HALF YEARLY EXAMINATION 2024-25

Class XI Mathematics (Code- 041)

Marking Scheme (SET-2)

SECTION A
(Multiple Choice Questions)
Each question carries 1 mark
1. D 1
2. B 1
3. D 1
4. C 1
5. B 1
6. A 1
7. B 1
8. A 1
9. C 1
10. B 1
11. D 1
12. C 1
13. B 1
14. C 1
15. C 1
16. A 1
17. A 1
18. C 1
19. D 1
20. B 1
SECTION B
This section comprises of very short answer type-questions (VSA) of 2 marks each
21. 𝐴 ∪ 𝐵 = {1,2,3,4,6} ½
′
(𝐴 ∪ 𝐵) = {5,7} ½
𝐵−𝐴 = {6} ½
(𝐵 − 𝐴)′ = {1,2,3,4,5,7} ½
22. 𝐹𝑜𝑟 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑔𝑟𝑎𝑝ℎ 1
Domain – R ½
Range - {−1, 0, 1} ½

23. 𝑡𝑎𝑛𝛼+𝑡𝑎𝑛𝛽 1
For using tan(𝛼 + 𝛽) = 1−𝑡𝑎𝑛𝛼 𝑡𝑎𝑛𝛽 2

1
For putting values 2
1
2
For finding tan(𝛼 + 𝛽) = 1
1
2
 𝜋
⇒𝛼 + 𝛽 =
4
OR 1
2𝑡𝑎𝑛𝑥
Using tan 2𝑥 = 2
1− 𝑡𝑎𝑛2 𝑥
𝜋
1
𝜋 𝜋 2𝑡𝑎𝑛 8
Putting 𝑥 = reducing above as tan 4 = 𝜋
2
8 1− 𝑡𝑎𝑛2
8

Reducing the above into a quadratic equation as 𝑥 2 + 2𝑥 − 1 = 0 1

2

𝜋 1
Solving the quadratic equation and finding tan 8 = −1 + √2
2
24. If (𝑥 – 𝑖𝑦)(3 + 5𝑖) is conjugate of −6 − 24𝑖
then (x – iy) (3 + 5i) = −6 + 24𝑖 1
2
−6 + 24𝑖
𝑥 – 𝑖𝑦 = 1
3 + 5𝑖 1
Solving we 𝑔𝑒𝑡 𝑥 = 3 𝑎𝑛𝑑 𝑦 = −3 2

25. 1 1
For finding 𝑎 = 3 𝑎𝑛𝑑 𝑟 = 2
2

𝑎( 1 − 𝑟 𝑛 )
𝑆𝑛 =
1−𝑟
1
3069 3 (1 − 2𝑛 ) 1
=
512 1 2
1− 2
3069 1
⇒ = 1−
512 2𝑛 1
For finding 𝑛 = 10

OR ½
𝑛
315 = 5(2 − 1) ½
𝑛=6 ½
𝑎𝑛 = 5 × 25 ½
𝑎𝑛 = 160
SECTION C
(This section comprises of short answer type questions (SA) of 3 marks each)
26. 𝐴∪𝐵 =𝐴∪𝐶
1
(𝐴 ∪ 𝐵) ∩ 𝐶 = (𝐴 ∪ 𝐶) ∩ 𝐶 2
For solving and simplifying as (𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶) = 𝐶
Again 1

1
(𝐴 ∪ 𝐵) ∩ 𝐵 = (𝐴 ∪ 𝐶) ∩ 𝐵
2
 For solving and simplifying as 𝐵 = (𝐴 ∩ 𝐵) ∪ (𝐵 ∩ 𝐶) and showing B = C
1

27. |𝑥−4| 𝑥−4 1

F(x)= 𝑥−4 = 𝑥−4=1 if x>4

|𝑥−4| (𝑥−4)
Also F(x)= =− =-1 if x<4
𝑥−4 𝑥−4 1
Thus the range of the function is {1,-1}
1
OR
For writing relation R = {(1,7), (2,5), (3,5)} 1
For writing Domain of R = {1,2,3} 1
For writing Range of R = {5,7} 1

28. 𝜋 9𝜋 𝜋 9𝜋 3𝜋 5𝜋 1
L.H.S. = cos (13 + 13 ) + cos (13 − 13 ) + cos 13 + cos 13 1
10𝜋 8𝜋 3𝜋 5𝜋
= cos ( 13 ) + cos ( 13 ) + cos 13 + cos 13
3𝜋 5𝜋 3𝜋 5𝜋
=cos (𝜋 − 13 ) + cos (𝜋 − 13 ) + cos 13 + cos 13 1
3𝜋 5𝜋 3𝜋 5𝜋 1
=− cos 13 − cos 13 +cos 13 + cos 13
=0=R.H.S.
1
29. Let 𝑥 𝑎𝑛𝑑 𝑥 + 2 𝑏𝑒 𝑡𝑤𝑜 𝑒𝑣𝑒𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠
𝑥 > 5 , 𝑥 + 2 > 5 𝑎𝑛𝑑 𝑡ℎ𝑒𝑖𝑟 𝑠𝑢𝑚 𝑥 + 𝑥 + 2 < 23
Solving above inequation to get 𝑥 < 10.5
1
Now for writing 5 < 𝑥 < 10.5 and finding values of 𝑥 𝑎𝑠 6 , 8 𝑎𝑛𝑑 10
For finding possible pairs of even integers as (6,8), (8,10)𝑎𝑛𝑑 (10,12)
1

1
𝑛 𝑛 𝑛 𝑛
30. writing 8 𝑎𝑠 (1 + 7) so 8 − 7𝑛 − 1= (1 + 7) − 7𝑛 − 1 applying 1
binomial theorem and showing that 8𝑛 − 7𝑛 − 1 𝑖𝑠 divisible by 49 2
OR
6
Expanding (√2 + 1) 1
6
Expanding (√2 − 1) 1
6 6
(√2 + 1) + (√2 − 1) = 2(8 + 60 + 30 + 1) = 198 1
31. Let 𝑎 𝑎𝑛𝑑 𝑏 𝑏𝑒 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑜𝑛 𝑋 𝑎𝑛𝑑 𝑌 𝑡ℎ𝑒 𝑎𝑥𝑒𝑠
𝑥 𝑦 1
Equation of line 𝑎 + 𝑏 = 1 , 𝑎 + 𝑏 = 1 , 𝑎𝑏 = −6
2

1
Solving and finding the values of 𝑎 = 3 , 𝑏 = −2 𝑜𝑟 𝑎 = −2 , 𝑏 = 3 12

For writing the equation of line 2𝑥 − 3𝑦 − 6 = 0 𝑎𝑛𝑑3𝑥 − 2𝑦 + 6 = 0 1

OR
 point on y axis = (0,6)
1
Slope of given line = -3/2 2
Sope of required line = 2/3 1
2
Hence equation of required line, 2x - 3y + 18 = 0
1
2
1
12
SECTION D
(This section comprises of long answer-type questions (LA) of 5 marks each)

32.
√3 1
For finding 𝑐𝑜𝑠𝑥 =
2 2
𝑥
𝑥 𝑙𝑖𝑒𝑠 𝑖𝑛 𝐼𝑉 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 ⇒ 𝑙𝑖𝑒𝑠 𝑖𝑛 𝐼𝐼 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡
2
𝑥 𝑥 𝑥 1
⇒ sin 2 > 0 , cos 2 < 0 𝑎𝑛𝑑 tan 2 < 0

𝑥 1 + 𝑐𝑜𝑠𝑥 𝑥 √2+ √3
Using formula cos 2 = ±√ and finding cos 2 = − 1
2 2 1
2

𝑥 1− 𝑐𝑜𝑠𝑥 𝑥 √2− √3 1
Using formula sin 2 = ±√ and finding sin 2 = 1
2 2
2

𝑥 √2− √3 1
For finding tan 2 = −
√2+ √3 2

33. 12! 1
Total number of arrangements = 3!4!2! = 1663200
11!
(i) Number of words beginning with P = 3!4!2! = 138600 1
(ii) Number of arrangements when all vowels occur together
8! 5!
= 3!2! × 4! ! = 16800 2
10!
(iii) Number of words = 3!4!2! = 12600
1
OR
(i) Number of ways of selecting team = 7𝐶5 = 21 1

(ii) the required number of ways = 7𝐶1 × 4𝐶4 + 7𝐶2 × 4𝐶3 + 7𝐶3 × 4𝐶2 + 7𝐶4 × 4𝐶1
= 7 + 84 + 210+140=441 1
1
(iii) the required number of ways
4𝐶3 × 7𝐶2 + 4𝐶4 × 7𝐶1
= 84 + 7 =91 1
 1
34.
𝑎 + 𝑏 = 3, 𝑎𝑏 = 𝑝, 𝑐 + 𝑑 = 12 𝑎𝑛𝑑 𝑐𝑑 = 𝑞
Let 𝑏 = 𝑎𝑟 , 𝑐 = 𝑎𝑟 2 𝑎𝑛𝑑 𝑑 = 𝑎𝑟 3 1

Then 𝑎(1 + 𝑟) = 3 𝑎𝑛𝑑 𝑎𝑟 2 (1 + 𝑟) = 12 1

Using above equation to find 𝑟 = 2 𝑎𝑛𝑑 𝑎 = 1. 1

Now 𝑝 = 𝑎𝑏 = 𝑎 . 𝑎𝑟 = 2 ,
𝑞 = 𝑐𝑑 = 𝑎𝑟 2 × 𝑎𝑟 3 = 25 = 32 1

𝑞+𝑝 32+2
Now the required ratio =
𝑞−𝑝 32−2

→ (𝑞 + 𝑝) ∶ (𝑞 − 𝑝) = 17 ∶ 15
OR
1
Let the numbers are a& b
a+b
Then A.M of a&b = A= 2

G.M of a & b= G = ab
According to question
a+b = 6G 1
2A = 6G
A = 3G
Quadratic equation whose roots are a &b is
1
x2- (a+b)x +ab=0
x2- 2Ax +G2=0
2 A  4 A2 − 4G 2 1
x=  x=A A2 − G 2
2
a= A+ A2 − G 2 , b = A − A2 − G 2
a A+ A2 − G 2
=
b A− A2 − G 2
a 3G + 9G 2 − G 2 a 3G + 8G 2
=  =
b 3G − 9G 2 − G 2 b 3G − 8G 2
a 3+2 2
=
b 3−2 2 1
 1
35. For finding point of intersection as (−1, −1) 1

3
For finding slope of given line 3𝑥 – 5𝑦 + 11 = 0 as 5 1

5
For finding slope of perpendicular line to 3𝑥 – 5𝑦 + 11 = 0 as − 3 1

For finding equation of required line as 5𝑥 + 3𝑦 + 8 = 0 2

SECTION E
(This section comprises of 3 case-study/passage-based questions of 4 marks each with
two sub-parts. First two case study questions have three sub parts (i), (ii), (iii) of marks 1,
1, 2 respectively. The third case study question has two sub parts of 2 marks each.)
36. (i) vowels will occupy even places in 2! Ways
Remaining three letters will occupy remaining places in 3! Ways
Total number of words with or without meaning are = 2! × 3! 1
(ii) No. of words start and end with vowels = 2 x 3! = 12
(iiI) No of words beginning with G = 4! = 24
Similarly no of words beginning with H are 24 and no of words beginning
with O are 24 . 1
After O words with T will start . Therefore number of words before the word
starting with T are =24+24+24 i.e. 72
OR
No of words beginning with TG = 3! = 6
No of words beginning with TH= 3! = 6
No of words beginning with TOG= 2! = 2
No of words beginning with TOH=2! = 2
2
Next come the words beginning with TOU is TOUGH
Rank of TOUGH = 24+24+24+6+6+2+2+1 = 89
37. (i) 256 1
(𝑖𝑖 ) 2046 1
(iii) 1023 2
OR
8
38. (i) 3𝑥 − 𝑦 = 8 2
(ii) 16 sq units 2