MCQ’s on Relations Functions and Binary Operations Class 12 OP Malhotra ISC Maths Solutions Ch-1,2&3. In this article you would learn full concept recap of Relation , Functions and Binary Operations. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Relation , Functions and Binary Operations Class 12 OP Malhotra MCQ and short answer ISC Maths Solutions Ch-1,2&3

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-1,2&3	Relation , Functions and Binary Operations
Writer	OP Malhotra
Exe	MCQ and short answer

MCQ’s on Relations Functions and Binary Operations

Relation , Functions and Binary Operations Class 12 OP Malhotra MCQ and short answer Solutions

Fill in the Blanks

1. If the relation R in a set A is reflexive, symmetric and transitive,then R is called an __________ relation.

Ans- If the relation R in a set A is reflexive, symmetric and transitive,then R is called an equivalence relation.

2. A relation R in set A is called transitive if (a, b) ∈ R and (b, c) ∈ R ⇒__________ for all a, b, c ∈ A.

Ans- A relation R in set A is called transitive if (a, b) ∈ R and (b, c) ∈ R ⇒ (a,c)∈R for all a, b, c ∈ A.

3. Let the relation R be defined in N by a R b, if 2a + 3b = 30, then R =__________.

Ans- Let the relation R be defined in N by a R b, if 2a + 3b = 30, then R ={(12,2), (9,4), (6,6), (3,8)}.
Sol : Given:
aRb if 2a + 3b = 30

2a = 30 − 3b

Checking natural values of b:

b = 2 ⇒ a = 12
b = 4 ⇒ a = 9
b = 6 ⇒ a = 6
b = 8 ⇒ a = 3

Therefore,

R = {(12,2), (9,4), (6,6), (3,8)}

4. Every function is __________ invertible. Only__________ functions are invertible.

Ans- Every function is not invertible. Only one-one(bijective) functions are invertible.

5. If f : R → R is defined by f(x) = 2x + 3, then f⁻¹(x) is given by__________.

Ans- If f : R → R is defined by f(x) = 2x + 3, then f⁻¹(x) is given by (x-3)/2
Sol: Given:
f(x) = 2x + 3

Let y = 2x + 3
x = (y − 3) / 2

Replacing y by x,
f⁻¹(x) = (x − 3) / 2

6. If f(x) = (x + 3), x ∈ R and g(x) = x − 3, x ∈ R, then fog(3) =__________.

Ans- If f(x) = (x + 3), x ∈ R and g(x) = x − 3, x ∈ R, then fog(3) = 3
Sol: Given:
f(x) = x + 3
g(x) = x − 3

First find g(3):
g(3) = 3 − 3 = 0

Now,
f(g(3)) = f(0) = 0 + 3 = 3
∴ fog(3) = 3

7. If f : A → B and g : B → C be the bijective functions, then (g∘f)⁻¹ is __________.

Ans- If f : A → B and g : B → C be the bijective functions, then (g∘f)⁻¹ is f⁻¹og⁻¹

8. Let the functions f, g, h be defined from R to R such that f(x) = x² − 1 , g(x) = √(x² + 1),
h(x) = { 0, if x<0 and x , if x>0   then ho(fog)(x) = ___________ .

Ans- Let the functions f, g, h be defined from R to R such that f(x) = x² − 1 , g(x) = √(x² + 1),
h(x) = { 0, if x<0 and x , if x>0   then ho(fog)(x) = x²
Sol: Given:
f(x) = x² − 1
g(x) = √(x² + 1)

First find f(g(x)):
f(g(x)) = (√(x² + 1))² − 1
= x² + 1 − 1
= x²

Now apply h:
Since x² ≥ 0 for all x,
h(x²) = x²
∴ h ∘ (f ∘ g)(x) = x²

9. Let * be a binary operation on N given by a * b = LCM(a, b) for all a, b ∈ N.Then 5 * 7 = __________.

Ans- Let * be a binary operation on N given by a * b = LCM(a, b) for all a, b ∈ N.Then 5 * 7 = 35
Sol: Since , 5 and 7 are prime numbers the lcm of 5 and 7 will be 5×7 = 35

10. If f be the greatest integer function defined as f(x) = [x] and g be the modulus function defined as g(x) = |x|, then the value of gof(−4/3)) is __________.

Ans- If f be the greatest integer function defined as f(x) = [x] and g be the modulus function defined as g(x) = |x|, then the value of gof(−4/3)) is 2
Sol: Given:
f(x) = [x] (Greatest Integer Function)
g(x) = |x|

First find f(−4/3):
−4/3 = −1.33…
Greatest integer ≤ −1.33 is −2
f(−4/3) = −2

Now apply g:
g(−2) = |−2| = 2
∴ g ∘ f(−4/3) = 2

 

Multiple Choice Questions

1. If R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} is a relation on the set A = {3, 6, 9, 12}, then the relation is:
(a) an equivalence relation
(b) reflexive and symmetric
(c) reflexive and transitive
(d) only reflexive

Ans- (c) reflexive and transitive
Given:
A = {3, 6, 9, 12}

R = {(3,3), (6,6), (9,9), (12,12), (6,12), (3,9), (3,12), (3,6)}

Since (3,3), (6,6), (9,9), (12,12) ∈ R,
R is reflexive.

(6,12) ∈ R but (12,6) ∉ R,
So R is not symmetric.

(3,6) and (6,12) ∈ R ⇒ (3,12) ∈ R,
Hence R is transitive.

∴ Relation is reflexive and transitive.

2. If R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a relation on the set A = {1, 2, 3, 4}, then the relation R is:
(a) a function
(b) transitive
(c) not symmetric
(d) reflexive

Ans- (c) not symmetric
Given:
A = {1, 2, 3, 4}

R = {(1,3), (4,2), (2,4), (2,3), (3,1)}

R is not reflexive since (1,1), (2,2), (3,3), (4,4) are not in R.

R is not a function since 2 is related to both 4 and 3.

(2,3) ∈ R but (3,2) ∉ R,
So R is not symmetric.

(1,3) and (3,1) ∈ R but (1,1) ∉ R,
So R is not transitive.

∴ R is (c) not symmetric.

3. The maximum number of equivalence relations on the set A = {1, 2, 3} are:
(a) 1
(b) 2
(c) 3
(d) 5

Ans- (d) 5
Given, set A = {1, 2, 3}

Now, the number of equivalence relations are as follows:
R₁ = {(1,1), (2,2), (3,3)}
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}
= A × A

Thus, maximum number of equivalence relations is 5.

4. If a relation R is defined by R = {aRb if a ≥ b}, then R is:
(a) an equivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric

Ans- (b) reflexive, transitive but not symmetric.
Given:
R = {(a, b) | a ≥ b}

Reflexive:
Since a ≥ a for all a, R is reflexive.

Symmetric:
If a ≥ b, it does not imply b ≥ a (unless a = b).
So R is not symmetric.

Transitive:
If a ≥ b and b ≥ c, then a ≥ c.
So R is transitive.

∴ R is (b) reflexive, transitive but not symmetric.

5. Let S denote the set of all straight lines in a plane. Let a relation R be defined by m R n ⇔ m perpendicular to n , m,n∈S. Then R is:
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) None of these

Ans- (d) None of these
Relation: m R n ⇔ m ⟂ n

Not reflexive (a line is not perpendicular to itself).
Symmetric (if m ⟂ n, then n ⟂ m).
Not transitive.

∴ R is not reflexive,symmetric and transitive .

6. If f(x) = 8x³ and g(x) = x^1/3, then find (g∘f)(x):
(a) x
(b) x²
(c) 2x
(d) None of these

Ans- (c) 2x
f(x) = 8x³, g(x) = x^1/3
(g ∘ f)(x) = g(8x³) = (8x³)^1/3 = 2x

7. If f(x) = (2x − 1)/(x + 5), x ≠ −5, then f⁻¹(x) is equal to:
(a) (x + 5)/(2x − 1), x ≠ 1/2
(b) (5x + 1)/(2 − x), x ≠ 2
(c) (x − 5)/(2x + 1), x ≠ 1/2
(d) (5x − 1)/(2 − x), x ≠ 2

Ans- (b) (5x + 1)/(2 − x), x ≠ 2
Let y = (2x − 1)/(x + 5)

yx + 5y = 2x − 1
x(y − 2) = −(1 + 5y)
x = (5y + 1)/(2 − y)

∴ f⁻¹(x) = (5x + 1)/(2 − x), x ≠ 2

8. If f : N → R be the function defined by f(x) = (2x − 1)/2 and g : Q → R be another function defined by g(x) = x + 2, then (g∘f)(3/2) is:
(a) 1
(b) 3
(c) 7/2
(d) None of these

Ans- (b) 3
f(3/2) = (2(3/2) − 1)/2 = (3 − 1)/2 = 1
g(1) = 1 + 2 = 3

9. If f(x) = x² − 1 and g(x) = (x + 1)², then (g∘f)(x) is equal to:
(a)(x + 1)⁴-1
(b)x⁴-1
(c)x⁴
(d) (x + 1)⁴

Ans- (c) x⁴
(g ∘ f)(x) = g(x² − 1)
= (x² − 1 + 1)²
= (x²)²
= x⁴

10. If f : R → R and g : R → R are defined by f(x) = x − 3 and g(x) = x² + 1, then the value of x for which g(f(x)) = 10 are:
(a) 0,-6
(b) 2,-2
(c) 1,-1
(d) 0,6

Ans- (d) 0,6
g(f(x)) = (x − 3)² + 1
(x − 3)² + 1 = 10
(x − 3)² = 9
x − 3 = ±3
x = 0, 6

11. If R is the set of real numbers and the functions f : R → R and g : R → R are defined by f(x) = x² + 2x − 3 and
g(x) = x + 1, then the value of which f(g(x)) = g(f(x)) is:
(a) −1
(b) 0
(c) 1
(d) 2

Ans- (a) -1
f(g(x)) = (x + 1)² + 2(x + 1) − 3
= x² + 4x

g(f(x)) = x² + 2x − 3 + 1
= x² + 2x − 2

x² + 4x = x² + 2x − 2
2x = −2
x = −1

12. If f : R → R is given by f(x) = tan x, then f⁻¹(1) is:
(a) π/4
(b) {nπ+π/4 , n∈Z}
(c) does not exist
(d) None of these

Ans- (a) π/4
f(x) = tan x
tan x = 1 ⇒ x = π/4 (principal value

13. The binary operation * : R × R → R is defined as a * b = 2a + b. Then (2 * 3) * 4 is:
(a) 24
(b) 9
(c) 56
(d) 18

Ans- (d) 18
a * b = 2a + b
2 * 3 = 2(2) + 3 = 7
7 * 4 = 2(7) + 4 = 18

14. If a * b denotes the larger of ‘a’ and ‘b’ and if a∘b = (a*b) + 3, then (5)∘(10) is:
(a) 18
(b) 53
(c) 13
(d) None of these

Ans- (c) 13
5 * 10 = 10 (larger number)
10 + 3 = 13

15. Let * be a binary operation on set Q of rational numbers defined as a * b = ab/5, the identity for * if any is:
(a) 0
(b) 1
(c) 5
(d) None of these

Ans- (c) 5
a * e = a ⇒ ae/5 = a ⇒ e = 5 

16. If * be binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R, then the operation ‘*’ is:
(a) commutative but not associative
(b) associative but not commutative
(c) neither commutative nor associative
(d) both commutative and associative

Ans- (a) commutative but not associative
a * b = 1 + ab
1 + ab = 1 + ba ⇒ commutative
Not associative

17. Let A be the non-void set of children in a family. The relation “a is brother of” on A is:
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these

Ans- (d) none of these
“Brother of” is not reflexive, not symmetric, not transitive.

18. If R is a relation on the set N defined by {(x, y) : 2x – y = 10}, then R is:
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these

Ans- (d) None of these
2x − y = 10
Not reflexive, not symmetric, not transitive.

19. The number of equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) is:
(a) 3
(b) 1
(c) 2
(d) None of these

Ans- (c) 2
(1,2) and (2,1) present ⇒ 1 and 2 in same class.
Possible partitions: { {1,2},{3} }, { {1,2,3} }
Total = 2

20. On set A = {1, 2, 3}, relations R and S are given by
R = {(1,1), (2,2), (3,3), (1,2) ,(2,1)}
S = {(1,1), (2,2), (3,3), (3,1)}.
(a) R ∪ S is an equivalence relation
(b) R ∪ S is reflexive and transitive but not symmetric
(c) R ∪ S is symmetric and transitive but not reflexive
(d) R ∪ S is reflexive and symmetric but not transitive

Ans-
R ∪ S is reflexive.
Not symmetric, not transitive.

 

21. If A = {1, 2, 3, 4}, then which one of the following is reflexive?
(a) {(1,1), (2,3), (3,3)}
(b) {(1,1), (2,2), (3,3), (4,4)}
(c) {(1,2), (2,1), (3,2), (2,3)}
(d) {(1,2), (1,3), (1,4)}

Ans- (b){(1,1), (2,2), (3,3), (4,4)}
Reflexive relation must contain (1,1),(2,2),(3,3),(4,4).

22. If f be the greatest integer function defined as f(x) = [x] and g be the modulus function defined as g(x) = |x|, then the value of g∘f(−5/4) is:
(a) 2
(b) −2
(c) 5/4
(d) 1

Ans- (b) 2
−5/4 = −1.25
[−1.25] = −2
|−2| = 2

23. If f : R → R defined as f(x) = (3x + 5)/2 is an invertible function, write f⁻¹:
(a) (2y − 3)/5
(b) (2y − 2)/5
(c) (2y + 5)/3
(d) (2y − 5)/3

Ans-  (d) (2y-5)/3
y = (3x+5)/2
2y = 3x+5
x = (2y−5)/3

24. If f(x) = (x + 1)/(x − 1), then the value of f(f(x)) is equal to:
(a) x
(b) 0
(c) −x
(d) 1

Ans- (a) x
f(x) = (x + 1)/(x − 1)

f(f(x)) = [ ( (x + 1)/(x − 1) + 1 ) ] / [ ( (x + 1)/(x − 1) − 1 ) ]

Simplify numerator:
(x + 1)/(x − 1) + 1
= (x + 1)/(x − 1) + (x − 1)/(x − 1)
= 2x/(x − 1)

Simplify denominator:
(x + 1)/(x − 1) − 1
= (x + 1)/(x − 1) − (x − 1)/(x − 1)
= 2/(x − 1)

Now divide:
f(f(x)) = [2x/(x − 1)] ÷ [2/(x − 1)]
= (2x/(x − 1)) × ((x − 1)/2)
= x

∴ f(f(x)) = x

25. Let X = {1, 2, 3, 4} and Y = {a, b, c}. Then the mapping f : X → Y defined by f(1) = a, f(2) = b, f(3) = a, f(4) = b is:
(a) one-one into
(b) one-one onto
(c) many-one into
(d) None of these

Ans- (c) many-one into
1,3 → a and 2,4 → b
Many-one, not onto

26. If f : R → R be given by f(x) = tan x, then f⁻¹(1) is:
(a) π/4
(b) nπ ± π/4, n ∈ Z
(c) Does not exist
(d) None of these

Ans- (a) π/4
tan⁻¹(1) = π/4

27. If f(x) = (2x − 3)/(3x + 4), then f⁻¹(−4/3) is:
(a) 0
(b) 3/4
(c) −2/3
(d) None of these

Ans- (d) none of these
Denominator zero at x = −4/3
So inverse does not exist there.

28. Domain of the function f(x) = 1 / √(x + 2) is:
(a) (−2, ∞)
(b) (0, ∞)
(c) (2, ∞)
(d) R

Ans- (a)(−2, ∞)
1/√(x+2) ⇒ x+2 > 0 ⇒ x > −2

29. The domain of the function sin⁻¹(3x) is:
(a) [−2/3, 2/3]
(b) [−1/3, 1/3]
(c) [−1, 1]
(d) [−3, 2]

Ans- (b) [−1/3, 1/3]
|3x| ≤ 1 ⇒ |x| ≤ 1/3
Domain = [−1/3, 1/3]

30. The domain and range of the function f(x) = x /(1-x) are respectively:
(a) R − {1}, R
(b) R − {1}, R − {-1}
(c) R − {−1, 1}, R
(d) R − {−1, 1}, R − {0}

Ans- (b) R − {1}, R − {-1}
Domain: x ≠ 1
Range: y ≠ −1

31. The range of the function f(x) = cos(x/3) is:
(a) [−1, 1]
(b) [−1/3, 1/3]
(c) [1, ∞)
(d) None of these

Ans- (a) [-1,1]
Range of cos(x/3) = [−1,1]

32. Let f : R → R be defined by f(x) = 1/x, x ∈ R, then f(x) is:
(a) onto
(b) not defined
(c) one-one
(d) bijective

Ans- (c) one-one
f(x) = 1/x
One-one but not onto R

33. If A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6} are two sets, and function f : A → B is defined by f(x) = (x + 2) ∀ x ∈ A, then the function is:
(a) bijective
(b) onto
(c) one-one
(d) many-one

Ans- (c) one-one
f(x) = x + 2
One-one, not onto

34. The function f : R → R given by f(x) = x³ − 1 is:
(a) a one-one function
(b) an onto function
(c) a bijection
(d) neither one-one nor onto

Ans- (c) a bijection
x³ − 1 is one-one and onto

35. If the binary operation * on the set of integers Z is defined by a * b = a + 3b², then find the value of 2 * 4:
(a) 8
(b) 6
(c) 16
(d) 50

Ans- (d) 50
2 * 4 = 2 + 3(4²)
= 2 + 48
= 50

Very Short Answer Type Questions

1. State the reasons for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive.

Sol: Given R = {(1,2), (2,1)} on the set {1,2,3}.

For a relation to be transitive, if (a,b) and (b,c) are in R, then (a,c) must also be in R.

Here, (1,2) and (2,1) are in R, so (1,1) should be in R.
But (1,1) is not in R.

Also, (2,1) and (1,2) are in R, so (2,2) should be in R.
But (2,2) is not in R.

Therefore, R is not transitive.

2. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. State whether f is one-one or not.

Sol: Given A = {1, 2, 3} and B = {4, 5, 6, 7}.

f = {(1,4), (2,5), (3,6)}

The images of the elements of A are:
1 → 4
2 → 5
3 → 6

All elements of A have different images in B.

Therefore, f is one-one (injective).

3. Write the range of the function f(x) = |x − 1| / (x − 1).

Sol: Given f(x) = |x − 1| / (x − 1)

Case 1: x > 1
|x − 1| = x − 1
f(x) = (x − 1) / (x − 1) = 1

Case 2: x < 1
|x − 1| = −(x − 1)
f(x) = −(x − 1) / (x − 1) = −1

At x = 1, the function is not defined.

Therefore, the range of the function is {−1, 1}.

4. Find the identity element in the set ‘Q’ of all positive rational numbers for the operation defined by a * b = 3ab / 2 for all a, b ∈ Q.

Sol: Let the identity element be e.

Given operation:
a * b = (3ab) / 2

For identity element,
a * e = a

So,
a * e = (3ae) / 2 = a

Multiply both sides by 2:
3ae = 2a

Divide both sides by a (a ≠ 0):
3e = 2
e = 2 / 3

Therefore, the identity element is 2/3.

5. If f : R → R be given by f(x) = (3 − x³)^1/3, then find f ∘ f (x).

Sol: Given f(x) = (3 − x³)^1/3

We have
f ∘ f (x) = f(f(x))

Substitute f(x) in place of x:
f(f(x)) = (3 − [(3 − x³)^1/3]³)^1/3

Since [(3 − x³)^1/3]³ = 3 − x³
f(f(x)) = (3 − (3 − x³))^1/3
= (x³)^1/3
= x

Therefore, f ∘ f (x) = x.

6. Let * be an operation defined as * : R × R → R such that a * b = 2a + b, a, b ∈ R. Check if * is a binary operation. Find if it is associative also.

Sol: Binary Operation:
Given a * b = 2a + b, where a, b ∈ R.

Since 2a + b is a real number for all real values of a and b,
the operation is closed in R.

Therefore, * is a binary operation on R.

Associativity:
a * b = 2a + b

(a * b) * c = (2a + b) * c
= 2(2a + b) + c
= 4a + 2b + c

b * c = 2b + c

a * (b * c) = a * (2b + c)
= 2a + (2b + c)
= 2a + 2b + c

Since 4a + 2b + c ≠ 2a + 2b + c,

Therefore, * is not associative.

7. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min {a, b}. Write the operation table of operation *.

Sol: Given a * b = min {a, b} on the set {1, 2, 3, 4, 5}

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

8. Let * be a binary operation on N given by a * b = HCF(a, b), ∀ a, b ∈ N. Write the value of 22 * 4.

Sol: Given a * b = HCF(a, b), where a, b ∈ N.

So,

22 * 4 = HCF(22, 4)

Factors of 22: 1, 2, 11, 22
Factors of 4: 1, 2, 4

Highest Common Factor = 2

Therefore, 22 * 4 = 2.

9. Let {a, b, c} and the relation R be defined as R = {(a, a), (b, c), (a, b)}.
Write the minimum number of ordered pairs which should be added to R to make R reflexive and transitive.

Sol: Given A = {a, b, c}

R = {(a, a), (b, c), (a, b)}

make R reflexive

For reflexive relation, we need:
(a, a), (b, b), (c, c)

(a, a) is already present.
Missing: (b, b) and (c, c)

make R transitive

(a, b) and (b, c) are in R,
So (a, c) must be in R.

(a, c) is missing.

Ordered pairs to be added:
(b, b), (c, c), (a, c)

Minimum number of ordered pairs = 3.

10. If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f⁻¹.

Sol: Given A = {a, b, c, d}

f = {(a, b), (b, d), (c, a), (d, c)}

To find f⁻¹, interchange each ordered pair:

(a, b) → (b, a)
(b, d) → (d, b)
(c, a) → (a, c)
(d, c) → (c, d)

f⁻¹ = {(b, a), (d, b), (a, c), (c, d)}

11. If f : R − {3/5} → R be defined as f(x) = (3x − 2) / (5x − 3),then show that f⁻¹(x) = f(x).

Sol: Given f(x) = (3x − 2) / (5x − 3), where x ≠ 3/5

Let y = (3x − 2) / (5x − 3)
y(5x − 3) = 3x − 2
5xy − 3y = 3x − 2
5xy − 3x = 3y − 2
x(5y − 3) = 3y − 2
x = (3y − 2) / (5y − 3)

Replace y by x:
x = (3x − 2) / (5x − 3)

Therefore, f⁻¹(x) =

12. If f(x) = 4 − (x − 7)³, then find f⁻¹(x).

Sol: Given f(x) = 4 − (x − 7)³

Let y = 4 − (x − 7)³
y − 4 = −(x − 7)³
4 − y = (x − 7)³

Taking cube root on both sides:
x − 7 = ∛(4 − y)
x = 7 + ∛(4 − y)

Replacing y by x:
f⁻¹(x) = 7 + ∛(4 − x)

–: End of MCQ’s on Relations Functions and Binary Operations Class 12 OP Malhotra ISC Maths Solutions :–

Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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