ISC Class 12 Sample Paper 2027 Mathematics

FOR ISC CLASS 12 EXAM PREPARATION

ISC Class 12 2027
Sample Paper ·
Mathematics
EXAM YEAR TYPE SUBJECT

ISC Class 12 2027 Sample Paper Mathematics

Notes · Sample Papers · Previous Year Papers · Mock Tests

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MATHEMATICS
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Maximum Marks: 80
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Time Allotted: Three Hours
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Reading Time: Additional Fifteen Minutes
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Instructions to Candidates
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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. It consists of twenty questions and three sections: Sections A, B, C and D.
5. All questions are compulsory.
6. Section A consists of very short answer questions of 1 mark each.
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7. Section B consists of short answer questions of 2 marks each.
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8. Section C consists of moderately long answer questions of 3 marks each.
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9. Section D consists of long answer questions of 5 marks each.
10. Internal choices have been provided in three questions of 2 marks each, three
questions of 3 marks each and three questions of 5 marks each.
11. While attempting Multiple Choice Questions in Section A, you are required
to write only ONE option as the answer.
12. The intended marks for questions or parts of questions are given in the
brackets [].
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13. All workings, including rough work, should be done on the same page as, and
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adjacent to, the rest of the answer.
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14. Mathematical tables and graph papers are provided.

las Instruction to Supervising Examiner
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ag Kindly read aloud the instructions given above to all the candidates present in the
examination hall.

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1 ISC SPECIMEN QUESTION PAPER 2027
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Page 1 of 32

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.

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) A relation 𝑅 is defined on ℤ as 𝑎𝑅𝑏 if and only if 𝑎2 − 7𝑎𝑏 + 6𝑏 2 = 0. [1]
Then R is: (Understand)
(a) reflexive and symmetric
(b) transitive but not reflexive
(c) symmetric but not reflexive
(d) reflexive but not symmetric
(ii) If 𝜃 = 𝑠𝑖𝑛−1 𝑥 + 𝑐𝑜𝑠 −1 𝑥 − 𝑡𝑎𝑛−1 𝑥, 𝑥 ≥ 0, then the smallest interval in which [1]
𝜃 lies is: (Analyse)
𝜋 3𝜋
(a) ≤ 𝜃 ≤
2 4

(b) 0 < 𝜃 <𝜋
−𝜋
(c) ≤ 𝜃 ≤0
4
𝜋 𝜋
(d) ≤ 𝜃 ≤
4 2

(iii) Let 𝐴 be the area of a triangle having vertices (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ) and (𝑥3 , 𝑦3 ). [1]
Which of the following is correct? (Understand)
(a) 𝑥1 𝑦1 1
|𝑥2 𝑦2 1| = +𝐴
𝑥3 𝑦3 1
(b) 𝑥1 𝑦1 1
|𝑥2 𝑦2 1| = ±2𝐴
𝑥3 𝑦3 1
(c) 𝑥1 𝑦1 1 𝐴
|𝑥2 𝑦2 1| = ±
𝑥3 𝑦3 1 2

(d) 𝑥1 𝑦1 12
|𝑥2 𝑦2 1| = 𝐴2
𝑥3 𝑦3 1

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(iv) A particle moves along the curve 6𝑦 = 𝑥 3 + 2. At what point(s) on the curve, is [1]
the y-coordinate changing 8 times as fast as the x-coordinate? (Apply)
(a) 5
(2, ) only
3

(b) (4,11) only
(c) −31
(4,11) and (−4, )
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3

(d)
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(8,86) and (−8, −85)
c If a satellite's altitude function ℎ(𝑡) has a positive derivative [1]se m .
(v)
em (ℎ′(𝑡) > 0), the satellite is moving away from the Earth.
Statement I:
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g la II: At the apogee (highest point), the instantaneous velocity ( ) of thea
a
𝑑ℎ
Statement
𝑑𝑡
satellite relative to its altitude is zero.
Which of the following is correct? (Understand)
(a) Statement I is true and Statement II is false.
(b) Statements I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.
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(vi) For what value(s) of 𝑘 do the tangents of. two curves 𝑥 = 𝑦 and 𝑥𝑦 = 𝑘 cut at
e m 2 [1]
right angles?
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(a) 𝑘 = ±
1
4 a g
(b) 𝑘 = ±1
(c) 𝑘=0
1
(d) 𝑘=±
2√2

(vii) Evaluate the nature of the point (0,0) for the curve 𝑦 = 𝑥 4 , given that [1]
𝑓 ′′ (0) = 0. (Understand)
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(a) The function at the point (0,0) is undefined because the second derivative

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test fails.
s em
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(b) It is a point of local maximum because the first derivative

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sign from positive to negative as 𝑥 increases through 0.
a (c) It is a local minimum because the second derivative is positive for all 𝑥 ≠ 0.
(d) It is a point of neither local maximum nor local minimum because the second
derivative is zero.

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(viii) The expression for finding the area of the shaded region is: [1]
Y

x=2
O X
(Apply)
8 𝑥
(a) ∫2 (4 − 2 − 2) 𝑑𝑥
8 𝑥
(b) ∫2 (2 + 4) 𝑑𝑥
8 𝑦
(c) ∫2 ( 2 + 4) 𝑑𝑦
8 𝑥
(d) ∫2 (4 − 2) 𝑑𝑥

(ix) Shown below is a solved anti-differentiation problem to obtain 𝑓(𝑥): [1]
𝑑 1
𝑓(𝑥) = such that 𝑓(𝑒) = −1
𝑑𝑥 𝑥(log 𝑥)2
Taking anti-derivative:
1
𝑓(𝑥) = ∫ 𝑑𝑥 + 𝐶
𝑥(log 𝑥)2
𝑑(log 𝑥)
Step 1 ⇒ 𝑓(𝑥) = ∫ (log +𝐶
𝑥)2
1
Step 2 ⇒ 𝑓(𝑥) = − +𝐶
log 𝑥
Given 𝑓(𝑒) = −1
⇒ 𝑓(𝑒) = −1 + 𝐶
⇒𝐶 =0
1
Step 3 𝑓(𝑥) = −
log 𝑥

In which step is there an error (if any) in the solution? (Understand)
(a) Step 1
(b) Step 2
(c) Step 3
(d) No error

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(x) On solving ∫ − √1−𝑥 2 𝑑𝑥, Anil’s answer was ∫ − √1−𝑥 2 𝑑𝑥 = 3 cos −1 𝑥 + 𝑐 and [1]
3 𝑑𝑥
Jaspreet’s answer was ∫ − √1−𝑥2 𝑑𝑥 = −3 ∫ √1−𝑥2 = −3 sin−1 𝑥 + 𝑐.
Whose answer was correct? (Recall)
(a) Only Anil’s answer was correct.
(b) Only Jaspreet’s answer was correct.
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(c)
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Both of them were correct.
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(d) Neither
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s em+ 3𝑗̂ − 𝑘̂, (𝑏⃗⃗) = -𝑖̂ + 2𝑗̂ − 4𝑘̂, (𝑐⃗) = 𝑖̂ + 𝑗̂ + 𝑘̂, then (𝑎⃗ × 𝑏⃗⃗). (𝑎⃗ × 𝑐⃗) is: gla[1]
(xi) If (𝑎⃗) = 2𝑖̂

g la (Recall)a
a (a) 60
(b) 64
(c) 74
(d) −74
𝑥−1 𝑦−2 3−𝑧 𝑥−1 2−𝑦 𝑧−3
(xii) The point of intersection of the lines = = and = = is: [1]
2 3 −4 5 −2 1

om
(Understand)
(a) (1, 2, −3)
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(b) (− 1, 2, 3)
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(c) (1, −2, 3)
(d) ( 1, 2, 3)
(xiii) An objective function 𝑍 = 𝑎𝑥 + 𝑏𝑦 is maximum at points (8,10) and (7,12). [1]
If 𝑎, 𝑏 ≥ 0 and ab = 72, then the maximum value of the function is equal to:
(Apply)
(a) 84
(b) 132

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(c) 156
m (d) 176
c. o(xiv) Statement I: Every Linear Programming Problem has at least onesoptimal
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em a solutions, then it
[1]
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Statement II: If a Linear Programming Problem has two optimal
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has infinitely many solutions.
Which of the statements is correct? (Understand)
(a) Statement I is true and Statement II is false.
(b) Statements I is false and Statement II is true.
(c) Both the statements are true.
(d) Both the statements are false.

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𝐵 𝐵 𝐴 [1]
(xv) If 𝑃(𝐴) = 𝑚, 𝑃 ( ̅ ) = 3𝑚, 𝑃 ( ) = 6𝑚, then what will be 𝑃 ( )? (Analyse)
𝐴 𝐴 𝐵

(a) 2𝑚
1+𝑚
(b) 4𝑚
1+𝑚
(c) 7𝑚
1+𝑚
(d) 9𝑚
1+𝑚
(xvi) Observe the given graph: [1]

Assertion: The curves given above can be considered as 𝑓(𝑥) and 𝑓 −1 (𝑥).
Reason: The curves are reflections of each other across the line 𝑥 = 𝑦.
Which one of the following is correct? (Understand)
(a) Both Assertion and Reason are true, and Reason is the correct explanation
of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
1
(xvii) Assertion: ∫ 𝑥 3 cos 𝑥 𝑑𝑥 = 0 [1]
−1
Reason: If 𝑓(𝑥) is a continuous function defined on [0, 𝑎], then
𝑎 𝑎
∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥)𝑑𝑥
Which one of the following is correct? (Analyse)
(a) Both Assertion and Reason are true, and Reason is the correct explanation
of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct
explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.
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(xviii) Using the properties of matrices, explain why the following concept is incorrect: [1]
Since (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 , therefore, (A + B) (A – B) = A2 − B 2 also must
be true if A and B are matrices. (Understand)
(xix) 𝑑𝑦 𝑎𝑥+3
If the solution of the differential equation = represents a circle, then find [1]
𝑑𝑥 2𝑦+5
the value of 𝑎. (Analyse)
(xx) The probability distribution of random variable X is given below. [1]
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omX 1 2 3 4 5 . co
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e m P(X) m 3m a 5m b
as
s (Apply) gl
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If 𝑃(𝑋 ≤ 2) = 0 · 28 and 𝑃(𝑋 ≥ 4) = 0 · 52, find 𝑃(𝑋 = 3).
a
a SECTION B – 14 MARKS

Question 2 [2]
A straight line 𝐿𝜃 has vector equation 𝑟⃗ = 5𝑖̂ + 𝜆(5𝑖̂ + 𝑠𝑖𝑛 𝜃 𝑗̂ + 𝑐𝑜𝑠 𝜃 𝑘̂) and a plane 𝜋𝑃
has equation 𝑥 = 𝑝, 𝑝𝜖ℝ.
Show that the angle between 𝐿𝜃 and 𝜋𝑃 is independent of both 𝜃 and p. (Understand)

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Question 3
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In a school, for a period of 8 working days,site is equally likely that Tanishka is present or
absent on any given day.
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a in school for at least 5 consecutive working days?
What is the probability that she is present
(Evaluate)

Question 4 [2]
Solve the following differential equation:
𝑥
𝑑𝑥 𝑥 𝑓( )

m
𝑦
= + (Apply)
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𝑥
𝑑𝑦 𝑦 𝑓′ ( )
m
𝑦

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s e Question 5
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(i) Determine the values of constants 𝑝 and 𝑞 such that the function [2]
a 𝑝 sin 𝑥 + 𝑞 , ∀𝑥 ≤ 0
𝑓(𝑥) = { 2 is differentiable at 𝑥 = 0. (Understand)
𝑥 + 2𝑥 + 1 , ∀𝑥 > 0
OR

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(ii) Observe the graph given below:

(a)State the value(s) of 𝑥 where the function has removable discontinuity. [1]
(Understand)
(b) State the value(s) of 𝑥 where the graph of the function is not differentiable. [1]
(Understand)

Question 6 [2]
(i) Parag is working on a school project on right-angled triangles. He draws a
right-angled triangle PQR with ∠R = 90°. The sides opposite angles P, Q, R are 𝑝, 𝑞, 𝑟
respectively.
𝑝 𝑞
Evaluate the expression: 𝑡𝑎𝑛−1 (𝑞+𝑟) + 𝑡𝑎𝑛−1 (𝑟+𝑝). (Apply)
OR
(ii) The equation given below has equal roots:
𝑎𝑥 2 + 𝑠𝑖𝑛−1 (𝑥 2 − 2𝑥 + 2) + 𝑐𝑜𝑠 −1 (𝑥 2 − 2𝑥 + 2) = 0
What is the value of ‘𝑎’? (Apply)

Question 7 [2]
(i) ⃗⃗ = 6𝑖̂ + 3𝑗̂ + 2𝑘̂ on 𝐴⃗ = 𝑖̂ − 2𝑗̂ − 2𝑘̂ and the scalar
Find the vector projection of 𝐵
component of 𝐵⃗⃗ on 𝐴⃗. (Apply)
OR
(ii) If the position vectors of three points A, B, C are respectively 𝑖̂ + 𝑗̂ + 𝑘̂,
2𝑖̂ + 3𝑗̂ − 4𝑘̂ and 7𝑖̂ + 4𝑗̂ + 9𝑘̂, find the unit vector perpendicular to the plane of
triangle ABC. (Apply)

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Question 8
Three landmarks of Dehradun are joined by straight roads. Clock tower of Dehradun is
considered as the 'origin'. IMA (I) is 3 km east and 9 km north of Clock Tower and
Rajpur (R) is 5 km east and 5 km south of IMA. A bus stop (S) is situated two thirds of the
way along the road from Clock Tower to IMA.
Considering 𝑖̂ as a 1 km vector pointing east and 𝑗̂ as a 1 km vector pointing north:
(i) Find the position vector of the bus stop (S) relative to the Clock Tower. (Understand) [1]
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(ii) Prove that the bus stop (S) is the closest point to Rajpur (R) on the Clock Tower to [1]
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IMA(I) Road.
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l as SECTION C – 21 MARKS
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a9
Question
The feasible region determined by some constraints is represented by the shaded region in
the graph given below:

12

10

m
(0,9)

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8

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C(1,7)
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6

l a s
gB(3,3)
4

2 a
A(1,1) (9/2,0)
(0,0)
2 4 6 8 10 12

(i) Formulate the constraints which represent the above feasible region. (Understand) [1]

(ii) Hence, maximise the objective function given by 𝑍 = 𝑥 + 𝑦.
o m (Understand) [1]
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c. o(iii) What change in the constraints will make the feasible region unbounded?
s
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e(Understand)
[1]

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ag Question 10 [3]
𝑑2 𝑦 𝑑2 𝑦
If 𝑥 = 3𝑧 + 1 and 𝑦 = 𝑓(𝑥), then prove that 9 = (Apply)
𝑑𝑥 2 𝑑𝑧 2

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Question 11
√1+𝑥 2 +𝑥 2𝑥 [2]
(i) Find the derivative of the function 𝑓(𝑥) = log [√1+𝑥 2 ] + tan−1 with respect
−𝑥 1−𝑥 2
to 𝑥. (Evaluate)
(ii) An analyst claims that the function 𝑓(𝑥) has a local maximum at some point 𝑥 = 𝑐. [1]
Evaluate this claim using the expression for 𝑓 ′ (𝑥). (Evaluate)

Question 12
(i) A vector 𝑛⃗⃗ of magnitude 8 units is inclined to 𝑥 - axis at 45o, 𝑦 - axis at 60o and at an [3]
acute angle with z-axis. A plane through the point (√2, −1,1) is normal to 𝑛⃗⃗.
Find the equation of the plane in vector form. (Analyse)
OR
(ii) The planes 𝜋1 and 𝜋2 have equations 2𝑥 + 6𝑦 − 2𝑧 = 5 and
51
3𝑥 + 9𝑦 + 𝑝𝑧 = − respectively.
2
1 [1]
(a) Verify that the point P (2, , 1) lies on the plane 𝜋1 . (Apply)
2

(b) Determine the value of 𝑝 if 𝜋2 is parallel to 𝜋1 . (Analyse) [1]

A line through P normal to 𝜋1 meets 𝜋2 at the point Q. Find the coordinates of [1]
(c)
Q. (Analyse)

Question 13 [3]
(i) If 𝑥, 𝑦, 𝑧 are distinct non zero real numbers, prove that
2
𝑥 𝑥 𝑦𝑧 2
𝑦 𝑧 𝑥 2 2
2
∆= |𝑦 𝑦 𝑧𝑥 | = |𝑦 3 𝑧 3 𝑥 3 |.
𝑧 𝑧 2 𝑥𝑦 1 1 1
Hence or otherwise evaluate ∆ in its simplest form if 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 = 1. (Analyse)
OR
(ii) A school is organising its Annual Day function and plans to decorate every chair with
a ribbon and every table with a cover. There are 150 chairs and 20 tables, all of which
need decorations. Company A charges ₹ 12 per chair ribbon and ₹ 180 per table cover.
Company B charges ₹ 10 per chair ribbon and ₹ 200 per table cover.
By setting up one matrix equation that include both companies, compare the overall
prices that would be charged by the two companies. (Analyse)

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Question 14
5
Let ∫1 3𝑓(𝑥)𝑑𝑥 = 12
1
(i) Show that ∫5 𝑓(𝑥)𝑑𝑥 = −4. (Understand) [1]
2 5
(ii) Find the value of ∫1 (𝑥 + 𝑓(𝑥)) 𝑑𝑥 + ∫2 (𝑥 + 𝑓(𝑥))𝑑𝑥. (Understand) [2]

o m
Question 15
m
o shows the graph of 𝑓(𝑥) = 2𝑥√𝑎 − 𝑥 , for −1 ≤ 𝑥 ≤ 𝑎, where [3] em . c
.
(i) The diagram belowc 2

s
2

𝑎 > 1.
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The linesL is the tangent to the graph of 𝑓(𝑥) at the origin O. l a
l a a g
ag that 𝑓 (𝑥) = ′ 2𝑎2 −4𝑥 2
Given , for −1 ≤ 𝑥 ≤ 𝑎.
√𝑎2 −𝑥 2
Using integration, find the area of ∆OPQ in terms of 𝑎. (Apply)

P(a,b)
L

f
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OR
(ii) A line with equation 𝑦 = −3𝑥 + 9 intersects the axes at the points P and Q.
A parabola of the form 𝑦 = 𝑎𝑥 2 + 𝑐 , where 𝑎 , 𝑐 ∈ ℤ, also passes through the points
P and Q as shown in the diagram below.
y
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P

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a Q
x

(a) Obtain the equation of the parabola. (Apply) [1]
(b) Using integration find the area of the shaded region. (Apply) [2]

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SECTION D – 25 MARKS

Question 16
The graph of 𝑓(𝑥) = 𝑥 − 6√𝑥 + 1 is shown below.
y
f(x) = x − 6√𝑥 + 1

9
x

(i) State the natural domain of 𝑓(𝑥). (Analyse) [1]
(ii) Does 𝑓(𝑥) have an inverse function? Explain your answer. (Analyse) [1]
(iii) Let g (𝑥) = 𝑥 − 6√𝑥 + 1, 0 ≤ 𝑥 ≤ 9. Find g −1 (𝑥) and verify (g o g −1 )(4) = 4. [1]
(Analyse)
(iv) Let h (𝑥) = 𝑥 − 6√𝑥 + 1, 𝑥 ≥ 9. Find h−1 (𝑥) and verify (h o h−1 )(16) = 16. [1]
(Analyse)
(v) Find the value of 𝑥 such that g −1 (𝑥) = h−1 (𝑥) (Analyse) [1]

Question 17
(i) An engineering team is designing a section of a new roller coaster track. The vertical
profile of the track for a specific horizontal stretch is modelled by the function:
1
𝑓(𝑥) = 𝑥 3 − 2𝑥 2 + 3𝑥 + 5
3
where 𝑥 represents the horizontal distance from the start of the section (in meters)
and 𝑓(𝑥) represents the height of the track (in meters). To ensure safety and a
thrilling experience, the engineering team must analyse the steepness and the peaks
of this track.

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(a) Identify the intervals of 𝑥 where the roller coaster is climbing (increasing [1]
height) and where it is descending (decreasing height). (Analyse)
(b) A support beam must be placed at the point where the track’s slope is exactly [1]
zero. Find the coordinates of the points where support beam must be placed.
(Evaluate)
(c) Determine the maximum and minimum heights reached by the roller coaster [2]
in the interval 𝑥 ∈ [0, 4]. (Evaluate)
m
om
(d) At the point 𝑥 = 1, a maintenance ladder must be placed perpendicular to [1]
. co
. c
the track, (along the normal). Find the equation of the maintenance ladder at
e m
𝑥 = 1.
e m (Evaluate)
l as
l as OR
ag
(ii)
agA large industrial water tank is shaped like an inverted right circular cone with a
semi-vertical angle of tan−1(0.5). Water is being drained out for a manufacturing
process at a constant rate of 5 m3 /min.

r

om
h=4m
tan α = 0.5

. c
e m
s
𝑑𝑣

a
= −5𝑚3 /𝑚𝑖𝑛
l 𝑑𝑡

agwater level is dropping when the height of the water
(a) Find the rate at which the [2]
is 4 meters. (Evaluate)
(b) Show that the wetted surface area of the tank 𝑆(ℎ) is a strictly increasing [1]
function of the water depth ℎ. (Analyse)
(c) A chemical additive must be added to the interior surface of the water tank. [2]
If the cost of the additive is proportional to the square of the surface area
(𝐶 = 𝐾𝑆 2 ), find the depth ℎ at which the cost is increasing most rapidly
relative to time. (Evaluate)
m
m
c. oQuestion 18 .co
m s em(Understand) [2]
s e (i) (a) Evaluate: ∫ ln 𝑥 𝑑𝑥.
g la
g la 1+𝑥
ln[ln( )] a
a
1−𝑥
(b) Hence, evaluate: ∫
1− 𝑥 2
𝑑𝑥 (Evaluate) [3]

OR
(ii) (a) Evaluate: ∫ tan 𝑥 𝑑𝑥 (Understand) [2]
𝑑𝑥
(b) Hence, evaluate: ∫ 𝑥 𝑥 𝑥 (Evaluate) [3]
cot cot cot
2 3 6

m . c
c. o
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Question 19
(i) In a game show, a contestant is shown three closed doors: Door A, Door B and
Door C. Behind one door is a laptop while the other two doors have nothing behind
them. The laptop is placed randomly, so each door is equally likely to contain the
laptop.
The contestant first selects Door A. The host, who knows where the laptop is and
never opens the door containing the laptop, opens Door B and reveals that there is
nothing behind the door.
The host then asks whether the contestant would like to choose to Door A or change
to Door C.
(a) Find the probability that the laptop is behind Door A, given that Door B has [2]
been opened. (Analyse)
(b) Find the probability that the laptop is behind Door C, given that Door B has [2]
been opened. (Analyse)
(c) Hence, state giving reasons, whether the contestant should change their [1]
selection or not. (Analyse)
OR
(ii) In a large organisation, only 1 out of every 1,500 emails contains a harmful
attachment. An automated filtering system is used to detect such emails.
The filtering system correctly marks a harmful email as dangerous 97% of the time.
It also correctly marks a safe email as safe 97% of the time.
(a) Find the probability that the email is safe. (Analyse) [1]
(b) Find the probability that given the email is harmful the filter marks it safe. [2]
(Analyse)
(c) One particular email has been flagged as dangerous by the system. Find the [2]
probability that this email is actually safe. (Analyse)

Question 20 [5]
A ball is thrown vertically downwards from the top of a cliff, and its position is tracked
from when it is first thrown until it hits the ground (at which point it may be assumed) that
the ball comes instantaneously to rest.
The height of the ball above the ground after 𝑡 seconds is given by the equation
𝑆(𝑡) = 𝑎𝑡 2 + 𝑏𝑡 + 𝑐, where 𝑎, 𝑏 and 𝑐 are real constants and the height 𝑠 is measured in
meters. It is observed that after 1 second, the ball is 285 m above the ground; after 2 seconds,
its height is 260 m and after 4 seconds it is 180 m.
Set up and solve a matrix equation to find the height of the cliff. (Evaluate)

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MATHEMATICS
ANSWER KEY

SECTION A – 20 MARKS

m
co
Question 1

om .
In answering Multiple Choice Questions, candidates have to write either the correct
. c
option number or the explanation against it. Please note that only ONE correct answer
e m
should be written.
e m a s
l [1]
(i)
l as
(d) reflexive but not symmetric
ag
g
Reflexive: ∀ 𝑎 ∈ ℤ, we have,
a 𝑎 − 7𝑎 ⋅ 𝑎 + 6𝑎 = 𝑎 − 7𝑎 + 6𝑎 = 0 ⇒ (𝑎, 𝑎) ∈ 𝑅
2 2 2 2 2

∴ Relation is reflexive
Symmetric:
For (6,1) as, 62 − 7 × 6 × 1 + 6 × 12 = 36 − 42 + 6 = 0 ⇒ (6,1) ∈ 𝑅.
But, for (1,6): 12 − 7.1.6 + 6. 62 = 1 − 42 + 216 ≠ 0 ⇒ (1,6) ∉ 𝑅
∴ Relation is not symmetric.
m
.co
𝝅 𝝅
(ii) (d) ≤ 𝜽 ≤ [1]
𝟒 𝟐

e m
𝑠𝑖𝑛 𝑥 and 𝑐𝑜𝑠 −1 𝑥 is defined for −1 ≤ 𝑥 ≤ 1
−1

𝑡𝑎𝑛 𝑥 is defined for all real 𝑥las
−1

ag is: 0 ≤ 𝑥 ≤ 1
Given, 𝑥 ≥ 0, the valid domain
𝜋
For all 𝑥 ∈ [−1, 1], 𝑠𝑖𝑛−1 𝑥 + 𝑐𝑜𝑠 −1 𝑥 = 2
𝜋
𝜃 = − 𝑡𝑎𝑛−1 𝑥
2
𝜋 𝜋
The principal value range of 𝑡𝑎𝑛−1 𝑥 is (− , )
2 2
𝜋
But since 𝑥 ≥ 0, 𝑡𝑎𝑛−1 𝑥 ∈ 0 [0, ] (∵ 𝑥 ≤ 1)
m
4

.co
−1 𝜋
0 ≤ 𝑡𝑎𝑛 𝑥 ≤
m
.co
4
𝜋
e m
e m ⇒−
4
≤ −𝑡𝑎𝑛−1 𝑥 ≤ 0
las
las 𝜋 𝜋
⇒ ≤ − 𝑡𝑎𝑛−1 𝑥 ≤
𝜋
ag
ag 4
𝜋
⇒4≤ 𝜃 ≤2
2
𝜋
2

m . c
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𝒙 𝟏 𝒚𝟏 𝟏 [1]
(iii) (b) |𝒙𝟐 𝒚𝟐 𝟏| = ±𝟐 A
𝒙 𝟑 𝒚𝟑 𝟏
Area of triangle formed by three points is:
𝑥1 𝑦1 1
1
𝐴 = |𝑥2 𝑦2 1|
2
𝑥3 𝑦3 1
Since determinant may be positive or negative
𝑥1 𝑦1 1
|𝑥2 𝑦2 1| = ±2 A
𝑥3 𝑦3 1
(iv) 𝟑𝟏 [1]
(c) (4,11) and (−𝟒, − )
𝟑
𝑑𝑦 𝑑𝑥
Differentiating w.r.t ‘t’ , 6 = 3𝑥 2 .
𝑑𝑡 𝑑𝑡
𝑑𝑦 𝑑𝑥
Substituting =8 gives the value of 𝑥 = ±4.
𝑑𝑡 𝑑𝑡
−31
When 𝑥 = 4, 𝑦 = 11 𝑎𝑛𝑑 𝑤ℎ𝑒𝑛 𝑥 = −4, 𝑦 =
3
31
∴ Points are (4,11) and (−4, − 3 )
(v) (c) Both the statements are true. [1]
Statement I: If ℎ′ (𝑡) > 0, it means altitude increases with time, so satellite is
moving away from earth
∴ Statement I is true.
Statement II: At highest point (apogee), altitude becomes maximum. At maximum
𝑑ℎ
point, = 0.
𝑑𝑡
This means change in altitude is zero.
∴ Statement II is true
(vi) 𝟏 [1]
(d) 𝒌 = ±
𝟐√𝟐
𝑑𝑦 𝑑𝑦 1
𝑥 = 𝑦 2 ⇒ 1 = 2𝑦 ⇒ =
𝑑𝑥 𝑑𝑥 2𝑦
𝑑𝑦 𝑑𝑦 −𝑦
𝑥𝑦 = 𝑘 ⇒ 𝑥 +𝑦 =0 ⇒ =
𝑑𝑥 𝑑𝑥 𝑥
Since tangents of the given two curves are at right angles, product of their slopes
= −1
1 −𝑦 1
⇒ . = −1 ⇒ 𝑥 = = 𝑦 2
2𝑦 𝑥 2
2 2 2 1
∴𝑘 =𝑥 𝑦 ⇒
8
1
∴𝑘=±
2√2

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(vii) (c) It is a local minimum because the second derivative is positive for all [1]
𝒙 ≠ 𝟎.
𝑑𝑦 𝑑2𝑦
= 4𝑥 ⇒ 2 = 12𝑥 2
3
𝑑𝑥 𝑑𝑥
𝑑2 𝑦 𝑑2 𝑦
Though = 0 when 𝑥 = 0, the fact that 2 = 12𝑥 2 is always positive for all
𝑑𝑥 2 𝑑𝑥
𝑥 ≠ 0 which confirms the curve is a concave up at 𝑥 = 0.
The function attains local minimum at (0,0).
m
(viii) 𝟖 𝒙
(d) ∫𝟐 (𝟒 − ) 𝒅𝒙
om [1] . co
. c𝟐
e m
e m
Given equation of line: 𝑥 + 2𝑦 = 8
l as
l as 𝑥
⇒ + =1
𝑦
8 4
ag
ag ⇒𝑦= 4 −
𝑥
2
𝑥
Required area is the area bounded by 𝑥 = 2; 𝑦 = 4 − and 𝑥-axis.
2
8 𝑥
∴ Required area = ∫2 (4 − ) 𝑑𝑥
2
(ix) (d) No error [1]
(x) (c) Both of them were correct. [1]
𝜋
Notice that: cos (− sin−1 𝑥 + )
m
2

.co
−1
= sin(sin 𝑥)
=𝑥
e m
So, the answers are equivalent.
l as
ag
(xi) (d) −74 [1]
𝑖̂ 𝑗̂ 𝑘̂
⃗⃗
𝑎⃗ × 𝑏 = | 2 3 −1| = −10𝑖̂ + 9𝑗̂ + 7𝑘 ̂
−1 2 −4
𝑖̂ 𝑗̂ 𝑘̂
𝑎⃗ × 𝑐⃗ = |2 3 −1| = 4𝑖̂ − 3𝑗̂ − 𝑘̂
1 1 1
∴ (𝑎⃗ × 𝑏⃗⃗). (𝑎⃗ × 𝑐⃗) = −40 − 27 − 7 = −74

om
(xii) (d) (𝟏, 𝟐, 𝟑) [1]

m
𝑥−1
=
𝑦−2
=
𝑧−3 𝑥−1
=
𝑦−2
=
𝑧−3
. c
c. o
Given lines are and
2 3 4 5 2 1
e m
em
∵ Both lines are passing through (1,2,3)
las
l as ∴ The point of intersection is (1,2,3).
ag
ag
(xiii) (c) 156 [1]
Maximum at points (8,10) and (7,12)
⟹ 8𝑎 + 10𝑏 = 7𝑎 + 12𝑏 ⟹ 𝑎 = 2𝑏
But as 𝑎𝑏 = 72 ⟹ 2𝑏 2 = 72 ⟹ 𝑏 2 = 36 ⟹ 𝑏 = 6.
∴ 𝑎 = 2𝑏 = 12
Maximum value = 8 × 12 + 10 × 6 = 156

m . c
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(xiv) (b) Statement I is false and Statement II is true. [1]
Statement I: Every Linear Programming Problem has at least one optimal Solution.
False - Some LPPs are infeasible
Statement II: If an Linear Programming Problem has two optimal solutions, then
it has infinitely many solutions.
True - If two points give same maximum value, then all points on the line joining
them also give same value.
Hence, infinitely many optimal solutions exist.
(xv) 𝟐𝒎 [1]
(a)
𝟏+𝒎
𝐵
𝑃 ( ) = 6𝑚 ⟹ 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 6𝑚 = 6𝑚2
𝐴
𝐵
𝑃 ( ̅ ) = 3𝑚 ⟹ 𝑃(𝐵 ∩ 𝐴̅) = 𝑃(𝐴̅) × 3𝑚 = 3𝑚(1 − 𝑚) = 3𝑚 − 3𝑚2
𝐴
𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 3𝑚 − 3𝑚 ⟹ 𝑃(𝐵) = 3𝑚 − 3𝑚 + 6𝑚 = 3𝑚 + 3𝑚2
2 2 2

𝐴 𝑃(𝐴∩𝐵) 6𝑚2 2𝑚
𝑃( ) = = 3𝑚+3𝑚2 =
𝐵 𝑃(𝐵) 1+𝑚
(xvi) (a) Both Assertion and Reason are true, and Reason is the correct explanation [1]
of Assertion.
The assertion states that the curves can be considered as 𝑓(𝑥) and 𝑓 −1 (𝑥)
Since the reflection property is satisfied
∴ Assertion is true.
The reason states that the curves are reflections of each other across the line
𝑥 = 𝑦.
The reflection of a curve across the line 𝑦 = 𝑥 yields the graph of the inverse
function because the geometric reflection swaps the 𝑥 and 𝑦 coordinates, which
corresponds to the algebraic definition of an inverse mapping.
∴ Reason is true and Reason correctly explains the Assertion.
(xvii) (b) Both Assertion and Reason are true, but Reason is not the correct [1]
explanation of Assertion.
Let 𝑓(𝑥) = 𝑥 3 cos 𝑥
𝑓(−𝑥) = −𝑥 3 cos 𝑥 = −𝑓(𝑥)
⇒ 𝑓(𝑥) is an odd function
1
⇒ ∫−1 𝑥 3 cos 𝑥 𝑑𝑥 = 0
𝑎
As ∫−𝑎 𝑓(𝑥)𝑑𝑥 = 0 when 𝑓(𝑥) is an odd function.
Hence assertion is true.
If 𝑓(𝑥) is a continuous function defined on [0, 𝑎], then
𝑎 𝑎
∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥)𝑑𝑥 is also true but not applicable for this assertion.
(xviii) Matrix Multiplication is non – commutative (AB ≠ BA) [1]
(A + B) (A – B) = A2 − AB + BA − B2
But AB ≠ BA
∴ (A + B) (A – B) ≠ A2 − B2
Hence, the concept is incorrect.

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(xix) 𝒂 = −𝟐 [1]
𝑑𝑦 𝑎𝑥+3
=
𝑑𝑥 2𝑦+5
⇒(2𝑦 + 5)𝑑𝑦 = (𝑎𝑥 + 3)𝑑𝑥
Integrating both sides:
⇒∫(2𝑦 + 5)𝑑𝑦 = ∫(𝑎𝑥 + 3)𝑑𝑥
𝑥2
⇒𝑦 2 + 5𝑦 + 𝑐 = 𝑎 2 + 3𝑥
m
𝑥2
om
⇒ −𝑎 2 − 3𝑥 + 𝑦 2 + 5𝑦 + 𝑐 = 0 . co
. c e m
m
Now if 𝑎 = −2, then the solution of the differential equation will be
e l as
l as
⇒𝑥 2 + 𝑦 2 − 3𝑥 + 5𝑦 + 𝑐 = 0, which is an equation of circle.
ag
(xx) a
g ∴ 𝑎 = −2
0·2 [1]
𝑃(𝑋 ≤ 2) = 0 · 28 ⟹ 4𝑚 = 0 · 28 ⟹ 𝑚 = 0 · 07
𝑃(𝑋 ≥ 4) = 0 · 52 ⟹ 5𝑚 + 𝑏 = 0 · 52 ⟹ 𝑏 = 0 · 52 − 0 · 35 = 0 · 17
9𝑚 + 𝑎 + 𝑏 = 1 ⟹ 𝑎 = 1 − 0 · 63 − 0 · 17 = 0 · 2
𝑃(𝑋 = 3) = 0 · 2

SECTION B – 𝟏𝟒 MARKS
m
.co
em
Question 2 [2]

l as
∵ 𝑟⃗ = 5𝑖̂ + 𝜆(5𝑖̂ + 𝑠𝑖𝑛 𝜃 𝑗̂ + 𝑐𝑜𝑠 𝜃 𝑘̂),

g
∴ Direction vector of above line 𝑑⃗ = 5𝑖̂ + 𝑠𝑖𝑛 𝜃 𝑗̂ + 𝑐𝑜𝑠 𝜃 𝑘̂ .
Now equation of plane 𝜋 is 𝑥 = 𝑝. a 𝑃
⇒ 𝑟⃗. 𝑛⃗⃗ = 𝑝 where n is the normal vector to the plane.
∴ 𝑛⃗⃗ = 𝑖̂ + 0𝑗̂ + 0𝑘̂
⇒ 𝑛⃗⃗ = 𝑖̂
|𝑑⃗.𝑛
⃗⃗| |5|
∴ 𝑠𝑖𝑛 𝜃 = |𝑑⃗|.|𝑛⃗⃗| =
√26.1
−1 5
𝜃 = 𝑠𝑖𝑛 ( )
√26
m
m
Hence proved.
c. oQuestion 3 m.co
s e
s em On any given day, Tanishka can be either present or absent. g la
[2]

g la Total possible outcomes = 2 8 a
a If we consider that she comes to school as P and she is absent as A,
Case 1: If Tanishka comes for at least the first 5 consecutive working days, then we just
need to consider the attendance for the remaining 3 days.
PPPPP _ _ _
3
This can be done in 2 ways.

m . c
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Case 2: If Tanishka comes for at least any other 5 consecutive working days, then she must
be absent on the preceding day. So, we just need to consider the attendance for the
remaining 2 days.
APPPPP_ _ or _ APPPPP _ or _ _ APPPPP
2
This can be done in 2 ways.
∴ Probability (comes to school for at least 5 consecutive working days)
23 +22 +22 +22 5
= =
28 64

Question 4 [2]
𝑥
𝑑𝑥 𝑥 𝑓( )
𝑑𝑦
= + ′ 𝑦𝑥
𝑦 𝑓 ( )
𝑦

Given differential equation is homogenous differential equation of type
𝑑𝑥 𝑥
= 𝑓( )
𝑑𝑦 𝑦
𝑥
Substituting, 𝑥 = 𝑣𝑦 i.e. =𝑣
𝑦
𝑑𝑥 𝑑𝑣
⇒ = 𝑣 + 𝑦.
𝑑𝑦 𝑑𝑦
𝑑𝑣
𝑓(𝑣)
⇒ 𝑣 + 𝑦. = 𝑣 + ′ (𝑣)
𝑑𝑦 𝑓
𝑑𝑣 𝑓(𝑣)
⇒ 𝑦. = ′ (𝑣)
𝑑𝑦 𝑓
𝑓 ′ (𝑣) 𝑑𝑦
⇒ 𝑑𝑣 =
𝑓(𝑣) 𝑦
Integrating both sides
𝑑(𝑓(𝑣)) 𝑑𝑦
⇒∫ =∫ + log 𝑐
𝑓(𝑣) 𝑦
⇒log|𝑓(𝑣)| = log 𝑦 + log 𝑐
⇒ 𝑓(𝑣) = 𝑐𝑦
𝑥
⇒ 𝑓 ( ) = 𝑐𝑦, which is the required solution.
𝑦

Question 5
(i) Since differentiability implies continuity, the limits from both sides at 𝑥 = 0 must [2]
be equal.
𝐿𝐻𝐿 = lim− 𝑝 sin 𝑥 + 𝑞 = 𝑞
𝑥→0
𝑅𝐻𝐿 = lim+ 𝑥 2 + 2𝑥 + 1 = 1
𝑥→0
At 𝑥 = 0, 𝑓(0) = 𝑞 = 𝐿𝐻𝐿 = 𝑅𝐻𝐿 ⟹ 𝑞 = 1
The Left-Hand Derivative (LHD) must equal the Right-Hand Derivative (RHD) at
𝑥 = 0.
𝑑
𝐿𝐻𝐷 = (𝑝 sin 𝑥 + 𝑞) = 𝑝 cos 𝑥,
𝑑𝑥
At 𝑥 = 0, 𝐿𝐻𝐷 = 𝑝 cos 𝑥 = 𝑝
𝑑
𝑅𝐻𝐷 = (𝑥 2 + 2𝑥 + 1) = 2𝑥 + 2
𝑑𝑥

6 ISC SPECIMEN ANSWER KEY 2027

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.
.co s e m
s em l a
a ag
At 𝑥 = 0, 𝑅𝐻𝐷 = 2𝑥 + 2 = 2
⟹𝑝=2
OR
(ii) (a) At 𝑥 = −2 the function has removable discontinuity. [1]
At 𝑥 = −2, LHL = RHL but 𝑓(−2) is missing.
(b) 𝑓(𝑥) is discontinuous at 𝑥 = −2, 𝑥 = 0 and 𝑥 = 2 hence not differentiable. [1]
m
om . co
. c [2]em
Question 6
e m l as
l as
(i) Given, In ∆PQR, ∠R = 90°
∴ 𝑟 2 = 𝑝2 + 𝑞 2 ….. (i) (Pythagoras ag
ag Now,
Theorem)

𝑝 𝑞
+
−1 𝑞+𝑟 𝑟+ 𝑝
𝑡𝑎𝑛 ( 𝑝𝑞 )
1− (𝑞
+ 𝑟) (𝑟 + 𝑝)
𝑝𝑟 + 𝑝2 + 𝑞 2 + 𝑞𝑟
= 𝑡𝑎𝑛−1 ( )
𝑞𝑟 + 𝑝𝑞 + 𝑟 2 + 𝑝𝑟 − 𝑝𝑞
𝑝𝑟 + 𝑟 2 + 𝑞𝑟
= 𝑡𝑎𝑛−1 ( ) … .. (from (i))
𝑝𝑟 + 𝑟 2 + 𝑞𝑟
= 𝑡𝑎𝑛−1 (1)
𝜋
o m
=
4
. c
s emOR
l a
Given, 𝑎𝑥 2 + 𝑠𝑖𝑛−1 (𝑥 2 − 2𝑥 + 2) + 𝑐𝑜𝑠 −1 (𝑥 2 − 2𝑥 + 2) = 0
ag
(ii)
2 𝜋 −1 −1 𝜋
∴ 𝑎𝑥 + = 0 …………………. (i) [∵ 𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 = ]
2 2
2
For the given inverse functions: −1 ≤ 𝑥 − 2𝑥 + 2 ≤ 1
Now, (𝑥 2 − 2𝑥 + 2) = (𝑥 − 1)2 + 1
Since (𝑥 − 1)2 ≥ 0
Minimum value occurs when 𝑥 = 1
∴ (𝑥 − 1)2 + 1 ≥ 1
To satisfy domain ≤ 1 we have,
m
.co
(𝑥 − 1)2 + 1 = 1
m
.co
∴ (𝑥 − 1)2 = 0
e m
e m ∴𝑥=1
las
las Thus, only one value of 𝑥 exists
ag
ag Hence, roots are equal at 𝑥 = 1
𝜋
∴ 𝑎 (1)2 + = 0
2
𝜋
∴𝑎=−
2

m . c
c. o
7 ISC SPECIMEN ANSWER KEY 2027
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em as
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Question 7 [2]
(i) Vector projection of B on A
⃗⃗ . 𝐴̂)𝐴̂
= (𝐵
𝐵.𝐴⃗⃗ ⃗
= ( |𝐴⃗| ) 𝐴̂
⃗⃗.𝐴⃗
𝐵
=( 2 ) 𝐴⃗
|𝐴⃗|
4
= − (𝑖̂ − 2𝑗̂ − 2𝑘̂)
9
⃗⃗ on 𝐴⃗
Scalar component of 𝐵
𝐴⃗
⃗⃗ .
=𝐵 |𝐴⃗|
𝑖̂ 2𝑗̂ 2
= (6𝑖̂ + 3𝑗̂ + 2𝑘̂). ( − − 𝑘̂)
3 3 3
4
=−
3
OR
(ii) A (𝑎⃗) = 𝑖̂ + 𝑗̂ + 𝑘̂, B (𝑏⃗⃗) = 2𝑖̂ + 3𝑗̂ − 4𝑘̂ , B (𝑐⃗) = 7𝑖̂ + 4𝑗̂ + 9𝑘̂
⃗⃗⃗⃗⃗⃗ = 𝑖̂ + 2𝑗̂ − 5𝑘̂ and 𝐴𝐶
∴ 𝐴𝐵 ⃗⃗⃗⃗⃗⃗ = 6𝑖̂ + 3𝑗̂ + 8𝑘̂
𝑖̂ 𝑗̂ 𝑘̂
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗
∴ 𝐴𝐵 × 𝐴𝐶 = |1 2 −5|
6 3 8
= 31𝑖̂ − 38𝑗̂ − 9𝑘̂
|31𝑖̂ − 38𝑗̂ − 9𝑘̂| = √2486
̂
31𝑖̂−38𝑗̂ −9𝑘
Unit vector =
√2486

Question 8
(i) Position vector of IMA(I) relative to Clock Tower (𝑂𝐼 ⃗⃗⃗⃗⃗ ) = 3𝑖̂ + 9𝑗̂ [1]
Position vector of Rajpur (R) relative to Clock Tower (𝑂𝑅 ⃗⃗⃗⃗⃗⃗ ) = 8𝑖̂ + 4𝑗̂
Position vector of Bus stop(S) relative to Clock Tower (𝑂𝑆 ⃗⃗⃗⃗⃗⃗) = 2 (3𝑖̂ + 9𝑗̂)
3
= 2𝑖̂ + 6𝑗̂
(ii) Now, ⃗⃗⃗⃗⃗⃗
𝑅𝑆 = ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ = 2𝑖̂ + 6𝑗̂ − 8𝑖̂ − 4𝑗̂ = −6𝑖̂ + 2𝑗̂.
𝑂𝑆 − 𝑂𝑅 [1]
⃗⃗⃗⃗⃗⃗. ⃗⃗⃗⃗⃗⃗
∴ 𝑂𝑆 𝑅𝑆 = −12 + 12 = 0.
⇒ Bus stop is the closest point to Rajpur on the Clock Tower to IMA Road.

SECTION C – 21 MARKS

Question 9
(i) 2𝑥 + 𝑦 ≤ 9 [1]
𝑦≥𝑥
𝑥≥1

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.co s e m
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a ag
(ii) 𝑍 = 8 𝑎𝑡 𝐶 (1,7) [1]
(iii) 2𝑥 + 𝑦 ≥ 9 [1]

Question 10 [3]
𝑑𝑥 𝑑𝑦 𝑑𝑦 𝑑𝑧 𝑑𝑦 𝑑𝑦
= 3, and = → =3
𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑥

m
co
𝑑2 𝑦 𝑑 𝑑𝑦
=3 ( )
om .
𝑑𝑧 2 𝑑𝑧 𝑑𝑥
𝑑 𝑑𝑦 𝑑𝑥
. c e m
=3
𝑑𝑥 𝑑𝑥
( ).
e 𝑑𝑧
m l as
=3
𝑑2 𝑦
l as
.3 ag
𝑑2 𝑦 a g
𝑑𝑥 2
𝑑2 𝑦
=9
𝑑𝑧 2 𝑑𝑥 2

Question 11
(i) Simplify the given function: [2]
√1+𝑥2 +𝑥 √1+𝑥2 +𝑥 2𝑥
𝑓(𝑥) = log [√1+𝑥2 . ] + tan−1
−𝑥 √1+𝑥2 +𝑥 1−𝑥 2
2
m
.co
(√1+𝑥 2 +𝑥)
𝑓(𝑥) = log [ 1+𝑥 2 −𝑥 2 ] + 2tan−1 𝑥

e m
𝑓(𝑥) = 2 log(√1 + 𝑥 2 + 𝑥) + 2tan−1 𝑥
s
+ 1) +la
ag
1 2𝑥 2
𝑓 ′ (𝑥) = 2 √1+𝑥 2 .(
+𝑥 2√1+𝑥 2 1+𝑥 2
2 2
𝑓 ′ (𝑥) = √1+𝑥 2 +
1+𝑥 2

(ii) The claim is incorrect. [1]
2 2
Since 𝑥 2 ≥ 0, both √1+𝑥2 and are always positive.
1+𝑥 2
Thus, 𝑓 ′ (𝑥) > 0 for all real 𝑥.
A local maximum requires 𝑓 ′ (𝑥) to be zero and change sign. Here, the function is
strictly increasing.
m
om 12
c. Question m.co
s e
s em (i) Let 𝛾 be the angle made by 𝑛⃗⃗ with z-axis, then direction cosines of 𝑛⃗⃗ are
g la [3]

la a
1 1
𝑙 = 𝑐𝑜𝑠 45° = , 𝑚 = 𝑐𝑜𝑠 60 ° = and n = cos𝛾
ag ∵ 𝑙 2 + 𝑚 2 + 𝑛2 = 1
√2 2

1
𝑛2 =
4
1 1
𝑛 = (neglecting 𝑛 = − as 𝛾 is acute)
2 2
∵ |𝑛⃗⃗| = 8
∴ 𝑛⃗⃗ = |𝑛⃗⃗|(𝑙𝑖̂ + 𝑚𝑗̂ + 𝑛𝑘̂)

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𝑖̂ 𝑗̂ ̂
𝑘
𝑛⃗⃗ = 8 ( + 2 + 2)
√2
= 4√2𝑖̂ + 4𝑗̂ + 4𝑘̂
∴ Required plane passes through the point (√2, −1,1) having position vector
𝑎⃗ = √2𝑖̂ − 𝑗̂ + 𝑘̂
∴ Vector equation is (𝑟⃗ − 𝑎⃗). 𝑛⃗⃗ = 0
𝑟⃗. (4√2𝑖̂ + 4𝑗̂ + 4𝑘̂ ) = (√2𝑖̂ − 𝑗̂ + 𝑘̂). (4√2𝑖̂ + 4𝑗̂ + 4𝑘̂)
⇒ 𝑟⃗ . (4√2𝑖̂ + 4𝑗̂ + 4𝑘̂) = 8
⇒ 𝑟⃗ . (√2𝑖̂ + 𝑗̂ + 𝑘̂ ) = 2
OR
(ii) (a) 𝜋1 : 2𝑥 + 6𝑦 − 2𝑧 = 5 [1]
1
Substituting the point P (2, , 1),
2
1
LHS = 2 × 2 + 6 × − 2 × 1 = 4 + 3 − 2 = 5 = 𝑅𝐻𝑆
2
51 [1]
(b) 𝜋2 : 3𝑥 + 9𝑦 + 𝑝𝑧 = −
2
3 9 𝑝
Given that 𝜋2 ∥ 𝜋1 ⇒ = =
2 6 −2
∴ 𝑝 = −3
(c) Equation of the line that is perpendicular to 𝜋1 is [1]
1
𝑥−2 𝑦− 𝑧−1
= 2
= =𝜆
1 3 −1
1
General points on the line are (𝜆 + 2, 3𝜆 + , −𝜆 + 1)
2
∵ 𝑄 𝑙𝑖𝑒𝑠 𝑜𝑛 𝜋2
1 51
⇒ 3(𝜆 + 2) + 9 (3𝜆 + ) − 3(1 − 𝜆) = −
2 2
3 17
⇒ 𝜆 + 2 + 9𝜆 + − 1 + 𝜆 = −
2 2
⇒ 11𝜆 = −11 ⇒ 𝜆 = −1
5
∴ Co-ordinates of Q (1, − , 2)
2

Question 13 [3]
(i) 𝑥 2 𝑥 3 𝑥𝑦𝑧 𝑥2 𝑥3 1
1 2 3
∆=
𝑥𝑦𝑧
|𝑦 𝑦 𝑥𝑦𝑧| = |𝑦 2 𝑦 3 1|
𝑧 2 𝑧 3 𝑥𝑦𝑧 𝑧2 𝑧3 1
𝑥2 𝑦2 𝑧2
= |𝑥 3 𝑦 3 𝑧 3 | Applied the property |𝐴| = |𝐴𝑡 |
1 1 1
𝑦2 𝑥2 𝑧2 𝑦2 𝑧2 𝑥2
= − |𝑦 3 𝑥 3 𝑧 3 | = |𝑦 3 𝑧 3 𝑥 3 | (Hence proved)
1 1 1 1 1 1

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.
.co s e m
s em l a
a ag
𝑥2 𝑥3 1
∆= |𝑦 2 𝑦 3 1|
𝑧2 𝑧3 1
𝑅1 → 𝑅1 − 𝑅2 and 𝑅2 → 𝑅2 − 𝑅3
𝑥2 − 𝑦2 𝑥3 − 𝑦3 0 𝑥 + 𝑦 𝑥 2 + 𝑦 2 + 𝑥𝑦 0
∆= | 𝑦 2 − 𝑧 2 𝑦 3 − 𝑧 3 0| = (𝑥 − 𝑦)(𝑦 − 𝑧) |𝑦 + 𝑧 𝑦 2 + 𝑧 2 + 𝑦𝑧 0|
𝑧2 𝑧3 𝑧2 𝑧3
1 1
m
Again 𝑅2 → 𝑅2 − 𝑅1
om . co
. c
𝑥+𝑦 𝑥 2 + 𝑦 2 + 𝑥𝑦 0
e m
m
= (𝑥 − 𝑦)(𝑦 − 𝑧) | 𝑧 − 𝑥 𝑧 2 − 𝑥 2 + 𝑦(𝑧 − 𝑥) 0|
e l as
l as 𝑧2 𝑧3 1
ag
ag
= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥) | 1
𝑥 + 𝑦 𝑥 2 + 𝑦 2 + 𝑥𝑦 0
𝑥+𝑦+𝑧 0|
2 3
𝑧 𝑧 1
= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥)
= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥) (As 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 = 1)
OR
(ii) Let 𝑄 be the quantities of chair ribbon and table cover needed
∴ 𝑄 = [150 20]
m
.co
Let P represents the prices of chair ribbon and table cover from each company.
12 10 ← 𝑐hair ribbons
∴𝑃=[ ]
e m
↑
180 200 ← table cover
↑
l as
Company Company
A B ag
∴ Total cost matrix is given by
C = QP
12 10
∴ 𝐶 = [150 20] [ ]
180 200
∴ 𝐶 = [150 × 12 + 20 × 180 150 × 10 + 20 × 200]
∴ 𝐶 = [1800 + 3600 1500 + 4000]
∴ 𝐶 = [5400 5500]
m
m Thus, the price of company A (₹ 5400) is ₹ 100 less than the price of the company B
.co
m .co (₹5500).
s e m
s e g la
g la Question 14
a
a (i) 5
Given, ∫1 3𝑓(𝑥)𝑑𝑥 = 12 [1]
5
⇒∫1 𝑓(𝑥)𝑑𝑥 = 4
5
⇒− ∫1 𝑓(𝑥)𝑑𝑥 = −4
1 𝑏 𝑎
⇒∫5 𝑓(𝑥)𝑑𝑥 = −4. [∵ ∫𝑎 𝑓(𝑥)𝑑𝑥 = − ∫𝑏 𝑓(𝑥)𝑑𝑥]
Hence proved.

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(ii) 2 5 [2]
∫(𝑥 + 𝑓(𝑥)) 𝑑𝑥 + ∫(𝑥 + 𝑓(𝑥))𝑑𝑥
1 2
5
=∫1 (𝑥 + 𝑓(𝑥)) 𝑑𝑥
5 5
= ∫1 𝑥 𝑑𝑥 + ∫1 𝑓(𝑥)𝑑𝑥
5
𝑥2
=[ ] + 4
2 2
25 1
= − +4
2 2
= 16

Question 15
(i) Given, L is the tangent to the graph of 𝑓 at the [3]
origin O.
Slope of the tangent at O = 𝑓 ′ (0)
2𝑎2 − 0
= = 2𝑎
√𝑎2 − 0
∴ Equation of L: 𝑦 − 0 = 2𝑎(𝑥 − 0)
⇒ 𝑦 = 2𝑎𝑥.
𝑎
∴∆OPQ =∫0 2𝑎𝑥 𝑑𝑥.
𝑎
𝑥2
= 2𝑎 × [ 2 ]
0
3
= 𝑎 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠.
OR

(ii) (a) Putting x=0 in 𝑦 = −3𝑥 + 9, y=9 [1]
⇒P (0,9)
and Putting y=0 in 𝑦 = −3𝑥 + 9, x=3
⇒ Q (3,0)
The parabola is of the form 𝑦 = 𝑎𝑥 2
+ 𝑐 , where 𝑎 , 𝑐 ∈ ℤ
Satisfying by, (0,9), we get 𝑐 = 9.
Satisfying by, (3,0), we get 𝑎 = −1.
∴ Equation of parabola is: 𝑦 = −𝑥 2 + 9

𝑥 𝑦
(b) Equation of line PQ is: + = 1 i.e. 𝑦 = 9 − 3𝑥 [2]
3 9
∴ The area of the shaded region
3
= ∫0 (9 − 𝑥 2 − 9 + 3𝑥) 𝑑𝑥
3
= ∫0 (3𝑥 − 𝑥 2 ) 𝑑𝑥

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.
.co s e m
s em l a
a 3
ag
3𝑥 2 𝑥3
=[ − ]
2 3 0
27
= −9
2
9
= sq units
2

SECTION D – 25 MARKS
m
om . co
m
Question 16
. c s e
m
(i) 𝑓(𝑥) = 𝑥 − 6√𝑥 + 1 [1]

s e
Natural domain of 𝑓(𝑥) is 𝑥 ≥ 0.
g l a
g l a
i.e. [0, ∞) a
a

m
c. o
(ii) As the graph fails the horizontal line test, hence it is not one-one. [1]
⇒ 𝑓(𝑥) is not invertible.
e m
𝑓(𝑥) = 𝑥 − 6√𝑥 + 1
l as
⇒ 𝑦 = 𝑥 − 6√𝑥 + 1
⇒ (√𝑥) − 6√𝑥 + (1 − 𝑦) =a0
2 g
6±√36−4(1−𝑦) 6±2√𝑦+8
∴ √𝑥 = = = 3 ± √𝑦 + 8
2 2
2
∴ 𝑥 = (3 ± √𝑦 + 8)
(iii) Given, g (𝑥) = 𝑥 − 6√𝑥 + 1, 0 ≤ 𝑥 ≤ 9. [1]
2
∴ 𝑔−1 (𝑥) = (3 − √𝑥 + 8)

m
2
∴ (g o g −1 )(𝑥) = (3 − √𝑥 + 8) − 6(3 − √𝑥 + 8) + 1
m .co
.co
= 9 − 6√𝑥 + 8 + 𝑥 + 8 − 18 + 6√𝑥 + 8 + 1 = 𝑥
e m
s
−1 )(4)
∴ (g o g =4
e m (iv) Given, h (𝑥) = 𝑥 − 6√𝑥 + 1, 𝑥 ≥ 9. la
las ag [1]

ag ∴ ℎ−1 (𝑥) = (3 + √𝑥 + 8)
2

2
∴ (h o ℎ−1 )(𝑥) = (3 + √𝑥 + 8) − 6(3 + √𝑥 + 8) + 1
= 9 + 6√𝑥 + 8 + 𝑥 + 8 − 18 − 6√𝑥 + 8 + 1 = 𝑥
∴ (h o ℎ−1 )(16) = 16

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(v) When g −1 (𝑥) = h−1 (𝑥) [1]
2 2
⇒ (3 − √𝑥 + 8) = (3 + √𝑥 + 8)
⇒ 9 − 6√𝑥 + 8 + 𝑥 + 8 = 9 + 6√𝑥 + 8 + 𝑥 + 8
⇒ √𝑥 + 8 = 0
⇒ 𝑥 = −8

Question 17
(i) (a) 1 [1]
𝑓(𝑥) = 𝑥 3 − 2𝑥 2 + 3𝑥 + 5
3
𝑓 ′ (𝑥) = 𝑥 2 − 4𝑥 + 3
𝑓 ′ (𝑥) = 0 ⟹ 𝑥 2 − 4𝑥 + 3 = 0 ⟹ (𝑥 − 1)(𝑥 − 3) = 0 ⟹ 𝑥 = 1 & 𝑥 = 3
Climbing up (increasing) ⟹ 𝑓 ′ (𝑥) > 0 on [0,1) ∪ (3, ∞)
Descending (decreasing) ⟹ 𝑓 ′ (𝑥) < 0 on (1, 3)
(b) Slope of the horizontal tangent 𝑓 ′ (𝑥) = 0, at 𝑥 = 1 & 𝑥 = 3 [1]
1
At 𝑥 = 1, 𝑓(1) = − 2 + 3 + 5 = 6 · 33
3
At 𝑥 = 3, 𝑓(3) = 9 − 18 + 9 + 5 = 5
Coordinates of the points on the curve: (1, 6 · 33) & (3, 5)
(c) Maxima & Minima in the interval [0, 4] [2]
Evaluate the function at the critical points 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 3 and end points of
the interval 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 4
𝑓(0) = 5
𝑓(1) = 6 · 33
𝑓(3) = 5
64
𝑓(4) = − 32 + 12 + 5 = 6.33
3
The absolute maximum is 6·33 at 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 4
The absolute minimum is 5 at 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 3
(d) The equation of normal at 𝑥 = 1 [1]
Slope of the tangent at 𝑥 = 1 ⟹ 𝑓 ′ (1) = 0
i.e. the tangent is a horizontal line ⟹ normal is a vertical line (𝑥 = 1)
OR
(ii) Let the radius, height, volume and semi vertical angle of the cone be 𝑟, ℎ, 𝑉 & 𝛼
𝑟
tan 𝛼 = = 0.5 ⟹ 𝑟 = 0.5 ℎ
ℎ
𝑑𝑉
Rate of flow (drained out) = = −5𝑚3 /𝑚𝑖𝑛
𝑑𝑡
1
Volume = 𝑉 = 𝜋𝑟 2 ℎ
3
1 0.25 𝜋
(a) 𝑉 = 𝜋(0.5ℎ)2 ℎ = 𝜋ℎ3 = ℎ3 [2]
3 3 12
𝑑𝑉
𝑑𝑉 𝜋 𝑑ℎ 𝜋 𝑑ℎ 𝑑ℎ −5
= 3ℎ2 ⟹ ℎ2 ⟹ = 𝜋𝑑𝑡2 = 𝜋 2
𝑑𝑡 12 𝑑𝑡 4 𝑑𝑡 𝑑𝑡 ℎ ℎ
4 4

14 ISC SPECIMEN ANSWER KEY 2027

For more Question Papers, Sample Papers, Notes & Syllabus visit Page 28 of 32

.
.co s e m
s em l a
a 𝑑𝑉
ag
Now substitute 𝑑𝑡 = −5 𝑎𝑛𝑑 ℎ = 4
𝜋 𝑑ℎ 𝑑ℎ 𝑑ℎ −5
−5 = 42 = 4𝜋 ⟹ = 𝑚/𝑚𝑖𝑛
4 𝑑𝑡 𝑑𝑡 𝑑𝑡 4𝜋
(b) √5
𝑙 = Slant height = √𝑟 2 + ℎ2 = √(0.5ℎ)2 + ℎ2 = √1.25ℎ2 = 2 ℎ [1]
Curved Surface area = 𝜋𝑟𝑙
√5 √5
𝑆 = 𝜋(0.5ℎ) 2 ℎ = 4 𝜋ℎ2
m
𝑑𝑆 √5
= 2 𝜋ℎ
om . co
c m
𝑑ℎ

.
Since ℎ > 0 for physical tank
m
𝑑𝑆
>0
s e
⟹s eSurface area 𝑆 is strictly increasing on(0, ∞) 𝑑ℎ

g l a
(c)gl
a 2
5𝑘𝜋 2 a [2]
a Given that Cost of the additive 𝐶 = 𝑘𝑆 = 𝑘 ( 𝜋ℎ ) = 2 √5 2
ℎ4
4 16
𝑑𝐶 𝑑𝐶 𝑑ℎ
Rate of change of cost = = .
𝑑𝑡 𝑑ℎ 𝑑𝑡
5𝑘𝜋 2 5
= ℎ3 . 𝜋 2 = 25𝑘𝜋ℎ
4 ℎ
4
𝑑𝐶
The rate of cost change is a linear function of ℎ.
𝑑𝑡
The rate of change increases as ℎ increases. Therefore, the cost is increasing
most rapidly at the maximum possible depth ℎ allowed by the tank.
m
Question 18
m .co
s e
a
(i) (a) ∫ ln 𝑥 𝑑𝑥 [2]
l
= ∫ ln 𝑥 . 1 𝑑𝑥 ag
I II
Applying by parts:
𝑑
= ln 𝑥 . ∫ 1 𝑑𝑥 − ∫ {∫ (ln 𝑥) ∫ 1𝑑𝑥} 𝑑𝑥
𝑑𝑥
1
= ln 𝑥 × 𝑥 − ∫ × 𝑥𝑑𝑥
𝑥
= 𝑥 ln 𝑥 − ∫ 1𝑑𝑥
= 𝑥(ln 𝑥 − 1) + 𝑐
m
om .co
1+𝑥
(b) 𝐼 = ∫ ln[ln(1−𝑥)] 𝑑𝑥 [3]
. c 1− 𝑥 2
e m
e m Let ln (
1+𝑥
)=𝑡
las
las 1−𝑥
Differentiating w.r.t’𝑡’: ag
ag 1−𝑥
⇒ 1+𝑥 ×
1−𝑥−(1+𝑥)(−1)
(1−𝑥)2
𝑑𝑥
× =1
𝑑𝑡
1
⇒2 × 𝑑𝑥 = 𝑑𝑡
1−𝑥 2
1+𝑥
ln[ln( )]
1−𝑥
∴𝐼=∫ 1− 𝑥 2 𝑑𝑥
1 1+𝑥 2
= 2 ∫ ln [ln ( )] × 𝑑𝑥
1−𝑥 1−𝑥 2

m . c
c. o
15 ISC SPECIMEN ANSWER KEY 2027
e m
em as
s
For more Question Papers, Sample Papers, Notes & Syllabus visit
gl
Page 29 of 32

1
= ∫ ln 𝑡 . 𝑑𝑡
2
1
= 𝑡(ln 𝑡 − 1) + 𝑐
2
1 1+𝑥 1+𝑥
= ln ( ) [ln [ln ( )] − 1] + 𝑐
2 1−𝑥 1−𝑥
OR
(ii) (a) ∫ tan 𝑥 𝑑𝑥 [2]
sin 𝑥
=∫ 𝑑𝑥
cos 𝑥
𝑑(cos 𝑥)
= −∫
cos 𝑥
= − ln|cos 𝑥| + 𝑐
(b) 𝑑𝑥
𝐼=∫ 𝑥 𝑥 𝑥 [3]
cot cot cot
2 3 6
𝑥 𝑥 𝑥
= ∫ tan tan tan 𝑑𝑥
2 3 6
𝑥 𝑥 𝑥
Now, = +
2 3 6
𝑥 𝑥 𝑥
⇒tan = tan ( + )
2 3 6
𝑥𝑥
𝑥 tan + tan
⇒ tan = 𝑥
36
𝑥
2 1− tan tan
3 6
𝑥 𝑥 𝑥 𝑥 𝑥 𝑥
⇒tan tan tan = tan −tan −tan
2 3 6 2 3 6
𝑥 𝑥 𝑥
∴𝐼 = ∫ tan 𝑑𝑥 − ∫ tan 𝑑𝑥 − ∫ tan 𝑑𝑥
2 3 6
𝑥 𝑥 𝑥
= −2 ln |cos | + 3 ln |cos | + 6 ln |cos | + 𝑐
2 3 6

Question 19
1
(i) P (laptop behind door A) = P (laptop behind door B) = P (laptop behind door C) =
3
If the laptop is behind door A, either door B or door C can be chosen to be opened.
1
⟹ P (Door B opened/ laptop behind A) =
2
Since the host will never reveal the laptop
⟹ P (Door B opened/ laptop behind B) = 0
If the laptop is behind door C, then only door B can be opened since door A is the
selection of the contestant and cannot be opened.
⟹P (Door B opened/ laptop behind C) = 1
1 1
(a) × 1 [2]
P (laptop is behind Door A/Door B has been opened) = 1 1
2 3
1 1 =
× + 0 × + 1× 3
2 3 3 3
1
(b) 1× 2 [2]
P (laptop is behind Door C/Door B has been opened) = 1 1
3
1 =
1
× +0× +1× 3
2 3 3 3

(c) The contestant should change their selection to Door C as the probability is [1]
more.
OR

16 ISC SPECIMEN ANSWER KEY 2027

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.
.co s e m
s em l a
a ag
(ii) A: Email is harmful
B: The filter marks email dangerous
1
Probability that the email is harmful = P(A) =
1500
1499
(a) Probability that the email is safe = 𝑃(𝐴̅) = [1]
1500
(b) Probability that given the email is harmful the filter marks it dangerous [2]
𝐵
= 𝑃 ( ) = 0 · 97
𝐴
m
co
𝐵̅
Probability that the given email is safe and the filter marks it safe = 𝑃 (𝐴̅) =
0.97
. c om e m .
e m
Probability that the given email is harmful and the filter marks it safe
l as
a s 𝐵 𝐵̅
= 𝑃 ( ̅ ) = 1 − 𝑃 (𝐴̅) = 0 · 03 g
a [2]
(c)glProbability that filter marks the email dangerous (Total probability)
𝐴

a = 𝑃(𝐵) = 𝑃 ( ) 𝑃(𝐴) + 𝑃 ( ) 𝑃(𝐴̅) = 0 · 97 × + 0 · 03 ×
𝐵 𝐵 1 1499
𝐴 𝐴̅ 1500 1500
= 0 · 0306
Probability that given the email is marked dangerous but it is safe =
𝐵 𝐵
𝐴̅ 𝑃(𝐴̅∩𝐵) 𝑃( ̅ )𝑃(𝐴̅) 𝑃( ̅ )𝑃(𝐴̅)
𝐴 𝐴
𝑃 (𝐵 ) = = = 𝐵 𝐵
𝑃(𝐵) 𝑃(𝐵) 𝑃( )𝑃(𝐴)+𝑃( ̅ )𝑃(𝐴̅)
𝐴 𝐴
1499
0·03 ×
= 1
1500
1499 = 0 · 9789 ≈ 0 · 98
m
0·97 × + 0·03 ×

.co
1500 1500

em
Question 20 [5]
Given, 𝑆(𝑡) = 𝑎𝑡 2 + 𝑏𝑡 + 𝑐
l as
ag
∴ 𝑆(1) = 𝑎(1)2 + 𝑏(1) + 𝑐 ⇒ 𝑎 + 𝑏 + 𝑐 = 285 → (𝑖)
𝑆(2) = 𝑎(2)2 + 𝑏(2) + 𝑐 ⇒ 4𝑎 + 2𝑏 + 𝑐 = 260 → (𝑖𝑖)
𝑆(4) = 𝑎(4)2 + 𝑏(4) + 𝑐 ⇒ 16𝑎 + 4𝑏 + 𝑐 = 180 → (𝑖𝑖𝑖)
The system can be written as 𝐴𝑋 = 𝐵
1 1 1 𝑎 285
[4 2 1] [𝑏 ] = [260]
16 4 1 𝑐 180
1 1 1 𝑎 285
where 𝐴 = [4 2 1] , 𝑋 = [𝑏 ] and B = [260]
m
.co
16 4 1 𝑐 180
m
.co
1 1 1
e m
e m
|𝐴| = |4 2 1| = 1(2 − 4) − 1(4 − 16) + 1(16 − 32)
16 4 1
las
las = −2 + 12 − 16 = −6 ≠ 0
ag
ag −1 −1 1
∴ 𝐴 exists and 𝐴 = |𝐴| . adj.A.
−2 3 −1
adj.A = [ 12 −15 3 ]
−16 12 −2
−2 3 −1
1 1
∴ 𝐴−1 = |𝐴| adj.A = − 6 [ 12 −15 3]
−16 12 −2

m . c
c. o
17 ISC SPECIMEN ANSWER KEY 2027
e m
em as
s
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Page 31 of 32

Now, 𝑋 = 𝐴−1 𝐵
𝑎 1 −2 3 −1 285
𝑏
∴ [ ] = − [ 12 −15 3 ] [260]
𝑐 6
−16 12 −2 180
𝑎 1 −570 + 780 − 180 1 30
∴ [𝑏 ] = − [ 3420 − 3900 + 540 ] = − [ 60 ]
𝑐 6 6
−4560 + 3120 − 360 −1800
∴ 𝑎 = −5, 𝑏 = −10 and 𝑐 = 300
∴ Height equation: 𝑆(𝑡) = −5𝑡 2 − 10𝑡 + 300
Height of the cliff:
At 𝑡 = 0, 𝑆(0) = −5(0)2 − 10(0) + 300
∴ Height of the cliff = 300 m.

18 ISC SPECIMEN ANSWER KEY 2027

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