Kerala Syllabus Class 10 Biology Solutions Chapter 3 Chemical Messages for Homeostasis

 

Question 1.
Show that the relation R in the set is given by A = {x ∈ Z, 0 = x = 12}
R = {(a,b): a – b is a multiple of 4}
Answer:
Given; R = {(a,b): a – b is a multiple of 4}
Clearly a – a = 0 is a multiple of 4. So
(a, a) ∈ R, hence Reflexive
(a,b) ∈ R ⇒ a  -b is a multiple of 4, imply that b – a is a multiple of 4 ⇒ (b, a) ∈ R So Symmetric.
(a, b), (b, c) ∈ R ⇒ a – b and b – c are multiple of 4.
a – b + b – c is a multiple of 4
a – c is a multiple of 4 ⇒ (a, c) ∈ R So Transitive.
Hence R is an equivalence Relation.

 

Question 2.
Let R be a Relation in the set A = {1, 2, 3, 4, 5, 6} defines as R = {(x,y): y = 2x – 1}

1. Write R in roster form and find it’s domain and range
2. Is R an equivalence relation ? Justify.

Answer:
1. R ={(1, 1), (2, 3), (3, 5)}
Domain = {1, 2, 3} ; Range ={1, 3, 5}

2. Since (2, 2) ∉ R, R is not reflexive
(2, 3) ∈ R but (3, 2) ∉ R
R is not symmetric.
(2, 3) ∈ R, (3, 5) ∉ R but, (2, 5) ∉ R
R is not transitive
∴ R is not an equivalence relation.

Question 3.

Answer:
Given;

 

Question 4.
Let ‘ * ’ is a binary operation on the set Q of rational numbers as follows;

1. a * b = a – b
2. a * b = a + ab
3. a * b = a2 + b2
4. a * b = (a – b)2

Find which of the binary operation are commutative and associative. Also find the identity element if exists.
Answer:
1. Given; a * b = a – b
b * a = b – a ≠ a – b = a * b
Hence not commutative.
Now; a * (b * c) = a * (b – c)
= a – (b – c) – a – b + c
(a * b) * c = (a – b) * c = a – b – c
a * (b * c) ≠ (a * b) * c
Hence not associative.
Now; a * e = a ⇒ a – e = a ⇒ e = 0
e * a = a ⇒ e – a = a ⇒ e = 2a
a * e ≠ e * a.
So identity element does not exist.

 

2. Given; a * b = a + ab
b * a = b + ab ≠ a + ab = a * b
Hence not commutative.
Now; a * (b * c) = a * (b + bc)
= a + a(b + be) = a + ab + abc
(a * b) * c = (a + ab) * c
= a + ab + c(a + ab)
= a + ab + ac + abc
a * (b * c) * (a * b) * c
Hence not associative.
Now; a * e = a ⇒ a + ae = a ⇒ e = 0
e * a = a ⇒ e + ea = a ⇒ e = a/1+a
a * e ≠ e * a.
So identity element does not exist.

3. Given; a * b = a2 + b2
b * a = b2 + a2 = a2 + b2 = a * b
Hence commutative.
Now; a * (b * c) = a* (b2 + c2) = a2 + (b2 + c2)2
(a * b) * c = (a2 + b2) * c
= (a2 + b2)2 + c2
a * (b * c) ≠ (a * b) * c
Hence not associative.
Now; a * e – a ⇒ a2 + e2 = a

So identity element does not exist.

4. Given; a * b = (a – b)2
b * a = (b – a)2 = (a – b)2 = a * b
Hence commutative.
Now; a * (b * c) = a * (b – c)2 (a – (b-c)2)2
(a * b) * c – (a – b)2 * c = ((a – b)2 – c)2
a * (b * c) ≠ (a * b) * c
Hence not associative.
Now; a * e = √a ⇒ e = a – √a ∉ Q
So identity element does not exist.

Question 5.
Show that the relation R on the set of natural numbers defined as R: { (x, y): y – x is a multiple of 2} is an equivalance relation
Answer:
Since x – x = 0 is multiple of 2, (x, x)∈ R
Therefore reflexive.
If y – x is a multiple of 2 then x – y is also a multiple of 2. Therefore (x, y) ∈ R ⇒ (y, x) ∈ R. Hence symmetric.
If y – x is a multiple of 2 and z-y is a multiple of 2, then their sum y – x + z – y = z – x is a multiple of 2. Therefore (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R
Hence transitive.
Therefore R is an equivalance relation.

 

Question 6.
Consider f : [3, ∞) → [1, ∞) given by f(x) = x2 – 6x + 10. Find f-1
Answer:
f(x) = (x – 3)2 + 1
y = (x – 3)2 + 1
⇒ (x – 3)2 = y – 1

Question 7.
‘*’ be a binary operation on N × N defined as (a, b) * (c, d) – (ac, bd)

1. Show that * is commutative.
2. Find the identity element of * if any
3. Write an element of N × N which has an inverse.

Answer:
1. (a, b) * (c, d) = (ac, bd) = (ca, db)
= (c, d) * (a, b)
Hence * is commutative.

2. (a, b) * (e, e) = (a, b)
⇒ (ae, be) = (a, b)
⇒ ae = a ⇒ e = 1
⇒ be = b ⇒ e = 1
Therefore (1,1) is the identity element.

3. Invertible element is (1,1).

Question 8.
Show that the relation R in the set R of Real numbers defined asR = {(a, b): a ≤ b2} is neither reflexive, nor symmetric nor transitive.
Answer:
a ≤ a2 is not true for all a. So (a, a) ∉ R hence not Reflexive.
(a, b) ∈ R ⇒ a ≤ b2 does not imply that b ≤ a2. So (b, a) ∉ R, hence not Symmetric. (a, b),(b, c) ∈ R ⇒ a ≤ b2 and b ≤ c2 does not imply that a ≤ c2. So (a, c) ∉ R, hence not Transitive.

Plus Two Maths Relations and Functions Four Mark Questions and Answers

Question 1.
The relation R defined in the set A = {-1, 0, 1} as R = {(a,b):a = b2}

1. Check whether R is reflexive, symmetric and transitive.
2. Is R an equivalence relation

Answer:
1. (-1, -1) ∉ R, R is not reflexive
(-1,1) ∈ R but (1, -1) ∉ R, R is not symmetric
(-1,1) ∈ R, (1, 1) ∈ R and (-1, 1) ∈ R,
R is transitive.

2. R is not reflexive, not symmetric and not transitive. So R is not an equivalence relation.

 

Question 2.
Let A = {1, 2, 3}. Give an example of a relation on A which is

1. Symmetric but neither reflexive nor transitive.
2. Transitive but neither reflexive nor symmetric.

Answer:
1. R = {(1, 2),(2, 1)}
(1, 1) ∉ R ⇒ R is not reflexive
(1, 2) ∈ R ⇒ (2, 1) ∈ R, R is symmetric
(1, 2) ∈ R, (2, 1) ∈ R but (1,1) ∉ R,
R is not transitive

2. R = {(1, 2),(1, 3),(2, 3)}
(1, 1) ∉ R ⇒ R is not reflexive
(1, 2) ∈ R ⇒ but (2,1) ∉ R, R is not symmetric
(1, 2) ∈ R, (2, 3) ∈R ⇒ (1, 3) ∈ R,
R is transitive.

Question 3.
Find fog and gof if

1. f(x) – |x| and g(x) – |3x + 4|
2. f(x) = 16x4 and g(x) = x ¼

Answer:
1. Given; f(x) = |x| and g(x) = |3x + 4|
fog(x) = f(g(x))
= f(|3x + 4|) = ||3x + 4|| = |3x + 4|
gof(x) = g(f(x)) = g(|x|) = |3|x| + 4|

2. fog(x) = f(g(x)) = g(x¼) = 16(x¼)4 = x
gof (x) = g(f(x)) = g(16x4) = (16x4)¼= 4x.

 

Question 4.
Consider the binary operation* : Q → Q where Q is the set of rational numbers is defined as a * b = a + b – ab

1. Find 2 * 3 (1)
2. Is identity for * exist? If yes, find the identity element. (2)
3. Are elements of Q invertible? Is yes, find the inverse of an element in Q (1)

Answer:
1. 2 * 3 = 2 + 3 – 6 = -1.

2. a * e = a ⇒ a + e – ae = a
⇒ e(1 – a) – 0 ⇒ e = 0
e * a = a ⇒ e + a – ea = a
⇒ e(1 – a) = 0 ⇒ e = 0 is the identity element.

3. a * b = 0 ⇒ a + b – ab = 0 ⇒ b = a/a−1

 

Question 5.
‘*’ is a binary operation on R defined as a * b = 2ab

1. Determine whether * is commutative and associative
2. Find the identity element if exists.
3. Find the inverse element, if exists

Answer:
1. a * b = 2ab = 2ba = b * a
Therefore commutative
a * (b * c) = a * (2bc) = 4abc
(a * b) * c = (2ab) * c = 4 abc
Therefore is associative.

2. a * e = a ⇒ 2ae = a ⇒ e = 1/2
e * a = a ⇒ 2ea = a ⇒ e = 1/2
Therefore identity element is 1/2.

3. a * b = 1/2 ⇒ 2ab = 1/2 ⇒ b = 1/4a, a ≠ 0.

 

Question 7.
If f :R → R is a function defined by f(x) = 3x – 2

1. Show that f is one-one.
2. Find fof(x).
3. Find the inverse of f if exists.

Answer:
1. f(x1) = f(x2)
⇒ 2x1 – 3 = 2x2 – 3 ⇒ x1 = x2
Therefore one-one.

2. fof(x) = f(f(x)) = f(3x – 2)
f (x) = 3(3x – 2) – 2 = 9x – 6 – 2 = 9x – 8.

Question 8.
Let A = N × N and be the binary operation. On A defined by (a, b) * (c, d) = (a+c, b+d). Show that ‘ * ’ is commutative and associative. Find the identity for ‘ * ’ on A if any.
Answer:
Given; A = N × N and
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b)
= (a + c, b + d) = (a, b) * (c, d))
Hence commutative.
Now; (a, b) * [(c, d) * (e, f)] = (a, b) * [c + e, d+f]
= (a + c + e, b + d + f)
[(a, b) * (c, d)] * (e, f) = [a + c, b + d] * (e, f)
= (a + c + e, b + d + f)
Hence associative.
(a, b) * (e, e) = (a, b) ⇒ (a + e, b + e) = (a, b)
⇒ a + e = a, b + e = b ⇒ e=0, e = 0 ⇒ (0,0) ∉ A
So identity element does not exist.

 

Question 10.
1. Let f be a function defined by f(x) = √x is a function if it defined from (1)
(a) f : N → N
(b) f : R → R
(c) f : R → R+
(d) f : R+ → R+
2. Check the injectivity and surjectivity of the following functions (3)
(a) f : N → N defined by f(x) = x3
(b) f : R → R given by f(x) = [x]
Answer:
1.
(d) f : R+ → R+

2.
(a) For x, y ∈ N,
f(x) = f(y) ⇒ x3 = y3 ⇒ x = y
Therefore, f is injective
For 2 ∈ N, there does not exist x in the domain N such that f(x) = x3 = 2.
∴ f is not surjective.

(b) f : R → R given by f(x) = [x]
It seen that f(1.1) = 1 and f(1.8) = 1;
But 1.1 ≠ 1.8;
∴ f is not injective
There does not exist any element x ∈ R
such that f(x) = 0.7
∴ f is not surjective.

 

Plus Two Maths Relations and Functions Six Mark Questions and Answers

 

Question 2.
1. Show that the function f: R → R defined by f(x) = 2x – 3 is one-one and onto. Find f-1
2. Which of the following figure represents the graph of a function on R which is onto but not one-one.


3. Write a function on R which is onto but not one-one
Answer:
1. f(x) = 2x – 3
f(x1) = f(x2)
⇒ 2x1 – 3 = 2x2 – 3 ⇒ x1 = x2
Therefore f is one-one.
Let f(x) = yeR, then
y = 2x – 3 ⇒ 2x = y + 3 ⇒ x = y+3 /2 ∈ R
Therefore f is onto. Then f-1(x) = x+3/2.

2. Option (c)
[(b),(c),(d) are not one-one since for different values of x, we have same value of y. ie; horizontal line meets at more than one point, (b) and (d) are not onto since range and codomain are different].

3. f(x) = x2, f : R → [0, ∞) (Any other function).

 

Question 3.
A = {1, 2, 3, 4, 6}, * is a binary operation on A is defined as a * b = HCF of a and b.

1. Represent * with the help of an operation table.
2. Find the identity element.
3. Write a commutative binary operation on A with 3 as the identity element. (Hint: Operation table may be used.

Answer:
1.

2. From the table it is clear that the identity element is 6.

3. We can write any operation table which is commutative with 3 as the identity element.

 

Question 4.
Let * be a binary operation on the set of all real numbers R defined by a * b = a + b + a2b for a, b R.

1. Find 2 * 6 and 6 * 2. (2)
2. Prove that * is neither commutative nor associative. (2)
3. Find the identity elements in R if exists. (2)

Answer:
1. a * b = a + b + a2b
2 * 6 = 2 + 6 + 22 × 6 = 8 + 24 = 32
6 * 2 = 6 + 2 + 62 × 2 = 8 + 72 = 80.

2. a * b = a + b + a2b a * b ≠ b * a
b * a = b + a + b2a
∴ is not commutative.
(ab)c = (a + b + a2b) c
= a + b + a2b + c + (a + b + a2b)2c
a * (b * c) = a * (b + c + b2c)
= a + b + c + b2c + a2(b + c + b2c)
∴ is not associative.

3. Let e is the identity element.
∴ a * e = a ⇒ a + e + a2e = a ⇒ e(1 + a2) = 0
∴ e = 0
∴ e * a = a ⇒ e + a + e2a = a ⇒ e + e2a = 0
⇒ e(1 + ae) = 0
⇒ e = 0; (1 + ae) = 0
ea = -1 ⇒ e = – 1a
Identity element does not exists.

 

Question 5.

1. Consider f: {3, 4, 5, 6} → {8, 10, 12, 13, 14} And f = { (3, 8), (4, 10), (5, 12), (6, 14)}. State whether f has inverse ? Give reason. (2)
2. Consider f : R → R given by f(x) = 3x + 2 Show that f is invertible. Find the inverse of f

Answer:
1. Distinct elements in set {3, 4, 5, 6} has distinct images, under f.
∴ f is one- one
But 14 in the codomain has no pre image.
∴ f is not onto.
∴ f has no inverse.

2. f(x) = 3x + 2; then
f(x1) = f(x2) = 3x2 + 2 ⇒ x1= x2
Hence F is one – one
For y ∈ R, let y = 3x + 2 ⇒ x = y−2/3 ∈ R

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