Class 12 Sample Paper 2023 Solutions Maths

Class 12 Sample Paper 2023 Solution

Practice Questions - Marking Scheme
Session 2022-23
Class XII
Mathematics (Code – 041)

SECTION A - Multiple Choice Questions - 1 Mark each

Q.No. Answer/Solution Marks
Q.1 C. sec-1 x 1
Q.2 B. P and Q must be square matrices of the same order. 1
Q.3 D. all - i), ii) and iii) 1
Q.4 A. -48 1
Q.5 1 1
C. 4
Q.6 1 1
B. −𝑡𝑎𝑛 𝑥 − B, where B is a constant.
Q.7 D. 4 1
Q.8 C. 9 sq units 1
Q.9 1

B.
Q.10 4
C. 3 𝑒 3𝑥 + 1 1
Q.11 A. only ii) 1
Q.12 1
C.
Q.13 D. 0 1
Q.14 B. 60° 1
Q.15 D. 8 1
Q.16 B. It has a unique solution. 1
Q.17 D. 0.08 1
Q.18 A. Minimise Z = x + y 1
Q.19 C. (A) is true but (R) is false. 1
Q.20 C. (A) is true but (R) is false. 1

SECTION B - VSA questions of 2 marks each

Q.21 2𝜋 0.5
Solves the RHS to obtain 3 as follows:

1 1
Equates the LHS to obtain x = − as follows:
√3

1 𝜋
Finds 𝑡𝑎𝑛−1 (𝑥) as − 3 as follows: 0.5

OR

2 4
i) Finds the domain as (−∞, 5] ∪ [5 , ∞)as follows: 1

ii) Finds the range as [−2, 3𝜋 − 2] as follows: 1

Q.22 1 0.5
Writes the expression for C as 2 (A − A′).

Finds A' as: 0.5

Finds C as:
1

Q.23 Differentiates y with respect to x using chain rule as: 1

𝑑𝑦 𝑑 sec 𝑥
= 4(𝑒 sec 𝑥 + 𝑥)3 (𝑒 + 𝑥)
𝑑𝑥 𝑑𝑥

Simplifies the above differential as:
1
𝑑𝑦
= 4(𝑒 sec 𝑥 + 𝑥)3 (𝑒 sec 𝑥 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 + 1)
𝑑𝑥

Q.24 0.5

0.5

1

OR

Uses the cross-product of vectors and writes: 0.5

Uses the cross-product of vectors and writes:
0.5

Simplifies RHS of the above equation as:
1

Concludes that the area of parallelogram QRST is equal to the area of
parallelogram QRTP.

Q.25 i) Expands the vector form to get the following: 0.5

Eliminates λ by equating the like coefficients of the position vectors of the 0.5
x, y and z axes to get the cartesian equation as follows:

ii) Assumes the coordinates of B as (𝑥2 , y2 , z2 ) and compares the cartesian
form of the equation from step 2 with the regular form of the cartesian 0.5
equation to find:

𝑥2 = 2𝑥1 , 𝑦2 = 3𝑦1 and 𝑧2 = 4𝑧1

Substitutes values 𝑥1 = (-2), 𝑦1 = 5 and 𝑧1 = (-3) in the equations from step 0.5
3 to get coordinates of B as (-4, 15, -12).

SECTION C - Short Answer Questions of 3 Marks each

Q.26 𝑑𝑢 1
Finds 𝑑𝜃 as:

𝑑𝑣
Finds 𝑑𝜃 as: 1

𝑑𝑢
Uses parametric differentiation and finds 𝑑𝑣 as: 0.5

𝜋
Concludes that the given statement is true as 𝑒 2 is a constant. 0.5

OR

Rewrites the given equation by taking logarithm on both sides as: 1

m(log x) − n(log y) = (m − n)(log x + log y)

Differentiates the above equation as: 1

𝑚 𝑛 1 1
𝑑𝑥 − 𝑑𝑦 = (𝑚 − 𝑛) ( 𝑑𝑥 + 𝑑𝑦)
𝑥 𝑦 𝑥 𝑦

Rearranges the above equation to get: 0.5
𝑛 𝑚
𝑑𝑥 = 𝑑𝑦
𝑥 𝑦

𝑑𝑦 𝑛𝑦
Finds 𝑑𝑥 to be 𝑚𝑥 . 0.5

Q.27 Interprets the question statement and writes it as: 0.5

Finds the derivative in the above step as: 1

(3x − 1)f(x) = 12𝑥 3 − 13𝑥 2 + 3x

1
Factorises the above cubic polynomial as:

(3x − 1)f(x) = (3x − 1)(4𝑥 2 − 3x)

and determines the value of f(x) as (4𝑥 2 − 3x).

Substitutes x = 6 in f(x) and evaluates f(6) as 126. 0.5

Q.28 Rewrites the integral using the identity cosec2 x = 1 + cot2 x as: 0.5

Substitutes cot x = u and hence cosec2 x dx = - du in the above step and 0.5
rewrites the integral as:

Integrates the above expression as: 0.5

0.5
Substitutes cot x in place of u in the above expression to get:

92
Substitutes the limits in the above expression to get 105 .
1

Q.29 Rearranges the given differential equation as: 0.5

Finds the integrating factor as follows as the equation obtained in the above step is of the 0.5
𝑑𝑦
form 𝑑𝑥 + y P(x) = Q(x).

Finds the solution as:
1.5

Substitutes x = y = 0 in the above equation and finds the value of C as 0.
0.5
Writes the particular solution as:

OR
𝑦
Rearranges the given equation in terms of 𝑥 as: 0.5

𝑑𝑦
Considers y = vx and finds 𝑑𝑥 in terms of v as:
0.5
𝑑𝑦 𝑑𝑦
= v + x
𝑑𝑥 𝑑𝑥

Equates the RHS obtained in steps 1 and 2 to get:
0.5

Rearranges the terms using the variable separable method as:
0.5

Integrates on both sides to find the general solution as:

sec −1 𝑣 = log |x| + C 1

or
𝑦
sec −1 𝑥 = log |x| + C, where C is the constant of integration.

Q.30 Takes the number of hens and cows to be x and y respectively and 1.5
formulates the linear programming problem as follows:

Maximise Z = 12x + 40y

subject to constraints,
25x + 75y ≤ 900
x + y ≤ 16
x ≤ 10
x ≥ 0
y ≥ 0

Graphs the constraints and marks the feasible region as: 1.5

Q.31 Takes E, F and G to be the events of taking out green marbles in the first, 0.5
second and third draws respectively, and writes:

8
𝑃(𝐸) = 𝑃(𝑔𝑟𝑒𝑒𝑛 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡 𝑑𝑟𝑎𝑤) =
14

Finds the probability that the second marble taken out is green provided 0.5
first is also green as:

7
𝑃(𝐹|𝐸) = 𝑃(𝑔𝑟𝑒𝑒𝑛 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑟𝑎𝑤) =
13

Finds the probability that the third marble taken out is green provided first 0.5
two are also green as:

6
𝑃(𝐺|𝐸𝐹) = 𝑃(𝑔𝑟𝑒𝑒𝑛 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑛 𝑡ℎ𝑖𝑟𝑑 𝑑𝑟𝑎𝑤) =
12

Finds the probability that all three marbles taken out are green in colour as: 1.5

𝑃(𝐸) × 𝑃(𝐹|𝐸) × 𝑃(𝐺|𝐸𝐹)

8 5 6
= × ×
14 13 12
10
=
91

OR

Assumes the number of students as a random variable X and writes that it
can take values of 0, 1 and 2.
0.5
Finds P(X = 0) as:

P(non-student and non-student)

10 9
= ×
18 17
90
=
306

Finds P(X = 1) as:

P(student and non-student) or P(non-student and student)
1.5
8 10 10 8
= × × ×
18 17 18 17

160
=
306

Finds P(X = 2) as: 0.5

P(student and student)

8 7
= ×
18 17
56
=
306
Writes the required probability distribution as:
0.5
X 0 1 2

P(X) 90 160 56
306 306 306

SECTION D - Long answer type questions (LA) of 5 marks each

Q.32 Assumes the number of litres of orange juice, beetroot juice and kiwi juice 0.5
as x, y and z, respectively to frame equations as follows:

500x + 20y + 800z = 1860
2x + 5y + 3z = 22
100x + 120y + 200z = 760

Writes the above system of equations in the matrix form using AX = B as 0.5

Finds |A| = 110000 ≠ 0 and hence writes that A is non-singular and has a 0.5
unique solution.

Finds adj A as:
1

Finds A-1 using |A| and adj A as: 1

Writes that X = A-1B and finds X as 1

Concludes that 2 litres of orange juice, 3 litres of beetroot juice and 1 litre 0.5
of kiwi juice should go into the mixture.

Q.33 Writes the endpoints of the ellipse as (−9, 0), (9, 0), (0, 6) and (0, −6) 1
respectively.

Expresses y in terms of x as:

Sets up the equation for the area of the shaded region as: 1

Evaluates the 1st part of the above equation as: 1

Applies the upper and the lower limit and finds the value of the integral as: 1

0.5
Evaluates the 2nd part of the equation from step 2 as:
1
Area of 2 triangles = 2 × 2 × 9 × 6 = 54 sq units

Adds the area obtained in step 3 and 4 to find the area of the shaded region in 0.5
terms of π as:

(27𝜋 + 54) sq units or 27(𝜋 + 2) sq units.

Q.34 1

Notes that the lines are skewed and writes the formula to find the shortest 1
distance between the lines (d) as follows:

1

1.5

10 0.5
Substitutes values from above steps to find distance as units.
√59

OR

0.5

Assumes P (x, y, z) to be the point of intersection of the two lines. Finds 0.5

x = λ + 4, y = 3λ + 2 and z = 2λ + 1.

Takes Tara’s position as T(2, -2, 1) to find the direction ratios of TP as
(𝜆 + 2), (3𝜆 + 4) and (2𝜆). 1

Notes that the dot product of the direction ratios of the given line and TP
will be 0, since they are perpendicular, and cos 90° = 0.
1.5
Writes that 1(𝜆 + 2) + 3(3𝜆 + 4) + 2(2𝜆) = 0.

Solves the above equation to find 𝜆 = (−1).

Substitutes the value of λ to find x = 3, y = 7 and z = 2. 0.5

Finds the length of TP as √83 units, using the following formula: 1

Q.35 Rewrites the given integral as: 0.5

1
Substitutes x4 = u and hence 𝑥 3 𝑑𝑥 = 4 du in the above integral to get: 1

Uses integration by parts to integrate the above expression as: 1

Integrates the above expression to get: 2

Substitutes x4 in place of u in the above expression to get: 0.5

OR

Expands the denominator using the identity (a3 - b3) as: 0.5

Rewrites the integral as a sum of two integrals using partial fractions as: 1

Solves the first integral as: 0.5

Rewrites the second integral as: 1

Solves the above integral as: 1.5

Concludes the final answer as: 0.5

SECTION E - Case Studies/Passage based questions of 4 Marks each

Q.36i) Lists all the elements of R as:
1
R = {(C, PB), (PB, C), (V, PB), (PB, V), (PB, SwD), (SwD, PB),
(PB, ShD), (ShD, PB), (SwD, ShD), (ShD, SwD)}

Q.36ii) Writes that the relation R is symmetric. 0.5

Gives a reason. For example, for every (x1, x2) ∈ R, (x2, x1) ∈ R as every 0.5
direct ship/direct ferry runs in both the directions.

Q.36iii) Writes that R is not transitive. 0.5

Gives a reason. For example,
1.5
(C, PB) ∈ R as there is a direct ship from Chennai to Port Blair.

(PB, SwD) ∈ R as there is a direct ferry from Port Blair to Swaraj Dweep.

But (C, SwD) ∉ R as there is no direct ship/ferry from Chennai to Swaraj Dweep.

OR

Writes that the function f is one-one. 0.5

Gives a reason. For example, no two elements of set Y are mapped to a common 0.5
element in set X.

Writes that the function f is not onto. 0.5

Gives a reason. For example, C ∈ X (co-domain of f) but it has no pre-image in 0.5
Y.

Q.37i) Finds the rate at which the amount of drug is changing in the blood stream 5 1
hours after the drug has been administered as:

𝐶 ′ (t) = −3𝑡 2 + 9t + 54
⇒ C′(5) = 24 mg/hr

Q.37ii) Equates the derivative 𝐶 ′ (t) to 0 and factorises 𝐶 ′ (t)as 3(3 + t)(6 − t). 0.5

Writes that for t ∈ (3, 4), 1.5

3 > 0,
(3 + t) > 0
and (6 − t) > 0
Therefore, C′(t) > 0.

Concludes that C(t) is strictly increasing in the interval (3, 4).
OR
0.5
Equates the derivative 𝐶 ′ (t) = −3𝑡 2 + 9t + 54 to 0 and finds the
critical points as t = 6 hours and t = (-3) hours.
0.5
Differentiate 𝐶 ′ (t) to get C"(t) as:

C"(t) = −6t + 9

Finds C"(6) as (−27) and writes that C(t) attains its maximum at t = 6 0.5
hours, as C"(6) = (−27) < 0.

Concludes that 6 hours after the drug is administered, Cmax is attained. 0.5

1
Q.37iii) Finds the value of C(t) at t = 6 hours as:

C(6) = −(6)3 + 4.5(6)2 + 54(6)

⇒ C(6) = 270

Writes the amount of drug in the bloodstream when the effect of the drug is
maximum as 270 mg.

Q.38i) Takes P(S), P(C) and P(T) as the probabilities that a person selected 1
randomly from the staff prefers sugar, coffee and tea respectively.

Finds P(T) = P(C′) = 1 − 0.6 = 0.4.

Finds P(S|T) = 1 − 0.2 = 0.8.

Uses theorem on total probability and finds the probability that a randomly 1
selected staff prefers a beverage with sugar as:

P(S) = P(C) × P(S|C) + P(T) × P(S|T)
86 43
= 0.6 × 0.9 + 0.4 × 0.8 = 0.86 or 100 𝑜𝑟 50

Q.38ii) Uses the sum of probabilities = 1 and finds the following probabilities: 0.5

♦ P(without sugar|coffee) = 1 − 0.9 = 0.1

♦ P(tea) = 1 − 0.6 = 0.4

Uses Bayes' theorem to find the probability that a staff selected at random
prefers coffee given that it is without sugar, P(coffee|without sugar) as: 1

0.6 × 0.1
=
0.6 × 0.1 + 0.4 × 0.2

(Award 0.5 marks if only the formula for Bayes′ theorem is written correctly. )

6 3
Simplifies the above expression and finds the required probability as 14 𝑜𝑟 7.
0.5