}}{2}$ (1 mark)
- (i) Differentiates to get $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ (½ mark)
- (ii) Puts $\frac{d V}{d r} = 0$ to get $S = 3 π r^{2}$ (1 mark)
- (ii) Shows $\frac{d ^{2} V}{d r ^{2}} = - 3 π r < 0$ , so V is maximum (½ mark)
- (ii) Concludes $h = r$ (½ mark)
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Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.