Marking Scheme

Class XII
Mathematics (Code – 041)
Section : A (Multiple Choice Questions- 1 Mark each)
Question Answer Hints/Solution
No
1. (c) In a skew-symmetric matrix, the (i, j)th element is negative of
the (j, i)th element. Hence, the (i, i)th element = 0
2. (a) |𝐴𝐴 | = |𝐴||𝐴′| = (−3)(−3) = 9
3. (b) The area of the parallelogram with adjacent sides AB and AC =
𝐴𝐵⃗ × 𝐴𝐶⃗ . Hence, the area of the triangle with vertices A, B, C
= 𝐴𝐵⃗ × 𝐴𝐶⃗
4. (c) The function f is continuous at x = 0 if 𝑙𝑖𝑚 → 𝑓(𝑥) = 𝑓(0)
We have f(0) = k and
𝑙𝑖𝑚 → 𝑓(𝑥) = 𝑙𝑖𝑚 → = 𝑙𝑖𝑚 → =𝑙𝑖𝑚 →

=𝑙𝑖𝑚 → =1
Hence, k =1
5. (b) 𝑥 1
+ log |𝑥| + 𝐶 ∵ 𝑓(𝑥) = 𝑥+ 𝑑𝑥
2 𝑥
6. (c)
The given differential equation is 4 = 0. Here, m = 2
and n = 1
Hence, m + n= 3
7. (b) The strict inequality represents an open half plane and it
contains the origin as (0, 0) satisfies it.
8. (a) Scalar Projection of 3𝚤̂ − 𝚥̂ − 2𝑘 𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑟 𝚤̂ + 2𝚥̂ − 3𝑘
̂ ̂ . ̂ ̂
= ̂ ̂
=
√
9. (c) ∫ = [𝑙 𝑜𝑔(𝑥 + 1)] = (𝑙𝑜𝑔10 − 𝑙𝑜𝑔5) = 𝑙𝑜𝑔
= 𝑙𝑜𝑔2
10. (c) (𝐴𝐵 ) = (𝐵 ) 𝐴 = 𝐵𝐴
11. (d) The minimum value of the objective function occurs at two
adjacent corner points (0.6, 1.6) and (3, 0) and there is no point
in the half plane 4𝑥 + 6𝑦 < 12 in common with the feasible
region. So, the minimum value occurs at every point of the line-
segment joining the two points.
12. (d) 2 − 20 = 2𝑥 − 24 ⟹ 2𝑥 = 6 ⟹ 𝑥 = 3 ⟹ 𝑥 = ±√3
13. (b) |𝑎𝑑𝑗𝐴| = |𝐴| ⟹ |𝑎𝑑𝑗𝐴| = 25
14. (c) P 𝐴 ∩ 𝐵 = 𝑃 𝐴′) × 𝑃(𝐵′) (As A and B are independent,
( ) (
𝐴 𝑎𝑛𝑑 𝐵′ are also independent.)
= 0.7 × 0.4 = 0.28
15. (c) 𝑑𝑦 𝑑𝑥
𝑦𝑑𝑥 − 𝑥𝑑𝑦 = 0 ⟹ 𝑦𝑑𝑥 − 𝑥𝑑𝑦 = 0 ⟹ =
𝑦 𝑥
𝑑𝑦 𝑑𝑥
⟹ = + 𝑙𝑜𝑔𝐾, 𝐾 > 0 ⟹ log|𝑦| = log|𝑥| + log 𝐾
𝑦 𝑥
⟹ log|𝑦| = log |𝑥|𝐾 ⟹ |𝑦| = |𝑥|𝐾 ⟹ y=±𝐾𝑥 ⟹ 𝑦 = 𝐶𝑥

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 16. (a) y = sin-1x
𝑑𝑦 1 𝑑𝑦
= ⟹ 1−𝑥 . =1
𝑑𝑥 √1 − 𝑥 𝑑𝑥
Again , differentiating both sides w. r. to x, we get
𝑑 𝑦 𝑑𝑦 −2𝑥
1−𝑥 + . =0
𝑑𝑥 𝑑𝑥 2√1 − 𝑥
Simplifying, we get (1 − 𝑥 )𝑦 = 𝑥𝑦
17. (b) 𝑎⃗ − 2𝑏⃗ = 𝑎⃗ − 2𝑏⃗ . 𝑎⃗ − 2𝑏⃗
𝑎⃗ − 2𝑏⃗ = 𝑎⃗. 𝑎⃗ − 4𝑎⃗. 𝑏⃗ + 4𝑏⃗ . 𝑏⃗
= |𝑎⃗| − 4𝑎⃗. 𝑏⃗+ 4 𝑏⃗
= 4 − 16 + 36 = 24
𝑎⃗ − 2𝑏⃗ = 24 ⟹ 𝑎⃗ − 2𝑏⃗ = 2√6
18. (b) The line through the points (0, 5, -2) and (3, -1, 2) is
= =
𝑥 𝑦−5 𝑧+2
𝑜𝑟, = =
3 −6 4
Any point on the line is (3𝑘, −6𝑘 + 5,4𝑘 − 2), where k is an
arbitrary scalar.
3𝑘 = 6 ⟹ 𝑘 = 2
The z-coordinate of the point P will be 4 × 2 − 2 = 6
19. (c) 𝑠𝑒𝑐 𝑥 is defined if 𝑥 ≤ −1 𝑜𝑟 𝑥 ≥ 1. Hence, 𝑠𝑒𝑐 2𝑥 will be
defined if 𝑥 ≤ − or 𝑥 ≥ .
Hence, A is true.
The range of the function 𝑠𝑒𝑐 𝑥 is [0, 𝜋] − { }
R is false.
20. (a) The equation of the x-axis may be written as 𝑟⃗ = 𝑡𝚤̂. Hence, the
acute angle 𝜃 between the given line and the x-axis is given by
|1 × 1 + (−1) × 0 + 0 × 0| 1 𝜋
𝑐𝑜𝑠𝜃 = = ⟹𝜃=
1 + (−1) + 0 × √1 + 0 + 0 √2 4

SECTION B (VSA questions of 2 marks each)

21. 13𝜋 𝜋 .1
𝑠𝑖𝑛 [𝑠𝑖𝑛 ] = 𝑠𝑖𝑛 [𝑠𝑖𝑛 2𝜋 − ]
7 7
= 𝑠𝑖𝑛 [𝑠𝑖𝑛 − ] = − 1
OR
Let 𝑦 ∈ 𝑁(codomain). Then ∃ 2𝑦 ∈ 𝑁(domain) such that
𝑓(2𝑦) = = 𝑦. Hence, f is surjective. 1
1, 2 ∈ 𝑁(domain) such that 𝑓(1) = 1 = 𝑓(2)
Hence, f is not injective. 1
22. Let AB represent the height of the street light from the ground. At
any time t seconds, let the man represented as ED of height 1.6 m
be at a distance of x m from AB and the length of his shadow EC
be y m.
Using similarity of triangles, we have = ⇒ 3𝑦 = 2𝑥 ½
.

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 Differentiating both sides w.r.to t, we get 3 = 2
𝑑𝑦 2 𝑑𝑦
= × 0.3 ⇒ = 0.2 ½
𝑑𝑡 3 𝑑𝑡
At any time t seconds, the tip of his shadow is at a distance of
(𝑥 + 𝑦) m from AB.
The rate at which the tip of his shadow moving
𝑑𝑥 𝑑𝑦
= + 𝑚/𝑠 = 0.5 𝑚/𝑠 ½
𝑑𝑡 𝑑𝑡
The rate at which his shadow is lengthening
𝑑𝑦
= 𝑚/𝑠 = 0.2 𝑚/𝑠 1/2
𝑑𝑡
23. 𝑎⃗ = 𝚤̂ − 𝚥̂ + 7𝑘 𝑎𝑛𝑑 𝑏⃗ = 5𝚤̂ − 𝚥̂ + 𝜆𝑘
Hence 𝑎⃗ + 𝑏⃗ = 6 𝚤̂ − 2𝚥̂ + (7 + 𝜆)𝑘 and 𝑎⃗ − 𝑏⃗ = −4𝚤̂ + (7 − 𝜆)𝑘
𝑘 ½
⃗ ⃗ ⃗ ⃗
𝑎⃗ + 𝑏 𝑎𝑛𝑑 𝑎⃗ − 𝑏 will be orthogonal if, 𝑎⃗ + 𝑏 . 𝑎⃗ − 𝑏 = 0 ½
i.e., if, −24 + (49 − 𝜆 ) = 0 ⟹ 𝜆 = 25
i.e., if, 𝜆 = ±5 1

OR
The equations of the line are 6𝑥 − 12 = 3𝑦 + 9 = 2𝑧 − 2, which,
when written in standard symmetric form, will be
( )
= = ½

Since, lines are parallel, we have = =

Hence, the required direction ratios are , , or (1,2,3) ½
and the required direction cosines are , , 1
√ √ √

24. 𝑦 1−𝑥 +𝑥 1−𝑦 = 1

Let 𝑠𝑖𝑛 𝑥 = 𝐴 and 𝑠𝑖𝑛 𝑦 = 𝐵.. Then x = sinA and y = sinB ½
𝑦 1 − 𝑥 + 𝑥 1 − 𝑦 = 1 ⟹ 𝑠𝑖𝑛𝐵𝑐𝑜𝑠𝐴 + 𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐵 = 1

𝜋
⟹ sin(𝐴 + 𝐵 ) = 1 ⟹ 𝐴 + 𝐵 = 𝑠𝑖𝑛 1 =
2
𝜋
⟹ 𝑠𝑖𝑛 𝑥 + 𝑠𝑖𝑛 𝑦 = ½
2
Differentiating w.r.to x, we obtain = − 1
25. Since 𝒂⃗ is a unit vector, ∴ |𝑎⃗| = 1 ½

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 (𝑥⃗ − 𝑎⃗). (𝑥⃗ + 𝑎⃗) = 12.

⟹ 𝑥⃗. 𝑥⃗ + 𝑥⃗. 𝑎⃗ − 𝑎⃗. 𝑥⃗ − 𝑎⃗. 𝑎⃗ = 12 ½

⟹ |𝑥⃗| − |𝑎⃗| = 12. ½

⟹ |𝑥⃗| − 1 = 12
⟹ |𝑥⃗| = 13 ⟹ |𝑥⃗| = √13 1/2
SECTION C
(Short Answer Questions of 3 Marks each)

26. 𝑑𝑥
√3 − 2𝑥 − 𝑥
2
=∫ =∫
( ) ( )

= 𝑠𝑖𝑛 +𝐶 [∫ = 𝑠𝑖𝑛 +𝐶] 1

√

27. P(not obtaining an odd person in a single round) = P(All three of

them throw tails or All three of them throw heads)
= × × ×2= 1+1/2
P(obtaining an odd person in a single round)
= 1 − P(not obtaining an odd person in a single round) = ½
The required probability
= P(‘In first round there is no odd person’ and ‘In second round
there is no odd person’ and ‘In third round there is an odd person’)
= × × = 1
OR
Let X denote the Random Variable defined by the number of
defective items.

P(X=0) = × =

P(X=1) = 2 × × =

2
P(X=2) = × =
𝑥 0 1 2
𝑝 2 8 1
5 15 15
𝑝𝑥 0 8 2 1/2
15 15

Mean = ∑ 𝑝 𝑥 = = 1/2
28. Let I = ∫ =∫
√
𝑑𝑥 ..(i)
√ √ √

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 𝒃 𝒃
Using ∫𝒂 𝒇(𝒙) 𝒅𝒙 = ∫𝒂 𝒇(𝒂 + 𝒃 − 𝒙)𝒅𝒙

I=∫ 𝑑𝑥

√
I=∫ dx ..(ii). 1
√ √
Adding (i) and (ii), we get
√ √
2I = ∫ 𝑑𝑥 + ∫ dx
√ √ √ √

1
2I = ∫ 𝑑𝑥

= [𝑥] = − =

Hence, I = ∫ = 1
√
OR
|𝑥 − 1| 𝑑𝑥 = (1 − 𝑥 )𝑑𝑥 + (𝑥 − 1)𝑑𝑥 1

= 𝑥−
1
+ −𝑥
4 1
0 1
= (1 − ) + (8 − 4) − ( − 1)
=5 1
29. 𝑦𝑑𝑥 + (𝑥 − 𝑦 )𝑑𝑦 = 0
𝒅𝒙
Reducing the given differential equation to the form + 𝑷𝒙 = 𝑸
𝒅𝒚
we get, + =𝑦 ½
∫
I.F = 𝑒 ∫ =𝑒 =𝑒 =𝑦 1
The general solution is given by
𝑥. 𝐼𝐹 = 𝑄. 𝐼𝐹𝑑𝑦 ⟹ 𝑥𝑦 = 𝑦 𝑑𝑦 1

⟹ 𝑥𝑦 = + 𝐶 , which is the required general solution ½

OR
𝑥𝑑𝑦 − 𝑦𝑑𝑥 = 𝑥 + 𝑦 𝑑𝑥
It is a Homogeneous Equation as
𝑑𝑦 𝑥 +𝑦 +𝑦 𝑦 𝑦 𝑦
= = 1+( ) + = 𝑓 .
𝑑𝑥 𝑥 𝑥 𝑥 𝑥 ½
Put 𝑦 = 𝑣𝑥

= 𝑣+𝑥

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 𝑑𝑣 ½
𝑣+𝑥 = 1+𝑣 +𝑣
𝑑𝑥
Separating variables, we get
𝑑𝑣 𝑑𝑥 ½
=
√1 + 𝑣 𝑥
Integrating, we get 𝑙𝑜𝑔 𝑣 + √1 + 𝑣 = 𝑙𝑜𝑔|𝑥| + 𝑙𝑜𝑔𝐾, 𝐾 > 0
𝑙𝑜𝑔 𝑦 + 𝑥 + 𝑦 = 𝑙𝑜𝑔𝑥 𝐾
⇒ 𝑦 + 𝑥 + 𝑦 = ±𝐾𝑥
⇒ 𝑦 + 𝑥 + 𝑦 = 𝐶𝑥 , which is the required general solution
1+1/2
30. We have Z= 400x +300y subject to
x + y ≤ 200,𝑥 ≤ 40, 𝑥 ≥ 20, 𝑦 ≥ 0
The corner points of the feasible region are C(20,0), D(40,0),
B(40,160), A(20,180)

Corner Point Z = 400x + 300y

C(20,0) 8000
D(40,0) 16000
B(40,160) 64000
A(20,180) 62000 1

Maximum profit occurs at x= 40, y=160

and the maximum profit =₹ 64,000 1
31. 𝑑𝑥 = ∫ 𝑥 + 𝑑𝑥 1
∫ ( ) ( )( )
Now resolving into partial fractions as
( )( )

2𝑥 + 1 𝐴 𝐵
= +
(𝑥 − 1)(𝑥 + 1) 𝑥−1 𝑥+1

We get = + 1
( )( ) ( ) ( )

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 Hence, ∫ ( )
𝑑𝑥 = ∫ 𝑥 + 𝑑𝑥
( )( )

=∫ 𝑥+ + 𝑑𝑥
( ) ( )

= + log|𝑥 − 1| + log|𝑥 + 1| + 𝐶
𝑥 1
= + (log|(𝑥 − 1) (𝑥 + 1)| + 𝐶 1
2 2

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

32.

(Correct
Fig: 1
Mark)

The points of intersection of the parabola 𝑦 = 𝑥 and the line

½
𝑦 = 𝑥 are (0, 0) and (1, 1).
Required Area = ∫ 𝑦 𝑑𝑥 + ∫ 𝑦 𝑑𝑥
Required Area = ∫ 𝑥 𝑑𝑥 + ∫ 𝑥 𝑑𝑥 2
1 2
= + = + = 1+1/2
0 1

33. Let (𝑎, 𝑏) ∈ 𝑁 × 𝑁. Then we have

ab = ba (by commutative property of multiplication of natural
numbers)
⟹ (𝑎, 𝑏)𝑅 (𝑎, 𝑏)
Hence, R is reflexive. 1
Let (𝑎, 𝑏), (𝑐, 𝑑) ∈ 𝑁 × 𝑁 such that (a, b) R (c, d). Then
ad = bc
⟹ 𝑐𝑏 = 𝑑𝑎 (by commutative property of multiplication of
natural numbers
⟹ (𝑐, 𝑑)𝑅(𝑎, 𝑏)
Hence, R is symmetric. 1+1/2
Let (𝑎, 𝑏), (𝑐, 𝑑), (𝑒, 𝑓) ∈ 𝑁 × 𝑁 such that

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 (a, b) R (c, d) and (c, d) R (e, f).
Then ad = bc, cf = de
⟹ 𝑎𝑑𝑐𝑓 = 𝑏𝑐𝑑𝑒
⟹ 𝑎𝑓 = 𝑏𝑒
⟹ (𝑎, 𝑏)𝑅(𝑒, 𝑓)
Hence, R is transitive. 2
Since, R is reflexive, symmetric and transitive, R is an
equivalence relation on 𝑁 × 𝑁. ½
OR
Let 𝐴 ∈ 𝑃 𝑋 . Then 𝐴 ⊂ 𝐴
( )
⟹ (𝐴, 𝐴) ∈ 𝑅
Hence, R is reflexive. 1
Let 𝐴, 𝐵, 𝐶 ∈ 𝑃(𝑋) such that
(𝐴, 𝐵), (𝐵, 𝐶) ∈ 𝑅
⟹ 𝐴 ⊂ 𝐵, 𝐵 ⊂ 𝐶
⟹𝐴⊂𝐶
⟹ (𝐴, 𝐶) ∈ 𝑅
Hence, R is transitive. 2
∅, 𝑋 ∈ 𝑃(𝑋) such that ∅ ⊂ 𝑋. Hence, (∅, 𝑋) ∈ 𝑅. But, 𝑋 ⊄ ∅,
which implies that (𝑋, ∅) ∉ 𝑅.
Thus, R is not symmetric. 2

34. The given lines are non-parallel lines. There is a unique line-
segment PQ (P lying on one and Q on the other, which is at right
angles to both the lines. PQ is the shortest distance between the
lines. Hence, the shortest possible distance between the insects =
PQ
The position vector of P lying on the line
𝑟⃗ = 6𝚤̂ + 2𝚥̂ + 2𝑘 + 𝜆 𝚤̂ − 2𝚥̂ + 2𝑘
is (6 + 𝜆)𝚤̂ + (2 − 2𝜆)𝚥̂ + (2 + 2𝜆)𝑘 for some 𝜆 ½
The position vector of Q lying on the line
𝑟⃗ = −4𝚤̂ − 𝑘 + 𝜇 3𝚤̂ − 2𝚥̂ − 2𝑘
½
is (−4 + 3𝜇)𝚤̂ + (−2𝜇)𝚥̂ + (−1 − 2𝜇)𝑘 for some 𝜇
𝑃𝑄⃗ = (−10 + 3𝜇 − 𝜆)𝚤̂ + (−2𝜇 − 2 + 2𝜆)𝚥̂ + (−3 − 2𝜇 − 2𝜆)𝑘 ½
Since, PQ is perpendicular to both the lines
(−10 + 3𝜇 − 𝜆) + (−2𝜇 − 2 + 2𝜆)(−2) + (−3 − 2𝜇 − 2𝜆)2
= 0, ½
𝑖. 𝑒. , 𝜇 − 3𝜆 = 4 …(i)
and (−10 + 3𝜇 − 𝜆)3 + (−2𝜇 − 2 + 2𝜆)(−2) + (−3 − 2𝜇 −
2𝜆)(−2) = 0, ½
𝑖. 𝑒. ,17𝜇 − 3𝜆 = 20 …(ii)
solving (i) and (ii) for 𝜆 𝑎𝑛𝑑 𝜇, we get 𝜇 = 1, 𝜆 = −1. 1
The position vector of the points, at which they should be so that
the distance between them is the shortest, are
5𝚤̂ + 4𝚥̂ and −𝚤̂ − 2𝚥̂ − 3𝑘 ½
𝑃𝑄⃗ = −6𝚤̂ − 6𝚥̂ − 3𝑘
The shortest distance = 𝑃𝑄⃗ = √6 + 6 + 3 = 9 1
OR

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 Eliminating t between the equations, we obtain the equation of the
path = = , which are the equations of the line passing
through the origin having direction ratios <2, -4, 4>. This line is
the path of the rocket. 1
When t = 10 seconds, the rocket will be at the point (20, -40, 40).
Hence, the required distance from the origin at 10 seconds = ½
20 + 40 + 40 𝑘𝑚 = 20 × 3 𝑘𝑚 = 60 𝑘𝑚 1
The distance of the point (20, -40, 40) from the given line
( ⃗ ⃗)× ⃗ ̂ ×( ̂ ̂ ) ̂
= = 𝑘𝑚 = 𝑘𝑚 2
⃗ ̂ ̂ ̂ ̂
√
= 𝑘𝑚 = 10√3 𝑘𝑚 1/2
√
35. 2 −3 5
A = 3 2 −4
1 1 −2
½
|A| = 2(0) + 3(−2) + 5(1) = −1
𝑎𝑑𝑗𝐴
𝐴 =
|𝐴|
0 −1 2 0 −1 2
𝑎𝑑𝑗𝐴 = 2 −9 23 , 𝐴 = 2 −9 23 3
( )
1 −5 13 1 −5 13
𝑥 0 −1 2 11
X=𝐴 𝐵⟹ 𝑦 = 2 −9 23 −5
( )
𝑧 1 −5 13 −3
0+5−6
= 22 + 45 − 69
( )
11 + 25 − 39
𝑥 1 −1
⟹ 𝑦 = −2 ⟹ 𝑥 = 1, 𝑦 = 2 , 𝑧 = 3. 1+1/2
𝑧 (−1)
−3

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

36. (i) f(𝑥 ) = −0.1𝑥 + 𝑚𝑥 + 98.6, being a polynomial function, is differentiable
everywhere, hence, differentiable in (0, 12) 1
(ii)𝑓 (𝑥 ) = − 0.2𝑥 + 𝑚
Since, 6 is the critical point,
𝑓 (6) = 0 ⟹ 𝑚 = 1.2 1

(iii) 𝑓(𝑥) = −0.1𝑥 + 1.2𝑥 + 98.6

𝑓 (𝑥 ) = − 0.2𝑥 + 1.2 = −0.2(𝑥 − 6)

In the Interval 𝒇 (𝒙) Conclusion

(0, 6) +ve f is strictly increasing
in [0, 6]
(6, 12) -ve f is strictly decreasing 1+1
in [6, 12]

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 OR
(iii) 𝑓(𝑥) = −0.1𝑥 + 1.2𝑥 + 98.6,
𝑓 (𝑥) = − 0.2𝑥 + 1.2, 𝑓 (6) = 0,
𝑓 ′(𝑥) = − 0.2
𝑓 (6) = − 0.2 < 0
Hence, by second derivative test 6 is a point of local maximum. The local
maximum value = 𝑓(6) = − 0.1 × 6 + 1.2 × 6 + 98.6 = 102.2 1
We have 𝑓(0) = 98.6, 𝑓(6) = 102.2, 𝑓(12) = 98.6
6 is the point of absolute maximum and the absolute maximum value of the
function = 102.2. 1/2
0 and 12 both are the points of absolute minimum and the absolute minimum value
of the function = 98.6. 1/2
37. (i)

Let (𝑥, 𝑦) = 𝑥, √𝑎 − 𝑥 be the upper right vertex of the rectangle.

The area function 𝐴 = 2𝑥 × 2 √𝑎 − 𝑥
= 𝑥√𝑎 − 𝑥 , 𝑥 ∈ (0, 𝑎). 1
(ii) = 𝑥× + √𝑎 − 𝑥
√

4𝑏 𝑎 − 2𝑥 4𝑏 2 𝑥 + √ (𝑥 − √ )
= × =− × ½
𝑎 √𝑎 − 𝑥 𝑎 √𝑎 − 𝑥
=0⇒𝑥= .
√
𝑥= is the critical point. 1/2
√
(iii)For the values of x less than and close to , >0
√ √
1
and for the values of x greater than and close to , < 0.
√ √
Hence, by the first derivative test, there is a local maximum at the critical point
𝑥 = . Since there is only one critical point, therefore, the area of the soccer field
√
is maximum at this critical point 𝑥 = 1/2
√
Thus, for maximum area of the soccer field, its length should be 𝑎√2 and its width
should be 𝑏√2. ½
OR

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 (iii) 𝐴 = 2𝑥 × 2 √𝑎 − 𝑥 , 𝑥 ∈ (0, 𝑎).
Squaring both sides, we get
𝑍=𝐴 = 𝑥 (𝑎 − 𝑥 ) = (𝑥 𝑎 − 𝑥 ), 𝑥 ∈ (0, 𝑎).
A is maximum when Z is maximum.
𝑑𝑍 16𝑏 32𝑏
= (2𝑥𝑎 − 4𝑥 ) = 𝑥 𝑎 + √2𝑥 (𝑎 − √2𝑥)
𝑑𝑥 𝑎 𝑎
=0⇒𝑥= .
√
𝑑 𝑍 32𝑏
= (𝑎 − 6𝑥 )
𝑑𝑥 𝑎
𝑑 𝑍 32𝑏
( ) = (𝑎 − 3𝑎 ) = −64𝑏 < 0
𝑑𝑥 √ 𝑎 1
Hence, by the second derivative test, there is a local maximum value of Z at the
critical point 𝑥 = . Since there is only one critical point, therefore, Z is
√
1/2
maximum at 𝑥 = , hence, A is maximum at 𝑥 = .
√ √
Thus, for maximum area of the soccer field, its length should be 𝑎√2 and its width
should be 𝑏√2. 1/2
38. (i)Let P be the event that the shell fired from A hits the plane and Q be the event
that the shell fired from B hits the plane. The following four hypotheses are
possible before the trial, with the guns operating independently:
𝐸 = 𝑃𝑄,𝐸 = 𝑃 𝑄,𝐸 = 𝑃 𝑄,𝐸 = 𝑃𝑄
Let E = The shell fired from exactly one of them hits the plane.
𝑃(𝐸 ) = 0.3 × 0.2 = 0.06, 𝑃 (𝐸 ) = 0.7 × 0.8 = 0.56, 𝑃(𝐸 ) = 0.7 × 0.2
= 0.14, 𝑃 (𝐸 ) = 0.3 × 0.8 = 0.24
𝐸 𝐸 𝐸 𝐸
𝑃 = 0, 𝑃 = 0, 𝑃 = 1, 𝑃 =1 1
𝐸 𝐸 𝐸 𝐸
𝐸 𝐸 𝐸 𝐸
𝑃(𝐸) = 𝑃(𝐸 ). 𝑃 + 𝑃(𝐸 ). 𝑃 + 𝑃(𝐸 ). 𝑃 + 𝑃(𝐸 ). 𝑃
𝐸 𝐸 𝐸 𝐸
= 0.14 + 0.24 = 0.38 1
( ).
(ii)By Bayes’ Theorem, P =
( ). ( ). ( ). ( ).

0.14 7 2
= =
0.38 19

NOTE: The four hypotheses form the partition of the sample space and it can be
seen that the sum of their probabilities is 1. The hypotheses 𝐸 𝑎𝑛𝑑 𝐸 are actually
eliminated as 𝑃 =𝑃 =0
Alternative way of writing the solution:
(i)P(Shell fired from exactly one of them hits the plane)
= P[(Shell from A hits the plane and Shell from B does not hit the plane) or (Shell
1
from A does not hit the plane and Shell from B hits the plane)]
= 0.3 × 0.8 + 0.7 × 0.2 = 0.38
1
(ii)P(Shell fired from B hit the plane/Exactly one of them hit the plane)
P(Shell ired from B hit the plane ∩ Exactly one of them hit the plane)
=
P(Exactly one of them hit the plane)

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 =
( ) 1
( )
0.14 7 1
= =
0.38 19

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 16.09.2022

To,
All the Heads of Schools, Teachers and Students

Subject: Practice Material recommendation based on the CBSE Sample Paper for
February 2022.

This is with reference to Board’s Circular No. Acad- CBSE/Academic/-

JS(PMS)/ 2022 dated 16.09.2022 regarding Sample Paper for Board Examina-
tion for Classes X for the Session 2022- 23.

It may be noted that the new pattern questions added by CBSE in their just
uploaded Sample Paper are incorporated strictly in large quantities within the
below recommended EDUCART book.

Class 10 - https://amzn.to/3xnrGp6

Class 12 - https://amzn.to/3ekcSkm

This Material is strictly based on the CBSE Sample Paper and available through
the above link publicly on Amazon.

Therefore, the schools are suggested to share the aforementioned

recommendation with all their teachers and students.