ISC Class 11 Specimen Paper Applied Maths

APPLIED 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 thirteen printed pages.
4. It consists of 20 questions and four sections: A, B, C and D. All questions
are compulsory.
5. Section A comprises very short answer questions of 1 mark each.
6. Section B consists of short answer questions of 2 marks each.
7. Section C consists of moderately long answer questions of 3 marks each.
8. Section D consists of long answer questions of 5 marks each.
9. Internal choices have been provided in three questions, each in Sections B,
C and D.
10. While attempting Multiple Choice Questions in Section A, 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 XI) SPECIMEN QUESTION PAPER 2026
1

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 – 20 MARKS
Question 1
In subparts (i) to (xvii) choose the correct options and in subparts (xviii) to (xx), answer
the questions as instructed.
(i) In a class of 60 Students, 25 play cricket, 20 play tennis and 10 students play [1]
both the games. The number of students who play neither of the games is:
(Application)
(a) 10
(b) 55
(c) 5
(d) 25
(ii) 1+𝑖 𝑛 [1]
Identify the least positive integral value of ‘n’ for which (1−𝑖) is real.
(Application)
(a) 0
(b) 1
(c) 2
(d) −1
(iii) Consider the parabola 3𝑥 2 = −8𝑦 [1]
Statement I: The equation of the directrix of the given parabola is 3𝑦 + 2 = 0.
Statement II: The equation of the latus rectum of the given parabola is
3𝑦 − 2 = 0.
Which one of the following is correct about the above statements?
(Understanding)
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement I is true and Statement II is false.
(d) Statements I is false and Statement II is true.

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
2

(iv) There are 5 items in Column A and 5 items in Column B. The number of ways in [1]
which each item in Column A matches with a unique item in Column B is:
(Understanding)
(a) 120
(b) 10
(c) 5
(d) 20
(v) A and B are two mutually exclusive events of an experiment. If P (not A) = 0∙57 [1]
and P (either A or B) = 0∙74 and P(B) = 𝑞, then the value of 𝑞 is:
(Application)
(a) 0∙17
(b) 0∙31
(c) 0∙69
(d) 0∙36
(vi) The amount of money today which is equal to series of payment in future is: [1]
(Recall)
(a) sinking value of annuity.
(b) nominal value of annuity.
(c) present value of annuity.
(d) future value of annuity.
(vii) Assertion: In a negatively skewed distribution, the mean is less than the [1]
median and the median is less than the mode.
Reason: In a negatively skewed distribution, the tail of the distribution
extends towards the negative side, indicating that most values are
clustered towards the higher end of the scale.
Choose the correct answer from the following options. (Understanding)
(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.

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
3

(viii) Considering the following statements. [1]
Statement I: CGST input tax credit cannot be claimed against IGST output tax
liability.
Statement II: SGST input tax credit cannot be claimed against IGST output tax
liability
Which one of the following is correct about the above statements? (Recall)
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement I is true and Statement II is false.
(d) Statements I is false and Statement II is true.
10 [1]
2𝑥 2 3
(ix) The middle term in the expansion of ( 3 + 2𝑥 2 ) is: (Application)

(a) 250
(b) 251
(c) 252
(d) 242
(x) Choose the suitable equation for the given curve: (Analysis) [1]

(a) 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, 𝑎 < 0 𝑎𝑛𝑑 𝐷 < 0
(b) 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, 𝑎 > 0 𝑎𝑛𝑑 𝐷 < 0
(c) 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, 𝑎 < 0 𝑎𝑛𝑑 𝐷 > 0
(d) 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, 𝑎 > 0 𝑎𝑛𝑑 𝐷 > 0
𝑥 5 −35
The value of lim 𝑥−3 + lim 𝑒 𝑥 is: (Understanding) [1]
(xi) 𝑥→3 𝑥→0

(a) 405
(b) 406
(c) 1
(d) 0

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
4

1 1 1
The value of 33 × 39 × 327 × … … … … … . upto infinite terms is: [1]
(xii)
(Understanding)
(a) 3
(b) √3
(c) 9
(d) ∞
(xiii) The algebraic equation of 2𝑙𝑜𝑔𝑥 + 3𝑙𝑜𝑔𝑦 − 𝑙𝑜𝑔𝑧 = 𝑙𝑜𝑔3 is: (Understanding) [1]
(a) 6𝑥𝑦
=3
𝑧
(b) 𝑥2 + 𝑦3 − 𝑧 = 3
(c) 𝑥2 3
=
𝑦3 𝑧
(d) 𝑥 2 𝑦 3 = 3𝑧
(xiv) Let 𝑎, 𝑏, 𝑐 are in Arithmetic Progression. 𝐴1 is the arithmetic mean between a and [1]
b. 𝐴2 is the arithmetic mean between b and c. The arithmetic mean between
𝐴1 𝑎𝑛𝑑 𝐴2 is: (Application)
(a) b
(b) 2b
(c) 𝑏
2
(d) b2
(xv) The multiplicative inverse of 3 + 2i is: (Understanding) [1]
(a) 3 – 2i
(b) 3 + 2𝑖
13
(c) 3 − 2𝑖
13
(d) 3 − 2𝑖
9

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
5

(xvi) Assertion: If nC12 = nC8 then 𝑛 = 20 [1]
n n
Reason: Cr = Cn-r (Understanding)
Choose the correct option.
(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.
(xvii) The quadratic equation with 3 + √2 i as one of the roots is: (Understanding) [1]
(a) 𝑥 2 + 6𝑥 + 7 = 0
(b) 𝑥 2 − 6𝑥 + 7 = 0
(c) 𝑥 2 − 6𝑥 + 5 = 0
(d) 𝑥 2 + 6𝑥 + 5 = 0
(xviii) In a code language TAPE is written as 4825, SMART is written as 91834 and [1]
BONE is written as 7605, then what is the code language for BASERA?
(Application)
(xix) Find the angle in radian through which a pendulum swings if its length is 75 cm [1]
and the tip describes an arc of length 21 cm. (Application)
(xx) What is the acute angle between the lines x + y = 0 and y = 0? (Understanding) [1]

SECTION B – 14 MARKS
Question 2 [2]
A committee of 4 persons is to be created from a group of 6 men and 7 women.
If the selection is made randomly, find the probability that there are equal numbers of
men and women in the committee. (Application)

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
6

Question 3 [2]
If a parabolic reflector is 20cm in diameter and 5cm deep. Find the coordinates of its
focus. (Application)

Question 4 [2]
If a relation 𝑅 = {(0,0), (2,4), (−1, −2), (3,6), (1,2)}, then (Recall)
(i) Write domain and range of R.
(ii) Write R in the set-builder form.

Question 5 [2]
(i) Find the nominal rate compounded quarterly equivalent to 5% effective rate of
interest. (Understanding)
OR
(ii) If the present worth of a sum of ₹29,160 due after 2 years is ₹ 25,000, find the rate
of interest compounded annually. (Understanding)

Question 6 [2]
(i) A company launches a marketing campaign for a new product. Online user
engagement (in thousands) at time ‘t’ seconds after the campaign stops is modelled
by the equation: I =𝐼0 × 2−0∙02𝑡 where:
𝐼 is the user engagement at time t,
𝐼0 is the initial user engagement when the campaign ends.
Calculate the time t (in seconds) at which the engagement drops to 90% from the
original level. (Application)
OR
(ii) Find the value of 𝑥 (𝑥 > 0 ) such that log 9 (log 3 𝑥) = log 3 (log 9 𝑥). (Application)

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
7

Question 7 [2]
(i) In how many ways can 5 chairpersons be arranged in a row such that
(a) President and Vice-president are always together?
(b) President and Vice-president are never together? (Application)
OR
(ii) If the 4th and 9th terms of a Geometric Progression be 54 and 13122 respectively.
Find the Geometric Progression. (Understanding)

Question 8 [2]
Find the derivative of 𝑦 = 3𝑠𝑖𝑛𝑥 − 5𝑐𝑜𝑠𝑥 + √𝑥 with respect to 𝑥. (Understanding)

SECTION C – 21 MARKS

Question 9 [3]
2𝑥 2 +𝑥
Solve for ‘x’: >3 (Evaluate)
𝑥 2 −𝑥+1

Question 10 [3]
Find all points on the line 𝑥 + 𝑦 = 4 that lie at a unit distance from the line 4𝑥 + 3𝑦 = 10
(Application)

Question 11 [3]
Let ‘𝑓’ be a function defined by 𝑓: 𝑅 → 𝑅, such that 𝑓(𝑥) = 5𝑥 2 + 2, 𝑥 ∈ 𝑅
(Understanding)
(i) Find the image of 3 under 𝑓.
(ii) Find the range of the 𝑓.
(iii) Find 𝑥 such that 𝑓(𝑥) = 1∙5.

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
8

Question 12 [3]
For any statement 𝑝 and 𝑞 construct the truth tables for: (~𝑝 ∨ 𝑞) ∧ (∼ 𝑝 ∨ ∼ 𝑞).
(Application)

Question 13 [3]
(i) The production of an item is given by the function 𝑃(𝑡) = 2𝑡 3 − 15𝑡 2 + 36𝑡,
where 𝑃(𝑡) is the number of units produced on the tth. day. (Analysis)
(a) Find the rate of production at any time 𝑡.
(b) Find the rate of production on the 2nd day.
(c) Interpret the result of part (b).
OR
𝑒 𝑥 𝑡𝑎𝑛𝑥 𝑑𝑦
(ii) If 𝑦 = , find (Application)
𝑙𝑜𝑔𝑥 𝑑𝑥

Question 14 [3]
(i) Mr Mohanty lives in Cuttack, Odisha. The reading of electric meter for his house is
6879 units. If the previous month’s reading was 5241 units and load is of 5kW,
calculate the electric bill for that month. Tariff plan is given below:
Energy charges
Number of units 0 – 50 51 – 200 201 – 400 > 400
Price per unit (₹) 3∙00 5∙00 6∙00 8∙00

Fixed charge ₹100 per kW/month
Surcharge is ₹0∙25 per unit
Energy tax is 5% of tariff rates. (Analysis)
OR
(ii) A man retires at the age of 60 years and his employer gives him a pension of
₹ 22,000 a year paid in half-yearly instalments for the rest of his life. Reckoning his
expectation of life to be 15 years and the rate of interest is 8% per annum
compounded half yearly, what single sum is equivalent to his pension?
(Analysis)

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
9

Question 15 [3]
(i) The first four central moments of a frequency distribution are 0, 2∙5, 0∙7 and 18∙75.
Answer the following questions. (Analysis)
(a) Find moment coefficient of kurtosis.

(b) Do you think it is negatively skewed? Justify your answer.

(c) From the point of view of peaked-ness, which type of curve is it? Justify your
answer.
OR
(ii) The following data given below shows the correlation co-efficient, mean and
Standard Deviation (s.d) of rainfall and yield of paddy in a certain area of the
country:
Yield per acre Annual rainfall (cm)
Mean 973∙5 18∙3
s.d 38∙4 2∙0
Coefficient of correlation = 0∙58
Estimate the most likely yield of paddy when the annual rainfall is 22cm, other
factors being assumed to remain constant. (Evaluate)

SECTION D – 25 MARKS
Question 16 [5]

(i) A company is evaluating two marketing strategies for the next 10 years.
Strategy A
• Investment in the first year is ₹6 lakhs.
• The investment increases by ₹1∙5 lakhs each year.
• At the end of 10 years, the company expects a profit of ₹108 lakhs from this
strategy.
Strategy B
• Investment in the first year is ₹3 lakhs.
• Each year, the investment becomes 1∙5 times the investment of the previous
year.
• The company expects to earn ₹272 lakhs profit on the total investment over
10 years. (Analysis)
Answer the following questions:
(a) Find the total investment made over 10 years under Strategy A and Strategy
B.
(b) Find the total profit percentage for Strategy A and Strategy B.
(c) Which strategy is more cost-effective? Why?

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
10

OR
(ii) Your school is preparing for annual function. Chairs are arranged in rows for the
guests. Every day, the number of rows and the number of chairs in each row
increases to accommodate more people.
On Day 1: 1 row with 2 chairs; each chair costs ₹3 → Cost = 1⋅2⋅3
On Day 2: 2 rows with 3 chairs; each chair costs ₹4 → Cost = 2⋅3⋅ 4
On Day 3: 3 rows with 4 chairs; each chair costs ₹5 → Cost = 3⋅4⋅5
and so on...
This pattern continues up to ‘n’ days. The total cost of chairs arranged over ‘n’
days is: 𝑆𝑛 = 1 ⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ 4 + 3 ⋅ 4 ⋅ 5 + ⋯ upto n terms (Application)
(a) Write the cost of chairs arranged on 𝑛𝑡ℎ day.
(b) Find a rule (formula) to calculate the total cost after ‘n’ days.
(c) Find the total cost of chairs for 5 days.
(d) If the total budget is ₹5040, for how many days can chairs be arranged in this
pattern?

Question 17 [5]
(i) A health team conducted a survey in a community to study the effect of mobile
phone usage on eye health. They collected data on the number of people who
reported eye problems in different age groups, possibly due to excessive screen
time. The collected data are tabulated bellow: (Analysis)

Age 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of persons 9 17 32 23 16 3

(a) Calculate the coefficient of variation (CV) for the age distribution of affected
individuals.
(b) Based on the coefficient of variation (CV), evaluate whether the age wise
distribution of eye problem is consistent or shows high variability.
(c) Suggest one age targeted awareness strategy that the health team could
implement based on your analysis.
OR

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
11

(ii) The internal and external assessment on mathematics were conducted for 10
students of class XII. Students obtained the following marks in the assessment.
Roll No of 1 2 3 4 5 6 7 8 9 10
the students:
Internal 16 18 15 14 12 10 14 9 12 20
marks (20)
External 72 78 64 68 65 45 68 54 60 78
Marks (80)
Calculate Spearman’s rank correlation coefficient and comment on the result.
(Analysis)

Question 18 [5]

(i) If the intercept on the line 𝑦 = 𝑥 by the circle 𝑥 2 + 𝑦 2 − 2𝑥 = 0 𝑖𝑠 𝐴𝐵, then find
the following: (Application)
(a) equation of the circle on 𝐴𝐵 as a diameter. [3]
(b) its centre and radius. [1]
(c) its equation in parametric form. [1]
OR
(ii) Refer to the diagram given below where C is the centre of the circle. (Analysis)

(i) The line through C and perpendicular to the chord AB is a diameter. Justify.
(a) [2]
Hence, find the equation of this diameter.
(b)
(ii) If the equation of the other diameter is 𝑥 − 3𝑦 − 11 = 0, find the [1]
coordinates of C.
(c)
(iii) Also, find the equation of the circle. [2]

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
12

Question 19 [5]
Consider the triangle given below: (Application)

(i) Write down the ratios tan 𝜑 and tan(𝜃 + 𝜑). [1]

(ii) Is there a way to calculate tan 𝜃 which involves these ratios? If yes, find tan 𝜃. [2]

(iii) Do 𝜃, 𝜑 and 𝜃 + 𝜑 lie in the same quadrant? Justify. Also state the quadrant(s). [2]

Question 20 [5]
In a financial year 2024-25, the gross salary of Mr. Ajay (aged 45 years) was ₹10,50,000
(excluding HRA) and income from interest on saving account was ₹15,500. He deposited
₹9200 per month in G.P.F. and paid ₹43,000 as life insurance premium.
He donated ₹25,000 in National Security Fund. He took a home loan of ₹ 24,00,000 from
state bank of India and paid ₹76000 as interest on home loan and ₹ 20000 as principal of
home loan.
Calculate his income tax under old tax regime at the end of the financial year if 4% health
and education cess is levied on the payable income tax.
Income tax slab for A.Y 2025-26. (Application)
Taxable income Income Tax
Upto ₹2,50,000 lakh NIL
₹2,50,001 lakh to ₹ 5 lakh 5% of taxable income exceeding ₹2,50,000
₹5.00,001 lakh to ₹ 10 lakh ₹12,500 + 20% taxable income exceeding ₹5 lakh
Above 10 Lakh ₹1,12,500 + 30% taxable income exceeding ₹10 lakh

ISC (CLASS XI) SPECIMEN QUESTION PAPER 2026
13

APPLIED MATHEMATICS
ANSWER KEY
SECTION A – 20 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 25 [1]
Let Tennis be represented by T and Cricket by C, sample space by S
students who play neither games is 𝑛(𝐶𝑈𝑇) = 𝑥
𝑛(𝑆) = 𝑛(𝑐) + 𝑛(𝑇) − 𝑛(𝐶 ∩ 𝑇) + 𝑛(𝐶𝑈𝑇)
60 = 25 + 20 − 10 + 𝑥
𝑥 = 25
(ii) (c) Or 2 [1]
1+𝑖 1+𝑖 𝑛 2𝑖 𝑛
(1−𝑖 × 1+𝑖) = ( 2 ) = 𝑖 𝑛 ; when 𝑛 = 2, 𝑖 2 = −1, which is real. n = 2
(iii) (b) Or Both the statements are false. [1]
2
The given parabola is 3𝑥 = −8𝑦 …..(i)
8
i.e. 𝑥 2 = − 3 𝑦
which is comparable with 𝑥 2 = −4𝑦 so (i) represents a standard (downward)
parabola
8 2
also 4𝑎= 3 ⇒ 𝑎 = 3
2
Therefore, the equation of directrix is y = a i.e. y = 3
3𝑦 − 2 = 0
∴ Statement I is false
2
The equation of latus rectum is 𝑦 + 𝑎 = 0 i.e. 𝑦 + 3 = 0
3𝑦 + 2 = 0
∴ Statement II is false
(iv) (a) Or 120 [1]
Number of ways = 5! = 120
(v) (b) Or 0∙31 [1]
A and B are two mutually exclusive events so 𝐴 ∩ 𝐵 = ∅
𝑃(𝑛𝑜𝑡𝐴) = 0 ∙ 57, 𝑠𝑜 𝑃(𝐴) = 0 ∙ 43 , 𝑃 (𝐴 ∪ 𝐵) = 0 ∙ 74
ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
1

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
⇒ 0 ∙ 74 = 0 ∙ 43 + 𝑞
∴ 𝑞 = 0 ∙ 31
(vi) (c) Or present value of annuity. [1]
(vii) (a) Or Both Assertion and Reason are true and Reason is the correct explanation [1]
for Assertion
Explanation: In a negatively skewed distribution (also known as left-skewed),
most values are clustered towards the higher end of the scale, and
there's a long tail extending towards the negative values
(left). This causes the mean to be pulled towards the lower values,
resulting in the order: Mean < Median < Mode.
(viii) (b) Both the statements are false. [1]
Explanation: Input tax credit of SGST can be utilised for payment of SGST first
and balance for payment of IGST on onward supply.
Input tax credit of CGST can be utilised for payment of CGST first
and balance for payment of IGST on onward supply.
(ix) (c) Or 252 [1]
2 5
2𝑥 3 5
𝑇6 = 𝑇1+5 = 10𝐶 5 ( ) ( 2 ) = 252
3 2𝑥
(x) (b) Or 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, 𝑎 > 0 𝑎𝑛𝑑 𝐷 < 0 [1]
Curve is facing upwards so a > 0 and curve is neither touching nor intersecting
the 𝑥 – axis hence D < 0
(xi) (b) Or 406 [1]
5 5
𝑥 −3
lim + lim 𝑒 𝑥 = 5(34 ) + 1 = 405 + 1 = 406
𝑥→3 𝑥 − 3 𝑥→0

(xii) (b) Or √3 [1]
1
1 1 1 1
+ 2 + 3 + ⋯.∞ = 3 1 =
3 3 3 1− 2
3
1 1 1 1
+ + +⋯.∞
33 32 33 = 3 2

(xiii) (d) Or 𝑥 2 𝑦 3 = 3𝑧 [1]
𝑥2𝑦3
log( 𝑧 ) = 𝑙𝑜𝑔3 ⟹ 𝑥 2 𝑦 3 = 3𝑧
(xiv) (a) Or b [1]
a, A1, b, A2, c

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
2

(xv) (c) Or
3−2𝑖 [1]
13
1 3 − 2𝑖 3 − 2𝑖
× =
3 + 2𝑖 3 − 2𝑖 13
(xvi) (a) Or Both Assertion and Reason are true and Reason is the correct explanation [1]
for Assertion.
n
𝐶𝑟 = n𝐶𝑛−𝑟 is true. ∴ n𝐶12 = n𝐶8 ⇒ 𝑟 = 12 𝑎𝑛𝑑 𝑛 = 12 + 8 = 20
(xvii) (b) Or 𝑥 2 − 6𝑥 + 7 = 0 [1]
One root = 3 + √2 ⟹ 3 − √2 𝑖𝑠 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑟𝑜𝑜𝑡
The quadratic equation is,
𝑥 2 − [(3 + √2) + (3 − √2)]𝑥 + (3 + √2)(3 − √2) = 0
⟹ 𝑥 2 − 6𝑥 + 7 = 0
(xviii) TAPE = 4825, SMART = 91834, BONE = 7605 [1]
From the above we get the code for B = 7, A = 8, S = 9, E =5, R = 3
So, code language for BASERA is 789538
(xix) The pendulum describes a circle of radius 75cm and its tip describes an arc of [1]
length 21 cm. Let θ radians be the angle through which the pendulum swings.
Here r = 75cm and l = 21 cm
𝑙 21 7
∴ θ = 𝑟= 75 = 25
(xx) Let the angle between the given lines be 𝜃 [1]
slope of 𝑥 + 𝑦 = 0 be 𝑚1 = −1
slope of 𝑦 = 0 be 𝑚2 = 0
1− 2𝑚 𝑚 −1
𝑡𝑎𝑛𝜃 = |1+𝑚 | =|1|=1
𝑚 1 2
𝜋
𝜃 = 45° =
4

SECTION B – 14 MARKS
Question 2 [2]

4 persons out of 13 can be selected in 13𝐶4 ways,
As per the data two gents and two ladies can be selected out of 6 gents and 7 ladies in
6𝐶 ×7𝐶 ways
2 2
6𝐶2 ×7𝐶2 63
Required probability: =
13𝐶4 143

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
3

Question 3 [2]

Let AOB the parabolic reflector which is 20cm in diameter and 5cm deep,
then AB = 20cm and OC =5cm
Where C is the midpoint of AB
The equation of the parabola can be taken as
y 2 = 4(𝑎𝑥)
Since the point A (5, 10) lies on the parabola
102 = 4 𝑎 × 5 ⇒ 𝑎 = 5
∴ The coordinate of the focus is (a,0) i.e (5,0)

Question 4 [2]
(i) Domain of R = {0,2,−1,3,1}, Range of R = {0,4,−2, 6, 2}
(ii) R in the builder form can be written as
𝑅 = {(𝑥, 𝑦): 𝑥 ∈ 𝐼, − 1 ≤ 𝑥 ≤ 3, 𝑦 = 2𝑥}

Question 5 [2]
(i) Let the nominal rate be 𝑟 %.
Number of conversions in a year, 𝑝 = 4
Effective rate of interest = 5% = 0∙05 per rupee.
𝑟 4
As effective rate per rupee= (1 + 100𝑝) − 1, 𝑤𝑒 𝑔𝑒𝑡
𝑟 4
0∙05 = (1 + 400) − 1
𝑟 4
1 ∙ 05 = (1 + )
400
𝑟
1+ = (1 ∙ 05)1/4
400
𝑟
= 1 ∙ 01227 − 1
400
𝑟 = 4 ∙ 908
Hence, the nominal rate compounded quarterly is 4∙908.
OR

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
4

(ii) Here, P.V = ₹25,000, 𝐴 = ₹29,160, 𝑛 = 2𝑦𝑟𝑠, 𝑖˙ = ?
𝐴 29160
Now, 𝑃, 𝑉 = (1+𝑖)𝑛 ⇒ 25,000 = (1+𝑖)2
2916
∴ (1 + 𝑖)2=
2500
∴ 1 + 𝑖= 1 ∙ 08
∴ 𝑖= 8%
∴ The rate of interest is 8% p.a.

Question 6 [2]
(i) Given, 𝐼 = 𝐼0 × 2−0∙02𝑡
We are to find the time when the user engagement becomes 90% of the original.
So, 𝐼 = 0 ∙ 9𝐼0
⟹ 𝐼0 × 2−0∙02𝑡 = 0 ∙ 9𝐼0 ⟹ log 0 ∙ 9 = −0 ∙ 02𝑡 𝑙𝑜𝑔2
𝑙𝑜𝑔0 ∙ 9 −0 ∙ 04576 −0 ∙ 04576
⟹𝑡= = = = 7.60 𝑠𝑒𝑐
−0 ∙ 02 𝑙𝑜𝑔2 −0 ∙ 02𝑙𝑜𝑔2 −0 ∙ 02 × 0 ∙ 3010

OR
(ii) 1 1
log 9 (log 3 𝑥) = log 3 (log 9 𝑥) ⟹ log 3 (log 3 𝑥) = log 3 ( log 3 𝑥)
2 2
2
1 1
⟹ log 3 (log 3 𝑥) = 2 log 3 ( log 3 𝑥) ⟹ log 3 (log 3 𝑥) = log 3 ( log 3 𝑥)
2 2
1
⟹ log 3 𝑥 = (log 3 𝑥)2 ⟹ 4 = log 3 𝑥 ⟹ 𝑥 = 34 ⟹ 𝑥 = 81
4

Question 7 [2]
(i) (a) Number of arrangements where P and V always together = 4!2! = 48 ways

(b) Number ways where P and V never be together =
Total number of arrangements − always be together = 5! – 48 = 72 ways
OR

(ii) 𝑎𝑟 3 = 54, 𝑎𝑟 8 = 13122 ⟹ 𝑟 5 = 243 ⟹ 𝑟 = 3 𝑎𝑛𝑑 𝑎 = 2
The GP is, 2, 6, 18, 54, …

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
5

Question 8 [2]
𝑑𝑦 1
= 3𝑐𝑜𝑠𝑥 + 5𝑠𝑖𝑛𝑥 +
𝑑𝑥 2 √𝑥

SECTION C – 21 MARKS
Question 9 [3]
2𝑥 2 + 𝑥 2𝑥 2 + 𝑥
> 3 ⟹ −3>0
𝑥2 − 𝑥 + 1 𝑥2 − 𝑥 + 1
𝑥 2 − 4𝑥 + 3
⟹ <0
𝑥2 − 𝑥 + 1
2 2
1 2 3
⟹ 𝑥 − 4𝑥 + 3 < 0 (∵ 𝑥 − 𝑥 + 1 = (𝑥 − ) + > 0, ∀𝑥 ∈ 𝑅)
2 4
⟹ (𝑥 − 1)(𝑥 − 3) < 0
⟹ 𝒙 ∈ (1,3)

Question 10 [3]
To take any point on the line 𝑥 + 𝑦 = 4
Let 𝑥 = 𝛼 , then from equation of line 𝑦 = 4 − 𝛼
Therefore, P (𝛼, 4 − 𝛼) is any point on the line
It will be required point if its perpendicular distance from the line
4𝑥 + 3𝑦 – 10 = 0 is 1 unit
| 4𝛼+3×(4−𝛼)−10|
 √42 +32
= 1⇒ |𝛼 + 2 | = 5
 𝛼 + 2 = ±5
 𝛼 = 3, −7
Hence, the required points are (3,1) and (−7,11)

Question 11 [3]
(i) Given 𝑓(𝑥) = 5𝑥 2 + 2, 𝑥 ∈ 𝑅
𝑓 (3) = 5 × 32 + 2 = 5× 9 + 2 = 47
(ii) 𝑓(𝑥) = 5𝑥 2 + 2
as 𝑥 2 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 ∈ 𝑅,
so 5𝑥 2 ≥ 0 hence 5𝑥 2 + 2 ≥ 2, Thus the minimum value of the 𝑓(𝑥) is 2 when
𝑥 = 0 and since the function increases without bound as |𝑥| → ∞, the range is
[2,∞)

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
6

(iii) 𝑓(𝑥) = 1 ∙ 5 does not belong to the range [2,∞). Therefore, no real solution exists.

Question 12 [3]

𝑝 𝑞 ~𝑝 ~𝑞 ~𝑝 ∨ 𝑞 ∼ 𝑝 ∨∼ 𝑞 (~𝑝 ∨ 𝑞) ∧ (∼ 𝑝 ∨ ∼ 𝑞)
T T F F T F F
T F F T F T F
F T T F T T T
F F T T T T T

Question 13 [3]
(i) (a) 𝑑𝑃
= 6(𝑡 2 − 5𝑡 + 6)
𝑑𝑡
(b) 𝑑𝑃
( 𝑑𝑡 ) = 6(22 − 5 × 2 + 6) = 0
𝑡=2

(c) There is no change in the production on the second day.
OR
(ii) 𝑥 2 𝑥 1 𝑥
𝑑𝑦 𝑙𝑜𝑔𝑥(𝑒 𝑠𝑒𝑐 𝑥 + 𝑡𝑎𝑛𝑥𝑒 ) − 𝑥 𝑒 𝑡𝑎𝑛𝑥
=
𝑑𝑥 (log 𝑥)2
𝑒𝑥 𝑡𝑎𝑛𝑥
= (log 𝑥)2 (𝑙𝑜𝑔𝑥 𝑠𝑒𝑐 2 𝑥 + 𝑙𝑜𝑔𝑥 𝑡𝑎𝑛𝑥 − )
𝑥

Question 14 [3]
(i) Given current month’s reading = 6879
Previous month’s reading = 5241
Number of unit consumed = 6878 – 5241 = 1578
Energy charges
Units Price Amount

0 – 50 ₹3.00 ₹75
51 – 200 ₹5.00 ₹750
201 – 400 ₹6.00 ₹1200
>400 ₹8.00 ₹9424
Total ₹11,449

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
7

Fixed charges = ₹(110 × 5) = ₹ 550
Surcharge = ₹(0∙25 × 1578) = ₹394∙50
Energy tax = 5% of ₹11999 = ₹599∙95
Total amount of electricity bill = ₹11,449 + ₹550 + ₹394∙50 + 599∙95
=₹12, 993∙45
Hence, electric bill for Mr Mohanty for the month is ₹12, 993∙45.
OR

(ii) Half yearly instalment = ₹ 11,000
Rate of interest = 4% half-yearly.
This single sum is evidently the present value of immediate annuity of ₹ 11∙000 for
15 years at 4% per half-yearly
A = ₹ 11,000 𝑖 =0.04, 𝑛= 30 V =?
𝐴
V = 𝑖 ( 1 − (1 + 𝑖)−𝑛 )
11,000
V = 0.04 ( 1 − (1 ∙ 04)−30 )
= 2,75,000 ( 1 − 0 ∙ 3083)
= 2,75,000 × 0 ∙ 6817
= ₹1,90,217 ∙ 50
Hence, the single sum equivalent to his pension is ₹ 1,90,217∙50

Question 15 [3]
(i) (a) We have 𝜇1 = 0, 𝜇2 = 2 ∙ 5, 𝜇3 = 0 ∙ 7 𝑎𝑛𝑑 𝜇4 = 18 ∙ 75
𝜇 2 (0∙7)2 0∙49
Moment coefficient of kurtosis 𝛽1 = 𝜇33 = (2∙5)3 = 15∙625 = 0 ∙ 031
2

(b) Since, 𝜇3 > 0, so, it is not negatively skewed. The distribution is positively
skewed.
(c) 𝛽2 = 𝜇4 = 18∙75 = 3
𝜇 2 (2∙5)2
2
∴ The value of 𝛽2 = 3
Hence, the curve is mesokurtic.
OR

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
8

(ii) Let yield of paddy per acre be 𝑥 and annual rainfall be 𝑦.
𝑥 = 973 ∙ 5, 𝑦 = 18 ∙ 3
𝜎𝑥 = 38 ∙ 4, 𝜎𝑦 = 2 ∙ 0 𝑎𝑛𝑑 𝑟 = 0 ∙ 58
𝜎𝑥
𝑏𝑥𝑦 = 𝑟
𝜎𝑦
38∙4
= 0 ∙ 58 × = 11∙136
2
Using, 𝑥 − 𝑥 = 𝑏𝑥𝑦 (𝑦 − 𝑦)
⇒ 𝑥 − 973 ∙ 5 = 11 ∙ 136(𝑦 − 18 ∙ 3)
⇒ 𝑥 − 973 ∙ 5 = 11 ∙ 136(22 − 18 ∙ 3)
⇒ 𝑥 = 41 ∙ 2032 + 973 ∙ 5 = 1014 ∙ 70.
Hence, when the annual rainfall is 22 cm, the most likely yield of paddy is 1014∙70
per acre.

SECTION D – 25 MARKS
Question 16 [5]
(i) Strategy A follows AP
6, 7.5, 9, 10∙5, …
𝑛 10
Total Investment 𝑆10 = 2 [2𝑎 + (𝑛 − 1)𝑑] = 2 [2(6) + 9(1 ∙ 5)]
= ₹127∙5 lakhs
Strategy B follows GP
3, 4.5, 6∙75, 10∙125…… having |𝑟|>1.
(a) Total Investment 𝑆 = 𝑎 (𝑟 𝑛−1) = 3 ((1.5)10−1) = ₹340 𝑙𝑎𝑘ℎ𝑠 (𝑎𝑝𝑝𝑟𝑜𝑥)
10 𝑟−1 1.5−1

(b) Profit for Strategy A = ₹108 lakhs
108
Profit % for Strategy A = 127∙5 × 100 = 84 ∙ 71%

Profit for Strategy B= ₹272 lakhs
272
Profit % for strategy B = 340 × 100 = 80%
Profit for Strategy B = 80%
(c) Strategy A. Although Strategy B offers a higher total profit, Strategy A is
more cost-efficient due to its higher profit percentage.
OR

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
9

(ii) (a) Cost for nth day = 𝑛(𝑛 + 1)(𝑛 + 2)

(b) Total cost after n days = ∑𝑛𝑘=1 𝑘(𝑘 + 1)(𝑘 + 2) =
𝑛(𝑛+1)(𝑛+2)(𝑛+3)
4
(c) Total cost for 5 days =
5.6.7.8
= ₹420
4

(d) 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3)
= 5040 ⟹ 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3) = 20160
4
𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3) = 2016 × 10 = 7 × 8 × 9 × 10⟹ 𝑛 = 7

Question 17 [5]
(i) (a) Since, the interval are of uniform width 10, we use step deviation method to
calculate mean with assumed mean 45
Class Class frequency 𝑥𝑖 − 𝐴 𝑓𝑡 𝑓𝑡 2
𝑡=
interval mark (𝑥𝑖 ) (𝑓) 𝑖
20 – 30 25 9 -2 -18 36
30 – 40 35 17 -1 -17 17
40 – 50 45 32 0 0 0
50 – 60 55 23 1 23 23
60 – 70 65 16 2 32 64
70 – 80 75 3 3 9 27
Total ∑ 𝑓 =100 ∑ 𝑓𝑡 =29 ∑ 𝑓𝑡 2 =167

∑ 𝑓𝑡 29
Mean = 𝐴 + ∑ 𝑓 × 𝑖 = 45 + 100 × 10 = 47 ∙ 9
∑ 𝑓𝑡 2 ∑ 𝑓𝑡 2
Standard deviation (𝜎) = √ ∑ 𝑓 − ( ∑ 𝑓 ) × 𝑖

167 29 2
= √ − ( ) × 10
100 100
= 10√1 ∙ 67 − 0 ∙ 0841
= 10 × 1 ∙ 2593
= 12 ∙ 59
Hence, mean of the given data is 47∙9 and Standard deviation is 12∙59
𝜎 12∙59
The coefficient of variation (CV) = 𝑥 × 100 = 47∙9 × 100 = 26 ∙ 28 %

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
10

(b) As C.V of 26.29% indicates a moderate variability in the age distribution of
people reporting eye problems. This means the eye problem are not highly
concentrated in one particular age group, but moderately spread out, with the
mean age of affected individuals being around 48 years
(c) Since the mean age is approximately 48 years and variability is moderate, the
health team could target awareness program at the 40 – 50 age group. This
group eye problem due to professional screen expose or reduce eye resistance.
Campaigns could be focus on promoting regular eye checkups and screen time
management and eye exercise for working adult.
OR
(ii)
Internal Rank External Rank Difference 𝑑 𝑑2
marks (𝑥) (R1) Marks (𝑦) (R2) = (R1−R2)
16 3 72 3 0 0
18 2 78 1∙5 0∙5 0∙25
15 4 64 7 −3 9
14 5∙5 68 4∙5 1 1
12 7∙5 65 6 1∙5 2∙25
10 9 45 10 −1 1
14 5∙5 68 4∙5 1 1
9 10 54 9 1 1
12 7∙5 60 8 −0 ∙ 5 0∙25
20 1 78 1∙5 −0 ∙ 5 0∙25
Total ∑ 𝑑 2 =16

1
6{∑ 𝑑2 + ∑ (𝑚3 −𝑚)}
Spearman’s rank correlation coefficient (r) = 1 − 12
𝑛3 − 𝑛
1 1 1 1
6{16+ (23 −2)+ (23 −2)+ (23 −2)+ (23 −2)}
Here, 𝑟 = 1 − 12 12 12 12
103 − 10
6{16+0.5+0.5+0.5+0.5}
𝑟= 1− 990
108
=1−
990
= 1 − 0 ∙ 109
= 0 ∙ 891.
Hence, rank correlation is 0∙891, which shows a high positive relation between the
internal marks and external marks.

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
11

Question 18 [5]
(i) (a) The equation of line is 𝑦 = 𝑥 ----------------------------- (i)
and the equation of the circle is
𝑥 2 + 𝑦 2 − 2𝑥= 0 ----------------------------- (ii)
Solving (i) and (ii) simultaneously for point of intersection we have
𝑥 2 +𝑦 2 − 2𝑥 = 0 ⇒ 2𝑥(𝑥 − 1) = 0
⇒ 𝑥 = 0,1
From (i), when 𝑥 = 0, 𝑦 = 0 when 𝑥 = 1, 𝑦 = 1
∴ The points of intersections are A (0,0) and B (1,1).
The equation of the circle on AB as diameter is
(𝑥 − 0) (𝑥 − 1) + (𝑦 − 0)(𝑦 − 1) = 0
2 2
i.e. 𝑥 +𝑦 −𝑥−𝑦 = 0
(b) Standard form
1 1 1 1
𝑥 2 +𝑦 2 − 𝑥 − 𝑦 + 4 + 4 = 4 + 4
1 2 1 2 1
(𝑥 − ) + (𝑦 − ) =
2 2 2
1 1 1
radius = , centre = ( 2 ,2)
√2

(c) 1 1 1 1
Parametric equation of circle: 𝑥 = 2 + cos 𝑡, 𝑦 = 2 + sin 𝑡, where 0 <
√2 √2
𝑡 ≤ 2𝜋
OR
(ii) (a) In a circle, the perpendicular from the centre to a chord bisects the chord.
If a line through the centre C is perpendicular to a chord AB it must pass
through the midpoint of the chord. Since it starts from the centre and goes
straight across the circle, it’s a diameter.
Therefore, CM is a diameter.
1
The midpoint M of the chord AB = (2 , 2)
−2 2
Slope of AB = −3 = 3
−1 −3
Slope of CM= 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝐴𝐵 = 2
−3 1
Equation of CM will be: 𝑦 − 2 = (𝑥 − 2)
2
−3 3
𝑦= 𝑥 + 4 +2
2
6𝑥 + 4𝑦 = 11

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
12

(b) If the equation of the other diameter is 𝑥 − 3𝑦 − 11 = 0
The intersection point of the two diameters will give centre C(𝑥, 𝑦), hence
solving the two equations of two diameters
6(3𝑦 + 11) + 4𝑦 = 11
7 5 7 5
𝑥= 𝑎𝑛𝑑 𝑦 = − 2 so centre 𝐶 (2 , − 2)
2

(c) 7 5
With Centre 𝐶 (2 , − 2) and using 𝐴 (2, 3) we find radius of the circle.

7 2 5 2 65
Radius of the circle 𝑟 = √(2 − 2) + (3 + 2) =√2

Equation of the given circle will be
7 2 5 2 65
(𝑥 − ) + (𝑦 + ) =
2 2 2

Question 19 [5]
Given two right angle triangles i.e. △ABC and △ABD,
(since BD=BC + CD = 2+3 = 5m)
𝐵𝐶 2
(i) In triangle ABC : tan ∅ = 𝐴𝐵 = 3
𝐵𝐷 5
In triangle ABD : tan ( 𝜃 + ∅ ) = 𝐴𝐵 =3
tan 𝜃+tan ∅
(ii) Yes, i.e. tan ( 𝜃 + ∅ ) = 1−tan 𝜃.tan ∅
Let tan 𝜃 = 𝑥 , then
2
5 𝑥+3
= 2
3 1 − 𝑥. 3
5 2𝑥 2
(1 − 3 ) = 𝑥 + 3
3
9 9
⇒ 𝑥 = 19 therefore tan 𝜃 = 19
(iii) Yes, 𝜃, ∅ and 𝜃 + ∅ lies in the same quadrant.
tan ∅,tan( 𝜃 + ∅ ) and tan 𝜃 , all are positive,
⇒ All angles will lie either in 1stor in 3rd. quadrant.
From the figure,∅ is angle of right triangle ABC, hence it is an acute angle.
Also, 𝜃 + ∅ is an angle of right triangle ABD, hence it is also an acute angle.
⇒ 𝜃 is also an acute angle.
Hence 𝜃, 𝜑 and 𝜃 + 𝜑 lies in the 1st quadrant.

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
13

Question 20 [5]
Calculation of Income tax for financial year 2024-25 (Old Tax Regime)
Income from salary = ₹ 10,50,000
Income from interest on saving account = + ₹ 15,500
∴ Gross income = ₹10,65,500
Less standard deduction = −₹ 50,000
Balance = ₹10, 15,500
Deduction under section 80G
Donation in National Security Fund = − ₹25,000
Deduction under section 80 TTA on interest from saving account = − ₹10000
Balance = ₹ 9,80,500
Deduction under section 24
Interest on home loan = −₹ 76000
Balance = ₹ 9,04,500
Deduction under section 80C
Deposited in G.P.F ₹ ( 12 × 9200) = ₹ 1,10,400
LIC premium = ₹ 43000
Principal of home loan = ₹ 20000
Total = ₹ 1,73,400
But deduction under section 80C is allowed up to ₹1,50,000
Less deduction under section 80C = − ₹1,50,000
Taxable income = ₹ 7, 54, 500
Since, taxable income is ₹ 7, 54, 500, so from income tax slab
Income tax = ₹ 12,500+ 20% of 2,54,500
= ₹ 12,500 + ₹50,900
= ₹ 63,400.
Since taxable income is more than 5 lakh, so Mr Ajay is not eligible for tax rebate under
section 87A
∴ Income tax = ₹ 63,400
Health and education cess 4% on ₹63,400 = + ₹2,536

Net Income tax = ₹ 65,936.

ISC (CLASS XI) SPECIMEN ANSWER KEY 2026
14