Solution
PERIODIC TEST -1
Class 12 - Mathematics
1.
(b) 2
Explanation: Total possible pairs {(1, 1) (1, 3), (1, 5), (3, 3), (3, 1), (3, 5) (5, 5), (5, 1), (5, 3)}
1st equivalence relation
R1 = {(1, 1,), (2, 2), (3, 3), (1, 3), (3, 1)}
2nd equivalence relation
R2 = {(1, 1,), (2, 2), (3, 3), (1, 3), (3, 1), (3, 5), (5, 3)}
no of possible equivalence relation
=2
∴
2.
(c) Reflexive, transitive but not symmetric
Explanation: Since n divides n, ∀ n ∈ N, R is reflexive.
R is not symmetric since for (3, 6) ∈ R but (6, 3)∉R.
R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will divide r.
3.
(d) bijection
Explanation: Given that f : [-1/2, 1/2] → [π /2, π /2] where f(x) = sin-1 (3x - 4x3)
Put x = sinθ in f(x) = sin-1 (3x - 4x3)
⇒ f (x = sin θ) = sin
−1
3
(3 sin θ − 4 sin θ )
f(x) = sin-1 (sin3θ)
⇒ f(x) = 3θ
⇒
⇒
f(x) = 3 sin-1x
If f(x) = f(y)
Then
3 sin-1x = 3 sin-1y
⇒ x = y
So, f is one-one.
y = 3 sin-1x
⇒ x = sin
y
3
x ∈ R also y ∈ R so f is onto.
Hence, f is bijection.
∵
4.
(b) equivalence
Explanation: Given that,
R be a relation on T defined as aRb if a is congruent to b ∀ a, b ∈ T
Now,
aRa ⇒ a is congruent to a, which is true since every triangle is congruent to itself.
⇒ (a,a) ∈ R ∀ a ∈ T
⇒ R is reflexive.
Let aRb ⇒ a is congruent to b
b is congruent to a
⇒ bRa
∴ (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a, b ∈ T
⇒ R is symmetric.
Let aRb ⇒ a is congruent to b and bRc ⇒ b is congruent to c
⇒ a is congruent to c
⇒
aRc
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a, b,c ∈ T
⇒ R is transitive.
Hence, R is an equivalence relation.
⇒
∴
1 / 10
5.
(b) a bijection
Explanation: Given that A = {x : −1 ≤ x ≤ 1} and f : A → A such that f(x) = x |x|.
For x < 0, f(x) < 0
⇒
y = -x2
−
−
−
⇒ x = √−y
, which is not possible for x > 0.
Hence, f is one-one and onto.
∴ the given function is bijective.
6. Let us take A = {2,4,6}
Define a relation R on A as:
A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}
Relation of R is reflexive as for every a ∈ A,
(a,a) ∈ R
⇒ (2,2), (4,4), (6,6) ∈ R,
Relation R is symmetric as (a,b) ∈ R
⇒ (b,a) ∈ R for all a ,b ∈ R
And Relation R is not transitive as (2,4), (4,6) ∈ R,
but (2,6) ∉ R
Therefore, relation R is reflexive and symmetric but not transitive.
7. Given: ∀ a, b ∈ Z, a R b if and only if a – b is divisible by 5.
Now,
R is Reflexive if (a, a) ∈ R ∀ a ∈ Z
a R a = (a-a) is divisible by 5.
a - a = 0 = 0 × 5 [since 0 is multiple of 5 it is divisible by 5]
= a - a is divisible by 5
= (a, a) ∈ R
Thus, R is reflexive on Z.
R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ Z
(a,b) ∈ R = (a-b) is divisible by 5
= (a - b) = 5z for some z ∈ Z
= -(b - a) = 5z
= b - a = 5(-z) [∵ z ∈ Z ⇒ -z ∈ Z]
= (b - a) is divisible by 5
= (b,a) ∈ R
Thus, R is symmetric on Z.
R is Transitive if (a,b) ∈ R and (b,c) ∈ R = (a,c) ∈ R ∀ a,b,c ∈ Z
(a,b) ∈ R = (a-b) is divisible by 5
= a-b = 5z1 for some z1∈ Z
(b,c) ∈ R = (b-c) is divisible by 5
= b-c = 5z2 for some z2∈ Z
Now,
a-b = 5z1 and b-c = 5z2
= (a-b) + (b-c) = 5z1 + 5z2
= a-c = 5(z1 + z2 ) = 5z3 where z1 + z2 = z3
= a-c = 5z3 [∵ z1,z2 ∈ Z ⇒ z3∈ Z]
= (a-c) is divisible by 5.
= (a, c) ∈ R
Thus, R is transitive on Z.
Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.
8. Given that,
R = {(1, 39), (2, 37), (3, 35) .... (19, 3), (20, 1)}
Domain = {1,2,3,.......,20}
Range = {1,3,5,7......,39}
2 / 10
R is not reflexive as (2, 2) ∉ R as
2 × 2 + 2 ≠ 41
R is not symmetric
as (1, 39) ∈ R but (39, 1) ∉ R
R is not transitive
as (11, 19) ∈ R, (19, 3) ∈ R
But (11, 3) ∉ R
Hence, R is neither reflexive, nor symmetric and nor transitive.
9.
i. (d) (X,Y) ∉ R
ii. (a) both (X,W) and (W,X) ∈ R
iii. (a) (F1, F2) ∈ R, (F2, F3) ∈ R and (F1, F3) ∈ R
iv. (c) Equivalence relation
v. (a) All those eligible voters who cast their votes
10.
(b) −
24
25
Explanation: put
cos
−1
sin 2θ = 2sinθcosθ =
11.
2.4
5
3
5
) = θ ⟹
. (−
3
5
) = (−
cos θ = −
24
25
3
5
therefore the given expression is
).
π
(a)
4
Explanation: sin
= sin
12.
(−
−1
(sin
π
4
−1
)=
(sin
π
3π
4
)
= sin
−1
(sin(π −
π
4
)
.
4
π
(d)
6
Explanation: Let the principle value be given by x
Now, let x = cosec-1(2)
⇒ cosec x = 2
⇒
cosec x = cosec (
(c) (−
(b) [
2
−π
2
) (∵ cos(
π
6
) = 2)
6
π
,
π
2
)
Explanation: (−
14.
6
π
⇒ x =
13.
π
,
π
2
]
π
2
,
π
2
)
- {0}
Explanation: To Find: The range of coses-1(x)
Here, the inverse function is given by y = f-1(x)
The graph of the function coses-1(x) can be obtained from the graph of
Y = coses-1(x) by interchanging x and y axes.i.e, if a, b is a point on Y = cosec x then b, a is the point on the function y = coses1(x)
Below is the Graph of the range of coses-1(x)
From the graph, it is clear that the range of coses-1(x) is restricted to interval
[
,
] - {0}
−π
2
15. Let tan
π
2
−1
–
(− √3) = y
–
⇒ tan y = − √3
3 / 10
π
⇒ tan y = − tan
3
π
⇒ tan y = tan(−
)
3
Since, the principal value branch of tan-1 is [−
Therefore, principal value of tan
16. sin
−1
∘
(sin(− 600 )) = sin
= sin
= sin
= sin
−1
−1
−1
10π
{sin(−
3
π
3
π
3
–
(− √3)
{sin(−600 ×
)} = sin
{− sin(3π +
(sin
−1
−1
−1
(− sin
)} = sin
−1
10π
3
π
is − .
3
π
180
)}
)
{− (− sin
⇒
(
2 cos x
2
sin
1− cos
2
2
=
) = tan
−1
[∵ 1 − cos
sin x
x
x
π
3
)}
3
−1
⇒ tan
.
π
) =
2 cos x
]
2
π
17. Here, we are required to find the value of x,
Now, the given equation is 2 tan (cos x) = tan
−1
π
,
2
(
2
2
sin x
) [∵ 2 tan
x = sin
2
−1
−1
(2cosecx)
x = tan
−1
(
2x
2
1−x
); −1 < x < 1]
x]
sinx cosx - sin2x = 0
⇒ sinx(cosx - sinx) = 0
⇒ sinx = 0 or cosx = sinx​​
⇒ sin x = sin 0 or cot x = 1 = cot π/4
⇒
π
∴ x = 0 or
4
But here at x = 0, the given equation does not exist.
π
Hence, x =
18.
is the only solution.
4
(b) 3 × 3
⎡
1
−1
Explanation: ⎢ 0
⎣
⎡
1
= ⎢0
⎣
⎡
= ⎢
⎣
19.
2
−1
2
3
2
2
3
⎤
⎤
−1
⎦
2
]
2
⎦
0
1
−
2×3
0
1
23
1
0
21
[
]
}
2×3
3×2
−1
−1
−21
1
0
−20
[
⎥
0
{[
⎥
]
2×3
3×2
−2
−1
2
0
1
−2
−1
⎤
−40 ⎥
−102
⎦
3×3
(d) B2 = B and A2 =A
Explanation: AB = A ...(i)
BA = B ....(ii)
From equation (ii)
B × (AB) = B
B2A= B
From equation (ii)
B2A = BA
B2 = B
From equation (i)
A × (BA) = A
A2B= A
From equation (i)
A2B = AB
A2 = A
Hence, A2 = A & B2 = B
20.
(c) A skew-symmetric matrix
Explanation: Given A and B are symmetric matrices
A'= A ...(i)
4 / 10
B'= B ...(ii)
Now (AB – BA)'= (AB)'– (BA)’
= B'A'– A’B’
[∵ (AB)' = B’A']
= BA – AB [Using (i) and (ii)]
∴ (AB – BA)' = - (AB - BA)
AB - BA is a skew symmetric matrix.
21.
(c) square matrix
Explanation: We know that as matrix is said to be square matrix if the number of rows is equal to the number of columns.
0
0
4
Therefore, the given matrix P = ⎢ 0
4
0⎥
⎡
⎣
4
0
0
⎤
is a square matrix.
⎦
(c) A + At is symmetric
Explanation: For every square matrix (A + A’ ) is always symmetric.
Since (A+A')'=A'+A=A+A'
22.
1
2
−1
3
1
2
23. We have A = [
A + (B + C) = [
5
0
1
2
2
24. Given: A = ⎢ 2
1
2⎥
2
1
2
2
2
1
6
0
]+ [
8
0
1
−2
]
]
3
2
0
7
]= [
2
5
2
],C = [
6
3
and (A + B) + C = [
⎣
0
1
]+ [
−1
⎡
4
],B = [
7
2
1
6
]= [
1
−2
]
= A + (B + C)
⎤
⎦
To prove A2 – 4A – 5 I2 = O2
Here I2 is the identity matrix of order 2 and O2 is the zero matrix of order 2.
Now, we will find the matrix for A2, we get
1
2
2
1
2
2
= A× A= ⎢2
1
2⎥⎢2
1
2⎥
2
1
2
1
⎡
2
A
⎣
⇒
2
⎤⎡
⎦⎣
2
⎤
⎦
A2
1 × 1 + 2 × 2 + 2 × 2
1 × 2 + 2 × 1 + 2 × 2
1 × 2 + 2 × 2 + 2 × 1
=⎢ 2 × 1 + 1 × 2 + 2 × 2
2 × 2 + 1 × 1 + 2 × 2
2 × 2 + 1 × 2 + 2 × 1⎥
2 × 2 + 2 × 1 + 1 × 2
2 × 2 + 2 × 2 + 1 × 1
⎡
⎣
2 × 1 + 2 × 2 + 1 × 2
⎤
⎦
[as cij = ai1b1j + ai2b2j + … + ainbnj]
⇒
A2 =
1 + 4 + 4
2 + 2 + 4
2 + 4 + 2
⎢2 + 2 + 4
4 + 1 + 4
4 + 2 + 2⎥
4 + 2 + 2
4 + 4 + 1
⎡
⎣
9
8
8
A2 =⎢ 8
9
8⎥
⎡
⇒
2 + 4 + 2
⎣
8
8
9
⎤
⎦
⎤
.........(1)
⎦
So, Substitute corresponding values from eqn(i) in equation
A2 – 4A – 5 I2, we get
9
8
8
1
2
2
1
0
0
= ⎢8
9
8⎥ − 4⎢2
1
2⎥ − 5⎢0
1
0⎥
8
8
9
2
1
0
1
9
8
8
4 × 1
4 × 2
4 × 2
5
0
0
= ⎢8
9
8⎥ − ⎢4 × 2
4 × 1
4 × 2⎥ − ⎢0
5
0⎥
8
8
9
4 × 2
4 × 1
0
5
9
8
8
4
8
8
5
0
0
= ⎢8
9
8⎥ − ⎢8
4
8⎥ − ⎢0
5
0⎥
8
9
8
4
0
5
⎡
⇒
⎣
⎡
⇒
⎣
⎡
⇒
⎣
8
⎤
⎦
⎤
⎡
⎣
⎡
⎦
⎣
⎤
⎡
⎦
⎣
2
4 × 2
8
⎤
⎡
⎦
⎤
⎦
⎣
⎡
⎣
0
0
⎤
⎦
⎤
⎦
⎡
⎣
0
⎤
⎦
⎤
⎦
5 / 10
9 − 4 − 5
8 − 8 − 0
8 − 8 − 0
= ⎢8 − 8 − 0
9 − 4 − 5
8 − 8 − 0⎥
8 − 8 − 0
9 − 4 − 5
⎡
⇒
⎣
8 − 8 − 0
⎤
⎦
[as rij = aij + bij + cij],
0
0
0
⇒= ⎢ 0
0
0⎥
⎡
⎣
0
0
= O2
⎦
0
A2 –
Hence the
⎤
4A – 5I = O2 (Proved)
25. Given equations are
2x + 3y = [
2
3
4
0
...... (i)
]
2
−2
−1
5
3x + 2y = [
]
.....(ii)
eqn. (i) × 3 - eqn. (ii) × 2
6x+9y
=
6x+4y
=
− −
9
0
]
4
−4
−2
10
[
]
=
5y
y =
6
12
[
−
=
1
5
2
13
14
−10
[
]
2
13
14
−10
[
]
eqn. (i) × 2 - eqn. (ii) × 3
4
6
8
0
4x+6y=[
]
6
−6
−3
15
9x+6y =[
−
]
− =
−5x
x =
−
−2
12
11
−15
=[
1
5
]
2
−12
−11
15
[
]
26. i. (a): 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 W omen Children
F=
F amily A
4
4
4
2
2
2
[
F amily B
]
Calories
M an
R=
⎡
woman
P roteins
2400
⎢ 1900
⎣
C hildren
1800
45
⎤
55 ⎥
33
⎦
ii. (b): The requirement of calories and proteins for each of the two families is given by the product matrix FR.
FR = [
4
4
⎡
2400
] ⎢ 1900
2
=[
4
2
2
⎣
1800
45
⎤
55 ⎥
33
⎦
4(2400 + 1900 + 1800)
4(45 + 55 + 33)
2(2400 + 1900 + 1800)
2(45 + 55 + 33)
]
Calories P roteins
FR = [
24400
532
F amily A
]
12200
266
F amily B
iii. (c)
iv. (c) 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
6 / 10
v. (a) A = (aij)m × n, B = (bij)n × p, C = (cij)p × q
BC = (bij)n × p × (Cij)p × q = (dij)n × q
(BC)A = (dij)n × q × (aij)m × M
Hence, (BC)A is possible only when m = q
27.
(c)
1
2
2
4
3
−5
[
]
Explanation: A = BX
B-1A = B-1BX
X = B-1A
Using Adjoint method of inverse
−1
B
1
=
2
2
0
0
1
[
]
X = B-1A
X =
X =
28.
1
2
1
2
2
0
[
1
2
3
−5
][
0
1
2
4
3
−5
[
]
]
(b) none of these
Explanation: 3A3 + 2A2 + 5A + I = 0 By pre-operating by A-1, we get
29.
⇒
3A-1 A3 + 2A-1 A2 + 5A-1 A + A-1 I = A-1 0
⇒
3A2 + 2A + 5 I + A-1 = 0, By making A-1 as the subject, we get
⇒
A-1 = -(3A2 + 2A + 5I)
(c) Idempotent
Explanation: clearly for given matrix A
Therefore idempotent
2
30.
(b)
= A
1
det(A)
Explanation: We know that, A-1 =
So, ∣∣A
−1
∣
∣
= ∣∣
=
=
1
|A|
1
| Adj(A)|
n
|A|
n−1
1
|A|
Adj (A)
∣
Adj(A)
∣
n
|A|
1
|A|
=
1
1
|A|
1
=
1
|A|
{since adj(A) is of order n and |Adj(A)| = |A|n-1}
31.
(d) A
Explanation: A2 = I
A-1A2 = A-1I
A = A-1
32. a
11
= 2, a12 = −3, a13 = 5
A31 = −12, A32 = 22, A33 = 18
L.H.S = a
11 A31
+ a12 A32 + a13 A33
= 2 (−12) + (−3) (22) + 5 (18)
= 0 Hence proved.
33. Given set of lines are:x+y-z=1
3x + y - 2z = 3
x-y-z=-1
Converting the following equations in matrix form,
AX = B
7 / 10
1
1
−1
⎢3
1
−2 ⎥ ⎢ y ⎥ = ⎢
⎡
⎣
1
−1
x
⎤⎡
⎦⎣
−1
z
⎤
1
⎡
⎦
⎣
⎤
⎡
⎤
3
⎥
−1
⎦
R2 - 3R1
R3 - R1
1
⎡
⎢0
⎣
0
1
−1
−2
1
−2
0
x
⎤⎡
1
⎥⎢y ⎥ = ⎢
⎦⎣
z
⎦
⎣
⎤
0
⎥
−2
⎦
Again converting into equations we get
x+y-z=1
-2y + z = 0
-2y = - 2
⇒ y = 1
-2 + z = 0
⇒z = 2
x+1-2=1
⇒ x = 2
∴ x = 2, y = 1, z = 2
∣
1
34. We have, ∣ cos C
1
∣ cos B
∣
∴ LHS =
∣
∣
cos B ∣
cos C
∣
∣
cos A
cos A
∣
1
1
cos C
cos C
1
cos A
cos A
1
∣ cos B
= 0
∣
cos B ∣
∣
∣
= 0
∣
Expanding along R1
2
= 1 (1 − cos A) − cos C (cos C − Cos A. cos B) + cos B (cos C . cos A − cos B)
2
2
2
2
2
= sin A − cos C + cos A. cos B. cos C + cos A. cos B. cos C − cos B
2
= sin A − cos B + 2 cos A. cos B. cos C − cos C
2
= − cos(A + B). cos(A − B) + 2 cos A. cos B. cos C − cos C
2
2
[∵ cos B − sin A = cos(A + B). cos(A − B)]
= − cos(−C ). cos(A − B) + cos C (2 cos A cos B − cos C ) [∵ cos(−θ = cos θ)]
= − cos C (cos A. cos B = sin A. sin B − 2 cos A. cos B + cos C )
= cos C (cos A. cos B − sin A. sin B − cos C )
= cos C (cos(A + B) − cos C )
= cos C (cos C − cos C ) = 0 = RHS
Hence proved.
35. x − y + z = 4
x − 2y − 2z = 9
2x + y + 3z = 1
1
−1
Let A = ⎢ 1
−2
⎡
⎣
2
1
⎤
⎡
x
⎤
⎡
4
⎤
−2 ⎥ X = ⎢ y ⎥ C = ⎢ 9 ⎥
1
3
⎦
⎣
z
⎦
⎣
1
⎦
Then, given system of equations can be rewritten as,
AX = C
1
−1
Now, AB = ⎢ 1
−2
⎡
⎣
2
8
0
0
= ⎢0
8
0⎥
0
8
⎡
⎣
0
1
1
−4
4
4
−2 ⎥ ⎢ −7
1
3
−3
−1
3
⎤⎡
⎦⎣
5
⎤
⎥
⎦
⎤
⎦
AB = 8I
−1
−1
A
=
1
8
B[
∵ A
−1
AB = 8A
−1
I
]
B = 8A
8 / 10
−1
1
1
2
2
2
⎢
⎢
⎢
−7
1
3
8
8
8
⎣
5
−3
−1
8
8
8
⎡
−1
⇒ A
=
1
8
⎤
⎥
⎥
⎥
⎦
Now, AX = C,
⇒
X = A-1C
⎡
−1
1
1
2
2
2
1
3
⎡
x
⎤
⎢
⇒ ⎢ y ⎥ = ⎢
⎢
⎣
⎦
z
⎣
−7
8
8
8
5
−3
−1
8
8
8
−4
⎡
⎢
= ⎢
⎢
⎣
2
−28
8
20
8
9
+
2
9
+
8
−27
+
8
1
+
+
4
⎡
⎤
⎥
⎥⎢9⎥
⎥
⎣
⎦
1
⎦
⎤
2
+
⎤
3
8
−1
3
⎡
⎤
⎥
⎥ = ⎢ −2 ⎥
⎥
⎣
⎦
−1
⎦
8
⇒ x = 3, y = −2, z = −1
36. 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
37. 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.
38. 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.
39. 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,
(1, 2), (2, 4) ∈ R
⇒ (1, 4) ∈ R
⇒ R is transitive.
Therefore, R is reflexive and transitive but not symmetric.
∣
40. Required Area = ∣
∣
∣0
1
∣
2
∣
3
0
1 ∣∣
–
√3
1
–
∣3
− √3
–
–
[1(−3√3 − 3√3)]∣
∣
=∣
=
1
2
6√3
2
–
= 3√3
∣∣
∣∣
1 ∣∣
sq units
41. 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.
42. Area of equilateral triangle =
–
∴ 3√3
=
√3
4
√3
4
(side)2
(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.
–
Then, × base × height = 3√3
⇒
1
2
9 / 10
or,
1
2
×
–
side × height = 3√3
or, height =
3√3×2
2√3
= 3 units.
43. 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
2
1
∣3
6
1∣
1
∣
2
∣
∣
∣
Since the three points are collinear, the area formed will be zero.
∣x
⇒
⇒
⇒
y
1∣
1
2
1
∣3
6
1∣
∣
∣
∣
∣
=0
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
⇒
10 / 10