ISC Class 12 Specimen Paper 2026 Maths

MATHEMATICS
Maximum Marks: 80
Time Allotted: Three Hours
Reading Time: Additional Fifteen minutes

Instructions to Candidates
1. You are allowed an additional fifteen minutes for only reading the paper.
2. You must NOT start writing during reading time.
3. The question paper has 14 printed pages.
4. The Question Paper is divided into three sections and has 22 questions in all.
5. Section A is compulsory and has fourteen questions.
6. You are required to attempt all questions either from Section B or Section C.
7. Section B and Section C have four questions each.
8. Internal choices have been provided in two questions of 2 marks, two
questions of 4 marks and two questions of 6 marks in Section A.
9. Internal choices have been provided in one question of 2 marks and one
question of 4 marks each in Section B and Section C.
10. While attempting Multiple Choice Questions in Section A, B and C, you
are required to write only ONE option as the answer.
11. The intended marks for questions or parts of questions are given in the
brackets [].
12. All workings, including rough work, should be done on the same page as, and
adjacent to, the rest of the answer.
13. Mathematical tables and graph papers are provided.
Instruction to Supervising Examiner
1. Kindly read aloud the instructions given above to all the candidates present in
the examination hall.

ISC (CLASS XII) SPECIMEN QUESTION PAPER 2026
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Note: The Specimen Question Paper in the subject provides a realistic format of the
Board Examination Question Paper and should be used as a practice tool. The questions for the
Board Examination can be set from any part of the syllabus, though the format of the Board
Examination Question Paper will remain the same as that of the Specimen Question Paper.
The weightage allocated to various topics, as given in the syllabus, will be strictly adhered to.

SECTION A - 65 MARKS

Question 1
In subparts (i) to (xi) choose the correct options and in subparts (xii) to (xv), answer the
questions as instructed.
(i) If A and B are square matrices of order 3, A is non-singular matrix and AB = O, [1]
then the matrix B is: (Understanding)
(a) unit matrix
(b) Scalar matrix
(c) non-singular matrix
(d) null matrix
(ii) If 𝑚 and 𝑛 are respectively the order and degree of the differential equation [1]
𝑑 𝑑𝑦 3
( ) = 0 then the value of (𝑚 − 𝑛) is: (Recall)
𝑑𝑥 𝑑𝑥

(a) 0
(b) 1
(c) 2
(d) 3
(iii) The derivative of 𝑥𝑦 = 𝑐 2 with respect to 𝑥 is: (Understanding) [1]
(a) 𝑑𝑦 2𝑐 − 𝑦
=
𝑑𝑥 𝑥
(b) 2
𝑑𝑦 𝑐
=
𝑑𝑥 𝑥 2
(c) 𝑑𝑦 −𝑦
=
𝑑𝑥 𝑥
(d) 𝑑𝑦 𝑦
=
𝑑𝑥 𝑥

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2𝑎 2𝑏 2𝑐 [1]
(iv) Consider ∆= |2𝑒 𝑓 𝑔|
2𝑖 𝑗 𝑘
𝑎 𝑏 𝑐
Assertion: The value of ∆ = 4 × | 𝑒 𝑓 𝑔|
𝑖 𝑗 𝑘
Reason: If all elements of one row or one column of a determinant are multiplied
by a scalar, 𝑘 then the value of the determinant is multiplied by 𝑘. (Analysis)
Which of the following is correct?
(a) Both Assertion and Reason are true, and Reason is the correct explanation for
Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation
for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
(v) Five numbers 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , 𝑥5 are randomly selected from the numbers 1, 2, 3, [1]
……., 18 and are arranged in the increasing order such that 𝑥1 < 𝑥2 < 𝑥3 < 𝑥4 < 𝑥5.
What is the probability that 𝑥2 = 7 and 𝑥4 = 11? (Application)
(a) 26
51

(b) 3
104

(c) 1
68

(d) 1
34

(vi) 𝑎 0 0 [1]
If 𝐴 = (0 𝑎 0), then 𝐴𝑛 equals to (Understanding)
0 0 𝑎
(a) 𝑎𝑛 0 0
( 0 𝑎𝑛 0 )
0 0 𝑎𝑛
(b) 𝑎 0 0
(0 𝑎𝑛 0)
0 0 𝑎
(c) 𝑎𝑛 0 0
(0 𝑎 0)
0 0 𝑎𝑛
(d) 𝑛𝑎 0 0
( 0 𝑛𝑎 0 )
0 0 𝑎𝑛

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(vii) Observe the following graphs (a), (b), (c), and (d), each representing different types [1]
of functions.

Statement 1: A function which is continuous at a point may not be differentiable at
that point.
Statement 2: Graph (c) is an example of a function that is continuous but not
differentiable at the origin. (Application)
Which of the following is correct?
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 2 is true and Statement 1 is false.
(c) Both the statements are true.
(d) Both the statements are false.
(viii) If 𝑓(𝑥) = 𝑘𝑥 2 + 7𝑥 − 4 and 𝑓 ′ (5) = 97 then what is the value of 𝑘? [1]
(Understanding)
(a) −4
(b) 0
(c) 4
(d) 9
(ix) Statement 1: If a relation 𝑅 on a set 𝐴 satisfies 𝑅 = 𝑅 −1, then 𝑅 is symmetric. [1]
Statement 2: For a relation 𝑅 to be symmetric, it is necessary that 𝑅 = 𝑅 −1
Which one of the following is correct? (Understanding)
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 2 is true and Statement 1 is false.
(c) Both the statements are true.
(d) Both the statements are false.

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(x) Assertion: The equality tan(cot −1 𝑥) = cot(tan−1 𝑥), is true for all 𝑥 ∈ 𝑅 [1]
𝜋
Reason: The identity tan−1 𝑥 + cot −1 𝑥 = 2 , is true for all 𝑥 ∈ 𝑅
Which of the following is correct? (Application)
(a) Both Assertion and Reason are true and Reason is the correct explanation for
Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation
for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
(xi) 𝐴
Given two events A and B such that P (𝐵)= 0.25 and 𝑃 (𝐴 ∩ 𝐵) = 0∙12. [1]

The value 𝑃 (𝐴’ ∩ 𝐵) is: (Understanding)
(a) 0·36
(b) 0·48
(c) 0·88
(d) 0·036
(xii) The value of the determinant of a matrix A of order 3 is 3. If C is the matrix of [1]
cofactors of the matrix A, then what is the value of determinant of C2? (Analysis)
(xiii) If a relation R on the set {𝑎, 𝑏, 𝑐} defined by R = {(𝑏, 𝑏)}, then classify the relation. [1]
(Understanding)
(xiv) [1]

The given function 𝑓: 𝑅 → 𝑅 is many to one function. Give reason.
(Understanding)
(xv) There are three machines and 2 of them are faulty. They are tested one by one in a [1]
random order till both the faulty machines are identified. What is the probability
that only two tests are needed to identify the faulty machines? (Application)

Question 2 [2]
𝑥
(i) 𝑑𝑦 𝑥−𝑦
(Understanding)
If 𝑥 = 𝑒 𝑦 , then prove that 𝑑𝑥 = 𝑥 𝑙𝑜𝑔 𝑥
OR

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(ii) Find the values of ′𝑎′ for which the function 𝑓(𝑥) = 𝑏 − 𝑎𝑥 + 𝑠𝑖𝑛 𝑥 is increasing on
R. (Understanding)

Question 3 [2]
Find the point on the curve 𝑦 = (𝑥 − 2)2 a t w h i c h t h e tangent is parallel to the chord joining the
end points (2,0) and (4,4). (Analysis)

Question 4 [2]
𝑑𝑦
Show that the general solution of the differential equation: = 𝑦𝑐𝑜𝑡2𝑥 is
𝑑𝑥
1
log 𝑦 = 2 log|𝑠𝑖𝑛2𝑥| + 𝐶 (Evaluate)

Question 5 [2]
(i) If ∫ 𝑥 5 cos( 𝑥 6 ) 𝑑𝑥 = 𝑘 sin(𝑥 6 ) + 𝐶, find the value of ‘𝑘’. (Evaluate)
OR
(ii) Evaluate: ∫0
5 √𝑥
𝑑𝑥 (Evaluate)
√5−𝑥 +√𝑥

Question 6 [2]
𝑥
A music streaming app uses the function: 𝑓(𝑥) = tan−1 (10) to assign a mood score based
on the number of hours a user listens to music per week.
Let the listening times (in hours/week) of two users, User A and User B, be 6 and 8
respectively.
Compute the combined mood score of user A and user B, that is, 𝑓(6) + 𝑓(8).
(Application)

Question 7 [4]
Solve: sin−1(𝑥) + sin−1(1 − 𝑥) = cos −1 𝑥. (Application)

Question 8 [4]
𝑑2 𝑦 𝑑𝑦
If 𝑦 = (𝐴 + 𝐵𝑥)𝑒 −2𝑥 , prove that: 𝑑𝑥 2 + 4 𝑑𝑥 + 4𝑦 = 0 (Application)

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Question 9 [4]
(i) Solve the following differential equation:

𝑑𝑦 𝑦
𝑥 𝑑𝑥 = 𝑦 − 𝑥 𝑡𝑎𝑛 ( 𝑥 ) (Application)
OR
(ii) Solve the following differential equation:
𝑑𝑦
(1 + 𝑥 2 ) + 2𝑥𝑦 = 4𝑥 2 (Application)
𝑑𝑥

Question 10 [4]
(i) Three friends go to a restaurant to have pizza. They decide who will pay for the pizza
by tossing a coin. It is decided that each one of them will toss a coin and if one person
gets a different result (heads or tails) than the other two, that person would pay. If all
three get the same result (all heads or all tails), they will toss again until they get a
different result. (Analysis)
(a) What is the probability that all three friends will get the same result (all heads
or all tails) in one round of tossing?
(b) What is the probability that they will get a different result in one round of
tossing?
(c) What is the probability that they will need exactly four rounds of tossing to
determine who would pay?
OR
(ii) A school offers students the choice of three modes for attending classes:
• Mode A: Offline (in-person) – 40% of students
• Mode B: Online (live virtual classes) – 35% of students
• Mode C: Recorded lectures – 25% of students
After a feedback survey:
• 20% of students from Mode A reported the class as “Excellent”
• 30% from Mode B rated it as “Excellent”
• 50% from Mode C rated it as “Excellent”
A student is selected at random from the entire group, and it is found that they rated
the class as “Excellent.” (Analysis)
(a) Represent the data in terms of probability. Define the events clearly.
(b) Using Bayes’ Theorem, find the probability that the student attended the
Recorded lectures (Mode C), given that they rated the class as “Excellent.”
(c) Interpret your result. Which mode has the highest likelihood of being chosen
if a student says “Excellent”?

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Question 11 [6]
To raise money for an orphanage, students of three schools A, B and C organised an
exhibition in their residential colony, where they sold paper bags, scrap books and pastel
sheets made by using recycled paper. Student of school A sold 30 paper bags, 20 scrap books
and 10 pastel sheets and raised ₹ 410. Student of school B sold 20 paper bags, 10 scrap books
and 20 pastel sheets and raised ₹ 290. Student of school C sold 20 paper bags, 20 scrap books
and 20 pastel sheets and raised ₹ 440.
Answer the following question:
(i) Translate the problem into a system of equations.
(ii) Solve the system of equation by using matrix method.
(iii) Hence, find the cost of one paper bag, one scrap book and one pastel sheet.
(Application)

Question 12 [6]
(i) 𝑥 2 +𝑥+1
Evaluate: ∫ (𝑥+2 )(𝑥 2 +1) 𝑑𝑥 (Evaluate)
OR
(ii) 3𝜋
𝑥 𝑑𝑥
Evaluate: ∫𝜋4 1+ sin 𝑥 𝑑𝑥 (Evaluate)
4

Question 13 [6]
(i) A person has manufactured a water tank in the shape of a closed right circular
539
cylinder. The volume of the cylinder is 2 cubic units. If the height and radius of the
cylinder be h and r. (Application)

(a) Express h in terms of radius r and given volume.
(b) Let the total surface area of the closed cylinder tank be S, express S in term of
radius r.
(c) 7
If the total surface area of the tank is minimum, then prove that radius r = 2
units.
(d) Find the height of the tank.
OR
ISC (CLASS XII) SPECIMEN QUESTION PAPER 2026
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(ii) A Dolphin jumps and taken a path given by the equation
1
ℎ(𝑡) = 2 (−7𝑡 2 + 3𝑡 + 2), (𝑡 ≥ 0) , ℎ(𝑡) is the height of the Dolphin at any point of
time. (Application)

(a) Is the function differentiable for 𝑡 ≥ 0? Justify.
1
(b) Find the instantaneous rate of change of height at 𝑡 = 14 .
3
(c) ℎ(𝑡) is increasing in (−∞, 14). Is this true or false? Justify.

(d) Find the time at which the Dolphin will attains the maximum height. Also find the
maximum height.

Question 14 [6]

In a school, three subject teachers English, Math, and Science sometimes give surprise tests
on the same day. Based on past records:

The English teacher gives a test 90% of the time
•
• The Math teacher gives a test 80% of the time
• The Science teacher gives a test 70% of the time
Each teacher decides independently. If the average number of surprise tests is less than 2.3
then the teachers should coordinate better to increase the performance of the students.
Otherwise, no action is needed.
Let 𝑋 be the number of surprise tests a student gets on a given day.
So, 𝑋 ∈ {0,1,2,3}. (Application)
(i) Find the probability for each possible number of surprise tests.
(ii) Use the probabilities to build a distribution table.
(iii) Calculate the average number of surprise tests per day.
(iv) Based on your calculations, decide: Should the teachers coordinate better? Or is the
current plan acceptable?

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SECTION B - 15 MARKS

Question 15 [5]
In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the
questions as instructed.
(i) Consider the following statements and choose the correct option:
Statement 1: If 𝑎⃗ and 𝑏⃗⃗ represents two adjacent sides of a parallelogram then the
diagonals are represented by 𝑎⃗ + 𝑏⃗⃗ and 𝑎⃗ − 𝑏⃗⃗ .
Statement 2: If 𝑎⃗ and 𝑏⃗⃗ represents two diagonals of a parallelogram then the adjacent
sides are represented by 2(𝑎⃗ + 𝑏⃗⃗ ) and 2(𝑎⃗ − 𝑏⃗⃗).
Which of the following is correct? (Recall)
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 2 is true and Statement 1 is false.
(c) Both the statements are true.
(d) Both the statements are false.
(ii) A plane passes through three points A, B and C with position vectors 𝑖̂ + 𝑗̂, 𝑗̂ + 𝑘̂ and
𝑘̂ + 𝑖̂ respectively. The equation of the line passing through the point P with position
vector 𝑖̂ + 2𝑗̂ + 2𝑘̂ and normal to the plane is (Application)
(a) 𝑟⃗ = (𝑖̂ + 2𝑗̂ + 2𝑘̂) + 𝜆(𝑖̂ + 𝑗̂ + 𝑘̂), 𝜆 ∈ 𝑅

(b) 𝑟⃗ = (𝑖̂ + 𝑗̂ + 𝑘̂) + 𝜆(𝑖̂ + 2𝑗̂ + 2𝑘̂), 𝜆 ∈ 𝑅

(c) 𝑟⃗ ⋅ (𝑖̂ − 𝑗̂ − 𝑘̂) = 𝑖̂ + 2𝑗̂ + 2𝑘̂

(d) 𝑥−1=𝑦 =𝑧
(iii) If the direction cosines of a line are < 1 , 1 , 1 > then (Understanding)
𝑐 𝑐 𝑐

(a) c>0
(b) 0<c<1
(c) c = ±√3
(d) 𝑐>2
(iv) If 𝑎⃗ and 𝑏⃗⃗ are unit vectors enclosing an angle 𝜃 and |𝑎⃗ + 𝑏⃗⃗| < 1, then find the values
between which 𝜃 lies. (Understanding)

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(v) Shown below is a cuboid. Find ⃗⃗⃗⃗⃗⃗
𝐵𝐴 . 𝐵𝐶⃗⃗⃗⃗⃗⃗

(Analysis)

Question 16 [2]
(i)

A building is to be constructed in the form of a triangular pyramid ABCD as shown
in the figure. Let the angular points be 𝐴 (0, 1, 2), 𝐵 (3, 0, 1), 𝐶 (4, 3, 6) and
𝐷 (2, 3, 2) and let G be the point of intersection of the medians of ∆BCD.
Using the above information, answer the following (Application)

(a) What will be the length of vector ⃗⃗⃗⃗⃗⃗
𝐴𝐺 ?
(b) Find the area of ∆ABC.
OR
(ii) What are the values of 𝑥 for which the angle between the vectors?
2𝑥 2 𝑖̂ + 3𝑥𝑗̂ + 𝑘̂ and 𝑖̂ − 2𝑗̂ + 𝑥 2 𝑘̂ is obtuse? (Application)

Question 17 [4]
(i)

𝑥+3 𝑦−1 𝑧+4
Given, B and C lie on the line = = and BC = 5 units find the area of
5 2 3
𝛥𝐴𝐵𝐶. (Application)
OR
ISC (CLASS XII) SPECIMEN QUESTION PAPER 2026
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(ii) Find the equation of the plane containing the line 𝑥 = 𝑦−1 = 1−𝑧 and the
−2 3 1
point (−1, 0, 2). (Application)

Question 18 [4]

Find the area bounded by the curve, 𝑦 = 𝑥(4 − 𝑥) and the 𝑥-axis from 𝑥 = 0 to 𝑥 = 5 as
shown in the figure given above. (Application)

SECTION C - 15 MARKS

Question 19 [5]
In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the
questions as instructed.
(i) Which condition is true if Average Cost (AC) is constant at all levels of output?
(Recall)
(a) MC > AC
(b) MC = AC
(c) MC < AC
(d) 1
MC = 2 AC
(ii) Which of the following statement(s) is/are correct with respect to regression
coefficients?
Statement 1: It measures the degree of linear relationship between two variables.
Statement 2: It gives the value by which one variable changes for a unit change in the
other variable.
Which of the following is correct? (Recall)
(a) Statement 1 is true and Statement 2 is false.
(b) Statement 2 is true and Statement 1 is false.
(c) Both the statements are true.
(d) Both the statements are false.

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(iii) Mean of 𝑥 = 53, mean of 𝑦 = 28 regression co-efficient 𝑦 𝑜𝑛 𝑥 = −1 · 2, regression
co-efficient 𝑥 𝑜𝑛 𝑦 = −0 · 3. Find coefficient of correlation (r). (Understanding)
(iv) The total revenue received from the sale of 𝑥 unit of a product is given by
𝑅(𝑥) = 3𝑥 2 + 36𝑥 + 5. Find the marginal revenue when 𝑥 = 5. (Understanding)
(v) A manufacturing company finds that the daily cost of producing 𝑥 item of product is
given by 𝐶(𝑥) = 210𝑥 + 7000. Find the minimum number that must be produced
and sold daily for break even, if each item is sold for ₹280. (Understanding)

Question 20 [2]
(i)

A real estate company is going to build a new residential complex. The land they have
purchased can hold at the most 500 apartments. Also, if they make 𝑥 apartments,
then the monthly maintenance cost for the whole complex would be as follows:
Fixed cost = ₹ 4000
Variable cost = ₹ (14𝑥 − 0 ∙ 04𝑥 2 )
How many apartments should the complex have in order to minimize the maintenance
costs? (Application)
OR
(ii) The demand function of a monopoly is given by 𝑥 = 100 − 4𝑝. Find the quantity at
which the Marginal Revenue will be zero. (Application)

Question 21 [4]
A survey of 50 families to study the relationships between expenditure on accommodation
in (₹ x) and expenditure on food and entertainment (₹ y) gave the following results:
∑ 𝑥 = 8500, ∑ 𝑦 = 9600, 𝜎𝑥 = 60, 𝜎𝑦 = 20, 𝑟 = 0.6
Estimate the expenditure on food and entertainment when expenditure on accommodation
is ₹200. (Application)

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Question 22 [4]
(i) A linear programming problem is given by 𝑍 = 𝑝𝑥 + 𝑞𝑦 where 𝑝, 𝑞 > 0 subject to
the constraints:
𝑥 + 𝑦 ≤ 60, 5𝑥 + 𝑦 ≤ 100, 𝑥 ≥ 0 𝑎𝑛𝑑 𝑦 ≥ 0 (Analysis)
(a) Solve graphically to find the corner points of the feasible region.
(b) If Z = 𝑝𝑥 + 𝑞𝑦 is maximum at (0,60) and (10, 50), find the relation of
𝑝 𝑎𝑛𝑑 𝑞. Also mention the number of optimal solution(s) in this case.
OR
(ii) The feasible region for an L.P.P. is shown in the adjoining figure:

(Analysis)
Based on the given graph, answer the following questions.
(a) Write the constraints for the L.P.P.
(b) Find the co-ordinates of the point B.
(c) Find the maximum value of the objective function 𝑍 = 𝑥 + 𝑦.

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MATHEMATICS
ANSWER KEY
SECTION A - 65 MARKS
Question 1
In answering Multiple Choice Questions, candidates have to write either the correct
option number or the explanation against it. Please note that only ONE correct answer
should be written.
(i) (d) or a null matrix [1]
(ii) (b) or 1 [1]
(iii) 𝑑𝑦
(c) or 𝑑𝑥 = 𝑥
−𝑦 [1]

(iv) (d) or Assertion is false, and Reason is true. [1]
(v) (c) or
1 [1]
68

(vi) 𝑎𝑛 0 0 [1]
(a) or ( 0 𝑎𝑛 0)
0 0 𝑎𝑛
(vii) (a) or Statement 1 is true and Statement 2 is false. [1]
(viii) (d) or 9 [1]
(ix) (c) or Both the statements are true. [1]
(x) (a) or Both Assertion and Reason are true and Reason is the correct explanation for [1]
Assertion.
(xi) (a) or 0·36 [1]
(xii) n=3 [1]
|𝐴| = 3.
𝐴𝑠 per question, C=(𝑎𝑑𝑗 𝐴)𝑇
∴|𝐶| = |(𝑎𝑑𝑗 𝐴)𝑇 | = |𝑎𝑑𝑗 𝐴|
∴ |𝐶| = |𝐴|𝑛−1 = 33−1 = 32 = 9
|𝐶 2 | = |𝐶| ⋅ |𝐶| = 9 × 9 = 81
Ans: 81
(xiii) The relation is symmetric, transitive but not reflexive. [1]
(xiv) The curve crosses 𝑥 -axis at three different points. That shows, for different value [1]
of 𝑥, the value of 𝑦 is zero. That is more than one domain, mapped to the same point
“zero”

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
1

(xv) Two tests will be required if first machine is faulty and second is good OR both [1]
machines are faulty.
2 1 2 1 2
Probability that only two tests are needed = 3 × 2 + 3 × 2 = 3
2
Ans: 3

Question 2 [2]
𝑥
(i) 𝑥 = 𝑒𝑦
𝑥 𝑥
log 𝑥 = log 𝑒 =
𝑦 𝑦
𝑦log 𝑥 = 𝑥
By differentiating on both sides,
1 𝑑𝑦
𝑦 ⋅ + log 𝑥 =1
𝑥 𝑑𝑥
𝑑𝑦 𝑦
log𝑥 = 1−
𝑑𝑥 𝑥
𝑑𝑦 𝑥−𝑦
=
𝑑𝑥 𝑥 log 𝑥

(ii) The values of ′𝑎′ for which the function 𝑓(𝑥) = 𝑏 − 𝑎𝑥 + 𝑠𝑖𝑛 𝑥 is increasing on R.
𝑓՚(𝑥) = −𝑎 + 𝑐𝑜𝑠 𝑥 ≥ 0
It implies 𝑎 ≤ 𝑐𝑜𝑠𝑥. But −1 ≤ 𝑐𝑜𝑠𝑥 ≤ 1
so 𝑎 ≤ −1.

Question 3 [2]
𝑦 = (𝑥 − 2)2
𝑑𝑦
= 2(𝑥 − 2)
𝑑𝑥
4−0
Slope of the chord = =2
4−2

Let 𝑃(𝑥1 , 𝑦1 ) be the point on the curve.
Since the tangent is parallel to the chord. Therefore, their slopes are equal.
∴ 2(𝑥1 − 2) = 2 ⟹ 𝑥1 =3 and 𝑦1 =1
So, the point 𝑃 is (3,1).

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
2

Question 4 [2]
𝑑𝑦
= 𝑦𝑐𝑜𝑡2𝑥 , by separation of variable
𝑑𝑥
𝑑𝑦
= 𝑐𝑜𝑡2𝑥 𝑑𝑥
𝑦

𝐼ntegrating both sides
𝑑𝑦
∫ = ∫ 𝑐𝑜𝑡2𝑥 𝑑𝑥
𝑦
1
⟹ log 𝑦 = 2 log|𝑠𝑖𝑛2𝑥| + 𝐶

Question 5 [2]
(i) ∫ 𝑥 5 cos( 𝑥 6 ) 𝑑𝑥
Let 𝑥 6 = 𝑡
6𝑥 5 𝑑𝑥 = 𝑑𝑡
𝑑𝑡 1 1
∫ cos( 𝑡) = sin 𝑡 + 𝑐 = sin(𝑥 6 ) + 𝑐
6 6 6
On comparing
1
𝑘 sin(𝑥 6 ) = sin(𝑥 6 )
6
1
𝑘=
6
OR
(ii) I = ∫0
5
𝑑𝑥
√𝑥
√5−𝑥 +√𝑥
Applying,
𝑎 𝑎

∫ 𝑓(𝑥) = ∫ 𝑓(𝑎 − 𝑥) 𝑑𝑥
0 0

5 √5−𝑥
𝐼 = ∫0 𝑑𝑥
√ +√5−𝑥
𝑥
5 √5−𝑥 𝑑𝑥 5 √𝑥 𝑑𝑥
2𝐼 = ∫0 + ∫0
√5−𝑥 +√𝑥 √5−𝑥 +√𝑥
5 5
2𝐼 = ∫0 1𝑑𝑥 = [𝑥]0
5
𝐼 = 2 = 2.5

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
3

Question 6 [2]
7
−1 6 −1 8 −1 5
tan ( ) + tan ( ) = tan 12
10 10 1−
25
35
= tan−1 ( )
13

Question 7 [4]
sin−1(𝑥) + sin−1(1 − 𝑥) = cos −1 𝑥
π −1
⇒ sin−1 (𝑥) + sin−1 (1 − 𝑥) = − sin 𝑥
2
π −1
⇒ sin−1 (1 − 𝑥) = − 2 sin 𝑥
2
π −1
⇒ (1 − 𝑥) = sin ( − 2 sin 𝑥)
2
−1
⇒ (1 − 𝑥) = cos( 2 sin 𝑥)
⇒ (1 − 𝑥) = cos(cos−1 (1 − 2𝑥2 ))
⇒ (1 − 𝑥) = 1 − 2𝑥2
⇒ 2𝑥 2 − 𝑥 = 0
1
∴ 𝑥 = 0,
2

Question 8

Given, 𝑦 = (𝐴 + 𝐵𝑥)𝑒 −2𝑥 [4]
Differentiating w.r.t ‘𝑥’
𝑑𝑦
⟹ 𝑑𝑥 = (𝐴 + 𝐵𝑥)(−2𝑒 −2𝑥 ) + 𝑒 −2𝑥 𝐵
𝑑𝑦
⟹ 𝑑𝑥 = (−2𝑦) + 𝑒 −2𝑥 𝐵
𝑑𝑦
⟹ 𝑑𝑥 + 2𝑦 = 𝑒 −2𝑥 𝐵 (i)
Differentiating w.r.t ‘𝑥’
𝑑2 𝑦 𝑑𝑦
⟹ 𝑑𝑥 2 + 2 𝑑𝑥 = −2𝑒 −2𝑥 𝐵 (ii)
(i) and (ii) ⟹
𝑑2𝑦 𝑑𝑦 𝑑𝑦
+ 2 𝑑𝑥 = −2 (𝑑𝑥 + 2𝑦)
𝑑𝑥 2
𝑑2 𝑦 𝑑𝑦
⟹ 𝑑𝑥 2 + 4 𝑑𝑥 + 4𝑦 = 0, Hence Proved.

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
4

Question 9 [4]
(i) Differential equation
𝑑𝑦 𝑦
𝑥 = 𝑦 − 𝑥 𝑡𝑎𝑛 ( )
𝑑𝑥 𝑥
𝑑𝑦 𝑦 𝑦
= − 𝑡𝑎𝑛 ( )
𝑑𝑥 𝑥 𝑥
𝑑𝑦 𝑑𝑣
Let 𝑦 = 𝑣𝑥 ⇒ 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥
𝑑𝑣
⇒ 𝑣 + 𝑥 𝑑𝑥 = 𝑣 − 𝑡𝑎𝑛𝑣
𝑑𝑣 𝑑𝑥
⇒ tan 𝑣 = − 𝑥
Integrating both sides
𝑑𝑥
⇒ ∫ cot 𝑣 𝑑𝑣 = − ∫ 𝑥
⇒ log| sin 𝑣| = −log𝑥 + log𝑐
𝑦 𝑐
⇒ log| sin 𝑥 | = −log𝑥 + log𝑐 = log 𝑥
𝑦
⇒ 𝑥 sin = 𝑐
𝑥

OR
𝑑𝑦
(ii) (1 + 𝑥 2 ) + 2𝑥𝑦 = 4𝑥 2
𝑑𝑥
𝑑𝑦 2𝑥𝑦 4𝑥 2
+ 2
(1+𝑥 )
= (1+𝑥 2 )
𝑑𝑥
𝑑𝑦
Comparing with 𝑑𝑥 + 𝑃𝑦 = 𝑄
2𝑥 4𝑥 2
𝑃 = (1+𝑥2) 𝑄 = (1+𝑥 2)
2𝑥
𝑑𝑥
I.F. = 𝑒 ∫1+𝑥2
2
I.F. = 𝑒 log (1+𝑥 )

I.F. = (1 + 𝑥 2 )
Thus, the solution of the differential equation is,
4𝑥 2
𝑦 (1 + 𝑥 2 ) =∫ (1+𝑥2) (1 + 𝑥 2 )𝑑𝑥
4𝑥 3
⟹ 𝑦(1 + 𝑥 2 ) = 3 + 𝑐

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
5

Question 10 [4]
1 1 1 1 1 1 1
(i) (a) P (no odd person) = P(HHH) + P(TTT) = 2 × 2 × 2 + 2 × 2 × 2 = 4
(b) 1 3
P (odd person) = 1 − 4 = 4
1 1 1 3 3
(c) P (odd person in 4th round) = 4 × 4 × 4 × 4 = 256
OR
(ii) (a) Let:
• A: student chose Mode A → P(A) = 0.40

• B: student chose Mode B → P(B) = 0.35

• C: student chose Mode C → P(C) = 0.25
Let E: student rated the class as Excellent
Given:
• P(E/A) = 0.20, P(E/B) = 0.30, P(E/C) = 0.50

(b) 𝑃(C/E)
P(C) × P(E/C)
=
P(A) × P(E/A) + P(B) × P(E/B) + P(C) × P(E/C)
0.25 ×0.50
= (0.40×0.20) + (0.35×0.30) + (0.25×0.50)
0.25 ×0.50
=
(0.08) + (0.105) + (0.125)

= 0.403
(c) There’s about a 40.3% chance that a student who rated the class as
“Excellent” attended Recorded lectures.
Now, numerator of, 𝑃(A/E) = P(A) × P(E/A) = 0.40 × 0.20 = 0.08
and
Numerator of 𝑃(B/E) = P(B) × P(E/B) = 0.35 × 0.30 = 0.105

By checking the numerators of 𝑃(C/E), 𝑃(B/E) and 𝑃(A/E) we observed
0.125 > 0.105 > 0.08. Therefore, recorded lectures (Mode C) have the highest
likelihood of being the chosen mode among students who gave an excellent
rating.

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
6

Question 11 [6]
(i) Let the cost of one paper bag, one scrap book and one pastel sheet be 𝑥, 𝑦 and 𝑧
respectively.
30𝑥 + 20𝑦 + 10𝑧 = 410 ⇒ 3𝑥 + 2𝑦 + 𝑧 = 41
20𝑥 + 10𝑦 + 20𝑧 = 290 ⇒ 2𝑥 + 𝑦 + 2𝑧 = 29
20𝑥 + 20𝑦 + 20𝑧 = 440 ⇒ 𝑥 + 𝑦 + 𝑧 = 22

(ii) Given system of equations is equivalent to AX = B
3 2 1 𝑥 41
Where 𝐴 = [2 1 2] , X = [𝑦] , 𝐵 = [29]
1 1 1 𝑧 22
−1
|𝐴| = −2 ≠ 0 ⇒𝐴 exists.
−1 −1 3
adj A = [ 0 2 −4]
1 −1 −1
−1 −1 3
1 1
Thus 𝐴−1 = |𝐴| 𝑎𝑑𝑗 𝐴 = − 2 [ 0 2 −4]
1 −1 −1
−1 −1 3 41 2
1
𝐴𝑋 = 𝐵 ⇒ 𝑋 = 𝐴−1 𝐵 = − 2 [ 0 2 −4] [29] = [15]
1 −1 −1 22 5
∴ 𝑥 = 2, 𝑦 = 15, 𝑧 = 5
(iii) The cost of one paper bag, one scrap book and one pastel sheet be Rs 2, Rs 15 and
Rs 5 respectively.

Question 12 [6]
(i) 𝑥2+ 𝑥 + 1
∫ (𝑥 + 2)(𝑥 2+ 1) dx
𝑥2+ 𝑥 + 1 𝐴 𝐵𝑥 + 𝐶
Let: (𝑥 + 2)(𝑥² + 1) = (𝑥 + 2) + (𝑥² + 1)
Equating the coefficients of 𝑥 2 , 𝑥 and constant respectively, we get
A+B=1
2B + C = 1
A + 2C = 1
Solving for A, B, C we get,
1 2 3
C = 5, B = 5 and A = 5
𝑥2+ 𝑥 + 1
∴∫ (𝑥 + 2)(𝑥 2+ 1) 𝑑𝑥

3 2𝑥 + 1
= ∫[ + ] 𝑑𝑥
5(𝑥 + 2) 5(𝑥² + 1)

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
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Integrating, we get
3 1
= 5 ln|𝑥 + 2| + 5 [ln|𝑥² + 1| + 𝑡𝑎𝑛⁻¹(𝑥)]+ C
OR
3𝜋
(ii) 𝑥
𝐼 = ∫𝜋4 𝑑𝑥
(1 + sin 𝑥)
4
𝜋 3𝜋
Here, 𝑎 = 4 and 𝑏 = 4 , so 𝑎 + 𝑏 = π.
𝑏 𝑏
Using, ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥)𝑑𝑥
3𝜋
𝜋−𝑥
𝐼 =∫𝜋 4
(1 + sin (𝜋 − 𝑥))
𝑑𝑥
4
3𝜋
𝜋−𝑥
=∫𝜋 (1 + sin 𝑥) 𝑑𝑥
4

4

Adding, both integrals,
3𝜋 3𝜋
𝑥 + (𝜋 − 𝑥) 1
𝐼 + 𝐼 = ∫𝜋4 (1 + sin 𝑥) 𝑑𝑥 = 𝜋 ∫𝜋 4 𝑑𝑥
4 4
(1 + sin 𝑥)
3𝜋 3𝜋
1−sin 𝑥
⟹2 𝐼 = 𝜋 ∫𝜋4 𝑑𝑥 = 𝜋 ∫𝜋4 (𝑠𝑒𝑐2 𝑥 − sec𝑥 tan𝑥 )𝑑𝑥
4 (1− sin2 𝑥) 4
3𝜋
⟹ 2 𝐼 = 𝜋[𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑐𝑥] 𝜋
4

4
𝐼 = (√2 − 1)𝜋

Question 13 [6]
(i) (a) express h in terms of radius r and given volume.
volume V = 𝜋𝑟 2 ℎ
539
=ℎ
2𝜋𝑟 2

(b) Let the total surface area of the closed cylinder tank
be S. Expressing S in term of radius r.
S=2𝜋𝑟ℎ + 2𝜋𝑟 2
539
S = 2𝜋𝑟 + 2𝜋𝑟 2
2𝜋𝑟 2
2 539
S = 2𝜋𝑟 + 𝑟
𝑑𝑠 539
(c) = 4𝜋𝑟 −
𝑑𝑟 𝑟2
𝑑𝑠
Setting, 𝑑𝑟 = 0 for stationary point, we get
539 = 4𝜋𝑟 3

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
8

7
𝑟 = 2 unit
𝑑2 𝑠 539×2 7
Now, 𝑑𝑟 2 = 4𝜋 + > 0, when r = 2 units.
𝑟3
7
Therefore, the total surface area (S) of the tank is minimum when r = 2 units.
(d) Finding the height of the tank:
539 539×4×7
ℎ = 2𝜋𝑟 2 = 2×22×7×7 = 7 𝑢𝑛𝑖𝑡s.

OR
1
(ii) (a) ℎ(𝑡) = 2 (−7𝑡 2 + 3𝑡 + 2), is a polynomial function, and all polynomial functions
are continuous and differentiable everywhere on R.
∴ The function is differentiable for 𝑡 ≥ 0.
1
(b) Given, ℎ(𝑡) = 2 (−7𝑡 2 + 3𝑡 + 2)
Differentiating w.r.t ‘t’ we get,
𝑑ℎ 1
𝑑𝑡
= (−14𝑡 + 3)
2
3
= −7 (𝑡 − 14).
1
∴ The instantaneous rate of change of height at t = 14
𝑑ℎ 3 1
= [ 𝑑𝑡 ] 1 = 2 − 2 =1 unit.
t=
14
𝑑ℎ 3
(c) 𝑑𝑡 = −7 (𝑡 − 14).
Given, t ≥ 0.
3 𝑑ℎ 3
In (0, 14) , 𝑑𝑡 > 0. i.e., ℎ(𝑡) is increasing in (0, 14 ).
3
∴ ℎ(𝑡) is increasing in (−∞, ). is false.
14
𝑑ℎ
(d) Setting 𝑑𝑡 = 0, for stationary point, we get
3
−7 (𝑡 − 14) = 0
3
⇒𝑡 = 14.
𝑑2 ℎ
Now, 𝑑𝑡 2 = −7 < 0.
3
∴ ℎ = 𝑓(𝑡) has a local maximum at t = 14..
∴ Maximum height =[ℎ(𝑡)] 3 = 1.161 units.
14

ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
9

Question 14 [6]
9 4 7
(i) Let 𝑃(𝐸) = 10 , 𝑃(𝑀) = 5 𝑎𝑛𝑑 𝑃(𝑆) = 10
1
⟹ 𝑃(𝐸̅ ) = , 𝑃(𝑀 ̅ ) = 1 𝑎𝑛𝑑 𝑃(𝑆̅) = 3
10 5 10
1 1 3 3
𝑃(𝑋 = 0) = 10 × 5 × 10 = 500
9 1 3 1 4 3 1 1 7 46
𝑃(𝑋 = 1) = (10 × 5 × 10) + (10 × 5 × 10) + (10 × 5 × 10) = 500
9 4 3 9 1 7 1 4 7 199
𝑃(𝑋 = 2) = (10 × 5 × 10) + (10 × 5 × 10) + (10 × 5 × 10) = 500
9 4 7 252
𝑃(𝑋 = 3) = 10 × 5 × 10 = 500

(ii) X 0 1 2 3
P(X) 3 46 199 252
500 500 500 500

(iii) Average number of surprise tests
46 199 252 1200
= 𝐸(𝑋) = 0 + (1 × 500) + (2 × 500) + (3 × 500) = 500 = 2.4

(iv) Average number of surprise tests = 2.4 > 2.3

Students are getting a good average of surprise tests. So, the current system is
balanced, and the teachers do not need to change anything.

SECTION B - 15 MARKS

Question 15 [5]
(i) (a) or Statement 1 is true and Statement 2 is false.
(ii) (a) or 𝑟⃗ = (𝑖̂ + 2𝑗̂ + 2𝑘̂) + 𝜆(𝑖̂ + 𝑗̂ + 𝑘̂), 𝜆 ∈ ℝ
𝐴𝐵 = (𝑗̂ + 𝑘̂) − (𝑖̂ + 𝑗̂) = −𝑖̂ + 𝑘̂
⃗⃗⃗⃗⃗⃗
𝐴𝐶 = (𝑘̂ + 𝑖̂) − (𝑖̂ + 𝑗̂) = −𝑗̂ + 𝑘̂
⃗⃗⃗⃗⃗⃗

𝑖̂ 𝑗̂ 𝑘̂
𝑛⃗⃗ = 𝐴𝐵 × 𝐴𝐶 = |−1 0 1| = 𝑖̂ + 𝑗̂ + 𝑘̂
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗
0 −1 1
𝑟⃗ = 𝑖̂ + 2𝑗̂ + 2𝑘̂ + 𝜆( 𝑖̂ + 𝑗̂ + 𝑘̂)
(iii) (c) or 𝑐 = ±√3
Using, 𝑙 2 + 𝑚2 + 𝑛2 = 1
1 1 1 3
⇒ 2+ 2+ 2=1⇒ 2=1
𝑐 𝑐 𝑐 𝑐
⇒𝑐 2 = 3
⇒ 𝑐 = ±√3
ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
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2
(iv) |𝑎⃗ + 𝑏⃗⃗| < 1 ⇒|𝑎⃗ + 𝑏⃗⃗ |2 < 1 ⇒ |𝑎⃗|2 + |𝑏⃗⃗| + 2 𝑎⃗. 𝑏⃗⃗ < 1
1 1
⇒1 + 1 + 2 𝑎⃗. 𝑏⃗⃗ < 1 ⇒𝑎⃗ ⋅ 𝑏⃗⃗ < − ⇒ |𝑎⃗||𝑏⃗⃗| cos 𝜃 < −
2 2
1 1 2𝜋 2𝜋
⇒ cos 𝜃 < − 2 ⇒ −1 ≤ cos 𝜃 < − 2 ⇒ 3 < 𝜃 ≤ 𝜋 i.e., 𝜃 ∈ ( 3 , 𝜋]
(v) Placing the coordinate axes as illustrated, coordinates of A is (2,0,0), B (0,4,3 ) and
C (1,4,0 ).

⃗⃗⃗⃗⃗⃗ = 2𝑖̂ - 4𝑗̂ -3 ̂𝑘 and 𝐵𝐶
∴𝐵𝐴 ⃗⃗⃗⃗⃗⃗ = 𝑖̂ -3 ̂𝑘
⃗⃗⃗⃗⃗⃗ . 𝐵𝐶
∴𝐵𝐴 ⃗⃗⃗⃗⃗⃗ = 2 + 9 = 11

Question 16 [2]
(i) (a) G is the centroid of 𝛥𝐵𝐶𝐷. The coordinates are

3+4+2 0+3+3 1+6+2
( , , ) = (3,2,3)
3 3 3
𝐴𝐺 = (3 − 0)𝑖̂ + (2 − 1)𝑗̂ + (3 − 2)𝑘̂ = 3𝑖̂ + 𝑗̂ + 𝑘̂
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ | = √32 + 12 + 12 = √11 units
|𝐴𝐺

(b) 𝐴𝐵 = (3 − 0)𝑖̂ + (0 − 1)𝑗̂ + (1 − 2)𝑘̂ = 3𝑖̂ − 𝑗̂ − 𝑘̂
⃗⃗⃗⃗⃗⃗
𝐴𝐶 = (4 − 0)𝑖̂ + (3 − 1)𝑗̂ + (6 − 2)𝑘̂ = 4𝑖̂ + 2𝑗̂ + 4𝑘̂
⃗⃗⃗⃗⃗⃗
1
Area of ∆ABC = |𝐴𝐵⃗⃗⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗⃗
𝐴𝐶 |
2
𝑖̂ 𝑗̂ ̂𝑘
𝐴𝐵 × 𝐴𝐶 = |3 −1 −1| = −2𝑖̂ − 16 𝑗̂ + 10 𝑘̂ and |𝐴𝐵
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗⃗
𝐴𝐶 | = 6√10
4 2 4
1
⃗⃗⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗⃗
Hence area of ∆ABC = 2 |𝐴𝐵 𝐴𝐶 | = 3√10 sq units
OR
(ii) Let 𝑎⃗ = 2𝑥 2 𝑖̂ + 3𝑥𝑗̂ + 𝑘̂ and 𝑏⃗⃗ = 𝑖̂ − 2𝑗̂ + 𝑥 2 𝑘̂
∵ angle between the vectors is obtuse
⇒ cos 𝜃 < 0
⃗⃗
𝑎⃗⃗ .𝑏
⇒ |𝑎⃗⃗ | | 𝑏⃗⃗| < 0
⇒ 𝑎⃗ . 𝑏⃗⃗ < 0
⇒ 2𝑥 2 − 6𝑥 + 𝑥 2 < 0

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⇒ 3𝑥(𝑥 − 2) < 0
⇒ 0 < 𝑥 < 2. i.e. 𝑥 ∈ (0,2)

Question 17 [4]
𝑥+3 𝑦−1 𝑧+4
(i) Let 5 = 2 = 3 = 𝜆.
∴𝐷(5𝜆 − 3,2𝜆 + 1,3𝜆 − 4)
Dr of AD ⟨5𝜆 − 3,2𝜆 − 1,3𝜆 − 7⟩ and Dr of given line L ⟨5,2,3⟩
AD ⊥ L, so 5(5𝜆 − 3) + 2(2𝜆 − 1) + 3(3𝜆 − 7) = 0 ⇒ 𝜆 = 1
Coordinates of D are (2, 3, −1)
AD = √(2 − 0)2 + (3 − 2)2 + (−1 − 3)2 = √21
1 1 5
Area of ∆ABC = 2 × 𝐵𝐶 × 𝐴𝐷 = 2 × 5 × √21 = 2 √21 sq. units

OR
(ii) Method-I
Let the equation of the plane be 𝑎(𝑥 − 𝑥1 ) + 𝑏(𝑦 − 𝑦1 ) + 𝑐(𝑧 − 𝑧1 ) = 0 is passing
through (−1, 0, 2)
⇒𝑎(𝑥 + 1) + 𝑏 (𝑦 – 0) + 𝑐 (𝑧 – 2) = 0………………………..(1)
𝑥−0 𝑦−1 𝑧−1
Given line −2 = 3 = −1 passing through (0, 1, 1) and having d.r. ⟨−2, 3, −1⟩
Since the plane contains the line and the point
⇒𝑎 (1) + 𝑏 (1) + 𝑐 (1 – 2) =0 ⇒ 𝑎 + 𝑏 – 𝑐 = 0……(2)
Also the line and normal to the plane are perpendicular
⇒– 2𝑎 + 3 𝑏 – 𝑐 = 0…………….(3)
𝑎 𝑏 𝑐
Solving (2) and (3) =3=5=𝑘
2
Hence required equation of the plane is
⇒2(𝑥 + 1) + 3(𝑦 − 0) + 5(𝑧 − 2) = 0
⇒ 2𝑥 + 3𝑦 + 5𝑧 − 8 = 0
Method –II
Let the equation of the plane be 𝑎(𝑥 − 𝑥1 ) + 𝑏(𝑦 − 𝑦1 ) + 𝑐(𝑧 − 𝑧1 ) = 0 is passing
through (−1, 0, 2)
⇒𝑎(𝑥 + 1) + 𝑏 (𝑦 – 0) + 𝑐 (𝑧 – 2) = 0………………………..(1)
𝑥−0 𝑦−1 𝑧−1
Given line −2 = 3 = −1 passing through (0, 1, 1) and having d.r. ⟨−2, 3, −1⟩
Since the plane contains the line and the point
⇒𝑎 (1) + 𝑏 (1) + 𝑐 (1 – 2) =0 ⇒ 𝑎 + 𝑏 – 𝑐 = 0……(2)
Also, the line and normal to the plane are perpendicular
⇒ – 2𝑎 + 3 𝑏 – c = 0…………….(3)
Hence required equation of the plane is
𝑥+1 𝑦 𝑧−2
| 1 1 −1 | = 0
−2 3 −1
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⇒2(𝑥 + 1) + 3(𝑦 − 0) + 5(𝑧 − 2) = 0
⇒ 2𝑥 + 3𝑦 + 5𝑧 − 8 = 0

Question 18 [4]
The required area is in two parts – one part above the 𝑥 -axis and the other below the 𝑥 -
axis.
4 5 4 5

𝐴 = ∫ 𝑦 𝑑𝑥 + ∫ −𝑦 𝑑𝑥 = ∫ 𝑥(4 − 𝑥) 𝑑𝑥 + ∫ 𝑥(𝑥 − 4) 𝑑𝑥
0 4 0 4
4 5
𝑥3 𝑥3 64 125 64
= [2𝑥 2 − 3 ] + [ 3 − 2𝑥 2 ] = (32 − 3 ) + ( 3 − 50) − ( 3 − 32)
0 4
32 25 32
= − + = 13 sq units
3 3 3

SECTION C - 15 MARKS

Question 19 [5]
(i) (b) or MC = AC
(ii) (b) or Both statements are true but Statement 2 is not the correct explanation of
Statement 1.
(iii) Given:𝑥̅ = 53, 𝑦̅ = 28, 𝑏𝑦𝑥 = −1.2, 𝑏𝑥𝑦 = −0.3
Using, 𝑟 2 = 𝑏𝑦𝑥 × 𝑏𝑥𝑦
= −1.2 × −0.3 = 0.36
⇒ 𝑟 = −0.6
Ans : − 0 ∙ 6
(iv) Given , 𝑅 (𝑥) = 36𝑥 + 3𝑥 2 + 5
𝑑
⇒ 𝑀𝑅 = (36𝑥 + 3𝑥 2 + 5)
𝑑𝑥
= 36 + 6 𝑥
∴𝑀𝑅 (𝑥 = 5) = 66
Ans: 66

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(v) Given 𝐶(𝑥) = 210𝑥 + 7000 , 𝑅(𝑥) = 280𝑥
Minimum number must be sold daily when 𝑅(𝑥) = 𝐶(𝑥)
⇒210𝑥 + 7000 = 280𝑥
⇒70𝑥 = 7000
⇒𝑥 = 100.
∴Minimum number that must be sold is 100.
Ans: 100

Question 20 [2]
(i) 𝐶(𝑥) = 4000 + 14𝑥 − 0.04𝑥 2
⇒𝐶 ′ (𝑥) = 14 − 0.08𝑥
Now, 𝐶 ′ (𝑥) = 0 ⇒ 14 = 0.08𝑥 ⇒ 𝑥 = 175
𝐶 ′′ (𝑥) = −0.08 < 0
∴C (𝑥) will be maximum at 𝑥 = 175.
As per question, we have to minimise the maintenance cost.
Since 𝑥 can be 0 to 500 apartments,
𝐶(0) = 4000 + 14 × 0 − 0.04 × (0)2 = 4000
𝐶(500) = 4000 + 14 × 500 − 0.04 × (500)2 = 1000
∴The complex must have 500 apartments to minimise the maintenance cost.
OR
(ii) Given, 𝑥 = 100 − 4𝑝
100 − 𝑥
⇒𝑝 =
4
100𝑥 − 𝑥 2
∴𝑅(𝑥) = 𝑝𝑥 =
4
𝑑 100𝑥−𝑥 2 100−2𝑥
⇒𝑀𝑅 = ( )=
𝑑𝑥 4 4
∵MR = 0
∴𝑥 = 50

Question 21 [4]
∑𝑥 ∑𝑦
𝑥̅ = = 170, 𝑦̅ = = 192,
𝑛 𝑛
𝜎𝑦 20
𝑏𝑦𝑥 = 𝑟 = 0.6 × = 0.2
𝜎𝑥 60
Regression equation 𝑦 on 𝑥 is
𝑦 − 192 = 0.2(𝑥 − 170)
ISC (CLASS XII) SPECIMEN ANSWER KEY 2026
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⇒𝑦 − 192 = 0.2𝑥 − 34
⇒ 𝑦 = 0.2𝑥 + 158
Putting, 𝑥 = 200
⇒𝑦 = 0.2(200) + 158 = 198
∴Expenditure on food and entertainment = ₹ 198.

Question 22 [4]
(i) (a) From graph, corner points are: A (0, 60), B (10, 50), C (20,0), D (0, 0)

(b) 𝑍 = 𝑝𝑥 + 𝑞𝑦
Given, Z is maximum at (0, 60) and (10, 50)
∴0. 𝑝 + 60. 𝑝 = 10. 𝑝 + 50. 𝑞
⇒10𝑝 = 10𝑞
⇒𝑝=𝑞
So, there can be infinite number of optimal solutions.
OR

(ii) (a) Equation of line AD:
𝑥 𝑦
+ = 1 ⇒ 2𝑥 + 𝑦 = 50
25 20
Equation of line EC:
𝑥 𝑦
+ = 1 ⇒ 𝑥 + 2𝑦 = 40
40 20
As origin lies in the region, 2𝑥 + 𝑦 ≤ 50 and 𝑥 + 2𝑦 ≤ 40
Therefore, the constraints are,
2𝑥 + 𝑦 ≤ 50
𝑥 + 2𝑦 ≤ 40
𝑥 ≥ 0, 𝑦 ≥ 0
(b) Solving simultaneously, the coordinate of B is (20, 10).

(c) The corner points are 𝑂(0,0), 𝐶(0,20), 𝐵(20,10) and 𝐴(25,0).
At O, Z = 0
At C, Z = 20
At B, Z = 30
At A, Z = 25
∴Maximum value of Z is 30.

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