MathsCBSE Class 12 · Sample paper, session 2024-25

CBSE Class 12 Mathematics: sample paper 2024-25

The sample question paper CBSE published for the 2024-25 session, 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 sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassXII_2024_25/Maths-SQP.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

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

Section A

MCQ 1-18, assertion-reason 19-20

Q1 · 1 mark · MCQDeterminants
If for a square matrix $A$ , $A \cdot (a d j A) = 202500020250002025$ , then the value of $∣ A ∣ + ∣ a d j A ∣$ is equal to:
1. (A) $1$
2. (B) $2025 + 1$
3. (C) $(2025)^{2} + 45$
4. (D) $2025 + (2025)^{2}$
Try it first, then show our solutionSolution
1. Since $A \cdot (a d j A) = ∣ A ∣ I_{3} = 2025 I_{3}$ , we get $∣ A ∣ = 2025$ .
2. $∣ a d j A ∣ = ∣ A ∣^{3 - 1} = (2025)^{2}$ , so $∣ A ∣ + ∣ a d j A ∣ = 2025 + (2025)^{2}$ . Option (D).
  1 mark
Answer: $2025 + (2025)^{2}$ , option (D)
Where the marks are
- Uses $A (a d j A) = ∣ A ∣ I$ and $∣ a d j A ∣ = ∣ A ∣^{n - 1}$ to get $2025 + 202 5^{2}$ (1 mark)
Q2 · 1 mark · MCQMatrices
Assume $X, Y, Z, W$ and $P$ are matrices of order $2 \times n$ , $3 \times k$ , $2 \times p$ , $n \times 3$ and $p \times k$ , respectively. Then the restriction on $n$ , $k$ and $p$ so that $P Y + W Y$ will be defined are:
1. (A) $k = 3, p = n$
2. (B) $k$ is arbitrary, $p = 2$
3. (C) $p$ is arbitrary, $k = 3$
4. (D) $k = 2, p = 3$
Try it first, then show our solutionSolution
1. $P Y$ : $(p \times k) (3 \times k)$ is defined only if $k = 3$ ; then $P Y$ is $p \times k$ .
2. $W Y$ : $(n \times 3) (3 \times k)$ is $n \times k$ . For the sum, orders must match, so $p = n$ . Option (A).
  1 mark
Answer: $k = 3, p = n$ , option (A)
Where the marks are
- Gets $k = 3$ from $P Y$ and $p = n$ from equal orders (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQApplication of Derivatives
The interval in which the function $f$ defined by $f (x) = e^{x}$ is strictly increasing, is
1. (A) $[1, \infty)$
2. (B) $(- \infty, 0)$
3. (C) $(- \infty, \infty)$
4. (D) $(0, \infty)$
Try it first, then show our solutionSolution
1. $f^{'} (x) = e^{x} > 0$ for all real $x$ , so $f$ is strictly increasing on $(- \infty, \infty)$ . Option (C).
  1 mark
Answer: $(- \infty, \infty)$ , option (C)
Where the marks are
- $f^{'} (x) = e^{x} > 0$ for all $x \in R$ (1 mark)
Q4 · 1 mark · MCQDeterminants
If $A$ and $B$ are non-singular matrices of same order with $det (A) = 5$ , then $det (B^{- 1} A B)^{2}$ is equal to
1. (A) $5$
2. (B) $5^{2}$
3. (C) $5^{4}$
4. (D) $5^{5}$
Try it first, then show our solutionSolution
1. $∣ B^{- 1} A B ∣ = ∣ B^{- 1} ∣∣ A ∣∣ B ∣ = \frac{1}{∣ B ∣} \cdot 5 \cdot ∣ B ∣ = 5$ .
2. Hence $(det (B^{- 1} A B))^{2} = 5^{2}$ . Option (B).
  1 mark
Answer: $5^{2}$ , option (B)
Where the marks are
- Uses $∣ X Y ∣ = ∣ X ∣∣ Y ∣$ and $∣ B^{- 1} ∣ = 1/∣ B ∣$ to get $5^{2}$ (1 mark)
Q5 · 1 mark · MCQDifferential Equations
The value of ' $n$ ', such that the differential equation $x^{n} \frac{d y}{d x} = y (lo g y - lo g x + 1)$ ; (where $x, y \in R^{+}$ ) is homogeneous, is
1. (A) $0$
2. (B) $1$
3. (C) $2$
4. (D) $3$
Try it first, then show our solutionSolution
1. $\frac{d y}{d x} = \frac{y}{x ^{n}} (lo g \frac{y}{x} + 1)$ ; the bracket is homogeneous of degree 0.
2. $\frac{y}{x ^{n}}$ has degree 0 only when $n = 1$ . Option (B).
  1 mark
Answer: $n = 1$ , option (B)
Where the marks are
- Writes RHS as $\frac{y}{x ^{n}} lo g (\frac{ey}{x})$ and requires degree 0, giving $n = 1$ (1 mark)
Q6 · 1 mark · MCQDeterminants
If the points $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ and $(x_{1} + x_{2}, y_{1} + y_{2})$ are collinear, then $x_{1} y_{2}$ is equal to
1. (A) $x_{2} y_{1}$
2. (B) $x_{1} y_{1}$
3. (C) $x_{2} y_{2}$
4. (D) $x_{1} x_{2}$
Try it first, then show our solutionSolution
1. Collinear ⇒ $x_{1} x_{2} x_{1} + x_{2} y_{1} y_{2} y_{1} + y_{2} 111 = 0$ .
2. Expanding gives $2 (x_{1} y_{2} - x_{2} y_{1}) = 0$ ⇒ $x_{1} y_{2} = x_{2} y_{1}$ . Option (A).
  1 mark
Answer: $x_{2} y_{1}$ , option (A)
Where the marks are
- Sets area determinant to zero and simplifies to $x_{1} y_{2} = x_{2} y_{1}$ (1 mark)
Q7 · 1 mark · MCQMatrices
If $A = 0 - 1 2 1 a 3 c - b 0$ is a skew-symmetric matrix then the value of $a + b + c =$
1. (A) $1$
2. (B) $2$
3. (C) $3$
4. (D) $4$
Try it first, then show our solutionSolution
1. Skew-symmetric: $a_{ij} = - a_{j i}$ , diagonal entries zero. So $a = 0$ , $c = - 2$ , $- b = - 3 \Rightarrow b = 3$ .
2. $a + b + c = 0 + 3 - 2 = 1$ . Option (A).
  1 mark
Answer: $1$ , option (A)
Where the marks are
- Finds $a = 0, b = 3, c = - 2$ , sum 1 (1 mark)
Q8 · 1 mark · MCQProbability
For any two events $A$ and $B$ , if $P (\overline{A}) = \frac{1}{2}$ , $P (\overline{B}) = \frac{2}{3}$ and $P (A \cap B) = \frac{1}{4}$ , then $P (\overset{ˉ}{A} / \overset{ˉ}{B})$ equals:
1. (A) $\frac{3}{8}$
2. (B) $\frac{8}{9}$
3. (C) $\frac{5}{8}$
4. (D) $\frac{1}{4}$
Try it first, then show our solutionSolution
1. $P (A) = \frac{1}{2}, P (B) = \frac{1}{3}$ , $P (A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$ .
2. $P (\overset{ˉ}{A} / \overset{ˉ}{B}) = \frac{1 - P ( A \cup B )}{P ( B ˉ )} = \frac{5/12}{2/3} = \frac{5}{8}$ . Option (C).
  1 mark
Answer: $\frac{5}{8}$ , option (C)
Where the marks are
- Uses $P (\overset{ˉ}{A} \cap \overset{ˉ}{B}) = 1 - P (A \cup B)$ to get $5/8$ (1 mark)
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Q9 · 1 mark · MCQVector Algebra
The value of $α$ if the angle between $p = 2 α^{2} \hat{i} - 3 α \hat{j} + \hat{k}$ and $q = \hat{i} + \hat{j} + α \hat{k}$ is obtuse, is
1. (A) $R - [0, 1]$
2. (B) $(0, 1)$
3. (C) $[0, \infty)$
4. (D) $[1, \infty)$
Try it first, then show our solutionSolution
1. Obtuse angle ⇒ $p \cdot q < 0$ : $2 α^{2} - 3 α + α < 0$ .
2. $2 α (α - 1) < 0 \Rightarrow α \in (0, 1)$ . Option (B).
  1 mark
Answer: $(0, 1)$ , option (B)
Where the marks are
- Sets dot product negative and solves to $(0, 1)$ (1 mark)
Q10 · 1 mark · MCQVector Algebra
If $∣ a ∣ = 3$ , $∣ b ∣ = 4$ and $∣ a + b ∣ = 5$ , then $∣ a - b ∣ =$
1. (A) $3$
2. (B) $4$
3. (C) $5$
4. (D) $8$
Try it first, then show our solutionSolution
1. $∣ a + b ∣^{2} + ∣ a - b ∣^{2} = 2 (∣ a ∣^{2} + ∣ b ∣^{2}) = 50$ .
2. $∣ a - b ∣^{2} = 50 - 25 = 25 \Rightarrow ∣ a - b ∣ = 5$ . Option (C).
  1 mark
Answer: $5$ , option (C)
Where the marks are
- Uses parallelogram law to get 5 (1 mark)
Q11 · 1 mark · MCQLinear Programming
For the linear programming problem (LPP), the objective function is $Z = 4 x + 3 y$ and the feasible region determined by a set of constraints is shown in the graph:
(Note: The figure is not to scale.)
Which of the following statements is true?
1. (A)Maximum value of $Z$ is at $R (40, 0)$ .
2. (B)Maximum value of $Z$ is at $Q (30, 20)$ .
3. (C)Value of $Z$ at $R (40, 0)$ is less than the value at $P (0, 40)$ .
4. (D)The value of $Z$ at $Q (30, 20)$ is less than the value at $R (40, 0)$ .
Figure in the paper: Shaded feasible region with vertices O(0,0), P(0,40), Q(30,20), R(40,0); two constraint lines intersect at Q, one passing through P and Q, the other through Q and R, with arrows showing the lines extending. See it in the official paper.
Try it first, then show our solutionSolution
1. $Z$ at corners: $O$ : 0, $R (40, 0)$ : 160, $Q (30, 20)$ : $120 + 60 = 180$ , $P (0, 40)$ : 120.
2. Bounded region, so max $Z = 180$ at $Q (30, 20)$ . Option (B).
  1 mark
Answer: Option (B): maximum at Q(30,20)
Where the marks are
- Evaluates Z at corners and identifies max 180 at Q (1 mark)
Q12 · 1 mark · MCQIntegrals
$\int \frac{d x}{x ^{3} ( 1 + x ^{4} ) ^{\frac{1}{2}}}$ equals
1. (A) $- \frac{1}{2 x ^{2}} 1 + x^{4} + c$
2. (B) $\frac{1}{2 x} 1 + x^{4} + c$
3. (C) $- \frac{1}{4 x} 1 + x^{4} + c$
4. (D) $\frac{1}{4 x ^{2}} 1 + x^{4} + c$
Try it first, then show our solutionSolution
1. Write integrand as $\frac{1}{x ^{5} ( 1 + x ^{- 4} ) ^{1/2}}$ ; put $t = 1 + x^{- 4}$ , $d t = - 4 x^{- 5} d x$ .
2. $- \frac{1}{4} \int t^{- 1/2} d t = - \frac{1}{2} t + c = - \frac{1}{2 x ^{2}} 1 + x^{4} + c$ . Option (A).
  1 mark
Answer: $- \frac{1}{2 x ^{2}} 1 + x^{4} + c$ , option (A)
Where the marks are
- Substitution $t = 1 + x^{- 4}$ and correct integration (1 mark)
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Q13 · 1 mark · MCQIntegrals
$\int_{0}^{2 π} cose c^{7} x d x =$
1. (A) $0$
2. (B) $1$
3. (C) $4$
4. (D) $2 π$
Try it first, then show our solutionSolution
1. $f (x) = csc^{7} x$ ; $f (2 π - x) = (- csc x)^{7} = - f (x)$ .
2. By $\int_{0}^{2 a} f = 0$ when $f (2 a - x) = - f (x)$ , the value is 0. Option (A).
  1 mark
Answer: $0$ , option (A)
Where the marks are
- Applies the property $\int_{0}^{2 a} f (x) d x = 0$ for $f (2 a - x) = - f (x)$ (1 mark)
Q14 · 1 mark · MCQDifferential Equations
What is the general solution of the differential equation $e^{y^{'}} = x$ ?
1. (A) $y = x lo g x + c$
2. (B) $y = x lo g x - x + c$
3. (C) $y = x lo g x + x + c$
4. (D) $y = x + c$
Try it first, then show our solutionSolution
1. $e^{y^{'}} = x \Rightarrow \frac{d y}{d x} = lo g x$ .
2. Integrating by parts: $y = x lo g x - x + c$ . Option (B).
  1 mark
Answer: $y = x lo g x - x + c$ , option (B)
Where the marks are
- Takes log and integrates $lo g x$ (1 mark)
Q15 · 1 mark · MCQInverse Trigonometric Functions
The graph drawn below depicts
1. (A) $y = sin^{- 1} x$
2. (B) $y = cos^{- 1} x$
3. (C) $y = cose c^{- 1} x$
4. (D) $y = cot^{- 1} x$
Figure in the paper: Red decreasing curve on x from -1 to 1, y-axis marked π/4, π/2, 3π/4, π; it starts at about π at x=-1, passes through (0, π/2), and ends at (1, 0). See it in the official paper.
Try it first, then show our solutionSolution
1. Domain $[- 1, 1]$ , range $[0, π]$ , decreasing with $y (0) = π /2$ , $y (1) = 0$ : this is $y = cos^{- 1} x$ . Option (B).
  1 mark
Answer: $y = cos^{- 1} x$ , option (B)
Where the marks are
- Identifies domain $[- 1, 1]$ , range $[0, π]$ ⇒ $cos^{- 1} x$ (1 mark)
Q16 · 1 mark · MCQLinear Programming
A linear programming problem (LPP) along with the graph of its constraints is shown below. The corresponding objective function is: $Z = 18 x + 10 y$ , which has to be minimized. The smallest value of the objective function Z is 134 and is obtained at the corner point $(3, 8)$ .
(Note: The figure is not to scale.)
The optimal solution of the above linear programming problem __________.
1. (A)does not exist as the feasible region is unbounded.
2. (B)does not exist as the inequality $18 x + 10 y < 134$ does not have any point in common with the feasible region.
3. (C)exists as the inequality $18 x + 10 y > 134$ has infinitely many points in common with the feasible region.
4. (D)exists as the inequality $18 x + 10 y < 134$ does not have any point in common with the feasible region.
Figure in the paper: Unbounded shaded feasible region with corner point P(3,8). Lines: 4x+y=20 (intercepts A1(5,0), B1(0,20)), 2x+3y=30 (intercepts A2(15,0), (0,10)), and dotted line 18x+10y=134 passing through P(3,8) and meeting y-axis at B2(0,10). See it in the official paper.
Try it first, then show our solutionSolution
1. For an unbounded region, the corner value 134 is the minimum if the half-plane $18 x + 10 y < 134$ has no point in common with the feasible region.
2. That is the case here, so the optimal solution exists: min $Z = 134$ at $(3, 8)$ . Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Applies the unbounded-region test $18 x + 10 y < 134$ (1 mark)
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Q17 · 1 mark · MCQContinuity and Differentiability
The function $f : R \to Z$ defined by $f (x) = [x]$ ; where $[.]$ denotes the greatest integer function, is
1. (A)Continuous at $x = 2.5$ but not differentiable at $x = 2.5$
2. (B)Not Continuous at $x = 2.5$ but differentiable at $x = 2.5$
3. (C)Not Continuous at $x = 2.5$ and not differentiable at $x = 2.5$
4. (D)Continuous as well as differentiable at $x = 2.5$
Try it first, then show our solutionSolution
1. Near $x = 2.5$ , $[x] = 2$ (constant), so $f$ is continuous and $f^{'} (2.5) = 0$ . Option (D).
  1 mark
Answer: Option (D)
Where the marks are
- Notes $[x] = 2$ on a neighbourhood of 2.5 ⇒ continuous and differentiable (1 mark)
Q18 · 1 mark · MCQCompetency-basedApplication of Integrals
A student observes an open-air Honeybee nest on the branch of a tree, whose plane figure is parabolic shape given by $x^{2} = 4 y$ . Then the area (in sq units) of the region bounded by parabola $x^{2} = 4 y$ and the line $y = 4$ is
1. (A) $\frac{32}{3}$
2. (B) $\frac{64}{3}$
3. (C) $\frac{128}{3}$
4. (D) $\frac{256}{3}$
Try it first, then show our solutionSolution
1. Region symmetric about y-axis: Area $= 2 \int_{0}^{4} 2 y d y$ .
2. $= 4 \cdot \frac{2}{3} [y^{3/2}]_{0}^{4} = \frac{8}{3} \cdot 8 = \frac{64}{3}$ . Option (B).
  1 mark
Answer: $\frac{64}{3}$ , option (B)
Where the marks are
- Sets up $2 \int_{0}^{4} 2 y d y$ and evaluates $64/3$ (1 mark)
Q19 · 1 mark · Assertion-reasonCompetency-basedContinuity and Differentiability
Assertion (A): Consider the function defined as $f (x) = ∣ x ∣ + ∣ x - 1∣$ , $x \in R$ . Then $f (x)$ is not differentiable at $x = 0$ and $x = 1$ .
Reason (R): Suppose $f$ be defined and continuous on $(a, b)$ and $c \in (a, b)$ , then $f (x)$ is not differentiable at $x = c$ if $lim_{h \to 0^{-}} \frac{f ( c + h ) - f ( c )}{h} \neq = lim_{h \to 0^{+}} \frac{f ( c + h ) - f ( c )}{h}$ .
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (C)(A) is true but (R) is false.
4. (D)(A) is false but (R) is true.
Try it first, then show our solutionSolution
1. At $x = 0$ : LHD $= - 2$ , RHD $= 0$ ; at $x = 1$ : LHD $= 0$ , RHD $= 2$ . So $f$ is not differentiable at 0 and 1; (A) true.
2. (R) is the definition-based criterion (unequal one-sided derivatives) used to prove (A). Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Shows unequal LHD and RHD at 0 and 1 and links to (R) (1 mark)
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Q20 · 1 mark · Assertion-reasonCompetency-basedRelations and Functions
Assertion (A): The function $f : R - {(2 n + 1) \frac{π}{2} : n \in Z} \to (- \infty, - 1] \cup [1, \infty)$ defined by $f (x) = sec x$ is not one-one function in its domain.
Reason (R): The line $y = 2$ meets the graph of the function at more than one point.
1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B)Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (C)(A) is true but (R) is false.
4. (D)(A) is false but (R) is true.
Try it first, then show our solutionSolution
1. $sec x = 2$ at $x = π /3$ and $x = - π /3$ (and more), so the line $y = 2$ cuts the graph at many points; (R) true.
2. By the horizontal line test, $sec x$ is not one-one; (A) true and (R) explains it. Option (A).
  1 mark
Answer: Option (A)
Where the marks are
- Uses horizontal line test with $y = 2$ to show not one-one (1 mark)
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Section B

VSA

Q21 · 2 marks · Very short answerInverse Trigonometric Functions
If $cot^{- 1} (3 x + 5) > \frac{π}{4}$ , then find the range of the values of $x$ .
Try it first, then show our solutionSolution
1. Write the right side as an inverse cotangent: $cot^{- 1} (3 x + 5) > \frac{π}{4} = cot^{- 1} 1$ .
  ½ mark
2. $cot^{- 1} x$ is strictly decreasing on its domain $R$ , so the inequality reverses: $3 x + 5 < 1$ .
  ½ mark
3. $3 x < - 4 \Rightarrow x < - \frac{4}{3}$ , so $x \in (- \infty, - \frac{4}{3})$ .
  1 mark
Answer: $x \in (- \infty, - \frac{4}{3})$
Where the marks are
- Writes $\frac{π}{4} = cot^{- 1} 1$ (½ mark)
- Uses that $cot^{- 1}$ is decreasing to get $3 x + 5 < 1$ (½ mark)
- Obtains $x < - \frac{4}{3}$ , i.e. $x \in (- \infty, - \frac{4}{3})$ (1 mark)
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Q22 · 2 marks · Very short answerCompetency-basedApplication of Derivatives
The cost (in rupees) of producing $x$ items in factory, each day is given by
$C (x) = 0.00013 x^{3} + 0.002 x^{2} + 5 x + 2200$
Find the marginal cost when 150 items are produced.
Try it first, then show our solutionSolution
1. Marginal cost is $C^{'} (x)$ : $C^{'} (x) = 3 (0.00013) x^{2} + 2 (0.002) x + 5 = 0.00039 x^{2} + 0.004 x + 5$ .
  1 mark
2. At $x = 150$ : $C^{'} (150) = 0.00039 (22500) + 0.004 (150) + 5 = 8.775 + 0.6 + 5 = 14.375$ . The marginal cost is ₹14.375.
  1 mark
Answer: Marginal cost = ₹14.375
Where the marks are
- Finds $C^{'} (x) = 0.00039 x^{2} + 0.004 x + 5$ (1 mark)
- Evaluates $C^{'} (150) = ₹14.375$ (1 mark)
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Answer one of these 2 alternatives.
Q23(a) · 2 marks · Very short answerContinuity and Differentiability
Find the derivative of $tan^{- 1} x$ with respect to $lo g x$ ; (where $x \in (1, \infty)$ ).
Try it first, then show our solutionSolution
1. Let $y = tan^{- 1} x$ and $z = lo g x$ . Then $\frac{d y}{d x} = \frac{1}{1 + x ^{2}}$ .
  ½ mark
2. $\frac{d z}{d x} = \frac{1}{x}$ .
  ½ mark
3. $\frac{d y}{d z} = \frac{d y / d x}{d z / d x}$ .
  ½ mark
4. $\frac{d y}{d z} = \frac{1/ ( 1 + x ^{2} )}{1/ x} = \frac{x}{1 + x ^{2}}$ .
  ½ mark
Answer: $\frac{x}{1 + x ^{2}}$
Where the marks are
- $\frac{d y}{d x} = \frac{1}{1 + x ^{2}}$ (½ mark)
- $\frac{d z}{d x} = \frac{1}{x}$ (½ mark)
- Uses $\frac{d y}{d z} = \frac{d y / d x}{d z / d x}$ (½ mark)
- Gets $\frac{x}{1 + x ^{2}}$ (½ mark)
Open Q23(a) on its own page
OR
Q23(b) · 2 marks · Very short answerContinuity and Differentiability
Differentiate the following function with respect to x : $(cos x)^{x}$ ; (where $x \in (0, \frac{π}{2})$ ).
Try it first, then show our solutionSolution
1. Let $y = (cos x)^{x} = e^{x l o g c o s x}$ . Differentiating, $\frac{d y}{d x} = e^{x l o g c o s x} \frac{d}{d x} (x lo g cos x)$ .
  ½ mark
2. By the product rule, $\frac{d y}{d x} = (cos x)^{x} {lo g cos x \cdot 1 + x \frac{d}{d x} (lo g cos x)}$ .
  ½ mark
3. $= (cos x)^{x} {lo g cos x + x \cdot \frac{1}{c o s x} (- sin x)} = (cos x)^{x} (lo g cos x - x tan x)$ .
  1 mark
Answer: $\frac{d y}{d x} = (cos x)^{x} (lo g cos x - x tan x)$
Where the marks are
- Writes $y = e^{x l o g c o s x}$ (or takes logs) and differentiates (½ mark)
- Applies product rule correctly (½ mark)
- Final derivative $(cos x)^{x} (lo g cos x - x tan x)$ (1 mark)
Open Q23(b) on its own page
Answer one of these 2 alternatives.
Q24(a) · 2 marks · Very short answerVector Algebra
If vectors $a = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $b = - \hat{i} + 2 \hat{j} + \hat{k}$ and $c = 3 \hat{i} + \hat{j}$ are such that $b + λ c$ is perpendicular to $a$ , then find the value of $λ$ .
Try it first, then show our solutionSolution
1. $b + λ c = (- 1 + 3 λ) \hat{i} + (2 + λ) \hat{j} + \hat{k}$ .
  ½ mark
2. Perpendicular to $a$ means $(b + λ c) \cdot a = 0$ : $2 (- 1 + 3 λ) + 2 (2 + λ) + 3 (1) = 0$ .
  1 mark
3. $8 λ + 5 = 0 \Rightarrow λ = - \frac{5}{8}$ .
  ½ mark
Answer: $λ = - \frac{5}{8}$
Where the marks are
- Writes $b + λ c$ in components (½ mark)
- Sets dot product with $a$ equal to zero (1 mark)
- Finds $λ = - \frac{5}{8}$ (½ mark)
Open Q24(a) on its own page
OR
Q24(b) · 2 marks · Very short answerCompetency-basedThree Dimensional Geometry
A person standing at $O (0, 0, 0)$ is watching an aeroplane which is at the coordinate point $A (4, 0, 3)$ . At the same time he saw a bird at the coordinate point $B (0, 0, 1)$ . Find the angles which $B A$ makes with the x,y and z axes.
Try it first, then show our solutionSolution
1. $B A = O A - O B = (4 \hat{i} + 3 \hat{k}) - \hat{k} = 4 \hat{i} + 0 \hat{j} + 2 \hat{k}$ .
  ½ mark
2. $∣ B A ∣ = 16 + 4 = 25$ , so the unit vector is $\frac{2}{5} \hat{i} + 0 \hat{j} + \frac{1}{5} \hat{k}$ ; direction cosines are $\frac{2}{5}, 0, \frac{1}{5}$ .
  ½ mark
3. Angles with the $x$ , $y$ , $z$ axes: $cos^{- 1} \frac{2}{5}$ , $\frac{π}{2}$ , $cos^{- 1} \frac{1}{5}$ .
  1 mark
Answer: $cos^{- 1} \frac{2}{5}, \frac{π}{2}, cos^{- 1} \frac{1}{5}$
Where the marks are
- Finds $B A = 4 \hat{i} + 2 \hat{k}$ (½ mark)
- Finds unit vector / direction cosines (½ mark)
- States the three angles (1 mark)
Open Q24(b) on its own page
Q25 · 2 marks · Very short answerVector Algebra
The two co-initial adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Find its diagonals and use them to find the area of the parallelogram.
Try it first, then show our solutionSolution
1. With $a = 2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ , $b = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , the diagonals are $d_{1} = a + b = 4 \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $d_{2} = a - b = - 6 \hat{j} - 8 \hat{k}$ .
  ½ mark
2. $d_{1} \times d_{2} = \hat{i} 40 \hat{j} - 2 - 6 \hat{k} - 2 - 8 = (16 - 12) \hat{i} - (- 32 - 0) \hat{j} + (- 24 - 0) \hat{k} = 4 \hat{i} + 32 \hat{j} - 24 \hat{k} = 4 (\hat{i} + 8 \hat{j} - 6 \hat{k})$ .
  1 mark
3. Area $= \frac{1}{2} ∣ d_{1} \times d_{2} ∣ = \frac{1}{2} \cdot 4 1 + 64 + 36 = 2101$ sq. units.
  ½ mark
Answer: Diagonals $4 \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $- 6 \hat{j} - 8 \hat{k}$ ; area $= 2101$ sq. units
Where the marks are
- Finds both diagonals (½ mark)
- Computes $d_{1} \times d_{2}$ correctly (1 mark)
- Area $= \frac{1}{2} ∣ d_{1} \times d_{2} ∣ = 2101$ sq units (½ mark)
Open Q25 on its own page

Section C

Q26 · 3 marks · Short answerCompetency-basedApplication of Derivatives
A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away horizontally at the rate of 200 cm/s, find the rate at which the string is being released.
Try it first, then show our solutionSolution
1. Figure: right triangle with vertical side 3 m (height of kite), horizontal side $x$ m and hypotenuse $y$ m (string).
  ½ mark
2. By Pythagoras, $x^{2} + 3^{2} = y^{2}$ .
  ½ mark
3. When $y = 5$ , $x = 25 - 9 = 4$ . Differentiating w.r.t. $t$ : $2 x \frac{d x}{d t} = 2 y \frac{d y}{d t}$ .
  1 mark
4. With $\frac{d x}{d t} = 200$ cm/s: $4 (200) = 5 \frac{d y}{d t} \Rightarrow \frac{d y}{d t} = 160$ cm/s. The string is released at 160 cm/s.
  1 mark
Answer: 160 cm/s
Where the marks are
- Correct figure (½ mark)
- Relation $x^{2} + 9 = y^{2}$ (½ mark)
- Finds $x = 4$ and differentiates (1 mark)
- $\frac{d y}{d t} = 160$ cm/s (1 mark)
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Q27 · 3 marks · Short answerCompetency-basedApplication of Derivatives
According to a psychologist, the ability of a person to understand spatial concepts is given by $A = \frac{1}{3} t$ , where $t$ is the age in years, $t \in [5, 18]$ . Show that the rate of increase of the ability to understand spatial concepts decreases with age in between 5 and 18.
Try it first, then show our solutionSolution
1. $A = \frac{1}{3} t \Rightarrow \frac{d A}{d t} = \frac{1}{3} \cdot \frac{1}{2} t^{- 1/2} = \frac{1}{6 t}$ , the rate of increase of ability.
  1 mark
2. $\frac{d ^{2} A}{d t ^{2}} = \frac{1}{6} (- \frac{1}{2}) t^{- 3/2} = - \frac{1}{12 t t}$ .
  1 mark
3. For $t \in (5, 18)$ , $t t > 0$ , so $\frac{d ^{2} A}{d t ^{2}} < 0$ .
  ½ mark
4. Hence $\frac{d A}{d t}$ is decreasing: the rate of increase of the ability to understand spatial concepts slows down with age between 5 and 18.
  ½ mark
Answer: $\frac{d ^{2} A}{d t ^{2}} = - \frac{1}{12 t t} < 0$ , so the rate of increase decreases with age.
Where the marks are
- $\frac{d A}{d t} = \frac{1}{6 t}$ (1 mark)
- $\frac{d ^{2} A}{d t ^{2}} = - \frac{1}{12 t t}$ (1 mark)
- Shows it is negative on $(5, 18)$ (½ mark)
- Conclusion (½ mark)
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Answer one of these 2 alternatives.
Q28(a) · 3 marks · Short answerCompetency-basedVector Algebra
An ant is moving along the vector $l_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ . Few sugar crystals are kept along the vector $l_{2} = 3 \hat{i} - 2 \hat{j} + \hat{k}$ which is inclined at an angle $θ$ with the vector $l_{1}$ . Then find the angle $θ$ . Also find the scalar projection of $l_{1}$ on $l_{2}$ .
Try it first, then show our solutionSolution
1. $cos θ = \frac{l _{1} \cdot l _{2}}{∣ l _{1} ∣∣ l _{2} ∣}$ , with $l_{1} \cdot l_{2} = (1) (3) + (- 2) (- 2) + (3) (1) = 10$ .
  1 mark
2. $∣ l_{1} ∣ = 14$ , $∣ l_{2} ∣ = 14$ , so $cos θ = \frac{10}{14} = \frac{5}{7}$ , $θ = cos^{- 1} \frac{5}{7}$ .
  ½ mark
3. Scalar projection of $l_{1}$ on $l_{2}$ $= \frac{l _{1} \cdot l _{2}}{∣ l _{2} ∣}$ .
  1 mark
4. $= \frac{10}{14}$ .
  ½ mark
Answer: $θ = cos^{- 1} \frac{5}{7}$ ; scalar projection $= \frac{10}{14}$
Where the marks are
- Uses dot-product formula for angle with $l_{1} \cdot l_{2} = 10$ (1 mark)
- $θ = cos^{- 1} (5/7)$ (½ mark)
- Formula for scalar projection (1 mark)
- Value $10/ 14$ (½ mark)
Open Q28(a) on its own page
OR
Q28(b) · 3 marks · Short answerThree Dimensional Geometry
Find the vector and the cartesian equation of the line that passes through $(- 1, 2, 7)$ and is perpendicular to the lines $r = 2 \hat{i} + \hat{j} - 3 \hat{k} + λ (\hat{i} + 2 \hat{j} + 5 \hat{k})$ and $r = 3 \hat{i} + 3 \hat{j} - 7 \hat{k} + μ (3 \hat{i} - 2 \hat{j} + 5 \hat{k})$ .
Try it first, then show our solutionSolution
1. The required line is parallel to $b_{1} \times b_{2} = \hat{i} 13 \hat{j} 2 - 2 \hat{k} 55 = (10 + 10) \hat{i} - (5 - 15) \hat{j} + (- 2 - 6) \hat{k} = 20 \hat{i} + 10 \hat{j} - 8 \hat{k}$ , i.e. parallel to $10 \hat{i} + 5 \hat{j} - 4 \hat{k}$ .
  1 mark
2. Vector equation through $(- 1, 2, 7)$ : $r = - \hat{i} + 2 \hat{j} + 7 \hat{k} + a (10 \hat{i} + 5 \hat{j} - 4 \hat{k})$ .
  1 mark
3. Cartesian equation: $\frac{x + 1}{10} = \frac{y - 2}{5} = \frac{z - 7}{- 4}$ .
  1 mark
Answer: $r = - \hat{i} + 2 \hat{j} + 7 \hat{k} + a (10 \hat{i} + 5 \hat{j} - 4 \hat{k})$ ; $\frac{x + 1}{10} = \frac{y - 2}{5} = \frac{z - 7}{- 4}$
Where the marks are
- Direction $b_{1} \times b_{2} = 20 \hat{i} + 10 \hat{j} - 8 \hat{k}$ (1 mark)
- Vector equation (1 mark)
- Cartesian equation (1 mark)
Open Q28(b) on its own page
Answer one of these 2 alternatives.
Q29(a) · 3 marks · Short answerIntegrals
Evaluate: $\int {\frac{1}{l o g x} - \frac{1}{( l o g x ) ^{2}}} d x$ ; (where $x > 1$ ).
Try it first, then show our solutionSolution
1. Split: $I = \int \frac{1}{l o g x} d x - \int \frac{1}{( l o g x ) ^{2}} d x$ . Integrate the first by parts with $\frac{1}{l o g x}$ as first function: $\int \frac{1}{l o g x} d x = \frac{1}{l o g x} \int d x - \int {\frac{d}{d x} (\frac{1}{l o g x}) \int d x} d x$ .
  1 mark
2. $= \frac{x}{l o g x} - \int (- \frac{1}{( l o g x ) ^{2}} \cdot \frac{1}{x}) x d x = \frac{x}{l o g x} + \int \frac{1}{( l o g x ) ^{2}} d x$ .
  1 mark
3. So $I = \frac{x}{l o g x} + \int \frac{d x}{( l o g x ) ^{2}} - \int \frac{d x}{( l o g x ) ^{2}} = \frac{x}{l o g x} + C$ .
  1 mark
Answer: $\frac{x}{lo g x} + C$
Where the marks are
- Splits and applies integration by parts to $\int \frac{d x}{l o g x}$ (1 mark)
- Correct derivative of $\frac{1}{l o g x}$ giving $+ \int \frac{d x}{( l o g x ) ^{2}}$ (1 mark)
- Cancellation to get $\frac{x}{l o g x} + C$ (1 mark)
Open Q29(a) on its own page
OR
Q29(b) · 3 marks · Short answerIntegrals
Evaluate : $\int_{0}^{1} x (1 - x)^{n} d x$ ; (where $n \in N$ ).
Try it first, then show our solutionSolution
1. Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{1} (1 - x) {1 - (1 - x)}^{n} d x = \int_{0}^{1} x^{n} (1 - x) d x$ .
  1 mark
2. $I = \int_{0}^{1} x^{n} d x - \int_{0}^{1} x^{n + 1} d x$ .
  ½ mark
3. $= \frac{1}{n + 1} [x^{n + 1}]_{0}^{1} - \frac{1}{n + 2} [x^{n + 2}]_{0}^{1}$ .
  ½ mark
4. $= \frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{( n + 1 ) ( n + 2 )}$ .
  1 mark
Answer: $\frac{1}{( n + 1 ) ( n + 2 )}$
Where the marks are
- Applies property $f (a - x)$ to get $\int_{0}^{1} x^{n} (1 - x) d x$ (1 mark)
- Splits into two integrals (½ mark)
- Integrates correctly (½ mark)
- Final value $\frac{1}{( n + 1 ) ( n + 2 )}$ (1 mark)
Open Q29(b) on its own page
Q30 · 3 marks · Short answerLinear Programming
Consider the following Linear Programming Problem:
Minimise $Z = x + 2 y$
Subject to $2 x + y \geq 3$ , $x + 2 y \geq 6$ , $x, y \geq 0$ .
Show graphically that the minimum of $Z$ occurs at more than two points
Try it first, then show our solutionSolution
1. Graph: draw $2 x + y = 3$ through $(\frac{3}{2}, 0)$ and $(0, 3)$ , and $x + 2 y = 6$ through $(6, 0)$ and $(0, 3)$ . In the first quadrant the region satisfying both $\geq$ constraints lies on and above $x + 2 y = 6$ (since $3 - \frac{x}{2} \geq 3 - 2 x$ for $x \geq 0$ ); it is unbounded with corner points $A (6, 0)$ and $B (0, 3)$ .
  1 mark
2. Values of $Z = x + 2 y$ at corners:
  Corner point $Z = x + 2 y$
  $A (6, 0)$ 6
  $B (0, 3)$ 6
  1 mark
3. Since the region is unbounded, draw the open half-plane $x + 2 y < 6$ ; it has no point in common with the feasible region, so the minimum value of $Z$ is 6.
  ½ mark
4. $Z = 6$ at every point of segment $A B$ on $x + 2 y = 6$ , e.g. at $(2, 2)$ : $Z = 2 + 4 = 6$ . Hence the minimum 6 occurs at more than two points (all points of segment $A B$ ).
  ½ mark
Answer: Minimum $Z = 6$ , attained at every point of segment joining $(6, 0)$ and $(0, 3)$ (e.g. (2,2)) — more than two points.
Where the marks are
- Correct graph of feasible region with corners $(6, 0)$ , $(0, 3)$ (1 mark)
- Corner-point table with $Z = 6$ at both (1 mark)
- Checks $x + 2 y < 6$ has no common point, so min $= 6$ (½ mark)
- Shows another point such as $(2, 2)$ gives 6, so more than two points (½ mark)
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Answer one of these 2 alternatives.
Q31(a) · 3 marks · Short answerCompetency-basedProbability
The probability that it rains today is 0.4. If it rains today, the probability that it will rain tomorrow is 0.8. If it does not rain today, the probability that it will rain tomorrow is 0.7. If
$P_{1}$ : denotes the probability that it does not rain today.
$P_{2}$ : denotes the probability that it will not rain tomorrow, if it rains today.
$P_{3}$ : denotes the probability that it will rain tomorrow, if it does not rain today.
$P_{4}$ : denotes the probability that it will not rain tomorrow, if it does not rain today.
(i) Find the value of $P_{1} \times P_{4} - P_{2} \times P_{3}$ . [2 Marks]
(ii) Calculate the probability of raining tomorrow. [1 Mark]
Try it first, then show our solutionSolution
1. $P_{1} = 1 - 0.4 = 0.6$ ; $P_{2} = 1 - 0.8 = 0.2$ ; $P_{3} = 0.7$ ; $P_{4} = 1 - 0.7 = 0.3$ .
  1 mark
2. (i) $P_{1} \times P_{4} - P_{2} \times P_{3} = 0.6 \times 0.3 - 0.2 \times 0.7 = 0.18 - 0.14 = 0.04$ .
  1 mark
3. (ii) Let $E_{1}$ : rains today, $E_{2}$ : does not rain today, $A$ : rains tomorrow. $P (E_{1}) = 0.4$ , $P (E_{2}) = 0.6$ , $P (A ∣ E_{1}) = 0.8$ , $P (A ∣ E_{2}) = 0.7$ .
  ½ mark
4. By total probability, $P (A) = 0.4 \times 0.8 + 0.6 \times 0.7 = 0.32 + 0.42 = 0.74$ .
  ½ mark
Answer: (i) 0.04; (ii) 0.74
Where the marks are
- Finds $P_{1} = 0.6$ , $P_{2} = 0.2$ , $P_{3} = 0.7$ , $P_{4} = 0.3$ (1 mark)
- Computes $P_{1} P_{4} - P_{2} P_{3} = 0.04$ (1 mark)
- Defines events and conditional probabilities (½ mark)
- Total probability gives 0.74 (½ mark)
Open Q31(a) on its own page
OR
Q31(b) · 3 marks · Short answerProbability
A random variable $X$ can take all non-negative integral values and the probability that $X$ takes the value $r$ is proportional to $5^{- r}$ . Find $P (X < 3)$ .
Try it first, then show our solutionSolution
1. $P (X = r) = k \cdot \frac{1}{5 ^{r}}$ , $r = 0, 1, 2, \dots$ , where $k$ is a non-zero constant.
  ½ mark
2. So $P (X = 0) = k$ , $P (X = 1) = \frac{k}{5}$ , $P (X = 2) = \frac{k}{5 ^{2}}$ , $P (X = 3) = \frac{k}{5 ^{3}}, \dots$
  ½ mark
3. Sum of probabilities is 1: $k (1 + \frac{1}{5} + \frac{1}{5 ^{2}} + \dots) = 1$ .
  ½ mark
4. Infinite G.P.: $k \cdot \frac{1}{1 - \frac{1}{5}} = 1 \Rightarrow k = \frac{4}{5}$ .
  ½ mark
5. $P (X < 3) = P (0) + P (1) + P (2) = \frac{4}{5} (1 + \frac{1}{5} + \frac{1}{25}) = \frac{4}{5} \cdot \frac{31}{25} = \frac{124}{125}$ .
  1 mark
Answer: $P (X < 3) = \frac{124}{125}$
Where the marks are
- Writes $P (X = r) = k 5^{- r}$ (½ mark)
- Lists probabilities (½ mark)
- Uses total probability $= 1$ (½ mark)
- Sums G.P. to get $k = \frac{4}{5}$ (½ mark)
- $P (X < 3) = \frac{124}{125}$ (1 mark)
Open Q31(b) on its own page

Corner point	$Z = x + 2 y$
$A (6, 0)$	6
$B (0, 3)$	6

Section D

Q32 · 5 marks · Long answerApplication of Integrals
Draw the rough sketch of the curve $y = 20 cos 2 x$ ; (where $\frac{π}{6} \leq x \leq \frac{π}{3}$ ).
Using integration, find the area of the region bounded by the curve $y = 20 cos 2 x$ from the ordinates $x = \frac{π}{6}$ to $x = \frac{π}{3}$ and the $x$ -axis.
Try it first, then show our solutionSolution
1. Sketch: $y = 20 cos 2 x$ on $[\frac{π}{6}, \frac{π}{3}]$ passes through $(\frac{π}{6}, 10)$ , crosses the $x$ -axis at $(\frac{π}{4}, 0)$ and ends at $(\frac{π}{3}, - 10)$ ; it is above the $x$ -axis on $[\frac{π}{6}, \frac{π}{4}]$ and below it on $[\frac{π}{4}, \frac{π}{3}]$ . Shade the regions between the curve and the $x$ -axis.
  1 mark
2. Required area $= \int_{π /6}^{π /4} 20 cos 2 x d x + \int_{π /4}^{π /3} 20 cos 2 x d x$ .
  2 marks
3. $= 20 [\frac{s i n 2 x}{2}]_{π /6}^{π /4} + 20 [\frac{s i n 2 x}{2}]_{π /4}^{π /3} = 10 (1 - \frac{3}{2}) + 10 (\frac{3}{2} - 1)$ .
  1 mark
4. $= 10 (1 - \frac{3}{2}) + 10 (1 - \frac{3}{2}) = 20 (1 - \frac{3}{2}) = (20 - 103)$ sq. units.
  1 mark
Answer: $20 (1 - \frac{3}{2}) = 20 - 103$ sq. units
Where the marks are
- Correct rough sketch showing crossing at $x = π /4$ (1 mark)
- Sets up area as sum of two integrals, with modulus for the part below the axis (1+1 marks)
- Integrates correctly (1 mark)
- Final area $20 (1 - \frac{3}{2})$ sq units (1 mark)
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Q33 · 5 marks · Long answerCompetency-basedMatrices
The equation of the path traversed by the ball headed by the footballer is $y = a x^{2} + b x + c$ ; (where $0 \leq x \leq 14$ and $a, b, c \in R$ and $a \neq = 0$ ) with respect to a XY-coordinate system in the vertical plane. The ball passes through the points $(2, 15)$ , $(4, 25)$ and $(14, 15)$ . Determine the values of a, b and c by solving the system of linear equations in a, b and c, using matrix method. Also find the equation of the path traversed by the ball.
Try it first, then show our solutionSolution
1. Substituting the points in $y = a x^{2} + b x + c$ : $4 a + 2 b + c = 15$ , $16 a + 4 b + c = 25$ , $196 a + 14 b + c = 15$ .
  1 mark
2. Matrix form $A X = B$ with $A = 4161962414111$ , $X = a b c$ , $B = 152515$ .
  ½ mark
3. $∣ A ∣ = 4 (4 - 14) - 2 (16 - 196) + 1 (224 - 784) = - 40 + 360 - 560 = - 240 \neq = 0$ , so $A^{- 1}$ exists.
  ½ mark
4. Cofactors: $A_{11} = - 10, A_{12} = 180, A_{13} = - 560, A_{21} = 12, A_{22} = - 192, A_{23} = 336, A_{31} = - 2, A_{32} = 12, A_{33} = - 16$ . So $adj A = - 10 180 - 560 12 - 192 336 - 2 12 - 16$ .
  1 mark
5. $X = A^{- 1} B = - \frac{1}{240} - 10 180 - 560 12 - 192 336 - 2 12 - 16 152515 = - \frac{1}{240} 120 - 1920 - 240$ .
  1 mark
6. Hence $a = - \frac{1}{2}$ , $b = 8$ , $c = 1$ .
  ½ mark
7. Path of the ball: $y = - \frac{1}{2} x^{2} + 8 x + 1$ .
  ½ mark
Answer: $a = - \frac{1}{2}, b = 8, c = 1$ ; path $y = - \frac{1}{2} x^{2} + 8 x + 1$
Where the marks are
- Forms the three linear equations (1 mark)
- Writes $A X = B$ (½ mark)
- $∣ A ∣ = - 240 \neq = 0$ (½ mark)
- Correct adj $A$ (1 mark)
- Computes $X = A^{- 1} B$ (1 mark)
- $a = - \frac{1}{2}, b = 8, c = 1$ (½ mark)
- Equation of path (½ mark)
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Answer one of these 2 alternatives.
Q34(a) · 5 marks · Long answerContinuity and Differentiability
If $f : R \to R$ is defined by $f (x) = ∣ x ∣^{3}$ , show that $f^{''} (x)$ exists for all real $x$ and find it.
Try it first, then show our solutionSolution
1. $f (x) = ∣ x ∣^{3} = {x^{3}, - x^{3}, x \geq 0 x < 0$ . For $x \neq = 0$ , $f$ is a polynomial on each side, hence differentiable. At $x = 0$ : LHD $= lim_{x \to 0^{-}} \frac{- x ^{3} - 0}{x} = lim_{x \to 0^{-}} (- x^{2}) = 0$ .
  ½ mark
2. RHD $= lim_{x \to 0^{+}} \frac{x ^{3} - 0}{x} = lim_{x \to 0^{+}} x^{2} = 0$ .
  ½ mark
3. LHD = RHD, so $f$ is differentiable at $x = 0$ with $f^{'} (0) = 0$ .
  ½ mark
4. Therefore $f^{'} (x) = {3 x^{2}, - 3 x^{2}, x \geq 0 x < 0$ .
  1 mark
5. For $f^{'}$ at $x = 0$ : LHD $= lim_{x \to 0^{-}} \frac{- 3 x ^{2} - 0}{x} = lim_{x \to 0^{-}} (- 3 x) = 0$ .
  ½ mark
6. RHD $= lim_{x \to 0^{+}} \frac{3 x ^{2} - 0}{x} = lim_{x \to 0^{+}} 3 x = 0$ .
  ½ mark
7. LHD = RHD, so $f^{'}$ is differentiable at $0$ (and elsewhere, being polynomial on each side).
  ½ mark
8. Hence $f^{''} (x)$ exists for all real $x$ and $f^{''} (x) = {6 x, - 6 x, x \geq 0 x < 0$ , i.e. $f^{''} (x) = 6∣ x ∣$ .
  1 mark
Answer: $f^{''} (x) = {6 x, - 6 x, x \geq 0 x < 0 = 6∣ x ∣$ , exists for all $x \in R$
Where the marks are
- Writes $f$ piecewise and finds LHD of $f$ at 0 (½ mark)
- RHD of $f$ at 0 (½ mark)
- Concludes $f$ differentiable at 0 (½ mark)
- Writes $f^{'} (x)$ piecewise (1 mark)
- LHD of $f^{'}$ at 0 (½ mark)
- RHD of $f^{'}$ at 0 (½ mark)
- Concludes $f^{'}$ differentiable at 0 (½ mark)
- States $f^{''} (x) = 6 x$ ( $x \geq 0$ ), $- 6 x$ ( $x < 0$ ) (1 mark)
Open Q34(a) on its own page
OR
Q34(b) · 5 marks · Long answerContinuity and Differentiability
If $(x - a)^{2} + (y - b)^{2} = c^{2}$ , for some $c > 0$ , prove that $\frac{[ 1 + ( \frac{d y}{d x} ) ^{2} ] ^{\frac{3}{2}}}{\frac{d ^{2} y}{d x ^{2}}}$ is a constant independent of $a$ and $b$ .
Try it first, then show our solutionSolution
1. The relation is a circle with centre $(a, b)$ and radius $c$ , so take the parametric form $x - a = c cos θ$ and $y - b = c sin θ$ .
  ½ mark
2. Differentiate with respect to $θ$ : $\frac{d x}{d θ} = - c sin θ$ and $\frac{d y}{d θ} = c cos θ$ .
  ½ mark
3. Divide the two: $\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ} = \frac{c cos θ}{- c sin θ} = - cot θ$ .
  1 mark
4. Differentiate again with respect to $x$ using the chain rule: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (- cot θ) \cdot \frac{d θ}{d x} = cosec^{2} θ \cdot \frac{1}{- c sin θ}$ .
  ½ mark
5. Simplify: $\frac{d ^{2} y}{d x ^{2}} = - \frac{cosec ^{3} θ}{c}$ .
  ½ mark
6. Find the numerator: $[1 + (\frac{d y}{d x})^{2}]^{3/2} = (1 + cot^{2} θ)^{3/2} = (cosec^{2} θ)^{3/2} = cosec^{3} θ$ .
  ½ mark
7. Form the ratio: $\frac{[ 1 + ( \frac{d y}{d x} ) ^{2} ] ^{3/2}}{\frac{d ^{2} y}{d x ^{2}}} = \frac{cosec ^{3} θ}{- \frac{cosec ^{3} θ}{c}} = - c$ .
  1 mark
8. The value $- c$ is a constant and contains neither $a$ nor $b$ . Hence the expression is a constant independent of $a$ and $b$ . Hence proved.
  ½ mark
Answer: The expression equals $- c$ , a constant independent of $a$ and $b$ .
Where the marks are
- Uses the parametric substitution $x - a = c cos θ$ , $y - b = c sin θ$ (1/2 mark)
- Finds $d x / d θ$ and $d y / d θ$ (1/2 mark)
- Obtains $d y / d x = - cot θ$ (1 mark)
- Differentiates again using the chain rule (1/2 mark)
- Obtains $d^{2} y / d x^{2} = - cosec^{3} θ / c$ (1/2 mark)
- Simplifies $(1 + cot^{2} θ)^{3/2}$ to $cosec^{3} θ$ (1/2 mark)
- Gets the ratio $= - c$ (1 mark)
- Concludes that the value is independent of $a$ and $b$ (1/2 mark)
Open Q34(b) on its own page
Answer one of these 2 alternatives.
Q35(a) · 5 marks · Long answerThree Dimensional Geometry
Find the shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equations are
$r = (- \hat{i} - \hat{j} - \hat{k}) + λ (7 \hat{i} - 6 \hat{j} + \hat{k})$ and $r = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + μ (\hat{i} - 2 \hat{j} + \hat{k})$
where $λ$ and $μ$ are parameters.
Try it first, then show our solutionSolution
1. Compare each line with $r = a + λ b$ : $a_{1} = - \hat{i} - \hat{j} - \hat{k}$ , $b_{1} = 7 \hat{i} - 6 \hat{j} + \hat{k}$ , $a_{2} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$ , $b_{2} = \hat{i} - 2 \hat{j} + \hat{k}$ . Since $b_{1}$ is not a multiple of $b_{2}$ , the lines are not parallel.
  1 mark
2. Find the vector joining the given points: $a_{2} - a_{1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k}$ .
  ½ mark
3. Find the cross product: $b_{1} \times b_{2} = \hat{i} 71 \hat{j} - 6 - 2 \hat{k} 11 = \hat{i} (- 6 + 2) - \hat{j} (7 - 1) + \hat{k} (- 14 + 6) = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}$ .
  1½ marks
4. Find its magnitude: $∣ b_{1} \times b_{2} ∣ = 16 + 36 + 64 = 116$ .
  ½ mark
5. Find the dot product: $(b_{1} \times b_{2}) \cdot (a_{2} - a_{1}) = - 16 - 36 - 64 = - 116$ .
  ½ mark
6. Apply the shortest distance formula: $d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{∣ b _{1} \times b _{2} ∣} = \frac{116}{116} = 116 = 229$ units.
  1 mark
Answer: Shortest distance $= 229$ units
Where the marks are
- Identifies $a_{1}, b_{1}, a_{2}, b_{2}$ and notes the lines are non-parallel (1 mark)
- Finds $a_{2} - a_{1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k}$ (1/2 mark)
- Computes $b_{1} \times b_{2} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}$ (1½ marks)
- Finds $∣ b_{1} \times b_{2} ∣ = 116$ (1/2 mark)
- Finds the dot product $= - 116$ (1/2 mark)
- Gets the shortest distance $229$ units (1 mark)
Open Q35(a) on its own page
OR
Q35(b) · 5 marks · Long answerThree Dimensional Geometry
Find the image of the point $(1, 2, 1)$ with respect to the line $\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 1}{3}$ . Also find the equation of the line joining the given point and its image.
Try it first, then show our solutionSolution
1. Let $P (1, 2, 1)$ be the given point and $L$ the foot of the perpendicular from $P$ to the line. Put $\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 1}{3} = λ$ . Any point on the line is $(λ + 3, 2 λ - 1, 3 λ + 1)$ , so take $L$ as this point.
  ½ mark
2. Direction ratios of $P L$ are $(λ + 3 - 1, 2 λ - 1 - 2, 3 λ + 1 - 1) = (λ + 2, 2 λ - 3, 3 λ)$ .
  ½ mark
3. $P L$ is perpendicular to the line, whose direction ratios are $1, 2, 3$ . So $(λ + 2) (1) + (2 λ - 3) (2) + 3 λ (3) = 0$ , which gives $14 λ - 4 = 0$ and $λ = \frac{2}{7}$ .
  ½ mark
4. Substitute $λ = \frac{2}{7}$ : the foot is $L (\frac{23}{7}, - \frac{3}{7}, \frac{13}{7})$ .
  ½ mark
5. Let $Q (x_{1}, y_{1}, z_{1})$ be the image. Then $L$ is the mid-point of $P Q$ : $\frac{1 + x _{1}}{2} = \frac{23}{7}$ , $\frac{2 + y _{1}}{2} = - \frac{3}{7}$ , $\frac{1 + z _{1}}{2} = \frac{13}{7}$ .
  1 mark
6. Solve these: $x_{1} = \frac{39}{7}$ , $y_{1} = - \frac{20}{7}$ , $z_{1} = \frac{19}{7}$ . So the image is $Q (\frac{39}{7}, - \frac{20}{7}, \frac{19}{7})$ .
  1 mark
7. Direction ratios of $P Q$ are $(\frac{32}{7}, - \frac{34}{7}, \frac{12}{7})$ , which are proportional to $(16, - 17, 6)$ . The line joining $P$ and $Q$ is $\frac{x - 1}{16} = \frac{y - 2}{- 17} = \frac{z - 1}{6}$ .
  1 mark
Answer: Image $(\frac{39}{7}, - \frac{20}{7}, \frac{19}{7})$ ; line: $\frac{x - 1}{16} = \frac{y - 2}{- 17} = \frac{z - 1}{6}$
Where the marks are
- Writes the general point $(λ + 3, 2 λ - 1, 3 λ + 1)$ for the foot $L$ (1/2 mark)
- Finds the direction ratios of $P L$ (1/2 mark)
- Uses the perpendicularity condition to get $λ = 2/7$ (1/2 mark)
- Finds the foot $L (23/7, - 3/7, 13/7)$ (1/2 mark)
- Uses $L$ as the mid-point of $P Q$ (1 mark)
- Finds the image $Q (39/7, - 20/7, 19/7)$ (1 mark)
- Writes the equation of line $P Q$ (1 mark)
Open Q35(b) on its own page

Section E

Case study

Q36 · 4 marks · Case-basedCompetency-basedApplication of Derivatives
Ramesh, the owner of a sweet selling shop, purchased some rectangular card board sheets of dimension $25 c m$ by $40 c m$ to make container packets without top. Let $x c m$ be the length of the side of the square to be cut out from each corner to give that sheet the shape of the container by folding up the flaps.
Based on the above information answer the following questions.
1. (i)
  Express the volume (V) of each container as function of $x$ only.
  1 mark
2. (ii)
  Find $\frac{d V}{d x}$
  1 mark
3. (iii)
  (a) For what value of $x$ , the volume of each container is maximum?
  2 marks
4. OR
  (iii)
  (b) Check whether V has a point of inflection at $x = \frac{65}{6}$ or not?
  2 marks
Try it first, then show our solutionSolution
1. (i) After cutting squares of side $x$ and folding, the base is $(40 - 2 x)$ by $(25 - 2 x)$ and the height is $x$ . So $V = x (40 - 2 x) (25 - 2 x) = 4 x^{3} - 130 x^{2} + 1000 x$ cm³, where $0 < x < 12.5$ .
  1 mark
2. (ii) Differentiate: $\frac{d V}{d x} = 12 x^{2} - 260 x + 1000 = 4 (3 x^{2} - 65 x + 250) = 4 (3 x - 50) (x - 5)$ .
  1 mark
3. (iii)(a) For extreme values put $\frac{d V}{d x} = 0$ , i.e. $4 (3 x - 50) (x - 5) = 0$ .
  ½ mark
4. (iii)(a) This gives $x = 5$ or $x = \frac{50}{3}$ . Reject $x = \frac{50}{3}$ , since then $25 - 2 x < 0$ ; so $x = 5$ .
  ½ mark
5. (iii)(a) Second derivative: $\frac{d ^{2} V}{d x ^{2}} = 24 x - 260$ .
  ½ mark
6. (iii)(a) At $x = 5$ : $\frac{d ^{2} V}{d x ^{2}} = 120 - 260 = - 140 < 0$ . So $V$ is maximum when $x = 5$ cm.
  ½ mark
7. (iii)(b) We have $\frac{d V}{d x} = 4 (3 x^{2} - 65 x + 250)$ .
  ½ mark
8. (iii)(b) So $\frac{d ^{2} V}{d x ^{2}} = 4 (6 x - 65)$ , which is $0$ at $x = \frac{65}{6}$ . This point lies in $(0, 12.5)$ .
  ½ mark
9. (iii)(b) For $x$ slightly less than $\frac{65}{6}$ , $6 x - 65 < 0$ , so $\frac{d ^{2} V}{d x ^{2}} < 0$ (concave down). For $x$ slightly more than $\frac{65}{6}$ , $\frac{d ^{2} V}{d x ^{2}} > 0$ (concave up).
  ½ mark
10. (iii)(b) The concavity changes sign at this point. Hence $V$ has a point of inflection at $x = \frac{65}{6}$ .
  ½ mark
Answer: (i) $V = x (40 - 2 x) (25 - 2 x)$ cm³; (ii) $\frac{d V}{d x} = 4 (3 x - 50) (x - 5)$ ; (iii)(a) $x = 5$ cm; (iii)(b) Yes, $x = \frac{65}{6}$ is a point of inflection
Where the marks are
- (i) Writes $V = x (40 - 2 x) (25 - 2 x)$ (1 mark)
- (ii) Gets $d V / d x = 4 (3 x - 50) (x - 5)$ (1 mark)
- (iii)(a) Sets $d V / d x = 0$ (1/2 mark)
- (iii)(a) Finds the critical points $x = 5$ and $50/3$ , rejecting $50/3$ (1/2 mark)
- (iii)(a) Finds $d^{2} V / d x^{2} = 24 x - 260$ (1/2 mark)
- (iii)(a) Shows $d^{2} V / d x^{2} = - 140 < 0$ at $x = 5$ , so the volume is maximum (1/2 mark)
- (iii)(b) Writes $d V / d x$ (1/2 mark)
- (iii)(b) Shows $d^{2} V / d x^{2} = 4 (6 x - 65) = 0$ at $x = 65/6$ (1/2 mark)
- (iii)(b) Shows the sign change of $d^{2} V / d x^{2}$ about $65/6$ (1/2 mark)
- (iii)(b) Concludes it is a point of inflection (1/2 mark)
Open Q36 on its own page
Q37 · 4 marks · Case-basedCompetency-basedRelations and Functions
An organization conducted bike race under 2 different categories-boys and girls. In all, there were 250 participants. Among all of them finally three from Category 1 and two from Category 2 were selected for the final race. Ravi forms two sets $B$ and $G$ with these participants for his college project.
Let $B = {b_{1}, b_{2}, b_{3}}$ , $G = {g_{1}, g_{2}}$ where $B$ represents the set of boys selected and $G$ the set of girls who were selected for the final race.
Ravi decides to explore these sets for various types of relations and functions.
On the basis of the above information, answer the following questions:
1. (i)
  Ravi wishes to form all the relations possible from $B$ to $G$ . How many such relations are possible?
  1 mark
2. (ii)
  Write the smallest equivalence relation on G.
  1 mark
3. (iii)
  (a) Ravi defines a relation from B to B as $R_{1} = {(b_{1}, b_{2}), (b_{2}, b_{1})}$ . Write the minimum ordered pairs to be added in $R_{1}$ so that it becomes (A) reflexive but not symmetric, (B) reflexive and symmetric but not transitive.
  2 marks
4. OR
  (iii)
  (b) If the track of the final race (for the biker $b_{1}$ ) follows the curve $x^{2} = 4 y$ ; (where $0 \leq x \leq 202$ & $0 \leq y \leq 200$ ), then state whether the track represents a one-one and onto function or not. (Justify).
  2 marks
Try it first, then show our solutionSolution
1. (i) A relation from $B$ to $G$ is any subset of $B \times G$ . Here $n (B \times G) = 3 \times 2 = 6$ , so the number of relations is $2^{6} = 64$ .
  1 mark
2. (ii) An equivalence relation must at least be reflexive. So the smallest equivalence relation on $G$ is ${(g_{1}, g_{1}), (g_{2}, g_{2})}$ , which is also symmetric and transitive.
  1 mark
3. (iii)(a)(A) For reflexivity, add $(b_{1}, b_{1}), (b_{2}, b_{2}), (b_{3}, b_{3})$ . To break symmetry, add one pair such as $(b_{2}, b_{3})$ without its reverse. So the minimum pairs to add are $(b_{1}, b_{1}), (b_{2}, b_{2}), (b_{3}, b_{3}), (b_{2}, b_{3})$ . Any one of $(b_{3}, b_{2}), (b_{1}, b_{3}), (b_{3}, b_{1})$ may be used instead of $(b_{2}, b_{3})$ .
  1 mark
4. (iii)(a)(B) Add $(b_{1}, b_{1}), (b_{2}, b_{2}), (b_{3}, b_{3}), (b_{2}, b_{3}), (b_{3}, b_{2})$ . The relation is now reflexive and symmetric. It is not transitive, because $(b_{1}, b_{2}) \in R$ and $(b_{2}, b_{3}) \in R$ but $(b_{1}, b_{3}) \in / R$ .
  1 mark
5. (iii)(b) One-one: $y = f (x) = \frac{x ^{2}}{4}$ , $f : [0, 202] \to [0, 200]$ . If $f (x_{1}) = f (x_{2})$ , then $x_{1}^{2} = x_{2}^{2}$ , so $(x_{1} - x_{2}) (x_{1} + x_{2}) = 0$ . Since $x_{1}, x_{2} \geq 0$ , this gives $x_{1} = x_{2}$ . So $f$ is one-one.
  1 mark
6. (iii)(b) Onto: for any $y \in [0, 200]$ , take $x = 2 y$ . Then $0 \leq x \leq 2200 = 202$ and $f (2 y) = \frac{4 y}{4} = y$ . Every $y$ has a pre-image, so $f$ is onto. Hence the track represents a one-one and onto function.
  1 mark
Answer: (i) $2^{6} = 64$ ; (ii) ${(g_{1}, g_{1}), (g_{2}, g_{2})}$ ; (iii)(a) (A) add $(b_{1}, b_{1}), (b_{2}, b_{2}), (b_{3}, b_{3}), (b_{2}, b_{3})$ ; (B) add $(b_{1}, b_{1}), (b_{2}, b_{2}), (b_{3}, b_{3}), (b_{2}, b_{3}), (b_{3}, b_{2})$ ; (iii)(b) Yes, it is one-one and onto
Where the marks are
- (i) Gets $2^{3 \times 2} = 64$ relations (1 mark)
- (ii) Writes ${(g_{1}, g_{1}), (g_{2}, g_{2})}$ (1 mark)
- (iii)(a)(A) Adds the three diagonal pairs plus one non-symmetric pair (1 mark)
- (iii)(a)(B) Adds the diagonal pairs plus $(b_{2}, b_{3}), (b_{3}, b_{2})$ (1 mark)
- (iii)(b) Proves one-one using $x \geq 0$ (1 mark)
- (iii)(b) Proves onto using the pre-image $2 y$ (1 mark)
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Q38 · 4 marks · Case-basedCompetency-basedProbability
Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage-II contains 6 parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously). Then two birds flew back from cage-II to cage-I (simultaneously).
Assume that all the birds have equal chances of flying.
On the basis of the above information, answer the following questions:-
1. (i)
  When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I then find the probability that the owl is still in Cage-I.
  2 marks
2. (ii)
  When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, the owl is still seen in Cage-I, what is the probability that one parrot and the owl flew from Cage-I to Cage-II?
  2 marks
Try it first, then show our solutionSolution
1. (i) Define the events: $E_{1}$ = one parrot and the owl fly from Cage-I, $E_{2}$ = two parrots fly from Cage-I, $A$ = the owl is still in Cage-I at the end. Two birds are chosen from 6 in $^{6} C_{2} = 15$ ways, so $P (E_{1}) = \frac{5 \times 1}{15} = \frac{1}{3}$ and $P (E_{2}) = \frac{^{5} C _{2}}{15} = \frac{10}{15} = \frac{2}{3}$ .
  ½ mark
2. (i) Under $E_{1}$ , Cage-II has 7 parrots and 1 owl. The owl is back in Cage-I only if it returns with one parrot, so $P (A ∣ E_{1}) = \frac{7 \times 1}{^{8} C _{2}} = \frac{7}{28} = \frac{1}{4}$ . Under $E_{2}$ , the owl never left, so $P (A ∣ E_{2}) = 1$ . By the theorem of total probability, $P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) = \frac{1}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot 1$ .
  1 mark
3. (i) Simplify: $P (A) = \frac{1}{12} + \frac{8}{12} = \frac{9}{12} = \frac{3}{4}$ . As a count check, $\frac{35 + 280}{420} = \frac{3}{4}$ .
  ½ mark
4. (ii) The required probability is $P (E_{1} ∣ A)$ .
  ½ mark
5. (ii) By Bayes' theorem, $P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} )}$ .
  ½ mark
6. (ii) Substitute: $P (E_{1} ∣ A) = \frac{1/12}{3/4} = \frac{1}{9}$ .
  1 mark
Answer: (i) $\frac{3}{4}$ ; (ii) $\frac{1}{9}$
Where the marks are
- (i) Defines the events and finds the probabilities of the two ways the birds can leave Cage-I (1/2 mark)
- (i) Applies total probability with the correct conditional probabilities (1 mark)
- (i) Gets $P (A) = 3/4$ (1/2 mark)
- (ii) Identifies the required probability as $P (E_{1} ∣ A)$ (1/2 mark)
- (ii) Writes Bayes' theorem (1/2 mark)
- (ii) Gets $1/9$ (1 mark)
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Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.