Category: Class 12 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Binary Operations”.

1. Let a binary operation ‘*’ be defined on a set A. The operation will be commutative if ________
a) a*b=b*a
b) (a*b)*c=a*(b*c)
c) (b ο c)*a=(b*a) ο (c*a)
d) a*b=a

Answer: a
Clarification: A binary operation ‘*’ defined on a set A is said to be commutative only if a*b=b*a, ∀a, b∈A.
If (a*b)*c=a*(b*c), then the operation is said to associative ∀ a, b∈ A.
If (b ο c)*a=(b*a) ο (c*a), then the operation is said to be distributive ∀ a, b, c ∈ A.

2. Let a*b=6a⁴-9b⁴ be a binary operation on R, then * is commutative.
a) True
b) False

Answer: b
Clarification: The given statement is false. The binary operation ‘*’ is commutative if a*b=b*a
Here, a*b=6a⁴-9b⁴ and b*a=6b⁴-9a⁴
⇒a*b≠b*a
Hence, the ‘*’ is not commutative.

3. Let ‘*’ be a binary operation on N defined by a*b=a-b+ab², then find 4*5.
a) 9
b) 88
c) 98
d) 99

Answer: d
Clarification: The binary operation is defined by a*b=a-b+ab².
∴4*5=4-5+4(5²)=-1+100=99.

4. Let ‘*’ be defined on the set N. Which of the following are both commutative and associative?
a) a*b=a+b
b) a*b=a-b
c) a*b=ab²
d) a*b=a^b

Answer: a
Clarification: The binary operation ‘*’ is both commutative and associative for a*b=a+b.
The operation is commutative on a*b=a+b because a+b=b+a.
The operation is associative on a*b=a+b because (a+b)+c=a+(b+c).

5. Let ‘&’ be a binary operation defined on the set N. Which of the following definitions is commutative but not associative?
a) a & b=a-b
b) a & b=a+b
c) a & b=ab – 8
d) a & b=ab

Answer: c
Clarification: The binary operation ‘&’ is commutative but not associative for a*b=ab-8.
For Commutative: a & b=ab-8 and b & a=ba-8
ab-8=ba-8. Hence, a & b=ab-8 is commutative.
For Associative: (a &b)& c=(ab-8)& c=(ab-8)c-8=abc-8c-8=abc-8c-8.
a& (b &c)=a&(bc-8)=a(bc-8)-8=abc-8a-8.
⇒(a&b) & c≠a& (b& c). Hence, the function is not associative.

6. Let ‘*’ be a binary operation defined by a*b=4ab. Find (a*b)*a.
a) 4a² b
b) 16a² b
c) 16ab²
d) 4ab²

Answer: b
Clarification: Given that, a*b=4ab.
Then, (a*b)*a=(4ab)*a
=4(4ab)(a)=16a² b.

7. Let ‘*’ and ‘^’ be two binary operations such that a*b=a² b and a ^ b = 2a+b. Find (2*3) ^ (6*7).
a) 256
b) 286
c) 276
d) 275

Answer: c
Clarification: Given that, a*b=a² b and a ^ b = 2a+b.
∴(2*3)^(6*7)=(2²×3)^(6²×7)
=12^252=2(12)+252=276.

8. An element is said to be invertible only if there is an identity element in that binary operation.
a) True
b) False

Answer: a
Clarification: The given statement is true. If there is a binary operation *:M×M → M with an identity element a∈ M is said to be invertible with respect to the binary operation * if there exists an element b ∈ M such that a*b = e = b*a, b is called inverse of a.

9. Let ‘*’ be a binary operation defined by a*b=3a^b+5. Find 8*3.
a) 1547
b) 1458
c) 1448
d) 1541

Answer: d
Clarification: It is given that a*b=3a^b+5.
Then, 8*3=3(8³)+5=3(512)+5=1536+5=1541.

10. Which of the following is not a type of binary operation?
a) Transitive
b) Commutative
c) Associative
d) Distributive

Answer: a
Clarification: Transitive is not a type of binary operation. It is a type of relation. Distributive, associative, commutative are different types of binary operations.

2. What will be the value of f(x) = (begin{vmatrix}p & 2 – i & i + 1 \2 + i & q & 3 + i \1 – i & 3 – i & r end {vmatrix})?
a) Real
b) Imaginary
c) Zero
d) Can’t be predicted
Answer: a
Clarification: Here, f(x) = f’(x)
=> f(x) is purely real

3. If, S_i = aⁱ + bⁱ + cⁱ then what is the value of (begin{vmatrix}S0 & S1 & S2 \S1 & S2 & S3 \S2 & S3 & S4 end {vmatrix})?
a) (a + b)²(b – c)²(c – a)²
b) (a – b)²(b – c)²(c + a)²
c) (a – b)²(b – c)²(c – a)²
d) (a – b)²(b + c)²(c – a)²
Answer: c
Clarification: We have, (begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix})
So, the value of the (begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix}) = (a – b)(b – c)(c – a)
Now, by circulant determinant,
(begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix}) X (begin{vmatrix}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2 end {vmatrix}) = (begin{vmatrix}S0 & S1 & S2 \S1 & S2 & S3 \S2 & S3 & S4 end {vmatrix})
Multiplying the determinant in row by row,
We get, (a – b)²(b – c)²(c – a)²

4. Let, α and β be real. Find the set of all values of β for which the system of equation βx + sin α*y + cosα*z = 0, x + cosα * y + sinα * z = 0 , -x + sinα*y – cosα * z = 0 has a non-trivial solution. For β = 1 what are all values of α?
a) 2α = 2nπ ± π/2 + π/2
b) 2α = 2nπ ± π/2 + π/4
c) 2α = 2nπ ± π/4 + π/4
d) 2α = 2nπ ± π/4 + π/2
Answer: c
Clarification: The given system have non-trivial solution if (begin{vmatrix}beta & sin alpha & cos alpha \1 & cos alpha & sin alpha \ -1 & sin alpha & -cos alpha end {vmatrix}) = 0
On opening the determinant we get β = sin 2α + cos 2 α
Therefore, -√2 ≤ β ≤ √2
Now, for β = 1,
sin 2α + cos 2 α = 1
=> (1/√2)sin 2α + (1/√2) cos 2α = (1/√2)
Or, cos(2α – π/4) = 1/√2 = cos(2nπ ± π/4)
=> 2α = 2nπ ± π/4 + π/4

5. Which one among the following is correct if x, y, z are eliminated from, ((frac{bx}{y+z}) = a, (frac{cy}{z+x}) = b, (frac{az}{x+y}) = c)?
a) a²b + b²c + c²a + abc = 0
b) a²b – b²c + c²a + abc = 0
c) a²b + b²c + c²a + 2abc = 0
d) a²b – b²c – c²a – abc = 0
Answer: a
Clarification: (frac{bx}{y+z}) = a  bx – ay – az = 0
(frac{cy}{z+x}) = b  bx – cy + bz = 0
(frac{az}{x+y}) = c  cx + cy – az = 0
(begin{vmatrix}b & -a & -a \b & -c & b \c & c & -a end {vmatrix}) = 0
Or, b(ca – bc) + a(-ab – bc) – a(bc + c²) = 0
or, abc – b²c – a²b – abc – abc – ac² = 0
or, a²b + b²c + c²a + abc = 0 which is the required eliminate.

6. The co-ordinates of the vertices of a triangle are [m(m + 1), (m + 1)], [(m + 1)(m + 2), (m + 2)] and [(m + 2)(m + 3), (m + 3)]. Then which one among the following is correct?
a) The area of the triangle is dependent on m
b) The area of the triangle is independent on m
c) Answer cannot be predicted
d) Data inadequate
Answer: b
Clarification: The area o the triangle with the given point as vertices is,
1/2 (begin{vmatrix}m(m+1) & (m+1) & 1 \(m+1)(m+2) & (m+2) & 1 \(m+2)(m+3) & (m+3) & 1 end {vmatrix})
= 1/2 (begin{vmatrix}m^2 + m & (m+1) & 1 \m^2 + 3m + 2 & (m+2) & 1 \m^2 + 5m + 6 & (m+3) & 1 end {vmatrix})
Now, by performing the row operation R₂ = R₂ – R₁ and R₃ = R₃ – R₂
= 1/2 (begin{vmatrix}m^2 + m & (m+1) & 1 \2m + 2 & 1 & 0 \2m + 4 & 1 & 0 end {vmatrix})
Now, breaking the determinant we get,
= 1/2 (2m + 2 – 2m – 4)
= -1
Thus, it is independent of m.

7. Which one is correct, the following system of linear equations 2x – 3y + 4z = 7, 3x – 4y + 5z = 8, 4x – 5y + 6z = 9 has?
a) No solutions
b) Infinitely many solutions
c) Unique Solution
d) Can’t be predicted
Answer: b
Clarification: Solving the given system of equation by Cramer’s rule, we get,
x = D₁/D, y = D₂/D, z = D₃/D where,
D = (begin{vmatrix}2 & -3 & 4 \3 & -4 & 5 \4 & -5 & 6 end {vmatrix})
D = –(begin{vmatrix}2 & 3 & 4 \3 & 4 & 5 \4 & 5 & 6 end {vmatrix})
Now, performing, C₃ = C₃ – C₂ and C₂ = C₂ – C₁ we get,
D = –(begin{vmatrix}2 & 1 & 1 \3 & 1 & 1 \4 & 1 & 1 end {vmatrix})
As two columns have identical values, so,
D = 0
Similarly,
D₁ = (begin{vmatrix}7 & -3 & 4 \8 & -4 & 5 \9 & -5 & 6 end {vmatrix})
Now, performing, C₁ = C₁ – C₃
D₁ = –(begin{vmatrix}3 & -3 & 4 \3 & -4 & 5 \3 & -5 & 6 end {vmatrix})
Now, performing, C₃ = C₃ – C₂
D₁ = -3(begin{vmatrix}1 & -3 & 1 \1 & -4 & 1 \1 & -5 & 1 end {vmatrix})
As two columns have identical values, so,
D₁ = 0
D₂ = (begin{vmatrix}2 & 7 & 4 \3 & 9 & 5 \4 & 8 & 6 end {vmatrix})
Now, performing,
D₂ = –(begin{vmatrix}2 & 3 & 2 \3 & 3 & 2 \4 & 3 & 2 end {vmatrix})
Now, performing, C₂ = C₂ – C₃ and C₃ = C₃ – C₁
D₂ = 6(begin{vmatrix}2 & 1 & 1 \3 & 1 & 1 \4 & 1 & 1 end {vmatrix})
As two columns have identical values, so,
D₂ = 0
D₃ = (begin{vmatrix}2 & -3 & 7 \3 & -4 & 9 \4 & -5 & 6 end {vmatrix})
D₃ = –(begin{vmatrix}2 & 3 & 4 \3 & 4 & 4 \4 & 5 & 4 end {vmatrix})
Now, performing, C₂ = C₂ – C₂ and C₃ = C₃ – C₂
D₃ = -4(begin{vmatrix}2 & 1 & 1 \3 & 1 & 1 \4 & 1 & 1 end {vmatrix})
As two columns have identical values, so,
D₃ = 0
Since, D = D₁ = D₂ = D₃ = 0, thus, it has infinitely many solutions.

8. What will be the value of x if (begin{vmatrix}2-x & 2 & 3 \2 & 5-x & 6 \3 & 4 & 10-x end {vmatrix}) = 0?
a) 8 ±√37
b) -8 ± √37
c) 8 ± √35
d) -8 ± √35
Answer: a
Clarification: Here, we have, (begin{vmatrix}2-x & 2 & 3 \2 & 5-x & 6 \3 & 4 & 10-x end {vmatrix}) = 0
Now, replacing C₃ = C₃ – 3C₁, we get,
(begin{vmatrix}2-x & 2 & 3-6 + 3x \2 & 5-x & 6-6 \3 & 4 & 10-x -9 end {vmatrix}) = 0
(begin{vmatrix}2-x & 2 & 3(x-1) \2 & 5-x & 0 \3 & 4 & -(x-1) end {vmatrix}) = 0
Or, (x – 1)(begin{vmatrix} 2-x & 2 & 3 \2 & 5-x & 0 \3 & 4 & -1 end {vmatrix}) = 0
Now, replacing R₁ = R₁ + 3R₃, we get,
Or, (x – 1)(begin{vmatrix}11-x & 14 & 0 \2 & 5-x & 0 \3 & 4 & -1 end {vmatrix}) = 0
Or, (x – 1)[-1 {(11 – x)(5 – x) – 28}] = 0
Or, -(x – 1)(55 – 11x – 5x + x² – 28)
Or, (x – 1)(x² – 16x + 27) = 0
Thus, either x – 1 = 0 i.e. x = 1 or x² – 16x + 27 = 0
Therefore, solving x² – 16x + 27 = 0 further, we get,
x = 8 ± √37

9. What is the relation between the two determinants f(x) = (begin{vmatrix}–a^2 & ab & ac \ab & -b^2 & bc \ac & bc & -c^2 end {vmatrix}) and g(x) = (begin{vmatrix}0 & c & b \c & 0 & a \b & a & 0 end {vmatrix})?
a) f(x) = g(x)
b) f(x) = (g(x))²
c) g(x) = 2f(x)
d) g(x) = (f(x))²
Answer: b
Clarification: Let, D = (begin{vmatrix}0 & c & b \c & 0 & a \b & a & 0 end {vmatrix})
Expanding D by the 1^st row we get,
D = – c(begin{vmatrix}c & a \b & 0 end {vmatrix}) + b(begin{vmatrix}c & 0 \b & a end {vmatrix})
= – c(0 – ab) + b(ac – 0)
= 2abc
Now, we have adjoint of D = D’
= (begin{vmatrix}
begin{vmatrix}0 & a\ a & 0\ end{vmatrix} & – begin{vmatrix}c & a\ b & 0\ end{vmatrix} & begin{vmatrix}c & 0\ b & a\ end{vmatrix}\
– begin{vmatrix}c & b\ a & 0\ end{vmatrix} & begin{vmatrix}0 & b\ b & 0\ end{vmatrix} & – begin{vmatrix}0 & c\ b & 0\ end{vmatrix} \
begin{vmatrix}c & b\ 0 & a\ end{vmatrix} & – begin{vmatrix}0 & b\ c & a\ end{vmatrix} & begin{vmatrix}0 & c\ c & 0\ end{vmatrix} \
end{vmatrix})
Or, D’ = (begin{vmatrix}–a^2 & ab & ac \ab & -b^2 & bc \ac & bc & -c^2 end {vmatrix})
Or, D’ = D²
Or, D’ = D² = (2abc)²

Mathematics MCQs for Class 12,