CBSE Class 12 Marking Scheme 2021 for Maths

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Class: XII Session: 2020-21
Subject: Mathematics
Marking Scheme (Theory)
Sr.No. Objective type Question Marks
Section I
1 Let 𝑓(𝑥1 ) = 𝑓(𝑥2 ) for 𝑠𝑜𝑚𝑒 𝑥1 , 𝑥2 ∈ 𝑅 1
 (𝑥1 )3 = (𝑥2 )3
 𝑥1 = 𝑥2 , Hence 𝑓(𝑥) is one − one

OR

26 reflexive relations 1

2 (1,2) 1

3 Since √𝑎 is not defined for 𝑎 ∈ (−∞, 0) 1
∴ √𝑎 = 𝑏 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

OR

𝐴1 ∪ 𝐴2 ∪ 𝐴3 = 𝐴 𝑎𝑛𝑑 𝐴1 ∩ 𝐴2 ∩ 𝐴3 =  1

4 3x5 1

5 0 1 0 1 0 1 1 0 1
𝐴=[ ]  𝐴2 = [ ][ ]=[ ]
1 0 1 0 1 0 0 1

OR

|adj A|=(-4)3-1=16

6 0 1

7 𝑒 𝑥 (1 − cot 𝑥) + C 1

OR

∵ 𝑓(𝑥) is an odd function
𝜋
2

∴ ∫ 𝑥 2 sin 𝑥 𝑑𝑥 = 0 1
−𝜋
2

8 1
1
2 31
𝐴 = 2 ∫ 𝑥 2 𝑑𝑥 = [𝑥 ]0
3
0
2
= 3 𝑠𝑞 𝑢𝑛𝑖𝑡

Page 1 of 14

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9 0 1

OR

3 1

10 1
𝐽̂

11 1 1 1
|2𝑖̂ × (−3𝑗̂)| = |−6𝑘̂ | = 3 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
2 2

12 2 1
|𝑎̂ + 𝑏̂| = 1
 𝑎̂2 + 𝑏̂ 2 + 2 𝑎̂. 𝑏̂ = 1
 2 𝑎̂. 𝑏̂ = 1 − 1 − 1
−1 −1 𝜋
 𝑎̂. 𝑏̂ =  |𝑎̂||𝑏̂| cos 𝜃 =  𝜃 = 𝜋 −
2 2 3
2𝜋
 𝜃= 3

13 1,0,0 1

14 (0,0,0) 1

15 2 3 1 1
1− × =
3 4 2

16 1 4 1 3 1 7 1
( ) ( ) =( )
2 2 2

Section II
17(i) (b) 1
17(ii) (a) 1
17(iii) (c) 1
17(iv) (a) 1
17(v) (d) 1
18(i) (b) 1
18(ii) (c) 1
18(iii) (b) 1
18(iv) (d) 1
18(v) (d) 1
Section III
19 𝜋
sin( −𝑥) 1
−1 𝑐𝑜𝑠𝑥 −1 2
𝑡𝑎𝑛 ( 1−𝑠𝑖𝑛𝑥
) = 𝑡𝑎𝑛 [ 𝜋 ] 2
1−cos( −𝑥)
2
𝜋 𝑥 𝜋 𝑥
2 sin( − ) cos( − )
𝑡𝑎𝑛−1 [ 4 2
2 𝜋 𝑥
4 2
]
2 𝑠𝑖𝑛 ( − )
4 2

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𝜋 𝑥 𝜋 𝜋 𝑥 1
𝑡𝑎𝑛−1 [𝑐𝑜𝑡 ( − )] = 𝑡𝑎𝑛−1 [tan − ( − )]
4 2 2 4 2

1
−1 𝜋 𝑥 𝜋 𝑥
𝑡𝑎𝑛 [𝑡𝑎𝑛 (4 + 2)] = 4 + 2 2

20
𝐴2 = 2𝐴
 |𝐴𝐴| = |2𝐴|
1
 |𝐴||𝐴| = 8|𝐴| (∵ |𝐴𝐵| = |𝐴| |𝐵| 𝑎𝑛𝑑 |2𝐴| = 23 |𝐴|) 2
 |𝐴| (|𝐴| − 8) = 0 1
1
 |𝐴| = 0 or 8
2

OR

3 1 3 1 8 5
𝐴2 = [ ][ ]=[ ]
−1 2 −1 2 −5 3

15 5 7 0
5𝐴 = [ ] , 7𝐼 = [ ]
−5 10 0 7
0 0 1
 𝐴2 − 5𝐴 + 7𝐼 = [ ]=O
0 0

 𝐴−1 (𝐴2 − 5𝐴 + 7𝐼) = 𝐴−1 O

 𝐴 − 5𝐼 + 7𝐴−1 = O

 7𝐴−1 = 5𝐼 − 𝐴

⇒ 𝐴−1 = 7 ([5 0] − [ 3 1
1
])
0 5 −1 2
1
1 2 −1
 𝐴 −1
= 7[ ]
1 3

21 2 𝑘𝑥
𝐿𝑡 1 − cos 𝑘 𝑥 𝐿𝑡 2 𝑠𝑖𝑛 ( 2 )
=
𝑥 → 0 𝑥 sin 𝑥 𝑥 → 0 𝑥 sin 𝑥

𝑘𝑥
2 𝑠𝑖𝑛2 ( 2 )
𝐿𝑡 𝑥2
=
𝑥→0 𝑥 sin 𝑥
𝑥 2

2 𝑘𝑥
𝐿𝑡 2 𝑠𝑖𝑛 ( 2 ) 𝑘 2
× ( 2)
𝑥→0 𝑘𝑥 2 𝑘2
(2) 2×1× 4 1
= = 12
𝐿𝑡 sin 𝑥 1
𝑥→0 𝑥
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∵𝑓(𝑥) 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0
𝐿𝑡
∴ 𝑓(𝑥) = 𝑓(0)
𝑥→0
𝑘2 1
 =  𝑘 2 = 1  𝑘 = ±1
2 2
1
2
22 1 𝑑𝑦 1
𝑦=𝑥+ ⇒ =1− 2
𝑥 𝑑𝑥 𝑥
∵𝑛𝑜𝑟𝑚𝑎𝑙 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 3𝑥 − 4𝑦 = 7, ∴ tangent is parallel to it
1 3 1
1− 2 = ⇒ 𝑥2 = 4 ⇒ 𝑥 = 2 (∵ 𝑥 > 0)
𝑥 4
1 5
𝑤ℎ𝑒𝑛 𝑥 = 2, 𝑦 = 2 + =
2 2
5 4 1
∴𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑁𝑜𝑟𝑚𝑎𝑙 ∶ 𝑦 − 2 = − 3 (𝑥 − 2) ⇒ 8𝑥 + 6𝑦 = 31

23 1
𝐼= ∫ 𝑑𝑥
𝑐𝑜𝑠 2 𝑥 (1 − tan 𝑥)2

Put, 1 − tan 𝑥 = 𝑦

So that, −𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = 𝑑𝑦 1

−1 𝑑𝑦
=∫ 𝑦2
= − ∫ 𝑦 −2 𝑑𝑦

1 1
=+ +𝑐 = +𝑐 1
𝑦 1 − tan 𝑥

OR
1
𝐼 = ∫0 𝑥 (1 − 𝑥)𝑛 𝑑𝑥
1 1
𝐼 = ∫0 (1 − 𝑥)[1 − (1 − 𝑥)]𝑛 𝑑𝑥
2
1 1
𝐼 = ∫0 (1 − 𝑥) 𝑥 𝑛 𝑑𝑥 = ∫0 (𝑥 𝑛 − 𝑥 𝑛+1 )𝑑𝑥

𝑥 𝑛+1 𝑥 𝑛+2
1 1
𝐼= [ − ]
𝑛+1 𝑛+2 0
1
1 1 1 2
𝐼 = [(𝑛+1 − 𝑛+2) − 0] = (𝑛+1) (𝑛+2)

24 2

𝐴𝑟𝑒𝑎 = 2 ∫ √8𝑥 𝑑𝑥 1
0
2
1
= 2 × 2√2 ∫ 𝑥 2 𝑑𝑥
0

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2 3 2
= 4√2 [ 𝑥 2 ]
3 0
1
8 3 8√2 2
= √2 [22 − 0] = × 2√2
3 3
1
32
= 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 2
3

25 𝑑𝑦
= 𝑥 3 𝑐𝑜𝑠𝑒𝑐 𝑦 ; 𝑦(0) = 0
𝑑𝑥
𝑑𝑦
∫ = ∫ 𝑥 3 𝑑𝑥 1
𝑐𝑜𝑠𝑒𝑐 𝑦
2
∫ sin 𝑦 𝑑𝑦 = ∫ 𝑥 3 𝑑𝑥

𝑥4
− cos 𝑦 = +𝑐 1
4

−1 = 𝑐 (∵ 𝑦 = 0, 𝑤ℎ𝑒𝑛 𝑥 = 0)

𝑥4 1
cos 𝑦 = 1 − 2
4

26 𝑎 = 𝑖̂ − 𝑗̂ + 𝑘̂
Let ⃗⃗⃗⃗
⃗⃗⃗⃗
𝑑 = 4 𝑖̂ + 5𝑘̂
1
𝑎 + ⃗⃗⃗⃗𝑏 = ⃗⃗⃗⃗
∵⃗⃗⃗⃗ 𝑑 ∴⃗⃗⃗⃗
𝑏 = ⃗⃗⃗⃗
𝑑 − ⃗⃗⃗⃗
𝑎 = 3 𝑖̂ + 𝑗̂ + 4 𝑘̂
2
𝑖̂ − 𝑗̂ + 𝑘̂
𝑎 x ⃗⃗⃗⃗𝑏
⃗⃗⃗⃗ = |1 − 1 1 |= −5𝑖̂ − 1𝑗̂ + 4 𝑘̂ 1
3 1 4
1
𝑎 x ⃗⃗⃗⃗
Area of parallelogram = |⃗⃗⃗⃗ 𝑏 | = √25 + 1 + 16 = √42 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 2

27 Let the normal vector to the plane be ⃗⃗⃗⃗𝑛
Equation of the plane passing through (1,0,0), i.e., 𝑖̂ is
𝑛 = 0 ………….(1)
(⃗⃗⃗𝑟 − 𝑖̂) ∙ ⃗⃗⃗⃗ 1
∵plane (1) contains the line𝑟⃗⃗ = 𝑜
⃗⃗⃗ + 𝜆 𝑗̂

∴𝑖̂ ∙ ⃗⃗⃗⃗
𝑛 = 0 and 𝑗̂ ∙ ⃗⃗⃗⃗
𝑛 =0 𝑛 = 𝑘̂
⇒ ⃗⃗⃗⃗
Hence equation of the plane is (⃗⃗⃗𝑟 − 𝑖̂) ∙ 𝑘̂ = 0
i.e., ⃗⃗⃗𝑟 ∙ 𝑘̂ = 0 1

28 Let x denote the number of milk chocolates drawn

X P(x)

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0 4 3 12
× =
6 5 30

1 2 4 16
( × )×2=
6 5 30
1
2 2 1 2 12
x =
6 5 30

1
Most likely outcome is getting one chocolate of each type
2

OR

(̄Ē ∩ F̄) ̅̅̅̅̅̅
(𝐸∪𝐹 ) 1−𝑃 (𝐸∪𝐹)
P (Ē | F̄) = P 𝑃 (F̄) = 𝑃 (F̄) = 1−𝑃(𝐹) -----------(1) 1

Now 𝑃 (𝐸 ∪ 𝐹) = P (E) + P (F) - P (E ∩ F)
= 0.8+0.7-0.6=0.9 1
2
Substituting value of 𝑃 (𝐸 ∪ 𝐹) in (1)
1−0.9 0.1
P (Ē | F̄) = 1−0.7 = 0.3 = 3
1 1
2

Section IV
29 (i) Reflexive :
Since, a+a=2a which is even ∴ (a,a) ∈ 𝑅 ∀𝑎 ∈ Z
Hence R is reflexive 1
2
(ii) Symmetric:
If (a,b) ∈R, then a+b = 2λ ⇒ b+a = 2 λ
⇒ (b,a) ∈R, Hence R is symmetric 1

(iii) Transitive:
If (a,b) ∈R and (b,c,) ∈R
then a+b = 2 λ---(1) and b+c =2 𝜇 ---- (2)
Adding (1) and (2) we get
a+2b+c=2(λ + 𝜇)
⇒ a+c=2 (λ + 𝜇 − 𝑏)
⇒ a+c=2k ,where λ + μ − b = k ⇒ (a,c) ∈R
Hence R is transitive
[0] = {...-4, -2, 0, 2, 4...} 1
1
2
2
30 Let u = 𝑒 𝑥 𝑠𝑖𝑛 𝑥 and v = (sin 𝑥) 𝑥
1
2

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𝑑𝑦 𝑑𝑢 𝑑𝑣
so that y = u + v ⇒ = + ----(1)
𝑑𝑥 𝑑𝑥 𝑑𝑥

2
Now, u = 𝑒 𝑥 𝑠𝑖𝑛 𝑥 , Differentiating both sides w.r.t. x, we get
1
𝑑𝑢 2
⇒ 𝑑𝑥 = 𝑒 𝑥 𝑠𝑖𝑛 𝑥 [𝑥(𝑠𝑖𝑛2𝑥) + 𝑠𝑖𝑛2 𝑥] ----- ( 2)

Also , v = (sin 𝑥) 𝑥

⇒ log v = 𝑥 log (sin 𝑥)

Differentiating both sides w.r.t. x, we get
1 𝑑𝑣
𝑣 𝑑𝑥
= 𝑥 cot 𝑥 + log (𝑠𝑖𝑛𝑥)
1
𝑑𝑣
𝑑𝑥
= (sin 𝑥) 𝑥 [𝑥 𝑐𝑜𝑡𝑥 + log(𝑠𝑖𝑛𝑥)] ------ (3)

Substituting from − (2), − (3) in − (1) we get 1
2
𝑑𝑦 2
= 𝑒 𝑥 𝑠𝑖𝑛 𝑥 [𝑥𝑠𝑖𝑛2𝑥 + 𝑠𝑖𝑛2 𝑥] + (sin 𝑥) 𝑥 [𝑥 𝑐𝑜𝑡𝑥 + log(𝑠𝑖𝑛𝑥)]
𝑑𝑥

31
𝐿𝑡 𝑓(1+ℎ)− 𝑓(1) 𝐿𝑡 [1+ℎ]−[1]
RHD = ℎ→0 ℎ
= ℎ→0 ℎ

𝐿𝑡 (1−1)
= ℎ→0 ℎ
=0 1

𝐿𝑡 𝑓(1−ℎ)− 𝑓(1) 𝐿𝑡 [1−ℎ]−[1] 𝐿𝑡 0−1
LHD = ℎ→0 −ℎ
= ℎ→0 −ℎ
= ℎ→0 −ℎ

𝐿𝑡 1
= ℎ→0 ℎ
=∞ 1

Since, RHD ≠LHD
1
Therefore f(x) is not differentiable at x = 1

OR
𝑑𝑦
𝑦 = 𝑏 tan 𝜃  = 𝑏 𝑠𝑒𝑐 2 𝜃 … (1)
𝑑𝜃
𝑑𝑥
𝑥 = 𝑎 sec 𝜃  = 𝑎 sec 𝜃 tan 𝜃 … (2)
𝑑𝜃

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𝑑𝑦
𝑑𝑦 𝑑𝜃 𝑏 𝑠𝑒𝑐 2 𝜃 𝑏 1
= = = 𝑐𝑜𝑠𝑒𝑐 𝜃 12
𝑑𝑥 𝑑𝑥 𝑎 sec 𝜃 tan 𝜃 𝑎
𝑑𝜃
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑤. 𝑟. 𝑡. 𝑥, 𝑤𝑒 𝑔𝑒𝑡
𝑑2 𝑦 −𝑏 𝑑𝜃
𝑑𝑥 2
= 𝑎 𝑐𝑜𝑠𝑒𝑐 𝜃 cot 𝜃 × 𝑑𝑥
−𝑏 1
= 𝑎 𝑐𝑜𝑠𝑒𝑐 𝜃 cot 𝜃 × 𝑎 sec 𝜃 tan 𝜃 [𝑢𝑠𝑖𝑛𝑔 (2)]

−𝑏 1
= 𝑎.𝑎 𝑐𝑜𝑡 3 𝜃

𝑑2 𝑦 −𝑏 𝜋 3 −𝑏 3 3√3𝑏 1
2
] = [cot ] = (√3) = −
𝑑𝑥 𝜃=𝜋 𝑎 6 𝑎 𝑎. 𝑎 2
6

32 𝑓 (𝑥) = tan 𝑥 − 4𝑥
1
𝑓 ′ (𝑥) = 𝑠𝑒𝑐 2 𝑥 − 4 2

a) For 𝑓 (𝑥) to be strictly increasing

𝑓 ′ (𝑥) > 0

⇒ 𝑠𝑒𝑐 2 𝑥 − 4 > 0

⇒ 𝑠𝑒𝑐 2 𝑥 > 4

1 1 2
⇒ 𝑐𝑜𝑠 2 𝑥 < ⇒ 𝑐𝑜𝑠 2 𝑥 < ( )
4 2

1 1 𝜋 𝜋 1
⇒ − 2 < cos 𝑥 < 2 ⇒ 3 < 𝑥 < 2 12

b) For 𝑓 (𝑥) to be strictly decreasing

𝑓 ′ (𝑥) < 0

⇒ 𝑠𝑒𝑐 2 𝑥 − 4 < 0

⇒ 𝑠𝑒𝑐 2 𝑥 < 4
1
⇒ 𝑐𝑜𝑠 2 𝑥 >
4

1 2
⇒ 𝑐𝑜𝑠 2 𝑥 > (2)

1 𝜋
⇒ cos 𝑥 > 2 [∵ 𝑥 ∈ (0, 2 )]

𝜋
⇒ 0 <𝑥 < 3

1

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33 Put 𝑥 2 = 𝑦 𝑡𝑜 𝑚𝑎𝑘𝑒 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠 1
2
𝑥2 + 1 𝑦+ 1 𝐴 𝐵
2 2
= = +
(𝑥 + 2)(𝑥 + 3) (𝑦 + 2)(𝑦 + 3) 𝑦 + 2 𝑦 + 3
1
⇒ 𝑦 + 1 = 𝐴(𝑦 + 3) + 𝐵 (𝑦 + 2)……………(1)
2
Comparing coefficients of y and constant terms on both sides of (1) we
get

A+B = 1 and 3A + 2B = 1

Solving, we get A = −1, B = 2 1

𝑥2+ 1 −1 1
∫ (𝑥2 +2)(𝑥2 +3) 𝑑𝑥 = ∫ 𝑥 2 +2 𝑑𝑥 + 2 ∫ 𝑥 2 +3 𝑑𝑥
1
1 𝑥 2 𝑥
= − 2 𝑡𝑎𝑛−1 ( 2) + 3 𝑡𝑎𝑛−1 ( 3) + 𝐶
√ √ √ √

34 Solving 𝑦 = √3𝑥 𝑎𝑛𝑑𝑥 2 + 𝑦 2 = 4
We get 𝑥 2 + 3𝑥 2 = 4
1
⇒ 𝑥2 = 1 ⇒ 𝑥 = 1
2

1
2

Required Area
1 2
1
= √3 ∫ 𝑥 𝑑𝑥 + ∫ √22 − 𝑥 2 𝑑𝑥 2
0 1

√3 2 1 𝑥 𝑥 2
= [𝑥 ]0 + [ √22 − 𝑥 2 + 2 sin−1 ( )] 1
2 2 2 1
√3 𝜋 √3 𝜋
= + [2 × − −2 × ]
2 2 2 6
2𝜋
𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
3 1
2
OR

Page 9 of 14

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4 6
Required Area = ∫0 √62 − 𝑥 2 𝑑𝑥
3 1
Y 2

x
1
2

4 𝑥 2 𝑥 6
= [ √6 − 𝑥 2 + 18 𝑠𝑖𝑛−1 ( )]
3 2 6 0
1
4 𝜋
= [18 × − 0] = 12𝜋 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠
3 2 1

35 The given differential equation can be written as

𝑑𝑦 𝑦 + 2𝑥 2 𝑑𝑦 1
= ⇒ − 𝑦 = 2𝑥
𝑑𝑥 𝑥 𝑑𝑥 𝑥
1
𝐻𝑒𝑟𝑒 𝑃 = − , 𝑄 = 2𝑥 1
𝑥
2
1
− ∫ 𝑑𝑥
IF = 𝑒 ∫ 𝑃𝑑𝑥 = 𝑒 = 𝑒 − log 𝑥 =
1
𝑥
𝑥 1

The solutions is :

1 1
𝑦× = ∫ (2𝑥 × ) 𝑑𝑥 1
𝑥 𝑥
𝑦
⇒ = 2𝑥 + 𝑐
𝑥 1
⇒ 𝑦 = 2𝑥 2 + 𝑐𝑥 2

36 |𝐴| = 1(−1 − 2) − 2(−2 − 0) = −3 + 4 = 1 1
2
A is nonsingular, therefore 𝐴−1 exists
−3 −2 −4
𝐴𝑑𝑗 𝐴 = [ 2 1 2]
2 1 3
−3 −2 −4
1 1
⇒ 𝐴−1 = |𝐴| (𝐴𝑑𝑗 𝐴) = [ 2 1 2] 12
2 1 3

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The given equations can be written as:
1 −2 0 𝑥 10 1
[2 −1 −1] [𝑦]=[ 8 ]
2
0 −2 1 𝑧 7

Which is of the form 𝐴′ 𝑋 = 𝐵

⇒ 𝑋 = (𝐴′ )−1 𝐵 = (𝐴−1 )′ 𝐵 1

𝑥 −3 2 2 10 0
⇒ [𝑦] = [−2 1 1] [ 8 ] = [−5]
𝑧 −4 2 3 7 −3
1
⇒ 𝑥 = 0, 𝑦 = −5, 𝑧 = −3 1
2

OR

1 −1 0 2 2 −4
𝐴𝐵 = [2 3 4] [−4 2 −4]
0 1 2 2 −1 5
6 0 0 1
= [0 6 0] 12
0 0 6

 𝐴𝐵 = 6𝐼
1 1
 𝐴 ( 𝐵) = 𝐼  𝐴−1 = (𝐵) 1
6 6

𝑇ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠

1 −1 0 𝑥 3
[2 3 4] [𝑦] = [17]
0 1 2 𝑧 7
3
𝐴𝑋 = 𝐷, 𝑤ℎ𝑒𝑟𝑒 𝐷 = [17]
7

 X = 𝐴−1 𝐷

𝑥 2 2 −4 3 12
1 1
 [𝑦] = 6 [−4 2 −4] [17] = 6 [−6]
𝑧 2 −1 5 7 24 1

𝑥 2
 [𝑦] = [−1]
𝑧 4

𝑥 = 2, 𝑦 = −1, 𝑧=4
1
1
2

37 We have 𝑎1 = 3𝑖̂ + 2𝑗̂ − 4𝑘̂ 𝑏1 = 𝑖̂ + 2𝑗̂ + 2𝑘̂

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𝑎2 = 5𝑖 − 2𝑗 𝑏2 = 3𝑖̂ + 2𝑗̂ + 6𝑘̂
⃗⃗⃗⃗ 𝑎1 = 2𝑖̂ − 4𝑗̂ + 4𝑘̂
𝑎2 − ⃗⃗⃗⃗ 1
𝑖̂ 𝑗̂ 𝑘̂
𝑏1 × 𝑏2 = |1 2 2| = 𝑖̂(12 − 4) − 𝑗̂(6 − 6) + 𝑘̂ (2 − 6)
⃗⃗⃗ ⃗⃗⃗⃗ 1
3 2 6

⃗⃗⃗⃗⃗ ⃗⃗⃗⃗2 = 8𝑖̂ + 0𝑗̂ − 4𝑘̂ = 8𝑖̂ − 4𝑘̂
𝑏1 × 𝑏

⃗⃗⃗1 × ⃗⃗⃗⃗
∵ (𝑏 𝑏2 ). (𝑎 𝑎1 ) = 16 − 16 = 0
⃗⃗⃗⃗2 − ⃗⃗⃗⃗ 1

∴ The lines are intersecting and the shortest distance between
the lines is 0.
Now for point of intersection
3𝑖̂ + 2𝑗̂ − 4𝑘̂ + 𝜆(𝑖̂ + 2𝑗̂ + 2𝑘̂ ) = 5𝑖̂ − 2𝑗̂ + 𝜇(3𝑖̂ + 2𝑗̂ + 6𝑘̂ )
⟹ 3 + 𝜆 = 5 + 3𝜇 −−−− (1)
2 + 2𝜆 = −2 + 2𝜇 −−−− (2)
−4 + 2𝜆 = 6𝜇 −−−− (3) 1

Solving (1) ad (2) we get, 𝜇 = −2 𝑎𝑛𝑑 𝜆 = −4
Substituting in equation of line we get
𝑟 = 5𝑖 − 2𝑗 + (−2)(3𝑖̂ + 2𝑗̂ − 6𝑘̂ ) = −𝑖̂ − 6𝑗̂ − 12𝑘̂
1
Point of intersection is (−1, −6, −12)

OR

Let P be the given point and Q be the foot of the perpendicular.
𝑥+1 𝑦−3 𝑧+6
Equation of PQ = = =𝜆 1
1
2 1 −2
2

P (−1,3, −6)

Let coordinates of Q be (2𝜆 − 1, 𝜆 + 3, −2𝜆 − 6)
Since Q lies in the plane 2𝑥 + 𝑦 − 2𝑧 + 5 = 0

∴ 2(2𝜆 − 1) + (𝜆 + 3) − 2(−2𝜆 − 6) + 5 = 0
1
⇒ 4𝜆 − 2 + 𝜆 + 3 + 4𝜆 + 12 + 5 = 0 2

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⇒ 9𝜆 + 18 = 0 ⟹ 𝜆 = −2

∴ 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑄 𝑎𝑟𝑒 (−5, 1, −2)

Length of the perpendicular = √(−5 + 1)2 + (1 − 3)2 + (−2 + 6)2 1

= 6 𝑢𝑛𝑖𝑡𝑠 1

1

38 Max 𝑍 = 3𝑥 + 𝑦
Subject to 𝑥 + 2𝑦 ≥ 100 ----- (1)
2𝑥 − 𝑦 ≤ 0 ----- (2)
2𝑥 + 𝑦 ≤ 200 ----- (3)
𝑥 ≥ 0, 𝑦≥0

3

Corner Points 𝑍 = 3𝑥 + 𝑦

A (0, 50) 50

B (0, 200) 200

C (50, 100) 250

D (20, 40) 100 1

𝑀𝑎𝑥 𝑧 = 250 𝑎𝑡 𝑥 = 50, 𝑦 = 100

1

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OR

(i)
Corner points 𝒁 = 𝟑𝒙 − 𝟒𝒚
O(0,0) 0
A(0,8) -32
B(4,10) -28
1
C(6,8) -14 1
2
D(6,5) -2
E(4,0) 12
𝑀𝑎𝑥 𝑍 = 12 𝑎𝑡 𝐸(4,0)
𝑀𝑖𝑛 𝑍 = −32 𝑎𝑡 𝐴(0,8)
1
(ii) Since maximum value of Z occurs at B(4,10) and C(6, 8)
∴ 4𝑝 + 10𝑞 = 6𝑝 + 8𝑞
2𝑞 = 2𝑝 2
𝑝 = 𝑞 1
Number of optimal solution are infinite 2

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