Solution

PRACTICE PAPER

Class 12 - Mathematics
Section A
1. i. a. Let (x, y) ∈ R
⇒ x − y is divisible by n.

⇒ x − y = kn for some k ∈ Z
⇒ y − x = ( − k)n

⇒ y − x is divisible by n.
⇒ (y, x) ∈ R

b. Let (x, y) ∈ R and (y, z) ∈ R

Now, (x, y) ∈ R ⇒ x − y is divisible by n.
⇒ x − y = kn for some k ∈ Z

Also, (y, z) ∈ R ⇒ y − z is divisible by n.

⇒ y − z = mn for some m ∈ Z.
⇒ (x − y) + (y − z) = kn + mn
⇒ x − z = (k + m)n
⇒ x - z is divisible by n.
⇒ (x, z) ∈ R
x2 − 9
ii. Here, f(x) = x−3
f(x) assume all real values of x except for x - 3 = 0
i.e., x = 3.
Thus, domain of f(x) = R - {3}.
Let f(x) = y
x2 − 9 (x+3) (x−3)
∴ y= x−3
= (x−3)
⇒ y=x+3
Since y takes all real values except 6.
Thus, range of f(x) = R - {6}.
x 2 + 3x + 5
iii. Here, f(x) =
x2 + x − 6
x 2 + 3x + 5
= (x+3) (x−2)
The function f(x) is defined for all values of x except for x + 3 = 0 and x − 2 = 0 i.e., x = − 3 and x = 2.
Thus, domain of f(x) = R - {-3, 2}.
2. R = {(a,b) = |a.b| is divisible by 2.
where a, b ∈ A = {1, 2, 3, 4, 5}
reflexivty
For any a ∈ A,|a−a|=0 Which is divisible by 2.
∴ (a, a) ∈ r for all a ∈ A

So ,R is Reflexive
Symmetric :
Let (a, b) ∈ R for all a, b ∈ R
|a−b| is divisible by 2
|b−a| is divisible by 2
(a, b) ∈ r ⇒ (b, a) ∈ R
So, R is symmetirc .
Transitive :
Let (a, b) ∈ R and (b, c) ∈ R then
(a, b) ∈ R and (b, c) ∈ R
|a−b| is divisible by 2

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 |b−c| is divisible by 2
Two cases :
Case 1:
When b is even
(a, b) ∈ R and (b, c) ∈ R
|a−c| is divisible by 2
|b−c| is divisible by 2
|a−c| is divisible by 2
∴ (a, c) ∈ R

Case 2:
When b is odd
(a, b) ∈ R and (b, c) ∈ R
|a−c| is divisible by 2
|b−c| is divisible by 2
|a−c| is divisible by 2
Thus, (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
So R is transitive.
Hence , R is an equivalence relation
3. i. R = {(x, y) : x is wife of y}
Reflexive: since x can not be wife of x.
∴ (x, x) ∉ R

⇒ R is not reflexive

Symmetric: Let (x, y) ∈ R

⇒ x is wife of y

⇒ Y is husband of x

⇒ (y, x) ∉ R

⇒ R is not symmetric

Transitive: Let (x, y) ∈ R and (y, z) ∈ R

⇒ x is wife of y and y is husband of z which is a contradiction

⇒ (x, z) ∉ R

⇒ R is not transitive

ii. A be the set of human beings

R = {(x, y) : x is father of y}
Reflexive: since x can not be father of x
∴ (x, x) ∉ R

⇒ R is not reflexive

Symmetric: Let (x, y) ∈ R

⇒ x is father of y
⇒ y can not be father of x

⇒ (y, x) ∉ R

⇒ R is not symmetric

Transitive:
Now, let (x,y), (y,z) ϵ R
⇒ x is the father of y and y is the father of z.
⇒ x is not the father of z.
⇒ Indeed x is the grandfather of z.
⇒ (x,z) ∉ R
⇒ R is not transitive.
4. A = R - {3}, B = R - {1}

f : A → B is defined as f(x) = ( )
x−2
x−3
.

Let x, y ∈ A such that f(x) = f(y).

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 x−2 y−2
⇒
x−3
= y−3
⇒ (x - 2) (y - 3) = (y - 2) (x - 3)
⇒ xy - 3x - 2y + 6 = xy - 3y - 2x + 6
⇒ -3x - 2y = -3y - 2x
⇒ 3x - 2x = 3y - 2y
⇒ x = y

Therefore, f is one-one.
Let y ∈ B = R - {1}
Then, y ≠ 1.
The function f is onto if there exists x ∈ A such that f(x) = y.
Now, f (x) = y
x−2
⇒
x−3
=y
⇒ x - 2 = xy - 3y
⇒ x(1 - y) = -3y + 2
2 − 3y
⇒ x= ∈ A [y ≠ 1]
1−y
2 − 3y
Thus, for any y ∈ B, there exists 1−y
∈ A such that

( ) 2 − 3y
−2

( )
1−y
2 − 3y 2 − 3y − 2 + 2y −y
f 1−y
= = 2 − 3y − 3 + 3y
= −1
=y
( ) 2 − 3y
1−y −3

f is onto.
∴

Hence, function f is one-one and onto.

[ ] []
1 3 2 1
5. We have, [ 1 x 1]
1×3
2 5 1 2 =0
15 3 2 3×3 x 3×1

[]
1
⇒ [ 1 + 2x + 15 3 + 5x + 3 2+x+2]
1×3
2 =0
x 3×1

[]
1
⇒ [ 16 + 2x 5x + 6 x+4]
1×3
2 =0
x 3×1

⇒ [16 + 2x + (5x + 6).2 + (x + 4)x] 1 × 1 = 0

⇒ [16 + 2x + 10x + 12 + x2 + 4x] = 0
⇒ [x2 + 16x + 28] = 0
⇒ [x2 + 2x + 14x + 28] = 0
⇒ (x + 2)(x + 14) = 0
∴ x = -2, -14

6. 1. Given: X + Y = [ ] 7
2
0
5
…..(i)

and X − Y =
[ ] 3
0
0
3
…..(ii)

Adding eq. (i) and (ii), we get

2X =
[ ][ ] [
7
2
0
5
+
3
0
0
3
=
7+3
2+0
0+0
5+3 ] [ ]
=
10
2
0
8

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 ⇒ X=
1
2
=
[ ] [ ]
10
2
0
8
=
5
1
0
4
Subtracting eq. (i) and (ii), we get

2Y =
[ ][7
2
0
5
−
3
0
0
3 ] [ =
7−3
2−0
0−0
5−3 ] [ ]
=
4
2
0
2

⇒ Y=
[ ]
1
2
=
4
2
0
2
=
[ ] 2
1
0
1

[
2. Given: 2X + 3Y =
2
4
3
0] …..(i)

[
and 3X + 2Y =
−2
−1
−2
5 ] …..(ii)

Multiplying eq. (i) by 2, 4X + 6Y = 2

[ ] [ ]
2
4
3
0
=
4
8
6
0
……….(iii)

Multiplying eq. (ii) by 3, 9X + 6Y = 3

[ ] [ ] 2
−1
−2
5
=
6
−3
−6
15
………(iv)

Eq. (iv) – Eq. (iii)

⇒ 5X =
[ 6
−3
−6
15 ] [ ]−
4
8
6
0

=
[ 6−4
−3 − 8
−6 − 6
15 − 0 ] [ ] =
− 11
2 − 12
15

[ ]
2 12
−
⇒ X=
1
5 [ − 11
2 − 12
15 ] =
−
5
11
5 3
5

Now, From eq. (i), 3Y =

[ ] 2
4
3
0
− 2X

[ ]
2 12
−
=
[ ] 2
4
3
0
−2
−
5
11
5 3
5

[ ][ ]
4 24
2 3 5
− 5
⇒ 3Y = − 22
4 0
− 5
6

[ ]
4 24
2− 5
3+ 5
3Y = 22
4+ 5 0−6

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 [ ]
6 39
1 5 5
⇒ Y= 3 42
5 −6

[ ]
2 13
5 5
⇒ Y= 14
5
−2

7. We are given that,

[ ] [ ][ ]
x+y
x−y
=
2
4
1
3 −2
1

We need to find the value of (x, y).

Multiply and find the product of the matrices on the right hand side of the equation,( which is possible because the columns of the
first matrix is equal to the rows of the second matrix.which will result as a matrix with the rows of the first matrix and the
columns of the second matrix ,Means the resultant matrix is of order 2x1.)

[ ][ ] [ ]
2 1 1 z 11
= (say)
4 3 −2 z 21

For z11: Dot multiply the 1st row of first matrix and 1st column of second matrix, then sum up.
(2 1)(1 -2) = 2 × 1 + 1 × -2
⇒ (2 1)(1 -2) = 2 – 2

⇒ (2 1)(1 -2) = 0

So,

[ ][ ] [ ]
2 1 1 0
=
4 3 −2 z 21

For z21: Dot multiply the 2nd row of first matrix and 1st column of second matrix, then sum up.
(4 3)(1 -2) = 4 × 1 + 3 × (-2)
⇒ (4 3)(1 -2) = 4 – 6

⇒ (4 3)(1 -2) = -2

So,

[ ][ ] [ ]
2
4
1
3 −2
1
=
−2
0

Equate the resulting matrix to the given matrix equation.

[ ] [ ][ ]
x+y
x−y
=
2
4
1
3 −2
1

[ ] [ ]
⇒
x+y
x−y
=
0
−2
We know by the property of matrices,

[ ][ ]
a 11 a 12 b 11 b 12
=
a 21 a 22 b 21 b 22

This implies,
a11 = b11, a12 = b12, a21 = b21 and a22 = b22
Therefore,
x+y=0
x – y = -2

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 Adding these two equations, we get
(x + y) + (x – y) = 0 + (-2)
⇒ x + y + x – y = -2

⇒ x + x + y – y = -2
⇒ 2x + 0 = -2
⇒ 2x = -2
2
⇒ x = -2
⇒ x = -1

Putting x = -1 in
x+y=0
⇒ (-1) + y = 0

⇒ -1 + y = 0
⇒ y = 1

So, putting values of x and y from above in (x, y), we get

(x, y) = (-1, 1)
Thus, (x, y) is (-1, 1).
1
8. i. Given: a ij = 2
|− 3i + j| ……….(i)
1 1 1
Putting i = 1, j = 1 in eq. (i) a 11 = 2
|− 3 + 1| = 2
|− 2| = 2
(2) = 1
1 1 1 1
Putting i = 1, j = 2 in eq. (i) a 12 = |− 3 + 2| = |− 1| = (1) =
2 2 2 2
1 1 1
Putting i = 1, j = 3 in eq. (i) a 13 = 2
|− 3 + 3| = 2
|0| = 2
(0) = 0
1 1 1 1
Putting i =1, j = 4 in eq. (i) a 14 = 2
|− 3 + 4| = 2
|1| = 2
(1) = 2
1 1 1 5
Putting i = 2, j = 1 in eq. (i) a 21 = 2
|− 6 + 1| = 2
|− 5| = 2
(5) = 2
1 1 1
Putting i = 2, j = 2 in eq. (i) a 22 = 2
|− 6 + 2| = 2
|− 4| = 2
(4) = 2
1 1 1 3
Putting i = 2, j = 3 in eq. (i) a 23 = 2
|− 6 + 3| = 2
|− 3| = 2
(3) = 2
1 1 1
Putting i = 2, j = 4 in eq. (i) a 24 = 2
|− 6 + 4| = 2
|− 2| = 2
(2) = 1
1 1 1
Putting i = 3, j = 1 in eq. (i) a 31 = 2
|− 9 + 1| = 2
|− 8| = 2
(8) = 4
1 1 1 7
Putting i = 3, j = 2 in eq. (i) a 32 = 2
|− 9 + 2| = 2
|− 7| = 2
(7) = 2
1 1 1
Putting i = 3, j = 3 in eq. (i) a 33 = |− 9 + 3| = |− 6| = (6) = 3
2 2 2
1 1 1 5
Putting i = 3, j = 4 in eq. (i) a 34 = 2
|− 9 + 4| = 2
|− 5| = 2
(5) = 2

[ ]
1

[ ]
1 0 1
a 11 a 12 a 13 a 14 2
2
5 3
∴ A 3 × 4 = [a 34] = a 21 a 22 a 23 a 24 = 2 1
2 2
a 31 a 32 a 33 a 34 5
7
4 2
3 2

ii. Given: a ij = 2i − j ……….(i)

Putting i = 1, j = 1 in eq. (i) a11 = 2 - 1 = 1
Putting i = 1, j = 2 in eq. (i) a12 = 2 - 2 = 0
Putting i = 1, j = 3 in eq. (i) a13 = 2 - 3 = -1
Putting i = 1, j = 4 in eq. (i) a14 = 2 - 4 = -2
Putting i = 2, j = 1 in eq. (i) a21 = 4 - 3 = 3
Putting i = 2, j = 2 in eq. (i) a22 = 4 - 2 = 2

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 Putting i = 2, j = 3 in eq. (i) a23 = 4 - 3 = 1
Putting i = 2, j = 4 in eq. (i) a24 = 4 - 4 = 0
Putting i = 3, j = 1 in eq. (i) a31 = 6 - 1 = 5
Putting i = 3, j = 2 in eq. (i) a32 = 6 - 2 = 4
Putting i = 3, j = 3 in eq. (i) a33 = 6 - 3 = 3
Putting i = 3, j = 4 in eq. (i) a34 = 6 - 4 = 2

[ ][ ]
a 11 a 12 a 13 a 14 1 0 −1 −2
∴ [ ]
A 3 × 4 = a 34 = a 21 a 22 a 23 a 24 3 2 1 0
a 31 a 32 a 33 a 34 5 4 3 2

9. We have A 2 =
[ ][ ] [ −1
2 3
2
2
−1
3
2
=
1
−4
12
1 ]
− 4A =
[ ] [ ]
−8
4
− 12
−8
and 7I =
7
0
0
7

[
Therefore, A 2 − 4A + 7I =
1−8+7
−4 + 4 + 0
12 − 12 + 0
1−8+7 ] [ ]
=
0
0
0
0
=0

⇒ A2 = 4A - 7I
Thus A3 = AA2 = A(4A - 7I) = 4(4A - 7I) - 7A
= 16A - 28I - 7A = 9A - 28I
And so A5 = A3A2
= (9A - 28I) (4A - 7I)
= 36A2- 63A - 112A + 196I
= 36 (4A - 7I) - 175A + 196I
= - 31A - 56I

= − 31
[ 2
−1
3
2 ] [ ]
− 56
1
0
0
1

=
[ − 118
31
− 93
− 118 ]
10. For the given system of equations, we have

| |
2 a 6
D= 1 2 b
1 1 3
⇒ D = 2(6 - b) -a(3 - b) + 6(1 - 2)
⇒ D = 12 - 2b - 3a + ab - 6 = 6 - 3a - 2b + ab = (b - 3)(a - 2)

| |
8 a 6
D1 = 5 2 b
4 1 3
⇒ D1 = 8(6 - b) - a(15 - 4b) + 6(5 - 8)
⇒ D1 = 48 − 8b − 15a + 4ab − 18 = 30 − 15a − 8b + 4ab = (a − 2)(4b − 15)

| |
2 8 6
D2 = 1 5 b
1 4 3

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 ⇒ D2 = 2(15 − 4b) − 8(3 − b) + 6(4 − 5) = 30 − 8b − 24 + 8b − 6 = 0

| |
2 a 8
and, D3 = 1 2 5 = 2(8 − 5) − a(4 − 5) + 8(1 − 2) = 6 + a − 8 = a − 2
1 1 4

i. For unique solution, we must have

D ≠ 0 ⇒ (a - 2) (b - 3) ≠ 0 ⇒ a ≠ 2, b ≠ 3
ii. For infinitely many solutions, we must have
D = D1 = D2 = D3 = 0
⇒ (a - 2)(b - 3) = 0, (a - 2)(4b - 15) = 0 and a - 2 = 0
⇒ a=2
Putting a = 2 in the given system of equations, we obtain
2x + 2y + 6z = 8
x + 2y + bz = 5
x + y + 3z = 4
This system is equivalent to the system
x + y + 3z = 4
x + 2y + bz = 5
Putting z = k, we get
x + y = 4 - 3k
x + 2y = 5 - bk
Solving these two equations, we get
x = 3 - 6k + bk, y = 1 - bk + 3k
Thus, the given system has infinitely many solutions given by
x = 3 - 6k + bk, y = 1 - bk + 3k, z = k, where k ∈ R.
Hence, the system has infinitely many solutions for a = 2
iii. For no solution, we must have
D = 0 and at least one of D1, D2 and D3 is non-zero.
Clearly, for b = 3, we have
D = 0 and D3 ≠ 0.
Hence, the system has no solution for b = 3.

[ ]
11. Given: A =
3
1
2
1

∴
[ ][ ]
A 2 = A. A =
3
1
2
1
3
1
2
1

=
[9+2
3+1] [ ] 6+2
2+1
=
11
4
8
3
∴ A2 + aA + bI 2 = 0

⇒
[ ] [ ] [ ]
11
4
8
3
+a
3
1
2
1
+b
1
0
0
1
=0

⇒
[ ][ ][ ] [ ]
11
4
8
3
+
3a
a
2a
a
+
b
0
0
b
=
0
0
0
0

⇒
[ ] [ ]
11 + 3a + b
4+a+0
8 + 2a + 0
3+a+b
=
0
0
0
0
∴ We have 11 + 3a + b = 0….(i)
8 + 2a + 0 = 0 ...(ii)
⇒ 2a = − 8

⇒ a = − 4

Here a = -4 satisfies 4 + a + 0 = 0 also, therefore a = -4

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 Putting a = -4 in eq. (i), 11 - 12 + b = 0 ⇒ b − 1 = 0 ⇒ b = 1
Here also b = 1 satisfies 3 + a + b = 0, therefore b = 1
Therefore, a = - 4 and b = 1

] [ ][
2 2 −4 1 −1 0
12. We have, A = −4 2 −4 , B = 2 3 4
2 −1 5 0 1 2

[ ][ ] [ ]
1 −1 0 2 2 −4 6 0 0
∴ BA = 2 3 4 −4 2 −4 = 0 6 0 = 6I
0 1 2 2 −1 5 0 0 6

[ ]
2 2 −4
A 1 1
∴ B-1 = = A= −4 2 −4
6 6 6
2 −1 5
Also, x - y = 3, 2x + 3y + 4z = 17 and y + 2z = 7

[ ][ ] [ ]
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
z 0 1 2 7

[ ][ ]
2 2 −4 3
1
= −4 2 −4 17 [using Eq. (i)]
6
2 −1 5 7

[ ] [][]
6 + 34 − 28 12 2
1 1
= − 12 + 34 − 28 = −6 = −1
6 6
6 − 17 + 35 24 4

∴ x = 2, y = -1 and z = 4
13. For the given system of equations, we have

| |
1 1 1
D= 1 2 3 = 1 × (10 - 9) - 1 × (5 - 3) +1 × (3 - 2) = 0,
1 3 5

| |
1 1 1
D1 = 4 2 3 = 1 × (10 - 9) - 1 × (20 - 21) + 1 × (12 - 14 ) = 0,
7 3 5

| |
1 1 1
D2 = 1 4 3 = 1 × (20 - 21) - 1 × (5 - 3) + 1 × (7 - 4) = 0,
1 7 5

| |
1 1 1
and, D3 = 1 2 4 = 1 × (14 -1 2 ) - 1 × ( 7 - 4 ) + 1 × ( 3 - 2 ) = 0.
1 3 7
Thus, we have D = D1 = D2 = D3 = 0.
So, either the system is consistent with infinitely many solutions or it is inconsistent. Consider the first two equations, these
equations can be written as

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 x+y=1-z
x + 2y = 4 - 3z
In order to solve these equations let us use Cramer's rule.

Here, D =
| |1
1
1
2
= 2 - 1 = 1, D1 =
| 1−z
4 − 3z
1
2 |
= 2 - 2z - 4 + 3z = z - 2

and, D2 =
| |1
1
1−z
4 − 3z
= 4 - 3z -1 + z = 3 - 2z.

D1 D2
∴ x= D
and y = D
⇒ x = z - 2, y = 3 - 2z.
Let z = k, where k is any real number. Then, we get
x = k - 2, y = 3 - 2k and z = k
Substituting these values in the LHS of third equation x + 3y + 5z = 7, we get
k - 2 + 3(3 - 2k) + 5k = k - 2 + 9 - 6k + 5k = 7 = RHS
Hence, these values satisfy the third equation.
Hence, x = k - 2, y = 3 - 2k , z = k is a solution of the given system of equation for every value of k.

| |
1 2 0
14. We have, A = − 2 −1 − 2 ...(i)
0 −1 1
∴|A| = 1(− 3) − 2(− 2) + 0 = 1 ≠ 0
Now, A 11 = − 3, A 12 = 2, A 13 = 2, A 21 = − 2, A 22 = 1, A 23 = 1, A 31 = − 4, A 32 = 2 and A 33 = 3

| || |
−3 2 2 T −3 −2 −4
∴ adj(A) = − 2 1 1 = 2 1 2
−4 2 3 2 1 3
adjA
∴ A −1 =
|A|

[ ]
−3 −2 −4
1
= 2 1 2
1
2 1 3

[ ]
−3 −2 −4
⇒ A −1 = 2 1 2 ...(i)
2 1 3
Also, we have the system of linear equations as
x − 2y = 10
2x − y − z = 8
and − 2y + z = 7
Now, the given system of equations can be rewritten in the form AX=B,

[ ] [] [ ]
1 2 0 x 10
where, A = 2 −1 −1 X = y and B = 8
0 −2 1 z 7
Since A is non singular , therefore given system of equations has a unique solution given by,
∴ X = A − 1B

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

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 [ ][]
− 30 + 60 + 14 0
= − 20 + 8 + 7 = −5
− 40 + 16 + 21 −3
∴ x = 0, y = − 5 and z = − 3

{
1 − sin 3x π
, if x <
3cos 2x 2
π π
15. Given, f(x) = p, if x = is continuous at x = 2
2
q(1 − sinx) π
, if x >
(π − 2x) 2 2

We shall use definition of continuity to find the values of constants a and b.

Since f(x) is continuous at x={tex}\frac{π}{2}{/tex},

Therefore,(LHL) x = π = (RHL) x = π = f
2 2 () π
2
........(i)

Now, LHL = lim f(x) = lim f

x→ 2
π− h→0
( )
π
2
−h

1 − sin 3
( ) π
2 −h

= lim
h→0
3cos 2 ( ) π
2 −h

1 − cos 3 h
= lim
h→0 3sin 2 h

( )
( 1 − cos h ) 1 2 + cos 2 h + 1 × cos h
= lim
h→0 3 ( 1 − cos h ) 2

( 1 − cos h ) ( 1 + cos h + cos h ) 2

= lim 3 ( 1 − cos h ) ( 1 + cos h )

h→0

( 1 + cos 2 h + cos h )
= lim
3 ( 1 + cos h )
h→0
2
1 + cos 0 + cos 0
= 3 ( 1 + cos 0 )
1+1+1 3 1
= 3(1+1)
= 3×2
= 2
.......(ii)

and RHL = lim f(x) = lim f

x→ 2
π h→0
( )
π
2
+h

[ ( )]
q 1 − sin
π
2 +h

= lim
h→0
[ ( )]
π−2
π
2 +h
2

q ( 1 − cos h ) q ( 1 − cos h )
= lim = lim
h→0 ( π − π − 2h ) 2 h→0 4h 2

(
q 2sin 2 2
h
)
= lim
h→0 4h 2 [ ∵ cosx = 1 − 2sin 2 2
x
]

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APS @ MUNDRA
 [ ] () h 2
sin 2
q
= 8
lim h
h→0 2

=
q
8
×1=
q
8 [ ∵ lim
x→0
sin x
x ]
= 1 ...........(iii)

On substituting the values from Eqs. (ii) and (iii) to Eq.(i), we get
1
2
=
q
8
=f ()
π
2

⇒
1
2
=
q
8
=p [ ()∵ f
π
2
= p( given ) ]
1 q 1
⇒
2
= 8
and 2
=p
1
∴ q = 4 and p = 2
16. When x < 1, we have f(x) = 1 - x.
We know that a polynomial function is everywhere continuous and differentiable. So, f (x) is
continuous and differentiable for all x < 1
Similarly, f (x) is continuous and differentiable for all x ∈ (1, 2) and x > 2.
Thus, the possible points where we have to check the continuity and differentiability of f (x) are
x = 1 and x = 2.
Continuity at x = 1:
We observe that:
lim f(x) = lim (1 − x) [ ∵ f(x) = 1 − x for x < 1]
x→1− x→1
=1-1=0
lim f(x) = lim (1 − x)(2 − x)=0 [ ∵ f(x) = (1 − x)(2 − x), for 1 ≤ x ≤ 2]
x→1+ x→1
and, f(1) = (1 - 1)(2 - 1) = 0
∴ lim f(x) = lim f(x) = f(1)
x→1− x→1+
So, f(x) is continuous at x = 1.
Continuity at x - 2: We observe that:
lim f(x) = lim (1 − x)(2 − x) [ ∵ f(x) = (1 − x)(2 − x) for 1 ≤ x ≤ 2]
x→2− x→2
and, lim f(x) = lim (3 − x) = 3 − 2 = 1 [f(x) = 3 − x for x > 2]
x→2+ x→2
∴ lim f(x) ≠ lim f(x)
x→2− x→2
So, f (x) is not continuous at x = 2.
Differentiability at x = 1:
we observe that:
f(x) −f(1)
(LHD at x = 1) = lim x−1
x→1−
(1−x) −0
⇒ (LHD at x = 1) = lim x−1
[Using definition of f(x)]
x→1
x−1
⇒ (LHD at x = 1) = − lim x−1
= −1
x→1
f(x) −f(1)
and (RHD at x = 1) = lim x−1
x→1+
(1−x) (2−x) −0
⇒ (RHD at x = 1) = lim [Using definition of f(x)[
x−1
x→1

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APS @ MUNDRA
 (x−1) (x−2)
⇒ (RHD at x = 1) = lim x−1
x→1
⇒ (RHD at x = 1) = lim x − 2 = 1 − 2 = − 1
x→1
Clearly, (LHD at x = 1) = (RHD at x = 1). So, f(x) is differentiable at x = 1.
Differentiability at x = 2:
Since, f (x) is not continuous at x = 2. So, it is not differentiable at x = 2.
17. It is given that f(x) = |x| - |x + 1|
The given function f is defined for real number and f can be written as the composition of two functions, as
f = goh, where g(x) = |x| and h(x) = |x + 1|
Then, f = g - h
First we have to prove that g(x) = |x| and h(x) = |x + 1| are continuous functions.
g(x) = |x| can be written as

g(x) =
{ − x, if x < 0
x, if x ≥ 0
Now, g is defined for all real number.
Let k be a real number.
Case I: If k < 0,
Then g(k) = -k
And lim g(x) = lim ( − x) = − k
x→k x→k
Thus, lim g(x) = g(k)
x→k
Therefore, g is continuous at all points x, i.e., x > 0|
Case II: If k > 0,
Then g(k) = k and
lim g(x) = lim x = k
x→k x→k
Thus lim g(x) = g(k)
x→k
Therefore, g is continuous at all points x, i.e., x < 0.
Case III: If k = 0,
Then, g(k) = g(0) = 0
lim g(x) = lim ( − x) = 0
x→0− x→0−
lim g(x) = lim (x) = 0
x→0+ x→0+
∴ lim g(x) = lim g(x) = g(0)
x→0− x→0+
Therefore, g is continuous at x = 0
From the above 3 cases, we get that g is continuous at all points.
g(x) = |x + 1| can be written as

g(x) =
{ − (x + 1), if x < − 1
x + 1, if x ⩾ −1
Now, h is defined for all real number.
Let k be a real number.
Case I: If k < -1,
Then h(k) = -(k + 1)
And lim h(x) = lim [ − (x + 1)] = − (k + 1)
x→k x→k
Thus, lim h(x) = h(k)
x→k
Therefore, h is continuous at all points x, i.e., x < -1
Case II: If k > -1,
Then h(k) = k + 1 and
lim h(x) = lim (x + 1) = k + 1
x→k x→k

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APS @ MUNDRA
 Thus, lim h(x) = h(k)
x→k
Therefore, h is continuous at all points x, i.e., x > -1.
Case III: If k = -1,
Then, h(k) = h(-1) = -1 + 1 = 0
lim h(x) = lim [-(x + 1)] = -(-1 + 1) = 0
x→1− x→1−
lim h(x) = lim (x + 1)= -(-1 + 1) = 0
x→1+ x→1+
∴ lim h(x) = lim h(x) = h( − 1)
x→1− x→1+
Therefore, g is continuous at x = -1
From the above 3 cases, we get that h is continuous at all points.
Hence, g and h are continuous function.
Therefore, f = g – h is also a continuous function.
18. The right hand limit of f at x = 0 is given by
1
ex −1
lim f(x) = lim 1
x→0+ x→0+ ex +1
1
1− 1
−1
ex 1−e x 1−0
= lim 1 = lim −1 = 1+0
=1
x→0+ 1+ 1 x→0+ 1+e x
ex

Also,f(0)=0
Thus lim f(x) ≠ f(0) . Hence f is discontinuous at x = 0.
x→0+
19. At x = -1
f(-1) = -2
lim f(x) = lim ( − 2)
x→ −1− x→ −1−
= lim −2=-2
x→ −1−
lim f(x) = lim 2x
x→ −1+ x→ −1+
lim 2( − 1 + h) = -2
h→0
Hence continuous at x = -1
x=1
f(1) = 2 × 1 = 2
lim f(x) = lim 2x
x→1− x→1−
lim 2(1 − h) = 2
h→0
lim f(x) = lim 2
x→1+ x→1+
lim 2 = 2
x→0
Hence it is a Continuous function.
20. According to the question, we have to show that the function defined is continuous at x = 2 but not differentiable at x = 2.

{
3x − 2, 0<x≤1
f(x) = 2x 2 − x, 1<x≤2
5x − 4, x>2

let us check its Continuity at x = 2

LHL = lim x → 2 − f(x) = lim x → 2 − (2x2 - x)

⇒ LHL = lim h → 0 [2(2 - h)2 - (2 - h)]

= lim h → 0[2(4 + h2 - 4h) - (2 - h)]

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APS @ MUNDRA
 = lim h → 0( 8 + 2h2 - 8h - 2 + h)
⇒ LHL = 8 - 2 = 6

and RHL = lim x → 2 + f(x) = lim x → 2 + (5x − 4)

⇒ RHL = lim h → 0[5(2 + h) - 4]
= lim h → 0(10 + 5h - 4) = = lim h → 0(5h + 6)
⇒ RHL = 6
Also, f(2) = 2(2)2 - 2 = 8 - 2 = 6
Since, LHL = RHL = f(2)
Therefore function f(x) is continuous at x = 2.
We will now check the differentiabililty of the given function at x = 2.
Differentiability at x = 2
f(2−h) −f(2)
LHD = lim h → 0
−h

[2(2−h) − (2−h) ] − [8−2]

2

⇒ LHD = lim h → 0 −h
2 ( 4 + h 2 − 4h ) − (2−h) −6
= lim h → 0 −h
8 + 2h 2 − 8h − 2 + h − 6
= lim h → 0 h
h ( 2h − 7 )
= lim h → 0 −h
= lim h → 0-(2h - 7)
⇒ LHD = 7
f(2+h) −f(2)
and RHD = lim h → 0 h
[5(2+h) −4] − [8−2]
= lim h → 0
h
( 6 + 5h ) − ( 6 ) 5h
= lim h → 0 h
= lim h → 0 h
⇒ RHD = 5

Since, LHD ≠ RHD

Therefore, function f(x) is not differentiable at x = 2.
Therefore, f(x) is continuous at x = 2 but not differentiable at x=2.
21. LHL = lim f(x)
x→0−
= lim f(0 − h)
h→0

= lim ( − h) msin −
h→0
( ) 1
h

= lim − ( − h) msin
h→0
() 1
h

= 0 × k [when − 1 ≤ k ≤ 1]
=0
RHL = lim f(x)
x→0+
= lim f(0 + h)
h→0

= lim (0 + h) msin
h→0
( ) 0+h
1

= lim (h) msin

h→0
()1
h

= 0 x k [when − 1 ≤ k ′ ≤ 1]
=0
LHL = RHL = f(0)
Since, f(x) is continuous at x = 0

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APS @ MUNDRA
 For Differentiability at x = 0
f(x) −f(0)
(LHD at x = 0) = lim x−0
x→0−
f(0−h) −f(0)
= lim
(0−h) −0
h→0

( − h ) msin ( )
−h
1

= lim h
h→0−

= lim
h→0−
− ( − h) m − 1sin ()
1
h

= Not Defined [since 0<m<1]

f(x) −f(0)
(RHD at x = 0) = lim x−0
x→0+
f(0+h) −f(0)
= lim (0+h) −0
h→0+

h msin ()1
h

= lim 0+h−0
h→0+

= lim (h) m − 1sin

h→0+
()1
h

=0
Since , (LHD at x = 0) ≠ (RHD at x = 0)
Hence, f(x) is continuous at x = 0 but not differentiable.
22. At x = - 3
f (-3) = |-3| + 3 = 3 + 3 = 6
lim f(x) = lim ( | x | + 3)
x→ −3− x→ −3−
= lim (|-3 - h| + 3) = |-3 - 0| + 3 = |3| + 3 = 3 + 3 = 6
h→0
lim f(x) = lim ( − 2x)
x→ −3+ x→ −3+
= lim − 2(− 3 + h) = 6
h→0
∴ lim f(x) = lim f(x) = f( − 3)
x→ −3− x→ −3+
Hence, f(x) is Continuous at x = -3.
At x = 3
f(3) = 6 × 3 + 2 = 20
lim f(x) = lim (− 2x)
x→3− x→3−
= lim − 2(3 − h)
h→0
= -6
lim f(x) = lim (6x + 2)
x→3+ x→3+
= lim [6(3 + h) + 2]
h→0
= 20
lim f(x) ≠ lim f(x)
x→3− x→3+
Hence f(x) is discontinuous at x = 3.
23. Let y = √cosx + √cosx + √cosx. . . . .
y= √cosx + y
Squaring both sides,

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APS @ MUNDRA
 y 2 = cosx + y
Differentiate
dy dy
2y dx = − sinx + dx
dy
dx
(2y − 1) = − sinx
dy
(1 − 2y) dx = + sinx
dy
(1 − 2y) dx = sinx
24. x√1 + y = − y√1 + x
Squaring both sides
x 2(1 + y) = y 2(1 + x)
x 2 + x 2y = y 2 + xy 2
x 2 − y 2 + x 2y − xy 2 = 0
(x − y)(x + y) + xy(x − y) = 0
(x − y)[x + y + xy] = 0
x + y + xy = 0
y(1 + x) = − x
−x
y=
1+x

dy
dx
= −
[ (1+x) (1) − (x) (1)
(1+x) 2 ]
= −

dy
[ ] 1+x−x
(1+x) 2

−1
=
dx (1+x) 2

√
25. y x 2 + 1 − log (√ x2 + 1 − x = 0 )
differentiating both sides w.r.t x

y.
2 √
1

x2 + 1
(2x) +
dy
√x 2 + 1. dx − √x + 1 − x 2
1

[ 1 ( 2x )

2 x2 + 1
√
−1
] =0

[ ] √x 2 + 1
x−
xy dy 1

√x 2 + 1
+ √ x2 + 1. dx
−
√x 2 + 1 − x √x 2 + 1
=0

xy + ( x 2 + 1 ) dy
− (√ x2 + 1 − x )
=
dx
√x 2 + 1 (√ x2 + 1 − x )√ x2 + 1

dy
xy + (x 2 + 1) dx = − 1
dy
(x 2 + 1) dx + xy + 1 = 0
26. We have,

√1 − x 2 + √1 − y 2 = a(x − y)
On putting x = sinα and y = sinβ we get

√1 − sin2α + √1 − sin2β = a(sinα − sinβ)

⇒ cosα + cosβ = a(sinα − sinβ)

⇒ 2cos
α+β

α−β
2
. cos
α−β
2
α−β
(
= a 2cos
α+β
2
. sin
α−β
2 )
⇒ cos 2
= asin 2

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APS @ MUNDRA
 α−β
⇒ cot 2
=a
α−β
⇒ = cot − 1a
2
⇒ α − β = 2cot − 1a
⇒ sin
− 1x − sin − 1y = 2cot − 1a [ ∵ x = sinα and y = sinβ]

On differentiating both sides w.r.t. x, we get

1 1 dy
− dx
=0
√ 1 − x2 √ 1 − y2

dy √1 − y 2
∴ = .
dx
√1 − x 2
Hence proved.
27. Let y = √cosx + √cosx + √cosx + . . . . .
y= √cosx + y
Squaring both sides
y 2 = cosx + y
Differentiating both sides
dy dy
2y dx = − sinx + dx
dy
dx
(2y − 1) = − sinx
dy
(1 − 2y) dx = sinx

28. Let u = sin − 1(4x 1 − 4x 2) √

put 2x = cosθ
⇒ u = sin − 1(2 × cosθ 1 − cos 2θ) √
−1
⇒ u = sin (2cosθsinθ)
⇒ u = sin-1(sin 2θ) ....(i)
Let v = √1 − 4x 2...(ii)
Here,

x ∈
( −
1
2√ 2 2√ 2
,
1
)
⇒ 2x ∈
( −
√2
1
,
1

√2 )
⇒ θ ∈
( ) π 3π
4
, 4

So, from equation (i),

[
u = π − 2θ…… since , sin − 1(sinθ) = π − θ, if θ ∈
( )]
π
2
,π

⇒ u = π − 2cos − 1(2x)[ since , 2x = cosθ]

Differentiating it with respect to x,

du
dx
=0−2

du
( √1 −
2
−1

( 2x ) 2 ) d
dx
(2x)

⇒
dx
= (2)
√1 − 4x 2
du 4
⇒
dx
= ...(iii)
√1 − 4x 2
from equation (ii)

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APS @ MUNDRA
 dv − 4x
dx
=
√1 − 4x 2

but, x

dv
∈
(
−4( −x)
1
− 2, −
1
2√ 2 )
=
dx
√1 − 4 ( − x ) 2
dv 4x
⇒
dx
= ...(iv)
√1 − 4x 2
Differentiating equation (ii) with respect to x,
dv 1 d
dx
=
2 √ 1 − 4x 2 dx (1 − 4x ) 2

dv 1
⇒
dx
= ( − 8x)
√
2 1 − 4x 2
dv − 4x
⇒ = ...(v)
dx
√1 − 4x 2
Dividing equation (iii) by (v)
du
dx 4 √1 − 4x 2
dv
= ×
− 4x
dx √1 − 4x 2
du 1
∴
dv
= − x
29. Let u = sin 1(2x
−
√1 − x 2 )
Put x = sin θ
⇒ u = sin − 1(2sinθ 1 − sin 2θ) √
⇒ u = sin − 1(2sinθcosθ)
⇒ u = sin − 1(sin2θ) ....(i)
and

Let, v = sec − 1
( ) 1

√1 − x 2

⇒ v = sec − 1
( ) 1

√1 − sin 2 θ

⇒ v = sec − 1
( ) 1
cos θ

⇒ v= sec − 1(secθ)

⇒
v = cos − 1

v = cos − 1(cosθ) ...(ii)

() 1
1
cos θ
[ since , sec − 1x = cos − 1 ( )]
1
x

Here,

x ∈
( ) 1

√2
,1

⇒
( )
sinθ ∈
1

√2
,1

⇒ θ ∈
( ) π π
,
4 2

19 / 54
APS @ MUNDRA
 So, from equation (i),

u = 2θ [ since , sin − 1(sinθ) = θ, if θ ∈

( − 2,
π π
2 )]
Let u = 2 sin-1x ...[Since, x = sin θ]
Differentiating it with respect to x,

du
dx
=2

du
( ) 1

√1 − x 2
2
⇒
dx
= ....(iii)
√1 − x 2
And, from equation (ii),

[
v = θ since, cos − 1(cosθ) = θ, if θ ∈ [0, π] ]
⇒ v= sin − 1x[
since , x = sinθ]
Differentiating it with respect to x,
dv 1
dx
= ...(iv)
√1 − x 2
dividing equation (iii) by (iv),
du 2 √1 − x 2
dx
= × 1
√1 − x 2
du
∴
dv
= 2.

30. Let u = tan − 1

( ) x

√1 − x 2
Put x = sinθ
⇒ θ = sin-1x

⇒ u = tan − 1
( sin θ

√1 − sin 2 θ )
⇒ u = tan − 1 ( ) sin θ
cos θ

⇒ u = tan-1(tan θ) ...(i)
And
√
Let v = sin − 1(2x 1 − x 2)

v = sin − 1(2sinθ√1 − sin 2θ)

v = sin-1(2 sin θ cos θ)

v = sin-1(sin 2θ) ...(ii)
Here,
1 1
− <x<
√2 √2
1 1
⇒ − < sinθ <
√2 √2
π π
⇒ − 4
<θ< 4
So, from equation (i)

[
u = θ since, tan − 1(tanθ) = θ, if θ ∈
( π π
− 2, 2 )]
⇒ u = sin-1x
Differentiating it with respect to x,

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APS @ MUNDRA
 du 1
dx
= ....(iii)
√1 − x 2
from equation (ii),

[
v = 2θ since, sin − 1(sinθ) = θ, if θ ∈
[ π π
− ,
2 2 ]]
⇒ v = 2sin-1x
Differentiating it with respect to x,
dv 2
dx
= ...(iv)
√1 − x 2
Dividing equation (iii) by (iv),

( )( )
du
dx 1 √1 − x 2
dv = 2
dx √1 − x 2
du 1
∴
dv
= 2
.

31. Let u = sin − 1

( ) 2x
1 + x2

Put x = tan θ

⇒ u = sin − 1
( ) 2tan θ
1 + tan 2 θ

⇒ u = sin-1(sin 2θ) ...(i)

Let v = cos − 1
( )
1 − x2
1 + x2

⇒ v = cos − 1
( )
1 − tan 2 θ
1 + tan 2 θ

⇒ v = cos-1(cos 2θ) ...(ii)

Here, 0 < x < 1
⇒ 0 < tanθ < 1

π
⇒ 0<θ< 4
So, from equation (i),

u = 2θ [ since , sin − 1(sinθ) = θ, if θ ∈

( − 2,
π π
2 )]
⇒ u = 2tan-1x ...[Since, x = tan θ]
Differentiating it with respect to x,
du 2
= ...(iii)
dx 1 + x2
from equation (ii),

v = 2θ [ since , cos − 1(cosθ) = θ, ifθ ∈ [0, π] ]

⇒ v = 2tan-1x ...[Since, x = tan θ]
Differentiating it with respect to x,
dv 2
= ...(iv)
dx 1 + x2
Dividing equation (iii) by (iv),
du
dx 2 1 + x2
dv = × =1
1 + x2 2
dx

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APS @ MUNDRA
32. Let u = sin − 1(4x 1 − 4x 2) √
Put 2x = cosθ
⇒ u = sin − 1(2 × cosθ 1 − cos 2θ) √
⇒ u= sin − 1(2cosθsinθ)
⇒ u = sin − 1(sin2θ) ...(i)
Let v = √1 − 4x 2 ....(ii)
Here,

x ∈
( − 2, −
1 1
2√ 2 )
⇒ 2x ∈
( − 1, −
√2
1
)
⇒ θ ∈
( ) 3π
4
,π

So, from equation (i),

[
u = π − 2θ since , sin − 1(sinθ) = π − θ, if θ ∈
( )]
π 3π
2
, 2

⇒ u = π − 2cos − 1(2x) [ since , 2x = cosθ]

Differentiate it with respect to x,

du
dx
=0−2

du
( √1 −
2
−1

( 2x ) 2 ) d
dx
(2x)

⇒
dx
= (2)
√1 − 4x 2
du 4
⇒
dx
= ....(iii)
√1 − 4x 2
from equation (ii),
dv − 4x
dx
=
√1 − 4x 2

but x

dv
∈
( − , −
1
2

−4( −x)
1
2√ 2 )
∴
dx
=
√1 − 4 ( − x ) 2
dv 4x
⇒
dx
= ....(iv)
√1 − 4x 2
Dividing equation (iii) by (iv)
du
dx 4 √1 − 4x 2
dv = × 4x
dx √1 − 4x 2
du 1
∴ = .
dv x
33. We have,
ex ex xe
y = ex + xe + ex
⇒ y=u+v+w
dy du dv dw
⇒ = + + ...(i)
dx dx dx dx
x x xe
where u = e , v = x xe ee and w = e x
ex
Now, u = e x ...(ii)
Taking log on both sides,

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APS @ MUNDRA
 ex
logu = loge x
x
⇒ logu = x e loge
x
⇒ logu = x e ....(iii)
taking log on both sides,
x
loglogu = logx e
⇒ log log u = ex log x
Differentiating with respect to x,

( )
1 d d d
⇒
log u dx
(logu) = e x dx (logx) + logx dx e x
1 1 du ex
⇒ = + e xlogx
log u u dx x

⇒
du
dx
= ulogu
[ ex
x
+ e xlogx
]
⇒
du
dx
ex
= ex × xe
x

[ ex
x
+ e xlogx …(A)
]
[Using equation (ii) and (iii)]
ex
Now, v = x e .... (iv)
Taking log on both sides,
ex
logv = logx e
x
⇒ logv = e e logx

⇒
1 dv
v dx
= ee
x d
dx
(logx) + logx dx e e
d
( ) x

⇒
1 dv
v dx
= ee () x 1
x
+ logxe e
x d
dx (e )x

⇒
dv
dx
= v ee[ () x 1
x
+ logxe e e x
x
]
⇒
dv
dx
ex
= xe × ee
[ x 1
x
+ e xlogx …(B)
]
{Using equation (iv)}
xe
Now, w = e x ...(v)
Taking log on both sides,
xe
logw = loge x
e
⇒ logw = x x loge
e
⇒ logw = x x ...(vi)
Taking log on both sides,
e
log log w = logx x
⇒ log log w = xe log x
1 d d d
⇒
log w dx
(logw) = x e dx (logx) + logx dx x e ( )
⇒
1
log w () 1
w
dw
dx
= xe ()
1
x
+ logxex e − 1

[ ]
dw
⇒ = wlogw x e − 1 + elogxx e − 1
dx
dw e
⇒
xx xe
= e x x e − 1(1 + elogx) ...(C)
dx
[using equation (v), vi)]

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APS @ MUNDRA
 Using equation (A), (B) and (C) in equation (i), we get

dy
dx
= ex xe
ex x

[ ex
x ]
+ e xlogx + x e

d 3y
ex

34. As we have to prove: = 2 cosx cosec3x

dx 2
We notice a third-order derivative in the expression to be proved so first take the step to find the third-order derivative.
Given, y = log (sin x)
d 2y
Let's find
dx 3

As
d 3y
dx 3
= dx
d
( ) ( ( ))
d 2y
dx 2
=
d
dx
d
dx
dy
dx

So lets first find dy/d $x$ and differentiate it again.

dy d
∴
dx
= dx
(log(sinx))
differentiating sin (logx) using the chain rule,
let, t = sinx and y = log t
dy dy dt
∵
dx
= dt
× dx
[using chain rule]
dy 1
dx
= cosx × t

[ ∵
d
dx
logx =
1 d
& (sinx) = cosx
x dx ]
dy cos x
dx
= sin x
= cotx
Differentiating again with respect to x:
d
dx () dy
dx
=
dx
d
(cotx)

d 2y
= − cosec 2x
dx 2

[2
∵
d
dx
cotx = − cosec 2x
]
d y
= − cosec 2x
dx 2
Differentiating again with respect to x:

d
dx ( ) d 2y
dx 2
=
d
dx (− cosec 2x

d
)
using the chain rule and dx
cosec x = -cosec x cot x
d 3y
= -2 cosec x (-cosec x cot x)
dx 3
cos x
= 2 cosec2x cot x = 2 cosec2x sin x
[ ∵ cot x = cos x/sin x]
d 3y
= 2 cosx cosec3x
dx 2
Hence proved.
35. Let u = y x, v = x y, w = x x
u + v + w = ab
du dw dv
Therefore + + = 0 ....(1)
dx dx dx
u = yx
Taking log both side
logu = logy x
logu = x. logy

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APS @ MUNDRA
 Differentiate both side w.r.t. to x
1 du 1 dy
u
. dx
= x. y . dx
+ logy.1

du
dx
=u
[ x dy
.
y dx
+ logy
]
du
dx
= yx
[ x dy
y dx
.
]
+ logy .... (2)

v = xy
Taking log both side
logv = logx y
logv = y. logx
1 dv 1 dy
.
v dx
= y. x
+ logx. dx

dv
dx
=v [ y
x
+ logx.
dy
dx ]
dv
dx
= xy [ y
x
+ logx.
dy
dx ] .... (3)

w = xx
Taking log both side
logw = logx x
logw = xlogx
1 dw 1
w
.dx
= x. x
+ logx.1
1 dw
.
w dx
= 1 + logx
dw
= w(1 + logx)
dx
dw
dx
= x x(1 + logx).... (4)
dy − x x ( 1 + log x ) − y . x y − 1 − y xlog y
= (by putting 2,3 and 4 in 1)
dx x . y x − 1 + x ylog x .
Section B

36. i. If a set P has m elements and set Q has n elements then the number of functions possible from P to Q is nm.
So, number of functions from A to B = 62
ii. As the total number of Relations that can be defined from a set P to Q is the number of possible subsets of P × Q.
If n(P) = m and n(Q) = n then n(P × Q) = mn and the number of subsets of P × Q = 2mn.
So number of relations possible from A to B = 22 × 6 = 212
If n(A) = P and n(B) = q then n(A × B) = pq and the number of subsets of A × B = 2pq.
iii. R = {(1, 2), (2, 2),(1, 3), (3, 4), (3, 1), (4, 3), (5, 5),}
R is not reflexive. (3, 3) ∉ R
R is not symmetric.
Because for (1, 2) ∈ R there
(2, 1) ∉ R.
R is not transitive.
Because for all element of B there does not exist,
(a, b) (b, c) ∈ R and (a, c) ∈ R.
OR
R is reflexive, since every element of B i.e,
B = {1, 2, 3, 4, 5, 6} is divisible by itself.
i.e, (1, 1), (2, 2), (3, 3), (4, 4,), (5, 5), (6, 6) ∈ R
Further, (1, 2) ∈ R
But (2, 1) ∉ R
Moreover,

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APS @ MUNDRA
 (1, 2), (2, 4) ∈ R
⇒ (1, 4) ∈ R

⇒ R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

37. i. Clearly, (1, 1), (2, 2), (3, 3), ∈ R. So, R is reflexive on A.
Since, (1, 2) ∈ R but (2, 1) ∉ R. So, R is not symmetric on A.
Since, (2, 3), ∈ R and (3, 1) ∈ R but (2, 1) ∉ R. So, R is not transitive on A.
So, R is only reflexive.
ii. Since, (1, 1), (2, 2) and (3, 3) are not in R. So, R is not reflexive on A.
Now, (1, 2) ∈ R ⇒ (2, 1) ∈ R
and (1, 3) ∈ R ⇒ (3, 1) ∈ R
So, R is symmetric
Clearly, (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∈ R.
So, R is not transitive on A.
iii. We have, R = {(x, y) : y = x + 5 and x < 4}, where x ,y ∈ N.
∴ R = {(1, 6), (2, 7), (3, 8)}

Clearly, (1, 1), (2, 2), etc. are not in R. So, R is not reflexive.
Since, (1, 6) ∈ R but (6, 1) ∉ R. So, R is not symmetric.
Since, (1, 6) ∈ R and there is no ordered pair in R which has 6 as the first element. The same is the case for (2, 7) and (3, 8).
So, R is transitive.
OR
We have, R = {(x, y) : 3x - y = 0}, where x, y ∈ A = {1, 2, ..... ,14}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Clearly, (1,1) ∉ R. So, R is not reflexive on A.

Since, (1, 3) ∈ R but (3, 1) ∉ R. So, R is not symmetric on A.
Since, (1, 3) ∈ R and (3, 9) ∈ R but (1, 9) ∉ R. So, R is not transitive on A.
38. i. B = {1, 2, 3, 4, 5, 6)
R = {(x, y) : y is divisible by x}
Now, since x is divisible by x
⇒ (x, x) ∈ R
So R is reflexive
Also, here (2, 4) ∈ R, as 4 is divisible by 2
But (4, 2) ∉ R as 2 is not divisible by 4
So R is not symmetric
Now, if (x, y) ∈ R & (y, z) ∈ R
Then (x,z) ∈ R
So R is transitive
Hence R is reflexive and transitive
ii. As A has 2 elements and B has 6 elements
So, number of functions from A to B = 62.
iii. ∵ (1, 1) ∉ R
So R is not reflexive
Now, here (1, 2) ∈ but (2, 1) ∉ R
So R is not symmetric
Also (1, 3) ∈ R and (3, 4) ∈ R
But (1, 4) ∉ R
So R is not transitive.
OR
Number of relations
×
= 2number of element in A number of element in B
×
= 22 6
= 212

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APS @ MUNDRA
39. i. For f(x) to be defined x - 2 ≠ 0 i.e., x ≠ 2
∴ Domain of f = R - {2}

x−1
ii. Let y = f(x), then y = x−2
2y − 1
⇒ xy - 2y = x - 1 ⇒ xy - x = 2y - 1 ⇒ x= y−1
Since, x ∈ R - {1}, therefore y ≠ 1
Hence, range of f(x)= R - {1}
iii. We have, g(x) = 2f(x) - 1

=2
( )
x−1
x−2
−1=
2x − 2 − x + 2
x−2
= x−2
x

OR
f(x1) = f{x2) ⇒ x1 = x2
40. i. Given that, A = {1, 2, 3}
Now, number of equivalence relations are as follows:
R1 = ((1, 1),(2, 2),(2, 3))
R2 = {(1, 1),(2, 2),(3, 3),(1, 2),(2, 1))
R3 = ((1, 1),(2, 2),(3, 3),(1, 3),(3, 1))
R4 = ((1, 1),(2, 2),(3, 3),(2, 3),(3, 2))

R5 = {(1, 2, 3) ⇒ A × A = A2}
∴ Maximum number of equivalence relations on the set A = {1, 2, 3} = 5

ii. aRb ⇒ a is brother of b.

This does not mean b is also a brother of a as b can be a sister of a.
Hence, R is not symmetric.
aRb ⇒ a is brother of b
and b Rc ⇒ b is a brother of c.
So, a is brother of c.
Hence, R is transitive.
iii. Given relation R1= {(x, y) | x2 = y2}

Reflexive: For all x ∈ R, x2 = x2, so, (x, x) ∈ R1

Hence, R1 is reflexive relation.
Symmetric: For all x, y ∈ R
Ifx2 = y2 then y2 n
= x2
Hence, R1 is symmetric relation.

Transitive: For all x, y ∈ R, x2 = y2 and for all y, z ∈ R

y2 = z2
∴ x2 = y2 = z2,or all x, y, z ∈ R
Hence, R1 is transitive.
Thus, R1 is an equivalence relation.
OR
Given, x greater than y, ∀ x, y ∈ N
⇒ x > y ∀ x, y ∈ N
Reflexive: Now, for (x, x) ∈ R
Therefore, x > x is not true for any x ∈ N
Thus, R is not reflexive.
Symmetric: Now, let (x, y) ∈ R, then x > y
If x > y, then y ≯ x for any x, y ∈ N
Thus, R is not symmetric.
Transitive: Now, let (x, y) ∈ R and (y, z) ∈ R
⇒ x > y and y > z

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APS @ MUNDRA
 Therefore, x > z ⇒ (x, z) ∈ R for all x, y, z ∈ N
Thus, R is transitive.
41. i. In △ABC,
BC
tan A = AB
10
tan A = 20
1
tan A =
2
1
∠ A = tan-1 2
1
∠ CAB = tan-1( 2 )
Now,
∠DAB = 2 × ∠CAB

1
= 2 × tan-1 2

Using 2tan-1x = tan − 1

( ) 2x
1 − x2

= tan − 1
( )1−
2× 2

() 1
2
1

= tan − 1
() 1
3
4
= tan − 1 ()4
3

∠ DAB = tan − 1
() 4
3

ii. ∠EAB = 3 × ∠CAB

= 3 × tan − 1
() 1
2

Using 3tan-1x = tan − 1

( ) 3x − x 3
1 − 3x 2

( ) ()
1 1 3
3× 2 − 2

∠ EAB = tan − 1
1−3 ()
1
2
2

()
11

∠ EAB = tan − 1
8
1
4
= tan − 1 ()
11
2

11
∠ EAB = tan-1 ( 2
)
iii. In △A'BC
BC
tan A' =
A ′B

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APS @ MUNDRA
 10
tan A' = 25
2
tan A' =
5
2
∠A
′ = tan-1 5
2
∠ CA ′ B = tan-1 5
Now, the difference

∠ CAB − ∠CA ′ B = tan − 1 () 1

2
− tan − 1 ()
2
5
x−y
Using tan-1x - tan-1y = tan-1 1 + xy

[ ]
1 2
2 −5
−1
= tan 1 2
1+ 2 × 5

[] ()
1
10 1
= tan − 1 6
= tan − 1
12
5

1
∠ CAB − ∠CA ′ B = tan-1 ( 12 )
OR
Domain of tan-1x is R i.e. all real numbers.

Range - R
( − 2,
π π
2 )
42. i. in △BDA
AB2 = AD2 + BD2
= (30√3) 2 + (30)2

= (60)2
AB = 60 m
BD
Now, sinα = AB
30 1
sinα = 60
= 2

α = sin − 1
() 1
2

α = sin-1
() 1
2

Again, In △BDA
AD 30√3 √3
cosα = AB
= 60
= 2

α = cos − 1
() √3
2

ii. DC = AC - AD
DC = 40√3 − 30√3
DC = 10√3 m
In △BDC,

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APS @ MUNDRA
 BD 30
tanβ = DC
=
10√3
= √3
β = tan-1(√3)
1
iii. ∵ sinα = 2
sinα = sin 30 ∘
α = 30 ∘
tanβ = √3
β = 60 ∘
Now, In △ABC
∠ ABC +∠BAC + ∠ACB = 180o
∠ ABC + 30o + 60o = 180o
∠ ABC = 90o
π
∠ ABC = 2
OR
Domain - [-1, 1] and range is [0, π]

43. i. Let f(x) = sin-1√x − 1

⇒ 0 ≤ x-1 ≤ 1
⇒ 1 ≤ x ≤ 2

∴ x ∈ [1, 2] [ ∵
√x − 1 ≥ 0 and − 1 ≤ √x − 1 ≤ 1]
ii. we know that, Domain of sin-1 is [-1, 1]
∴ Domain of sin-1[x] is {x : -1 ≤ [x] ≤ 1}

{
− 1∀ −1 ≤ x < 0
But {x} = 0∀ 0≤x<1
1∀ 1≤x<2

∴ Domain of sin-1[x] is [-1, 2)

iii. sin
[ π
3
− sin − 1 −
( )] 1
2

= sin
[ π
3
+ sin − 1 −
( )] 1
2
[ ∵ sin-1(-θ) = -sinθ]

= sin [ π
3
+ sin − 1 sin 6 ( )] π

= sin [ ] π
3
+
π
6

= sin
[ ] 2π + π
6

= sin
() 3π
6

= sin
() π
2
=1

OR
Let,
2π 2π
y = cos-1(cos 3
) + sin-1(sin 3
)
2π π
= 3
+ sin-1(sin(π − 3 ))
2π π
= 3
+ sin-1(sin 3 )

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APS @ MUNDRA
 2π π
= 3
+ 3
=π
2π 2π
∴ Principal value of cos-1(cos ) + sin-1(sin ) = π.
3 3

44. i. Let cosec-1 (-1) = y

Then, cosec y = -1

Since the range of principal value branch of y = cosec-1x is

π
[ − 2,
π π
2 ] - {0} and cosec
( )
−
π
2
= -1,

we have y = − 2 .

Since, − 2
π
∈
[ − 2,
π π
2 ] - {0}.

π
Thus, the principal value of cosec-1(-1) is − 2 .

ii. Let sec-1(-2) = y

Then, sec y = -2
1
⇒ cos y = − 2
1
Now, cos y = −
2
π
= -cos 3

= cos π − ( ) π
3
2π
= cos 3
2π
Therefore, y = 3
⋅

Since
2π
3
∈ [0, π] -
{} π
2
, the principal value of sec-1(-2) is
2π
3
[ ∵ The range of the principal value branch of sec-1 x is [0,π]

-
{}
π
2
]

iii. We have,

tan − 1
( ) x+1
x−1
+ tan − 1
( ) x−1
x
= tan-1(-7)

[ ]
x+1 x−1

( )
x−1 + x A+B
⇒ tan − 1 x+1 x−1 = tan-1(-7) [ ∵ tan-1A + tan-1B = 1 − AB
]
1− x−1 × x

x ( x + 1 ) + ( x − 1 )2
⇒ = -7
(
x ( x − 1 ) − x2 − 1 )
⇒ 2x2 - 8x + 8 = 0
⇒ x2 - 4x + 4 = 0
⇒ (x- 2)2 = 0
Hence, x = 2.
OR
Let x = sinθ
then, θ = sin-1x

L.H.S. = sin-1 2x 1 − x 2 ( √ )
⇒
(
sin-1 2sinθ 1 − sin 2θ √ )
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APS @ MUNDRA
 ⇒ sin-1(2 sinθ ⋅ cos = θ)
⇒ sin-1(sin 2θ) = 2θ
⇒ = 2 sin-1x = R.H.S.
45. i. Total number of toilets that can be expected in each village is given by the following matrix.

[ ][ ]
X 400 300 100 2 / 100
Y 300 250 75 4 / 100
Z 500 400 150 20 / 100

[ ] []
X 8 + 12 + 20 X 40
Y 6 + 10 + 15 = Y 31
Z 10 + 16 + 30 Z 56
X = 40, Y = 31, Z = 56

[ ][ ]
X 400 300 100 2 / 100
ii. Y 300 250 75 4 / 100
Z 500 400 150 20 / 100

[ ] []
X 8 + 12 + 20 X 40
Y 6 + 10 + 15 = Y 31
Z 10 + 16 + 30 Z 56
Total attempt made in all the villages = 2475
Total number of toilets that can be expected after the promotion in all the three-villages = 40 + 31 + 56 = 127
127
The percentage of toilets that can be expected after the promotion in all the three-villages = 2475
× 100 = 5.13%
iii. Let ₹A, ₹B and ₹C be the cost incurred by the organization for villages X, Y and Z respectively. Then A, B, C will be given by
the following matrix equation.

[ ][ ] [ ]
400 300 100 50 A
300 250 75 20 = B
500 400 150 40 C

[][ ]
A 400 × 50 + 300 × 20 + 100 × 40
⇒ B = 300 × 50 + 250 × 20 + 75 × 40
C 500 × 50 + 400 × 20 + 150 × 40

[ ][ ]
20000 + 6000 + 4000 30000
= 15000 + 5000 + 3000 = 23000
25000 + 8000 + 6000 39000
Cost is ₹30,000.
OR

[ ][ ] [ ]
400 300 100 50 A
300 250 75 20 = B
500 400 150 40 C

[][ ]
A 400 × 50 + 300 × 20 + 100 × 40
⇒ B = 300 × 50 + 250 × 20 + 75 × 40
C 500 × 50 + 400 × 20 + 150 × 40

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APS @ MUNDRA
 [][ ][ ]
A 20000 + 6000 + 4000 30000
⇒ B = 15000 + 5000 + 3000 = 23000
C 25000 + 8000 + 6000 39000
Hence total cost is = ₹92000
46. i. 5x - 4y = 40
5x - 8y = -80

[ ][ ] [ ]
5
5
−4
−8
x
y
=
− 80
40

[ ] []
ii. A =
5
5
−4
−8
,X=
x
y
and B =
[ ] 40
− 80
|A| = -40 + 20 = -20 ≠ 0

[ ] [
Cofactor matrix A =
−8
4
−5
5
adj A =
−8
−5
4
5 ]
iii. A =
[ ] [] [ ]
5
5
−4
−8
,X=
x
y
and B =
40
− 80
|A| = -40 + 20 = -20 ≠ 0

Cofactor matrix A =
[ −8
4
−5
5 ] , adj A =
[ −8
−5
4
5 ]
X = A-1 B ...(i)
1
A-1 = |A|
⋅ adjA

A-1 =
− 20
1
⋅
[ ] −8
−5
4
5
From (i)

[ ] [ ][ ]
x
y
=
1
− 20
⋅
−8
−5
4
5 − 80
40

⇒
[] [ x
y] [ ] = − 20
1 − 320 − 320
− 200 − 400
=
32
30
x = 32 and y = 30
OR
There are 32 Children, and each child is given ₹30.
Total money spent by Seema = 32 × 30 = ₹960
Hence Seema spends ₹960 in distributing the money to all the students of the Orphanage.

47. i. AB =
[ ][ ] 2
3
−1
4
5
7
2
4

=
[ ][
10 − 7
15 + 28
4−4
6 + 16
=
3
43 22
0
]
ii. We have, CD - AB = 0

⇒
[ ][ ] [ ] [ ]
2
3
5
8
a
c
b
d
−
3
43
0
22
=
0
0
0
0

⇒
[ ][ ] [ ]
2a + 5c
3a + 8c
2b + 5d
3b + 8d
−
3
43
0
22
=
0
0
0
0

⇒
[ ] [ ]
2a + 5c − 3
3a + 8c − 43
2b + 5d
3b + 8d − 22
=
0
0
0
0
By equality of matrices, we get 2a + 5c - 3 = 0 ...(i)

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APS @ MUNDRA
 3a + 8c - 43 = 0 ...(ii)
2b + 5d = 0 ...(iii)
3b + 8d - 22 = 0 ...(iv)
Solving (i) and (ii), we get a = -191, c = 77
iii. We have, CD - AB = 0

⇒
[ ][ ] [ ] [ ]
2
3
5
8
a
c
b
d
−
3
43
0
22
=
0
0
0
0

⇒
[ ][ ] [ ]
2a + 5c
3a + 8c
2b + 5d
3b + 8d
−
3
43
0
22
=
0
0
0
0

⇒
[ ] [ ]
2a + 5c − 3
3a + 8c − 43
2b + 5d
3b + 8d − 22
=
0
0
0
0
By equality of matrices, we get 2a + 5c - 3 = 0 ...(i)
3a + 8c - 43 = 0 ...(ii)
2b + 5d = 0 ...(iii)
3b + 8d - 22 = 0 ...(iv)
Solving (iii) and (iv), we get b = -110, d = 44
OR

We have, B + D =
[ ] [
5
7
2
4
+
− 191
77
− 110
44 ]
[ − 186
84
− 108
48 ]
48. i. A + B =
[ 10000
50000
20000
30000
30000
10000 ][
+
5000
20000
10000
10000
6000
10000 ] [ =
15000
70000
30000
40000
36000
20000 ]
The combined sales of Masoor in September and October, for farmer Girish ₹40000.

[ ][ ]
10000 20000 30000 5000 10000 6000
ii. A + B = +
50000 30000 10000 20000 10000 10000

[ ]
15000 30000 36000
=
70000 40000 20000
The combined sales of Urad in September and October, for farmer Ankit is ₹15000.

iii. A - B =
[ 10, 000
50, 000
20, 000
30, 000
30, 000
10, 000 ][ -
5000
20, 000
10, 000
10, 000
6000
10, 000 ]
=
[ 10, 000 − 5000
50, 000 − 20, 000
20, 000 − 10, 000
30, 000 − 10, 000
30, 000 − 6000
10, 000 − 10, 000 ]
A-B=
[ 5000
30, 000
10, 000
20, 000
24, 000
0 ] Ankit
Girish
OR
Profit = 2% × sales on october
2
= 100
×B

= 0.02 × [ 5000
20, 000
10, 000
10, 000
6000
10, 000 ]
=
[ 0.02 × 5000
0.02 × 20, 000
0.02 × 10, 000
0.02 × 10, 000
0.02 × 6000
0.02 × 10, 000 ]
=
[ 100
400
200
200
120
200 ] Ankit
Girish

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49. i. If ₹ 15000 is invested in bond X, then the amount invested in bond Y = ₹ (35000 - 15000) = ₹ 20000.

A = Investment
[ 15000
X Y
20000 ]
Interest rate Interest rate

and B = X
Y [ ] [ ]
10%
8%
= X
Y
0.08
0.1

ii. The amount of interest received on each bond is given by

AB = [ 15000 20000 ] ×
[ ]
0.08
0.1

= [15000 × 0.1 + 20000 × 0.08] = [1500 + 1600] = 3100

iii. Let ₹ x be invested in bond X and then ₹ (35000 - x) will be invested in bond Y.
Now, total amount of interest is given by

[x 35000 − x ]
[ ] 0.1
0.08
= [0.1x + (35000 - x)0.08]

But, it is given that total amount of interest = ₹ 3200

∴ 0.1x + 2800 - 0.08x = 3200
⇒ 0.02x = 400 ⇒ x = 20000

Thus, ₹ 20000 invested in bond X and ₹ 35000 - ₹ 20000 = ₹ 15000 invested in bond Y.
OR
Let ₹ x invested in bond X, then we have
10
x× 100
= 500 ⇒ x = 5000
Thus, amount invested in bond X is ₹ 5000 and so investment in bond Y be ₹ (35000 - 5000) = ₹ 30000
Hatchback Sedan SUV

[ ]
A 120 50 10
50. i. B 100 30 5
C 90 40 2
In 2019, dealer A sold 120 Hatchbacks, 50 Sedans and 10 SUV; dealer B sold 100 Hatchbacks, 30 Sedans and 5 SUVs and
dealer C sold 90 Hatchbacks, 40 Sedans and 2 SUVs.
∴ Required matrix, say P, is given by

Hatchback Sedan SUV

[ ]
A 120 50 10
P= B 100 30 5
C 90 40 2
In 2020, dealer A sold 300 Hatchbacks, 150 Sedans, 20 SUVs dealer B sold 200 Hatchbacks, 50 sedans, 6 SUVs dealer C sold
100 Hatchbacks, 60 sedans, 5 SUVs.
∴ Required matrix, say Q, is given by

Hatchback Sedan SUV

[ ]
A 300 150 20
Q= B 200 50 6
C 100 60 5
Hatchback Sedan SUV

[ ]
A 300 150 20
ii. B 200 50 6
C 100 60 5

In 2020, dealer A sold 300 Hatchback, 150 Sedan, 20 SUV dealer B sold 200 Hatchback, 50 sedan, 6 SUV dealer C sold 100
Hatchback, 60 sedan, 5 SUV.

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APS @ MUNDRA
 ∴ Required matrix, say Q, is given by
Hatchback Sedan SUV

[ ]
A 300 150 20
Q= B 200 50 6
C 100 60 5

iii. Total number of cars sold in two given years, by each dealer, is given by
Hatchback Sedan SUV

[ ]
A 120 + 300 50 + 150 10 + 20
P + Q = B 100 + 200 30 + 50 5+6
C 90 + 100 40 + 60 2+5
Hatchback Sedan SUV

[ ]
A 420 200 30
= B 300 80 11
C 190 100 7
OR
The amount of profit in 2020 received by each dealer is given by the matrix
Hatchback Sedan SUV

[ ][ ]
A 300 150 20 50000
B 200 50 6 100000
C 100 60 5 200000

[ ]
A 15000000 + 15000000 + 4000000
= B 10000000 + 5000000 + 1200000
C 5000000 + 6000000 + 1000000

[ ]
A 34000000
= B 16200000
C 12000000

51. i. Let A be the 2 × 3 matrix representing the annual sales of products in two markets.
x y z

A=
[ 10000
6000
2000
20000
18000
8000 ] Market I
Market II
Now, revenue = sale price × number of items sold

][ ]
2.5
AB =
[ 10000
6000
2000
20000
18000
8000
1.5
1

⇒ AB =
[ 25000 + 3000 + 18000
15000 + 30000 + 8000 ] [ ]
=
46000
53000
Therefore, the revenue collected from Market I = ₹46000 and the revenue collected from Market II = ₹53000.
ii. Let C be the column matrix representing cost price of each unit of products x, y, z.

[]
2
C= 1
0.5

][ ]
2
AC =
[ 10000
6000
2000
20000
18000
8000
1
0.5

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APS @ MUNDRA
 ⇒ AC =
[ 20000 + 2000 + 9000
12000 + 20000 + 4000 ] [ ]
=
31000
36000
Cost price in Market I is ₹31000 and in Market is ₹36000.
iii. Now, Profit matrix = Revenue matrix - Cost matrix
⇒ AB - AC

⇒
[ ] [ ] [ ]
46000
53000
−
31000
36000
=
15000
17000
Therefore, the gross profit from both the markets = ₹15000 + ₹17000 = ₹32000
OR

A=
[ ]
0
1
1
0

A2 = [ ][ ] [
0
1
1
0
0
1
1
0
=
0+1
0+0
0+0
0+1 ] [ ]
=
0
1
1
0

⇒ A2 = I
52. i. Combined sales in September and October for each farmer in each variety is given by
Urad Masoor Mung

A+B=
[ 15000
70000
30000
40000
36000
]
20000 Balwan Singh
Shyam

Combined sales of Masoor in September and October for farmer Balwan Singh = ₹ 40000
ii. Combined sales in September and October for each farmer in each variety is given by
Urad Masoor Mung

A+B=
[ 15000
70000
30000
40000
36000
]
20000 Balwan Singh
Shyam

Combined sales of Urad in September and October for farmer Shyam = ₹ 15000
iii. Change in sales from September to October is given by
Urad Masoor Mung

A-B=
[ 5000
30000
10000
20000
24000
0 ] Shyam
Balwan Singh
∴ Decrease in sales of Mung from September to October for farmer Shyam = ₹ 24000.
OR
Required profit is given by
2
2% of B = 100
× B = 0.02 × B
Urad Masoor Mung

= 0.02
[ 5000
20, 000
10, 000
10, 000
6, 000
]
10, 000 Balwan Singh
Shyam

Urad Masoor Mung

= [ 100
400
200
200
120
] Shyam
200 Balwan Singh
Thus, in October Shyam receives ₹ 100, ₹ 200 and ₹ 120 as profit in the sale of each variety of pulses, respectively and
Balwan Singh receives a profit of ₹ 400, ₹ 200 and ₹ 200 in the sale of each variety of pulses respectively.
53. i. Let number of children = x
Amount distributed by Shanti for one child = ₹ y
Now, Total money = xy
and Total money will remain the same.
Given that, if there were 8 children less, everyone would have got ₹ 10 more.
Total money now = Total money before
(x - 8) × (y + 10) = xy
⇒ x(y + 10) - 8(y + 10) = xy

⇒ xy + 10x - 8y - 80 = xy

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APS @ MUNDRA
 ⇒ 10x - 8y - 80 = 0
⇒ 10x - 8y = 80
⇒ 5x - 4y = 40
Also, if there were 16 children more, everyone would have got ₹ 10 less.
Total money now = Total money before
(x + 16) × (y - 10) = xy
⇒ x(y - 10) + 16(y - 10) = xy

⇒ xy - 10x + 16y - 160 = xy

⇒ -10x + 16y - 160 = 0

⇒ 10x - 16y + 160 = 0
⇒ 5x - 8y = -80

Thus, required equations are:

5x - 4y = 40 ...(i)
5x - 8y = - 80 ...(ii)
ii. On solving eqs. (i) & (ii), we get x = 32 and y = 30.
Hence, the number of children = 32
The amount is given to each child by Shanti = Rs. 30.
iii. Writing eq. (i) & eq. (ii) in matrix form, we get

[ ][ ] [ ]
5
5
−4
−8
x
y
=
− 80
40

OR
Total amount = xy = 32 × 30 = ₹ 960.
54. i. 2x - 4y + 5z = 1500
3x + y - 2z = 100
-x + 3y + z = 400

[ ][ ] [ ]
2 −4 5 x 1500
3 1 −2 y = 100
−1 3 1 z 400

[ ] [] [ ]
2 −4 5 x 1500
ii. A = 3 1 −2 , X = y , B = 100
−1 3 1 z 400

X = A-1B ...(i)
|A| = 2(1 + 6) + 4(3 - 2) + 5(9 + 1) = 68 ≠ 0

[ ] [ ]
7 −1 10 7 19 3
co-factor matrix A = 19 7 − 2 , adj A = −1 7 19
3 19 14 10 −2 14
1
A-1 = |A|
⋅ adjA

[ ]
7 19 3
1
A-1 = −1 7 19
68
10 −2 14

From (i)

[] [ ][ ]
x 7 19 3 1500
1
⇒ y = −1 7 19 100
68
z 10 −2 14 400

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APS @ MUNDRA
 [] [ ]
x 10500 + 1900 + 1200
1
⇒ y = − 1500 + 700 + 7600
68
z 15000 − 200 + 5600

[] [ ]
x 13600
1
⇒ y = 6800
68
z 20400

x = 200, y = 100 and z = 300

Total price of one dozen of pens and one dozen of notebooks = 200 + 100 = ₹300
iii. The sale amount of Ravi = 2x + 5z = 2 × 200 + 5 × 300 = 400 + 1500 = ₹1900
OR
The amount of purchases made by all three friends
4y + 2z + x = 4 × 100 + 2 × 300 + 200 = ₹1200
The amount of sales made by all three friends
2x + 5z + 3x + y + 3y + z = 5x + 4y + 6z = 5 × 200 + 4 × 100 + 6 × 300 = ₹3200
55. i. In factory A, number of units of types I, II and III for boys are 80, 70, 65 respectively and for girls number of units of types I,
II and III are 80, 75,90 respectively.
Boys Girls

[ ]
I 80 80
∴ P = II 70 75
III 65 90
In factory B, number of units of types I, II and III for boys are 85, 65, 72 respectively and for girls number of units of types I,
II and III are 50, 55, 80 respectively.
Boys Girls

[ ]
I 85 50
Q = II 65 55
III 72 80
ii. Let X be the matrix that represent the number of units of each type produced by factory A for boys and Y be the matrix that
represents the number of units of each type produced by factory B for boys.
I II III I II III

Then, X = [ 80 70 65 ] and Y = [ 85 65 72 ]
Now, required matrix = X + Y = [80 70 65] + [80 65 72]
= [165 135 137]
iii. Let X be the matrix that represent the number of units of each type produced by factory A for girls and Y be the matrix that
represents the number of units of each type produced by factory B for girls.
I II III I II III

X = [ 80 75 90 ] and Y = [ 50 55 80 ]
Required matrix = [80 75 90] + [50 55 80]
= [130 130 137]
OR
Clearly, R = P + Q

[ ][ ][ ]
80 80 85 50 165 130
= 70 75 + 65 55 = 135 130
65 90 72 80 137 170

∴ R' =
[ 165
130
135
130
137
170 ]
56. i. In factory A, number of units of type I, II and III for boys are 80, 70, 65 respectively and for girls number of units of type I, II
and III are 80, 75, 90 respectively.

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APS @ MUNDRA
 Boys Girls

[ ]
I 80 80
∴ P = II 70 75
III 65 90
ii. In factory B, number of units of type I, II and III for boys are 85, 65, 72 respectively and for girls number of units of type I, II
and III are 50, 55, 80 respectively.
Boys Girls

[ ]
I 85 50
∴ Q = II 65 55
III 72 80
iii. Let matrix X represent the number of units of each type produced by factory A for boys and matrix Y represents the number
of units of each type produced by factory B for boys.
I II III

∴ X = [ 80 70 65 ]
I II III

Y = [ 85 65 72 ]
Now, total production of sports clothes of each type for boys = X + Y
= [80 70 65] + [85 65 72]
= [165 135 137]
OR
For girls, let matrix S represents the number of units of each type produced by factory A and matrix T represents the number
of units of each type produced by factory B.
I II III

∴ S = [ 80 75 90 ]
I II III

T = [ 50 55 80 ]
Now, required matrix = S + T
= [80 75 90] + [50 55 80]
= [130 130 170]
57. i. If ₹15000 is invested in bond X, then the amount invested in bond Y = ₹(35000 - 15000) = ₹20000
Invest rate
X Y

Investment A = [ 15000 20000 ]; B =

X
[ ]
0.1
Y 0.08
ii. The amount of interest received on each bond is given by

AB = [15000 20000] ×
[ ]
0.1
0.08
= [15000 × 0.1 + 20000 × 0.08] = [1500 + 1600] = 3100
iii. Let ₹x be invested in bond X and then ₹(35000 - x) will be invested in bond Y.
Now, total amount of interest is given by

[x 35000 - x]
[ ] 0.1
0.08
= [0.1x + (35000 - x) 0.08]

But, it is given that total amount of interest = ₹3200

∴ 0.1x + 2800 - 0.08x = 3200

⇒ 0.02 x = 400 ⇒ x = 20000

Thus, ₹20000 invested in bond X and ₹35000 - ₹20000

= ₹15000 invested in bond Y.
OR
Let ₹x invested in bond X, then we have
10
x × 100
= 500 ⇒ x = 5000
Thus, the amount invested in bond X is ₹5000 and so investment in bond Y be ₹(35000 - 5000) = ₹30000

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APS @ MUNDRA
58. i. Let F be the matrix representing the number of family members and R be the matrix representing the requirement of calories
and proteins for each person. Then
Men Women Children

F=
Family A
Family B [ 4
2
4
2
4
2 ]
Calories Proteins

[ ]
Man 2400 45
R = woman 1900 55
Children 1800 33
ii. The requirement of calories and proteins for each of the two families is given by the product matrix FR.

][ ]
2400 45
FR =
[ 4
2
4
2
4
2
1900
1800
55
33

=
[ 4(2400 + 1900 + 1800)
2(2400 + 1900 + 1800)
4(45 + 55 + 33)
2(45 + 55 + 33) ]
Calories Proteins

FR =
[ 24400
12200 ]
532 Family A
266 Family B

[ ]
2400 − 2400 × 5% 45 + 45 × 5%
iii. R' = 1900 − 1900 × 5% 55 + 55 × 5%
1800 − 1800 × 5% 33 + 33 × 5%

[ ]
2400 − 120 45 + 2.25
⇒ R' = 1900 − 95 55 + 2.75
1800 − 90 33 + 1.65
Calories Proteins

[ ]
Man 2280 45.25
⇒ R' = Woman 1805 55.75
Children 1710 34.65
OR
Since, AB = B ...(i) and BA = A ..(ii)
∴ A2 + B2 = A ⋅ A + B ⋅ B
= A(BA) + B(AB) [using (i) and (ii)]
- (AB)A + (BA)B [Associative law]
= BA + AB [using (i) and (ii)]
=A+B
Fans Mats Plates

[ ]
A 40 50 20
59. i. P = B 25 40 30
C 35 50 40

[]
25 Fans
Q = 100 Mats
50 Plates

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APS @ MUNDRA
 ii. Clearly, total funds collected by each school is given by the matrix

[ ][ ]
40 50 20 25
PQ = 25 40 30 100
35 50 40 50

[ ][ ]
1000 + 5000 + 1000 7000
= 625 + 4000 + 1500 = 6125
875 + 5000 + 2000 7875
∴ Funds collected by school A is ₹7000.

Funds collected by school B is ₹6125.

Funds collected by school C is ₹7875.

[]
25 Fans
New price matrix Q = 20% × 100 Mats
50 Plates

[ ]
25 + 25 × 0.20 Fans
iii. ⇒ Q= 100 + 100 × 0.20 Mats
50 + 50 × 0.20 Plates

[]
30 Fans
Q= 120 Mats
60 Plates

[]
25 Fans
New price matrix Q = 20% × 100 Mats
50 Plates

[ ]
25 + 25 × 0.20 Fans
⇒ Q= 100 + 100 × 0.20 Mats
50 + 50 × 0.20 Plates

[]
30 Fans
Q = 120 Mats
60 Plates

OR

[ ][ ]
40 50 20 30
PQ = 25 40 30 120
35 50 40 60

[ ][ ]
1200 + 6000 + 1200 8400
PQ = 750 + 4800 + 1800 = 7350
1050 + 6000 + 2400 9450
Total fund collected = 8400 + 7350 + 9450 = ₹25,200
60. i. Number of items purchased by shopkeepers A, B and C can be written in matrix form as
Notebooks pens pencils

[ ]
144 60 72 A
X= 120 72 84 B
132 156 96 C

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APS @ MUNDRA
 []
40 Note book
ii. Since, Y = 12 Pen
3 Pencil

[ ][ ]
144 60 72 40
∴ XY = 120 72 84 12
132 156 96 3

[ ][ ]
5760 + 720 + 216 6696
= 4800 + 864 + 252 = 5916
5280 + 1872 + 288 7440

iii. (A + I)2 = A2 + 2A + I = 3A + I
⇒ (A + I)3 = (3A + I) (A + I)
= 3A2 + 4A + I = 7A + I
∴ (A + I)3 - 7A = I
OR
A2 - B2 = (A - B) (A + B) = A2 + AB - BA - B2
∴ AB = BA

61. i. Let A be the 2 × 3 matrix representing the annual sales of products in two markets.

[ ]
x y z
∴ A = 10000 2000 18000
6000 20000 8000
Let B be the column matrix representing the sale price of each unit of products x, y, z.

[]
2.5
∴ B= 1.5
1
Now, revenue = sale price × number of items sold

][ ]
2.5
=
[ 10000
6000
2000
20000
18000
8000
1.5
1

=
[ 25000 + 3000 + 18000
15000 + 30000 + 8000 ][ ]
=
46000
53000
Therefore, the revenue collected from Market I = ₹ 46000
ii. Let A be the 2 × 3 matrix representing the annual sales of products in two markets.

[ ]
x y z
∴ A = 10000 2000 18000
6000 20000 8000
Let B be the column matrix representing the sale price of each unit of products x, y, z.

[]
2.5
∴ B= 1.5
1

Now, revenue = sale price × number of items sold

][ ]
2.5
=
[ 10000
6000
2000
20000
18000
8000
1.5
1

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APS @ MUNDRA
 =
[ 25000 + 3000 + 18000
15000 + 30000 + 8000 ][ ]=
46000
53000
The revenue collected from Market II = ₹ 53000.
iii. Let C be the column matrix representing cost price of each unit of products x, y, z.

[]
2
Then, C = 1
0.5
∴ Total cost in each market is given by

][ ]
2
AC =
[ 10000
6000
2000
20000
18000
8000
1
0.5

=
[ 20000 + 2000 + 9000
12000 + 20000 + 4000 ][ ]=
31000
36000
Now, Profit matrix = Revenue matrix - Cost matrix

=
[ ][ ][ ]
46000
53000
−
31000
36000
=
15000
17000
Therefore, the gross profit from both the markets = ₹ 15000 + ₹ 17000 = ₹ 32000
OR

We have, A =
[ ] 0
1
1
0

∴ A2 =
[ ][ ] [ ]
0
1
1
0
0
1
1
0
=
1
0
0
1
=I

62. i. From question:

For Rajat x + y + z = 21
For Raj 4x + 3y + 2z = 60
For Raman 6x + 2y + 3z = 70
Therefore, Algebraical representation is
x + y + z = 21
4x + 3 y + 2z = 60
6x + 2y + 3z = 70
ii. We have AX = B
Pre multiplying by A-1 on both sides, we have
A-1 AX = A-1B
⇒ (A-1 A)X = A-1B
⇒ IX = A-1 B [A -1A = I (Identity matrix]
⇒ X = A-1 B [IX = X]

| |
1 1 1
iii. |A|= 4 3 2 = 1 (9 - 4) -1 (12 - 12) + (8 -18)
6 2 3
= 5 - 0 - 10 = - 5 ≠ 0
Therefore, A-1 exists.
Now, A11 = 9 - 4 = 5 ; A21 = - ( 3 - 2 ) = - 1 ; A31 = 2 - 3 = -1
A12 = - (12 -12 ) = 0; A22 =(3 -6) = -3; A32 = - (2 - 4) = 2
A13 = (8 -18 ) = -10; A23 = -(2 - 6) = 4; A33 = (3 - 4) = -1

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APS @ MUNDRA
 | || |
5 0 − 10 T 5 −1 −1
∴ Adj A= − 1 −3 4 = 0 −3 2
−1 2 −1 − 10 4 −1
1
∵ A −1 = |A|
AdjA

| |
5 −1 −1
1
=− 0 −3 2
5
− 10 4 −1
OR
We have X = A-1B

|| | | |
x 5 −1 −1 21
1
⇒ y =− 0 −3 2 60
5
z − 10 4 −1 70

|| | |
x − 25
1
⇒ y = − − 40
5
z − 40

||||
x 5
⇒ y = 8
z 8
⇒ x = 5, y = 8, z = 8
⇒ Cost of potato,onion and brinjal are ₹ 5, ₹ 8 and ₹ 8
63. i. Monthly savings of Rita
= Monthly income of Rita - Monthly Expenditure of Rita
15000 = 3x - 5y
Similarly for Ritika
15000 = 4x - 7y
Here required system of linear equation
3x - 5y = 15000
4x - 7y = 15000
ii. We have system of linear equations
3x - 5y = 15000
4x -7y = 15000
This may be written in matrix system as

AX = B, where A =
[ ] [] [ ]
3
4
−5
−7
,X=
x
y
,B=
15000
15000
Because,

| 3
4 |[ ] [ ]
−5
−7
x
y
=
15000
15000

⇒
[ ][ ] 3x − 5y
4x − 7y
=
15000
15000

iii. We have |A| =| | 3

4
−5
−7
∴ |A| = -21 + 20 = -1

A11 = -7 A12 = -4
A21 = 5 A22 = 3

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 Adj A =
[ −7
5
−4
3 ] [ ]
T
=
−7
−4
5
3
1
∴ A-1 = |A|
adj A

=
1
−1 [ ][ −7
−4
5
3
=
7
4
−5
−3 ]
OR
We have X = A-1 B

⇒
[ ] [ ][ ]
x
y
=
7
4
−5
−3
15000
15000

⇒
[] [x
y ]
=
105000 − 75000
60000 − 45000

⇒
[] [ ]
x
y
=
30000
15000
⇒ x = 30,000, y = 15,000
∴ Rita's monthly income = 3x = 3 × 30000 = ₹ 90000
And Ritika's income = 4x = 4 × 30000 = ₹ 120000
64. i. Let Δ be the area of the triangle then,

| |
−2 6 1
1
Δ= 3 −6 1
2
1 5 1
1
= 2
|-2(-6 - 5) - 6(3 - 1) + 1(15 + 6)| [Expanding along R1]
1
⇒ Δ= 2
|43 - 12| = 15.5 sq. units
ii. The given points are collinear.

| |
2 −3 1
1
∴ k −1 1 =0
2
0 4 1
Expanding along R1, we get
2(-1 - 4) + 3(k) + 1(4k) = 0
10 40
⇒ 7k - 10 = 0 ⇒ k= ⇒ 4k =
7 7
iii. Area of △ABC = 3 sq. units [Given]

| | | |
1 3 1 1 3 1
1
⇒ 0 0 1 = ±3 ⇒ 0 0 1 = ±6
2
k 0 1 k 0 1
⇒ 1(0 - 0) - 3(0 - k) + 1(0 - 0) = ± 6
⇒ 3k = ± 6 ⇒ k = ± 2
OR
Let Q(x, y) be any point on the line joining A(1, 2) and B(3, 6). Then, area of △ABQ = 0

| |
1 2 1
1
⇒ 3 6 1 =0
2
x y 1
⇒ 1(6 - y) - 2(3 - x) + 1(3y - 6x) = 0
⇒ 6 - y - 6 + 2x + 3y - 6x = 0
⇒ -4x = -2y ⇒ 2x = y

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APS @ MUNDRA
 || |
x1 y1 1
1
65. i. Area of triangle is given by x2 y2 1
2
x3 y3 1

|| |
0 0 1
1 3
∴ Required area = √3 1
2
3 − √3 1

= | 1
2 |
[1( − 3√3 − 3√3)] [Expanding along R1]

= 3√3 sq. units.

ii. Since a face of the Pyramid consist of 25 smaller equilateral triangles.
∴ Required area = 25 × 3√3 = 75√3 sq. units.
√3
iii. Area of equilateral triangle = 4
a2
√3
∴ 3√3 = 4
a2 ⇒ a2 = 12 ⇒ a = 2√3
Let h be the length of altitude of a smaller equilateral triangle. Then,
1
2
× base × height = 3√3
1
⇒
2
× 2√3 × h = 3√3 ⇒ h = 3 units
OR

| |
a 0 1
1
Area of ΔABC = 0 b 1
2
1 1 1
1 1 1 1
= 2
[a (b - 1) - 0 + 1 (0 - b)] = 2
(ab - a - b) = 0 [ ∵
a
+ b
=1 ⇒ b + a = ab]
∴ Points A, B and C are collinear.

|| |
0 0 1
1
66. i. Required Area =
3 √3 1
2
3 − √3 1
1
= ∣ 2 [1( − 3√3 − 3√3)]∣
6√ 3
= 2
= 3√3 sq units
ii. Since, a face of the Pyтamid consists of 25 smaller equilateral triangles.
∴ Area of a face of the Pyramid = 25 × 3√3 = 75√3 sq. units.
√3
iii. Area of equilateral triangle = 4
(side)2
√3
∴ 3√3 = 4
(side)2 [As calculated above area of equilateral triangle is 3√3 sq. units]

⇒ (side)2 = 12
⇒ side = 2√3 (units)
Let h be the length of the altitude of a smaller equilateral triangle.
1
Then, 2
× base × height = 3√3
1
or, 2
× side × height = 3√3
3√ 3 × 2
or, height = = 3 units.
2√ 3

OR

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 Let the third point on the line be (x, y).
The area of triangle with vertices (x, y), (1, 2), (3, 6)

| |
x y 1
1
= 1 2 1
2
3 6 1
Since the three points are collinear, the area formed will be zero.

| |
x y 1
⇒ 1 2 1 =0
3 6 1
⇒ x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0
⇒ -4x + 2y = 0
⇒ 2x - y = 0
Hence, the equation of line joining (1, 2) and (3, 6) is 2x - y = 0
67. i. (Constraints are x + 2y ≥ 10
x+y≥6
3x + y ≥ 8
x≥0
y≥0

ii. Corner points Value of Z = 16x + 20y

A(10, 0) 160

B(2, 4) 112

C(1, 5) 116

D(0, 8) 160
The minimum cost is ₹ 112
68. i. According to given information, the system of linear equations is:
x + y + z = 45
z=x+8
x + z = 2y
or, x + y + z = 45
-x + z = 8
x - 2y + z = 0
In matrix form,

[ ][ ] [ ]
1 1 1 x 45
−1 0 1 y = 8
1 −2 1 z 0

[ ]
1 1 1
ii. Let A = 1 0 −2
1 −1 1

[ ]
2 2 2
1
Then A-1 = 6
3 0 −3
1 −2 1
We know that,
(A')-1 = (A-1)'
or, (A-1)' = (A')'-1

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 [ ]
1 1 1
3 3 3
1 1
∴
(A ) −1 ′ = 2
0 −2
1 −1 1
6 3 6

[ ]
1 1 1
3 2 6
1 −1
= 3
0 3
1 −1 1
3 2 6

[ ] [ ]
1 1 1
3 2 6
1 1 1 −1
1 −1
∴ 1 0 −1 = 0
3 3
1 −2 1 1 −1 1
3 2 6

[ ] [ ][ ]
x 2 −3 1 45
1
iii. y = 6
2 0 −2 8
z 2 3 1 0

[] [ ]
x 90 − 24 + 0
1
⇒ y = 90 + 0 + 0
6
z 90 + 24 + 0

[] [ ]
x 66
1
⇒ y = 90
6
z 114

∴ x = 11, y = 15 and z = 19
Required Ratio is 11 : 15 : 19.
OR
Given, P3 = Q3 and P2Q = Q2P
On subtracting, we get
P3 - P2Q = Q3 - Q2P
⇒ P2(P - Q) + Q2(P - Q) = 0
⇒ (P2 + Q2) (P - Q) = 0
If |P2 + Q2| ≠ 0, then P2 + Q2 is invertible.
⇒ P - Q = 0 contradiction

∴ |P2 + Q2| = 0

69. i. We have, A =
[ 3
1
−4
−1 ]
∴ A2 =
[ 3
1
−4
−1 ][ 3
1
−4
−1 ] [ ]
=
5
2
−8
−3
, which can be obtained from An =
[ 1 + 2n
n
− 4n
1 − 2n ] for n = 2.

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 ii. We have, A =
[ ] 1
0
2
1

∴ |A| =
| |
1
0
2
1
=1-0=1

Also, |An| = |A ⋅ A ...A (n times)| = |A|n = 1n = 1

[ ]
a 0 0
iii. We have, A = 0 a 0
0 0 a

[ ][ ]
a 0 0 a 0 0
∴ A2 = A ⋅ A= 0 a 0 0 a 0
0 0 a 0 0 a

[ ]
a2 0 0
0 2 0
= a
0 0 a2

[ ]
an 0 0
Similarly, An = 0 an 0
0 0 an

Now, cofactor of a13 = (-1)1+3

| |
0
0
an
0
=0

OR
We have, |A| = 2
and |An| = |A ⋅ A ...A (n-times)|
= |A| |A| ... |A| (n-times) = |A|n = 2n

70. i. Let Δ =
| 1
4
−2
3 |
Cofactor of 1 = 3, cofactor of -2 = -4, Cofactor of 4 = 2, cofactor of 3 = 1
∴ Required sum = 3 - 4 + 2 + 1 = 2

| |
5 6 −3
ii. Let Δ = − 4 3 2
−4 −7 3

Minor of a21 =
| 6
−7
−3
3 | = 18 - 21 = -3

| |
2 −3 5
iii. Let Δ = 6 0 4
1 5 −7
Clearly, a32 = 5

and A32 = cofactor of a32 in Δ = (-1)3+2

| |
2
6
5
4
= (-1) (8 - 30) = 22
∴ a32 ⋅ A32 = 5 × 22 = 110

OR

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APS @ MUNDRA
 | |
5 3 8
Here, Δ = 2 0 1
1 2 3

∴ Minor of a23 =
| | 5
1
3
2
= 10 - 3 = 7

71. i. We have, X = [3 2] [ ][ ]
−2
1 2
−1
3
2

= [3 2]
[ ]
7
−8
= [21 - 16] = [5]

∴ |X| = 5
ii. a22 in U is -1 and its minor is 1

iii. We have, A =
[ ] 1
2
0
1

[]
Let U1 =
[]a
b
then AU1 =
1
0

⇒
[ ][ ] [ ] [ ] [ ]
1
2
0
1
a
b
=
1
0
⇒
a
2a + b
=
1
0
⇒ a = 1 and 2a + b = 0 ⇒ a = 1 and b = -2

[]
Let U2 =
[]c
d
then AU2 =
2
3

⇒
[ ][ ] [ ] [ ] [ ]
1
2
0
1
c
d
=
2
3
⇒
c
2c + d
=
2
3
⇒ c = 2 and 2c + d = 3
⇒ c = 2 and d = 3 - 4 = -1

Thus, U1 + U2 =
[ ][ ] [ ]
1
−2
+
2
−1
=
3
−3
OR

Clearly, U =
[ 1
−2
2
−1 ]
∴ |U| =
| 1
−2
2
−1 | = -1 + 4 = 3

72. i. (x - 50)(y + 50) = xy ⇒ 50x - 50y = 2500 ⇒ x - y = 50

Also (x - 10)(y - 20) = xy - 5300
⇒ xy - 20x - 10y + 200 = xy - 5300
⇒ - 20x - 10y = -5300 - 200
⇒ 2x + y = 550

Hence, represented system of linear equations is

x - y = 50
2x + y = 550
ii. We can write matrix system as

[ ] [] [ ]
1 −1 x 50
AX = B, where A = ,X= ,B=
2 1 y 550

[ ][ ] [ ]
1 −1 x 50
Because =
2 1 y 550

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 [ ][ ]
x−y 50
⇒ =
2x + y 550
⇒ x - y = 50

2x + y = 550
iii. We have

A=
[ ]1
2
−1
1
⇒ |A| = 1 + 2 = 3 ≠ 0

Now, A11 = 1, A12 = -2

A21 = 1, A22 = 1

∴ adj A = [ 1
2
−1
1 ] [ ]
T
=
1
2
−1
1

∴ A-1 = |A|
1
adj A =
[ ]1
3
1
2
−1
1
OR
We have

⇒ X = A-1 ⇒
[] [x
y
=
1
3
1
−2
1
1 ]| | [ ] [ ]
50
550
⇒
x
y
=
1
3
600
450

⇒ [] [ ]
x
y = 150
200
⇒ x = 200, y = 150
Hence dimension is
200 m and 150 m
73. i. Matrix equation is AX = B, where

[ ] [] [ ]
5 3 1 x 160
A= 2 1 3 , X = y , B = 190
1 2 4 z 250
where x is the number of pens bought, y the number of bags and z the number of instrument boxes.
ii. |A| = 5(4 - 6) - 3(8 - 3) + 1(4 - 1) = -22

[ ][ ]
−2 −5 3 ′ −2 − 10 8
iii. adj (A) = − 10 19 −7 = −5 19 − 13
8 − 13 −1 3 −7 −1

[ ]
−2 − 10 8
1
⇒ A-1 = ( − 22 )
−5 19 − 13
3 −7 −1
OR

[ ][ ]
32 20 18 25 15 5
P= A2 - 5A = 15 13 17 - 10 5 15
13 13 23 5 10 20

[ ]
7 5 13
= 5 8 2
8 3 3

74. i. f(3) = 4
x2 − 9 (x+3) (x−3)
lim f(x) = lim x−3
= lim (x−3)
x→3 x→3 x→3

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 = lim (x + 3) = 6 ∵ f(x) ≠ f(3)
x→3
∴ f(x) has removable discontinuity at x = 3.
ii. lim f(x) = lim (x + 2) = 4 + 2 = 6
x→4− x→4
lim f(x) = lim (x + 4) = 4 + 4 = 8
x→4+ x→4
∴ lim f(x) ≠ lim f(x)
x→4− x→4+
∴ f(x ) has an irremovable discontinuity at x = 4.
(x −4 )
2

iii. lim f(x) = lim = lim (x + 2) = 4

(x−2)
x→2 x→2 x→2
and f(2) = 5(given) ∴ lim f(x) ≠ f(2)
x→2
∴ f(x) has removable discontinuity at x = 2
OR
f(0) = 2
x+x
lim f(x) = lim x
=2
x→0− x→0
x−x
∵ lim f(x) = lim x
=0
x→0+ x→0
∴ f(x ) has an irremovable discontinuity at x = 0.
75. i. We have,
x3 + x2y + xy2 + y3 = 81
Differentiating both sides w.r.t. x, we get

( dy
3x 2 + 2xy + x 2 dx + y 2 + 2xy dx
)( dy
dy
) + 3y 2 dx = 0
dy

⇒ (x2 + 2xy + 3y2) dx (3x2 + 2xy + y2) = 0

dy
⇒ (x2 + 2xy + 3y2) dx = - (3x2 + 2xy + y2)

dy (
− 3x 2 + 2xy + y 2)
⇒
dx
=
( x + 2xy + 3y )
2 2

dy
ii. To find slope of curve y, we find dx
.

We have, y = x2 + 6y2 + xy

dy ( )
d x2 ( )
d 6y 2
d ( xy )
∴
dx
= dx
+ dx
+ dx
dy dy dy
⇒
dx
= 2x + 12y dx
+ y + x dx
dy
⇒ (1 - x - 12y) dx
= 2x + y
dy 2x + y
⇒
dx
= ( 1 − x − 12y )
iii. From the given equation

2 sin y cos y.
dy
dx
- sin xy. x ⋅
[ dy
dx
+y⋅1 =0
]
dy ysin xy
or dx
= sin 2y − xsin ( xy )
dy ysin xy
dx
= sin 2y − xsin xy
π 1
π π

()
⋅
4 sin 1⋅ 4
dy 4 √3
dx π = π π = 1
x=1, y = 4 sin 2 4 − 1sin 1 ⋅ 4 1−
√3

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 π
=
4 ( √2 − 1 )

() dy π π ( √2 + 1 )
∴
dx π = = 4
x=1, y = 4 4 ( √2 − 1 )

OR
We have,
y = (√x) y
⇒ log y = y log (√x) [Taking by both sides]
Now, differentiating both sides w.r.t. x, we get
1 dy d d
y
⋅
dx
= y dx (log√x) + log√x dx (y)

⇒
1
y
⋅
dy
dx
=y
( 1

√x
⋅
2√ x
1
) + log√x dx
dy

⇒
1
y
⋅
dy
dx
=
y
2x
+
dy
dx ( ) 1
2
logx

⇒
dy
dx ( 1
y
1
− logx =
2 ) 2x
y

dy y 2y
⇒
dx
= 2x
× 2 − ylog x
dy y2
⇒
dx
= x ( 2 − ylog x )

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