KENDRIYA VIDYALAYA SANGATHAN LUCKNOW REGION
Second Pre Board Exam 2022-23
Marking Scheme
Class XII Subject-mathematics
1. (b) 1
2. (d) 1
3. (d) 1
4. (a) 1
5. (d) 1
6. (c) 1
7. (b) 1
8. (b) 1
9. (b) 1
10. (a) 1
11. (d) 1
12. (c) 1
13. (c) 1
14. (b) 1
15. (d) 1
16. (c) 1
17. (c) 1
18. (b) 1
19. (a) 1
20. (d) 1
𝑒 𝑥 (𝑥−3) (𝑥−1)−2
21. ∫ (𝑥−1)3
= ∫ 𝑒 𝑥 [ (𝑥−1)3 ] ½
1 −2
= ∫ 𝑒 𝑥 [(𝑥−1)2 + ((𝑥−1)3 )] ½
𝑒𝑥
= (𝑥−1)2 + 𝑐 1
22. Since 𝑙 2 + 𝑚2 + 𝑛2 = 1 ⇒ 𝑐𝑜𝑠 2 𝛼 + 𝑐𝑜𝑠 2 𝛽 + 𝑐𝑜𝑠 2 𝛾 = 1 1
⇒ 1 − 𝑠𝑖𝑛2 𝛼 + 1 − 𝑠𝑖𝑛2 𝛽 + 1 − 𝑠𝑖𝑛2 𝛾 = 1
⇒ 𝑠𝑖𝑛2 𝛼 + 𝑠𝑖𝑛2 𝛽 + 𝑠𝑖𝑛2 𝛾 = 2 1
OR
For getting direction ratios 1,2,3 1
1 2 3
For getting direction cosines , , 1
√14 √14 √14
𝐸 𝑃(𝐸∩𝐹) 4
23. (i) 𝑃 (𝐹 ) = 𝑃(𝐹)
=9 1
𝐸𝑐 𝑃(𝐸 𝑐∩𝐹) 𝑃(𝐹)−𝑃(𝐸∩𝐹) 5
(ii) 𝑃 ( 𝐹 ) = 𝑃(𝐹) = 𝑃(𝐹)
=9 1
24. For reflexivity 1
For transitivity 1
25. (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗). (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗) = 0
⇒ 𝑎⃗. 𝑎⃗ + 𝑎⃗. 𝑏⃗⃗ + 𝑎⃗. 𝑐⃗ + 𝑏⃗⃗. 𝑎⃗ + 𝑏⃗⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ + 𝑐⃗. 𝑏⃗⃗ + 𝑐⃗. 𝑐⃗ = 0 1
2
⇒ |𝑎⃗|2 + |𝑏⃗⃗| + |𝑐⃗|2 + 2(𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗) = 0
⇒ 9 + 25 + 49 + 2(𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗) = 0 ½
−83
⇒ (𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗) = 2 1/2
OR
𝑖̂ 𝑗̂ 𝑘̂
𝑎⃗ × 𝑏⃗⃗ = |2 −3 4| = −2𝑖̂ + 4𝑗̂ + 4𝑘̂ 1
2 −1 2
|𝑎⃗ × 𝑏⃗⃗| = √4 + 16 + 16 = 6 1/2
1
Area of parallelogram = |𝑎⃗ × 𝑏⃗⃗| = 3 sq.units ½
2
4
26. ∫1 {|𝑥 − 1| + |𝑥 − 2| + |𝑥 − 4|}𝑑𝑥
4 2 4 4
= ∫1 (𝑥 − 1)𝑑𝑥 − ∫1 (𝑥 − 2)𝑑𝑥 + ∫2 (𝑥 − 1)𝑑𝑥 − ∫1 (𝑥 − 1)𝑑𝑥 1
(𝑥−1)2 4 (𝑥−2)2 2 (𝑥−2)2 4 (𝑥−4)2 4
=[ 2
] −[ 2
] +[ 2
] −[ 2
] 1
1 1 2 1
9 1 9 23
= + +2+ = 1
2 2 2 2
𝑥
27. (𝑥 − 𝑦)𝑒 (𝑥−𝑦) = 𝑎
Taking log on both sides
𝑥
log(𝑥 − 𝑦) + (𝑥−𝑦) = log 𝑎 1
𝑑𝑦 𝑑𝑦
1− 1.(𝑥−𝑦)−𝑥(1− )
Differentiating both sides w.r.t.x, we get 𝑑𝑥
+ (𝑥−𝑦)2
𝑑𝑥
=0 1
𝑥−𝑦
𝑑𝑦 𝑑𝑦
⇒ (𝑥 − 𝑦) (1 − 𝑑𝑥 ) + 𝑥 − 𝑦 − 𝑥 + 𝑥 𝑑𝑥 = 0 ½
𝑑𝑦
⇒ 𝑦 𝑑𝑥 + 𝑥 − 2𝑦 = 0
𝑑𝑦
⇒ 𝑦 𝑑𝑥 + 𝑥 = 2𝑦 ½
OR
𝑦 = log(1 + 2𝑡 2 + 𝑡 4 ) = log(1 + 𝑡2 )2 = 2 log(1 + 𝑡2 )
𝑑𝑦 2(2𝑡) 4𝑡
= = 1
𝑑𝑡 1+𝑡 2 1+𝑡 2
𝑑𝑥 1
𝑥 = tan−1 𝑡 ⇒ = 1+𝑡 2 ½
𝑑𝑡
𝑑𝑦
𝑑𝑦
𝑑𝑡
= 𝑑𝑡
𝑑𝑥 = 4𝑡 ½
𝑑𝑡
𝑑2 𝑦 𝑑 𝑑𝑡
𝑑𝑥 2
= 𝑑𝑡 (4𝑡) × 𝑑𝑥 = 4(1 + 𝑡 2 ) 1
28. Let 𝑃(2,4, −1) be the given point and line
𝑥+5 𝑦+3 𝑧−6
1
= 4
= −9
= 𝜆 ⇒ direction ratios of this line are 1,4, −9
⇒ Any point Q on this line is given by 𝑄(𝜆 − 5,4𝜆 − 3,6 − 9𝜆) 1
Then direction ratios of PQ are (𝜆 − 7,4𝜆 − 7,7 − 9𝜆)
Since PQ is perpendicular to the given line
So (𝜆 − 7)1 + (4𝜆 − 7)4 + (7 − 9𝜆)(−9) = 0 ⇒ 𝜆 = 1 1
So coordinates of Q are (−4,1, −3)
Also PQ = 7 units 1
29. Let 𝐸1 : Problem is solved by A
𝐸2 : Problem is solved by B
1 1
So 𝑃(𝐸1 ) = 2 and 𝑃(𝐸2 ) = 3
1 2
⇒ 𝑃(𝐸1𝑐 ) = 2 and 𝑃(𝐸2𝑐 ) = 3 1
1 2 2
(𝑖) 𝑃(Problem is solved) = 1 − 𝑃(𝐸1𝑐 )𝑃(𝐸2𝑐 )=1 − × = 1
2 3 3
1
(𝑖𝑖)𝑃(one of them solves the problem)= 𝑃(𝐸1 )𝑃(𝐸2𝑐 ) + 𝑃(𝐸2 )𝑃(𝐸1𝑐 ) =2 1
OR
Let 𝐸1 : Red ball is transferred from Bag I to Bag II
𝐸2 : Black ball is transferred from Bag I to Bag II
𝐴: Ball drawn from bag II is black 1
3 5
𝑃(𝐸1 ) = and 𝑃(𝐸2 ) =
8 8
3 4
𝑃(𝐴/𝐸1 ) = and 𝑃(𝐸2 ) = 1
8 8
𝑨
𝑬 𝑷(𝑬𝟐 )𝑷( ) 20
𝑬𝟐
𝑷 ( 𝑨𝟐) = 𝐴 𝐴 = 29 1
𝑃(𝐸1 )𝑃( )+𝑃(𝐸2 )𝑃( )
𝐸1 𝐸2
30. Let 𝑄(𝑥, 𝑦) be a point on the curve 𝑦 2 = 4𝑥, which is nearest to the point 𝑃(2,1)
So (𝑃𝑄)2 = (𝑥 − 2)2 + (𝑦 − 1)2
2
𝑦2
(𝑃𝑄)2 = ( − 2) + (𝑦 − 1)2 = 𝑆(𝑠𝑎𝑦) 1
4
𝑑𝑆 𝑦 3 −8
=
𝑑𝑦 4
𝑑𝑆
𝑑𝑦
= 0 gives 𝑦 = 2 and so 𝑥 = 1 1
𝑑2 𝑆 3𝑦 2
Now 𝑑𝑦2 = 4 > 0
Therefore (1,2) will be at minimum distance from (2,1) 1
𝑑𝑥 𝑦 tan 𝑦−𝑥 tan 𝑦−𝑥𝑦 𝑑𝑥 1 1
31. = ⇒ +( + )𝑥 = 1
𝑑𝑦 𝑦 tan 𝑦 𝑑𝑦 𝑦 tan 𝑦
1
∫( +cot 𝑦)𝑑𝑦
IF = 𝑒 𝑦 = 𝑒 log 𝑦+log sin 𝑦 = 𝑒 log(𝑦 sin 𝑦) = 𝑦 sin 𝑦 1
Solution of the given differential equation is
𝑥(IF) = ∫ Q. IF 𝑑𝑦 + 𝑐
𝑥(𝑦 sin 𝑦) = ∫ 𝑦 sin 𝑦 𝑑𝑦 + 𝑐 1
𝑥(𝑦 sin 𝑦) = −𝑦 cos 𝑦 + sin 𝑦 + 𝑐
sin 𝑦−𝑦 cos 𝑦+𝑐
𝑥= 1
𝑦 sin 𝑦
OR
cos 𝑦 𝑑𝑥 + (1 + 2𝑒 −𝑥 ) sin 𝑦 𝑑𝑦 = 0
𝑑𝑥 − sin 𝑦 𝑑𝑦
∫ (1+2𝑒 −𝑥 ) = ∫ 1
cos 𝑦
𝑒𝑥 − sin 𝑦 𝑑𝑦
∫ 2+𝑒 𝑥 𝑑𝑥 = ∫ cos 𝑦
log(𝑒 𝑥 + 2) = log|cos 𝑦| + log 𝑐
log(𝑒 𝑥 + 2) = log|cos 𝑦|𝑐 1
𝑒 𝑥 + 2 = |cos 𝑦|𝑐
𝜋
Substituting 𝑥 = 0 𝑎𝑛𝑑 𝑦 = 4 we get
1 + 2 = 𝑐(1/√2)
𝑐 = 3√2 ½
𝑒 𝑥 + 2 = 3√2|cos 𝑦| is the particular solution ½
32. |𝐴| = 10 1
4 −5 1
1
𝑎𝑑𝑗. 𝐴 = [2 0 −2] 12
2 5 3
4 −5 1
1
So 𝐴−1= 10 [2 0 −2] 1
2 5 3
Given equations can be written in the form of 𝐴𝑋 = 𝐵 𝑎𝑠
1 2 1 𝑥 4
[−1 1 1] [𝑦] = [0]
1 −3 1 𝑧 4
Since 𝐴𝑋 = 𝐵
⇒ 𝐴−1 𝐴𝑋=𝐴−1 𝐵
⇒ 𝐼𝑋 = 𝐴−1 𝐵
⇒ 𝑋 = 𝐴−1 𝐵 1/2
𝑥 4 −5 1 4
1
⇒ [𝑦] = 10 [2 0 −2] [0]
𝑧 2 5 3 4
𝑥 20
1
⇒ [𝑦] = 10 [ 0 ]
𝑧 20
⇒ 𝑥 = 2, 𝑦 = 0, 𝑧 = 2 1
OR
To Prove 𝐵𝐴 = 6𝐼 2
Given equations can be written in the form of 𝐴𝑋 = 𝐵 𝑎𝑠
1 −1 0 𝑥 3
[2 3 4] [𝑦] = [17]
0 1 2 𝑧 7
𝑥 1 −1 0 −1 3
⇒ [𝑦] = [2 3 4] [17] 1
𝑧 0 1 2 7
1
Since 𝐵𝐴 = 6𝐼 ⇒ 𝐴−1 = 6 𝐵
𝑥 2 2 −4 3 12 2
1 1
So [𝑦] = 6 [−4 2 −4] [17] = 6 [−6] = [−1]
𝑧 2 −1 5 7 24 4
⇒ 𝑥 = 2, 𝑦 = −1, 𝑧 = 4 2
33. For Correct figure 1
√3 𝑥 2
Area= ∫0 √3 𝑑𝑥 + ∫√3 √22 − 𝑥 2 𝑑𝑥 1
1 𝑥 2 √3 𝑥 4 𝑥 2
= [2] + [2 √22 − 𝑥 2 + 2 sin−1 2] 1
√3 0 √3
𝜋
= 3 sq.units 2
OR
For Correct figure 2
The point of intersection of the parabola 𝑦 = 𝑥 2 and the line 𝑦 = 𝑥 are
(0,0) and (1,1) ½
1 2
Required Area= ∫0 𝑥 2 𝑑𝑥 + ∫1 𝑥 𝑑𝑥 1
11 1
= sq. units 12
6
3𝑥−2 𝐴 𝐵 𝐶
34.
(1+𝑥)2 (𝑥+3)
= (𝑥+1) + (𝑥+1)2 + 1
(𝑥+3)
11 −5 −11 1
Getting values 𝐴 = ,𝐵 = ,𝐶 = 12
4 2 4
11 5 11
𝐼 = ∫ (4(𝑥+1) − 2(𝑥+1)2 − 4(𝑥+3)) 𝑑𝑥 1
11 𝑥+1 5 1
= log |𝑥+3| + 2(𝑥+1) + 𝑐 12
4
OR
If 𝑥 = 𝑡 then
2
𝑡 𝐴 𝐵
= (𝑡+𝑎2 ) + (𝑡+𝑏2 ) 1
(𝑡+𝑎2 )(𝑡+𝑏2 )
−𝑎2 𝑏2 1
Getting values 𝐴 = 𝑏2 −𝑎2 𝑎𝑛𝑑 𝐵 = 𝑏2 −𝑎2 12
−𝑎2 1 𝑏2 1
𝐼 = 𝑏2 −𝑎2 ∫ 𝑥 2 +𝑎2 𝑑𝑥 + 𝑏2 −𝑎2 ∫ 𝑥 2 +𝑏2 𝑑𝑥 1
1 𝑥 𝑥 1
= 𝑎2 −𝑏2 [𝑎 tan−1 𝑎 − 𝑏 tan−1 𝑏] + 𝑐 12
35. For correct graph 2
For getting correct corner points (5,0),(6,0),(4,4),(0,6),(0,4) 1
Points 𝑍 = 600𝑥 + 400𝑦
(5,0) 3000
(6,0) 3600
(4,4) 4000
(0,6) 2400
(0,4) 1600
1
12
1
𝑍 is maximum for C(4,4) i.e. at 𝑥 = 4; 𝑦 = 4 2
𝑥 𝑦 𝑧
36. (𝑖) = = −1 1
1 2
2 1 1
(𝑖𝑖) , , 1
√6 √6 √6
(𝑖𝑖𝑖) ⃗⃗⃗⃗⃗
𝑎2 − ⃗⃗⃗⃗⃗
𝑎1 = 3𝑖̂ + 3𝑗̂ ½
⃗⃗⃗⃗1 × 𝑏
𝑏 ⃗⃗⃗⃗⃗2 = 3𝑖̂ − 3𝑗̂ − 3𝑘̂ 1
( ⃗⃗⃗⃗⃗
𝑎2 − ⃗⃗⃗⃗⃗).
𝑎1 (𝑏 ⃗⃗⃗⃗1 × ⃗⃗⃗⃗⃗
𝑏2 ) = 0
So shortest distance between the given lines = 0 ½
OR
Any point on line (i) is 𝑃(𝜆, 2𝜆, −𝜆)
Any point on line (ii) is 𝑄(2𝜇 + 3, 𝜇 + 3, 𝜇) 1
To meet the lines 𝜆 = 2𝜇 + 3 𝑎𝑛𝑑 −𝜆 = 𝜇
We get 𝜇 = −1 𝑎𝑛𝑑 𝜆 = 1
So, motorcycles meet with an accident at (1,2, −1) 1
37. (i) Combined sales in September and October
Basmati Permal Naura
15000 30000 36000 Ramakishan
𝐴+𝐵 =[ ] 1
70000 40000 20000 Gurucharan
(ii) Change in sales from September to October
Basmati Permal Naura
5000 10000 24000 Ramakishan
𝐴−𝐵 =[ ] 1
30000 20000 0 Gurucharan
2
(iii) 2% of B = 100 × 𝐵 = 0.02 𝐵 ½
Basmati Permal Naura
100 200 120 Ramakishan
𝐶=[ ] 1
400 200 200 Gurucharan
Thus in October Ramakishan receives Rs.100, Rs.200, Rs.120 as profit in the sale of each
variety of rice respectively ½
OR
2
2% of A = 100 × 𝐴 = 0.02 𝐴 ½
Basmati Permal Naura
200 400 600 Ramakishan
𝐷=[ ] 1
1000 600 200 Gurucharan
Thus in September Gurucharan receives Rs.1000, Rs.600, Rs.200 as profit in the sale of each
variety of rice respectively ½
38. (𝑖)𝑓(𝑥) = −0.1𝑥 2 + 𝑚𝑥 + 98.6, being a polynomial function, is differentiable everywhere,
Hence differentiable in (0,12). 1
′
𝐴𝑙𝑠𝑜 𝑓 (𝑥) = −0.2𝑥 + 𝑚
At critical point 𝑓 ′ (6) = 0 gives 𝑚 = 1.2 1
(𝑖𝑖) 𝑓(𝑥) = −0.1𝑥 2 + 1.2𝑥 + 98.6
𝑓 ′ (𝑥) = −0.2𝑥 + 1.2 = −0.2(𝑥 − 6) 1
Interval Sign of 𝑓 (𝑥)
′ Conclusion
(0,6) +ve 𝑓 is strictly increasing
(6,12) -ve 𝑓 is strictly decreasing
1
OR
𝑓(𝑥) = −0.1𝑥 2 + 1.2𝑥 + 98.6
𝑓 ′ (𝑥) = −0.2𝑥 + 1.2
𝑓 ′′ (𝑥) = −0.2 < 0
Hence by second derivative test 6 is a point of local maxima
Local Maximum Value= −0.1 × 62 + 1.2 × 6 + 98.6 = 102.2 1`
Now
We have 𝑓(0) = 98.6, 𝑓(6) = 102.2, 𝑓(12) = 98.6
So, 6 is the point of absolute maximum and absolute maximum value=102.2
And 0 and 12 both are the points of absolute minimum and absolute minimum value = 98.6 1
