4 Marks

Question 514 Marks

Let Z be the set of all integers and Z₀ be the set of all non-zero integers. Let a relation R on Z × Z₀ be defined as (a, b)R(c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ Z × Z₀, Prove that R is an equivalence relation on Z × Z₀.

Answer

We have, Z be the set of integers and Z₀ be the set of non-zero integers.
R = {(a, b), (c, d): ad = bc} be a relation on Z × Z₀
Now,
Reflexivity: (a, b) ∈ Z × Z₀
⇒ ab = ba
⇒ {(a, b), (a, b)} ∈ R
⇒ R is reflexive.
Symmetric: Let {(a, b), (c, d)} ∈ R
⇒ ad = bc
⇒ cd = da
⇒ {(c, d), (a, b)} ∈ R
⇒ R is symmetric.
Transitive: Let (a, b), (c, d) ∈ R and (c, d), (e, f) ∈ R
⇒ ad = bc and cf = de
$\Rightarrow\ \frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}$ and $\frac{\text{c}}{\text{d}}=\frac{\text{e}}{\text{f}}$
$\Rightarrow\ \frac{\text{a}}{\text{b}}=\frac{\text{e}}{\text{f}}$
$\Rightarrow\ \text{af}=\text{be}$
$\Rightarrow\ (\text{a, b})(\text{e, f})\in\text{R}$
⇒ R is transitive.
Hence, R is an equivalence relation on Z × Z₀

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Question 524 Marks

Let f : N → N be defined by $\text{f(n)}=\begin{cases}\text{n}+1,&\text{if n is odd}\\\text{n}-1,&\text{if n is even}\end{cases}$ Show that f is a bijection.

Answer

We have, $\text{f(n)}=\begin{cases}\text{n}+1,&\text{if n is odd}\\\text{n}-1,&\text{if n is even}\end{cases}$

Injection test:

Case I: If n is odd,

Let $\text{x, y}\in\text{N}$ such that f(x) = f(y)

As, f(x) = f(y)

⇒ x + 1 = y + 1

⇒ x = y

Case II: If n is even,

Let $\text{x, y}\in\text{N}$ such that f(x) = f(y)

As, f(x) = f(y)

⇒ x - 1 = y - 1

⇒ x = y

So, f is injective.

Surjection test:

Case I: If n is odd,

As, for every $\text{n}\in\text{N},$ there exists y = n - 1 in N such that

f(y) = f(n - 1) = n - 1 + 1 = n

Case II: If n is even,

As, for every $\text{n}\in\text{N},$ there exists y = n + 1 in N such that

f(y) = f(n + 1) = n + 1 - 1 = n

So, f is surjective.

So, f is bijection.

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Question 534 Marks

Let R be the relation defined on the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b): both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other and all the elements of the subset {2, 4, 6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

Answer

We observe the following properties of R.
Reflexivity: Let a be an arbitrary element of R. Then,
$\text{a}\in\text{R}$
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ Both a and b are either even or odd.
⇒ Both b and a are either even or odd.
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{A}$
So, R is symmetric on A.
Transitivity: Let (a, b) and (b, c) $\in\text{R}$
⇒ Both a and b are either even or odd and both b and c are either even or odd.
⇒ a, b and c are either even or odd.
⇒ a and c both are either even or odd.
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a, c}\in\text{A}$
So, R is transitive on A.
Thus, R is an equivalence relation on A.
We observe that all the elements of the subset {1, 3, 5, 7} are odd. Thus, they are related to each other.
This is because the relation R on A is an equivalence relation.
Similarly, the elements of the subset {2, 4, 6} are even. Thus, they are related to each other because every element is even.
Hence proved.

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Question 544 Marks

Classify the following functions as injection, surjection or bijection:

f : R → R, defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$

Answer

f : R → R, defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$

xy² + x = x²y + y

xy² - x²y + x - y = 0

-xy(-y + x) + 1(x - y) = 0

(x - y)(1 - xy) = 0

x = y or $\text{x}=\frac{1}{\text{y}}$

So, f is not an injection.

Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$\frac{\text{x}}{\text{x}^2+1}=\text{y}$

yx² - x + y = 0

$\text{x}=\frac{-(-1)\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ if $\text{y}\neq0$

$=\frac{1\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ which may not be in R

For example, if y = 1, then

$\text{x}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\text{i}\sqrt{3}}{2},$ which is not in R

So, f is not surjection and f is not bijection.

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Question 554 Marks

Let C be the set of all complex numbers and C₀be the set of all no-zero complex numbers. Let a relation R on C₀be defined as $\text{z}_1\text{R z}_2\Leftrightarrow\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}$ is real for all $\text{z}_1,\ \text{z}_2\in\text{C}_0.$ Show that R is an equivalence relation.

Answer

Test for reflexivity: Since, $\frac{\text{z}_1-\text{z}_1}{\text{z}_1+\text{z}_1}=0,$ which is a real number.

So, $(\text{z}_1,\text{ z}_1)\in\text{R}$

Hence, R is relexive relation.

Test for symmetric: Let $(\text{z}_1,\text{ z}_2)\in\text{R.}$

Then, $\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x,}$ where x is real

$\Rightarrow\ -\Big(\frac{\text{z}_1-\text{ z}_2}{\text{z}_1+\text{ z}_2}\Big)=-\text{x}$

$\Rightarrow\ \Big(\frac{\text{z}_2-\text{ z}_1}{\text{z}_2+\text{ z}_1}\Big)=-\text{x},$ is also a real number

So, $(\text{z}_2,\text{ z}_1)\in\text{R}$

Hence, R is symmetric relation.

Test for transivity: Let $(\text{z}_1,\text{ z}_2)\in\text{R}$ and $(\text{z}_2,\text{ z}_3)\in\text{R}.$

Then,

$\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x},$ where x is a real number.

$\Rightarrow\ \text{z}_1-\text{z}_2=\text{xz}_1+\text{xz}_2$

$\Rightarrow\ \text{z}_1-\text{xz}_1=\text{z}_2+\text{xz}_2$

$\Rightarrow\ \text{z}_1(1-\text{x})=\text{z}_2(1+\text{x})$

$\Rightarrow\ \frac{\text{z}_1}{\text{z}_2}=\frac{(1+\text{x})}{(1-\text{x})}\ .....(1)$

Also,

$\frac{\text{z}_2-\text{z}_3}{\text{z}_2+\text{z}_3}=\text{y},$ where y is a real number.

$\Rightarrow\ \text{z}_2-\text{z}_3=\text{yz}_2+\text{yz}_3$

$\Rightarrow\ \text{z}_2-\text{yz}_2=\text{z}_3+\text{yz}_3$

$\Rightarrow\ \text{z}_2(1-\text{y})=\text{z}_3(1+\text{y})$

$\Rightarrow\ \frac{\text{z}_2}{\text{z}_3}=\frac{(1+\text{y})}{(1-\text{y})}\ .....(2)$

Dividing (1) and (2), we get

$\frac{\text{z}_1}{\text{z}_3}=\Big(\frac{1+\text{x}}{1-\text{x}}\Big)\times\Big(\frac{1-\text{y}}{1+\text{y}}\Big)=\text{z,}$ where z is a real number.

$\Rightarrow\ \frac{\text{z}_1-\text{z}_3}{\text{z}_1+\text{z}_3}=\frac{\text{z}-1}{\text{z}+1},$ which is real

$\Rightarrow\ (\text{z}_1,\text{ z}_3)\in\text{R}$

Hence, R is transitive relation.

From (i), (ii) and (iii), R is an equivalenve relation.

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Question 564 Marks

Let A = {1, 2, 3, ... 9} and R be the relation in A × A defined by (a, b)R(c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer

Given that, A = {1, 2, 3, ... 9} and (a, b)R(c, d) if a + d = b + c for (a, b) ∈ A × A and (c, d) ∈ A × A.

Let (a, b)R(a, b)

⇒ a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.

Since, the sum of two numbers doesn’t depend on the order.

Hence, R is reflexive.

Let (a, b)R(c, d)

Now, a + d = b + c

Also, c + b = d + a

⇒ (c, d)R(a, b)

Hence, R is symmetric.

Let (a, b)R(c, d) and (c, d)R(e, f)

$\therefore$ a + d = b + c and c + f = d + e

⇒ a + d = b + c and d + e = c + f

⇒ (a + d) - (d + e) = (b + c) - (c + f)

⇒ a - e = b - f

⇒ a + f = b + e

⇒ (a, b)R(e, f)

Hence, R is transitive.

Since, R is symmetric, reflexive and transitive.

Hence, R is an equivalence relation.

The equivalence class containing [(2, 5)] is given by {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

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Question 574 Marks

If R and S are relations on a set A, then prove that:

R and S are symmetric $\Rightarrow\ \text{R}\cap\text{S}$ and $\text{R}\cup\text{S}$ are symmetric
R is reflexive and S is any relation $\Rightarrow\ \text{R}\cup\text{S}$ is reflexive.

Answer

R and S are symmetric relations on the set A.

$\Rightarrow\ \text{R}\subset\text{A}\times\text{A}$ and $\text{S}\subset\text{A}\times\text{A}$

$\Rightarrow\ \text{R}\cap\text{S}\subset\text{A}\times\text{A}$

Thus, $\text{R}\cap\text{S}$ is a relation on A.

Let $\text{a, b}\in\text{A}$ such that $(\text{a, b})\in\text{R}\cap\text{S.}$ Then,

$(\text{a, b})\in\text{R}\cap\text{S}$

$\Rightarrow\ (\text{a, b})\in\text{R}$ and $(\text{a, b})\in\text{S}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ and $(\text{b, a})\in\text{S}$ [Since R and S are symmetric]

$\Rightarrow\ (\text{b, a})\in\text{R}\cap\text{S}$

Thus,

$(\text{a, b})\in\text{R}\cap\text{S}$

$\Rightarrow\ (\text{b, a})\in\text{R}\cap\text{S}$ for all $\text{a, b}\in\text{A}$

So, $\text{R}\cap\text{S}$ is symmetric on A.

Also,

Let $\text{a, b}\in\text{A}$ such that $(\text{a, b})\in\text{R}\cup\text{S}$

$\Rightarrow\ (\text{a, b})\in\text{R}$ or $(\text{a, b})\in\text{S}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ or $(\text{b, a})\in\text{S}$ [Since R and S are symmetric]

$\Rightarrow\ (\text{b, a})\in\text{R}\cup\text{S}$

So, $\text{R}\cup\text{S}$ is symmetric on A.

R is reflexive and S is any relation.

Suppose $\text{a}\in\text{A.}$ Then,

$(\text{a, a})\in\text{R}$ [Since R is reflexive]

$\Rightarrow\ (\text{a, a})\in\text{R}\cup\text{S}$

$\Rightarrow\ \text{R}\cup\text{S}$ is reflexive on A.

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Question 584 Marks

Consider $\text{f}:\text{R}_+\rightarrow[-5,\infty)$ given by f(x) = 9x² + 6x - 5. Show that f is invertible with $\text{f}^{-1}(\text{x})=\frac{\sqrt{\text{x}+6}-1}{3}.$

Answer

Injectivity of f: Let x and y be two elements of domain (R₊), such that

f(x) = f(y)

⇒ 9x² + 6x - 5 = 9y² + 6y - 5

⇒ 9x² + 6x = 9y² + 6y

⇒ x = y (As, $\text{x, y}\in\text{R}_+$)

So, f is one-one.

Surjectivity of f: Let y is in the co domain (Q) such that f(x) = y

⇒ 9x² + 6x - 5 = y

⇒ 9x² + 6x = y + 5

⇒ 9x² + 6x + 1 = y + 6 (Adding 1 on both sides)

⇒ (3x + 1)² = y + 6

$\Rightarrow\ \text{3x}+1=\sqrt{\text{y}+6}$

$\Rightarrow\ 3\text{x}=\sqrt{\text{y}+6}-1$

$\Rightarrow\ \text{x}=\frac{\sqrt{\text{y}+6}-1}{3}\in\text{R}^+$ (domain)

⇒ f is onto.

So, f is a bijection and hence, it is invertible.

Finding f^-1: Let f^-1(x) = y .....(1)

⇒ x = f(y)

⇒ x = 9y² + 6y - 5

⇒ x + 5 = 9y² + 6y

⇒ x + 6 = 9y² + 6y + 1 (adding 1 on both sides)

⇒ x + 6 = (3y + 1)²

$\Rightarrow\ 3\text{y}+1=\sqrt{\text{x}+6}$

$\Rightarrow\ 3\text{y}=\sqrt{\text{x}+6}-1$

$\Rightarrow\ \text{y}=\frac{\sqrt{\text{x}+6}-1}{3}$

So, $\text{f}^{-1}(\text{x})=\frac{\sqrt{\text{x}+6}-1}{3}$ [from (1)]

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Question 594 Marks

If $\text{f(x)}=\sqrt{1-\text{x}}$ and $\text{g(x)}=\log_\text{e}\text{x}$ are two real functions, then describe, functions fog and gof.

Answer

$\text{f(x)}={1-\text{x}}$

For domain,

$1-\text{x}\geq0$

$\text{x}\leq1$

⇒ domain of g

$\text{f}:(-\infty,1]\rightarrow0,\infty=\log_\text{e}\text{x}$

Clearly, $\text{g}:0,\infty\rightarrow\text{R}$

Computation of fog: Clearly, the range of g is not a subset of the domain of f.

Therefore,

We need to compute the domain of fog.

⇒ Domain fog = x : x $\in$ Domain g and $\text{g}(\text{x})\in$ Domain of f

⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\log_\text{e}\text{x}\in(-\infty,1]$

⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\text{x}\in(0,\text{e}]$

⇒ Domain fog = x : x $\in(0,\text{e}]$

⇒ Domain fog = (0, e]

⇒ fog : 0, e → R

Therefore,

(fog)(x) = f(g(x))

$=\text{f}(\log_\text{e}\text{x})$

$=\sqrt{1-\log_\text{e}\text{x}}$

Computation of gof: Clearly, the range of f is a subset of the domain of g.

$\Rightarrow\ \text{gof}:(-\infty,1]\rightarrow\text{R}$

$(\text{gof})(\text{x})=\text{g(f(x))}$

$=\text{g}(\sqrt{1-\text{x}})$

$\Rightarrow\ \log_\text{e}\sqrt{1-\text{x}}$

$=\log_\text{e}(1-\text{x})^\frac{1}{2}$

$=\frac{1}{2}\log_\text{e}(1-\text{x})$

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Question 604 Marks

Let n be a fixed positive integer. Define a relation R on Z as follows:
$(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by n. Show that R is an equivalence relation on Z.

Answer

We observe the following properties of R.

Reflexivity: Consider $\text{a}\in\text{N}$

Here, a - a = 0 = 0 × n

Implies that a - a is divisible by n

Implies that $\text{a, a}\in\text{R}$

Implies that $\text{a, a}\in\text{R}$ for all $\text{a}\in\text{Z}.$

So, R is reflexive on Z.

Symmetry: Consider $\text{a, b}\in\text{R}$

Here a - b is divisible by n

Implies that a - b = np for some $\text{p}\in\text{Z}$

Implies that b - a = n - p.

Implies that b - a is divisible by n $\big[\text{p}\in\text{Z}$ implies that $-\text{p}\in\text{Z}\big]$

implies that $\text{b, a}\in\text{R}$

So, R is symmetric on Z.

Transitivity: Consider a, b and b, c $\in\text{R}$

Here, a - b is divisible by n and b - c is divisible by n.

implies that a - b = np for some $\text{p}\in\text{Z}$ and b - c = nq for some $\text{q}\in\text{Z}$

Adding the above two

we get a - b + b - c = np + nq

Implies that a - c = n(p + q).

Here, $\text{p}+\text{q}\in\text{Z}$

Implies that $\text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z.}$

So, R is transitive on Z.

Hence, R is an equivalence relation on Z.

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Question 614 Marks

Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as:
f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Show that f and g both are bijections and find fog and gof.

Answer

Proving f is a bijection:

f = {(a, v), (b, u), (c, w)} and f : A → B

Injectivity of f: No two elements of A have the same image in B.

So, f is one-one.

Surjectivity of f: Co-domain of f = {u v, w}

Range of f = {u v, w}

Both are same.

So, f is onto.

Hence, f is a bijection.

Proving g is a bijection:

g = {(u, b), (v, a), (w, c)} and g : B → A

Injectivity of g: No two elements of B have the same image in A.

So, g is one-one.

Surjectivity of g: Co-domain of g = {a, b, c}

Range of g = {a, b, c}

Both are the same.

So, g is onto.

Hence, g is a bijection.

Finding fog: Co-domain of g is same as the domain of f.

So, fog exists and fog: {u v, w} → {u v, w}

(fog)(u) = f(g(u)) = f(b) = u

(fog)(v) = f(g(v)) = f(a) = v

(fog)(w) = f(g(w)) = f(c) = w

So, fog = {(u, u), (v, v), (w, w)}

Finding gof.

Co-domain of f is same as the domain of g.

So, fog exists and gof: {a, b, c} → {a, b, c}

(gof)(a) = g(f(a)) = g(v) = a

(gof)(b) = g(f(b)) = g(u) = b

(gof)(c) = g(f(c)) = g(w) = c

So, gof = {(a, a), (b, b), (c, c)}

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Question 624 Marks

Consider the function $\text{f}:\text{R}^{+}\rightarrow[-9,\infty]$ given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with $\text{f}^{-1}\text{(y)}=\frac{\sqrt{54+5\text{y}}-3}{5}.$

Answer

$\text{f}:\text{R}^{+}\rightarrow\ [-9,\infty)$ given by f(x) = 5x² + 6x - 9

For any $\text{x, y}\in\text{R}^{+}$

f(x) = f(y)

⇒ 5x² + 6x - 9 = 5y² + 6y - 9

⇒ 5(x² - y²) + 6(x - y) = 0

⇒ (x - y)[5(x + y) + 6] = 0

⇒ (x - y) = 0 $[\because5(\text{x}+\text{y})+6\neq0\text{ as x, y}\in\text{R}^{+}]$

⇒ x = y

So, f is an injection.

Let y be an arbitrary element of $[-9,\infty).$

f(x) = y

⇒ 5x² + 6x - 9 = y

⇒ 25x² + 30x - 45 = 5y

⇒ 25x² + 30x + 9 - 54 = 5y

⇒ (5x + 3)² = 5y + 54

$\Rightarrow(5\text{x}+3)=\sqrt{5\text{y}+54}$

$\Rightarrow\ \text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}$

Now, $\text{y}\in[-9,\infty)$

$\Rightarrow\ \text{y}\geq-9$

$\Rightarrow\ 5\text{y}+54\geq9$

$\Rightarrow\ \sqrt{5\text{y}+54}\geq3$

$\Rightarrow\ \sqrt{5\text{y}+54}-3\geq0$

$\Rightarrow\ \frac{\sqrt{5\text{y}+54}-3}{5}\geq0$

$\Rightarrow\ \text{x}\geq0\Rightarrow\ \text{x}\in\text{R}^{+}$

Thus, for every $\text{y}\in[-9,\infty)$ there exist $\text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}\in\text{R}^{+}$ such that f(x) = y.

So, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is onto.

Thus, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is a bijection and hence invertible.

Let f^-1 denotes the inverse of f.

Then,

(fof^-1)(y) = y for all $\text{y}\in[-9,\infty)$

f(f^-1(y)) = y for all $\text{y}\in[-9,\infty)$

⇒ 5(f^-1(y))² + 6(f^-1(y)) - 9 = y for all $\text{y}\in[-9,\infty)$

⇒ 25(f^-1(y))² + 30(f^-1(y)) - 45 = 5y for all $\text{y}\in[-9,\infty)$

⇒ 25(f^-1(y))² + 30(f^-1(y)) + 9 = 5y + 54 for all $\text{y}\in[-9,\infty)$

⇒ {5f^-1(y) + 3}² = 5y + 54 for all $\text{y}\in[-9,\infty)$

⇒ 5f^-1(y) + 3 $=\sqrt{5\text{y}+54}$ for all $\text{y}\in[-9,\infty)$

$\Rightarrow\ \text{f}^{-1}(\text{y})=\frac{\sqrt{5\text{y}+54}-3}{5}$

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Question 634 Marks

Let $\text{f}:[-1,\infty)\rightarrow[-1,\infty)$ be given by f(x) = (x + 1)² - 1, $\text{x}\geq-1.$ Show that f is invertible. Also, find the set S = {x : f(x) = f^-1(x)}.

Answer

Injectivity: Let x and $\text{y}\in[-1,\infty),$ such that
f(x) = f(y)
⇒ (x + 1)² - 1 = (y + 1)² - 1
⇒ (x + 1)² = (y + 1)²
⇒ (x + 1) = (y + 1)
⇒ x = y
So, f is a injection.
Surjectivity: Let $\text{y}\in[-1,\infty)$
Then, f(x) = y
⇒ (x + 1)² - 1 = y
$\Rightarrow\ \text{x}+1=\sqrt{\text{y}+1}$
$\Rightarrow\ \text{x}=\sqrt{\text{y}+1}-1$
Clearly, $\text{x}=\sqrt{\text{y}+1}-1$ is real for all $\text{y}\geq-1.$
Thus, every element $\text{y}\in[-1,\infty)$ has its pre-image $\text{x}\in[-1,\infty)$ given by $\text{x}=\sqrt{\text{y}+1}-1$
⇒ f is a surjection.
So, f is a bijection.
Hence, f is invertible.
Let f^-1(x) = y ......(1)
⇒ f(y) = x
⇒ (y + 1)² - 1 = x
⇒ (y + 1)² = x + 1
$\Rightarrow\ \text{y}+1=\sqrt{\text{x}+1}$
$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+1}-1$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\pm\sqrt{\text{x}+1}-1$ [from (1)]
$\text{f}(\text{x})=\text{f}^{-1}(\text{x})$
$\Rightarrow\ (\text{x}+1)^2-1=\pm\sqrt{\text{x}+1}-1$
$\Rightarrow\ (\text{x}+1)^2=\pm\sqrt{\text{x}+1}$
⇒ (x + 1)⁴ = x + 1
⇒ (x + 1)[(x + 1)³ - 1] = 0
⇒ x + 1 = 0 or (x + 1)³ - 1 = 0
⇒ x = -1 or (x + 1)³ = 1
⇒ x = -1 or x + 1 = 1
⇒ x = -1 or x = 0
⇒ S = {0, -1}

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Question 644 Marks

m is said to be related to n if m and n are integers and m - n is divisible by 13. Does this define an equivalence relation?

Answer

We observe the following properties of relation R.

$\text{Let R} = \big\{(\text{m, n}): \text{(m, n)}\in\text{Z}:\text{m}-\text{n}\ \text{is divisible by} 13\big\}$$$$$

Relexivity: Let m be an arbitrary element of Z. Then, $\text{m}\in\text{R}$

⇒ m - m = 0 = 0 × 13

⇒ m - m is divisible by 13

⇒ (m, m) is reflexive on Z.

Symmetry: Let $(\text{m, n})\in\text{R.}$ Then,

⇒ m - n is divisible by 13

⇒ m - n = 13p

Here, $\text{p}\in\text{Z}$

⇒ n - m = 13(-p)

Here, $-\text{p}\in\text{Z}$

⇒ n - m is divisible by 13

$\Rightarrow\ (\text{n, m})\in\text{R}$ for all $\text{m, n}\in\text{Z}$

So, R is symmetric on Z.

Transitivity: Let (m, n) and (n, o) $\in\text{R}$

⇒ m - n and n - o are divisible by 13

⇒ m - n = 13p and n - o = 13q for some $\text{p, q}\in\text{Z}$

Adding the above two, we get

⇒ m - n + n - o = 13p + 13q

⇒ m - o = 13(p + q)

Here, $\text{p}+\text{q}\in\text{Z}$

⇒ m - o is divisible by 13

$\Rightarrow\ (\text{m, o})\in\text{R}$ for all $\text{m, o}\in\text{Z}$

So, R is transitive on Z.

Hence, R is an equivalence relation on Z.

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Question 654 Marks

On R − {1}, a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.

Answer

Commutativity:

Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,

a * b = a + b - ab

= b + a - ba

= b * a

Therefore,

a * b = b * a, $\forall\ \text{a},\text{b}\in\text{R}-\{1\}$

Thus, * is commutative on R - {1}.

Associativity:

Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,

a * (b * c) = a * (b + c - bc)

= a + b + c - bc - a(b + c - bc)

= a + b + c - bc - ab - ac - abc

(a * b) * c = (a + b - ab) * c

= a + b - ab + c - (a + b - ab)c

= a + b + c - ab - ac - bc + abc

Therefore,

a * (b * c) = (a * b) * c, $\forall\text{ a},\text{b},\text{c}\in\text{R}-\{1\}$

Thus, * is associative on R - {1}.

Finding identity element:

Let e be the element in R - {1} with respect to * such that

a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$

a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$

⇒ a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{R}-\{1\}$

e(1 - a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$

$\text{e}=0\in\forall\text{ a}\in\text{R}-\{1\},\forall\text{ a}\in\text{R}-\{1\}$ $[\because\ \text{a}\neq1]$

Thus, 0 is the identity element in R - {1} with respect to *.

Finding inverse:

Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

⇒ a + b - ab = 0 and b + a - ba = 0

⇒ a = ab - b

⇒ a = b(a - 1)

$\Rightarrow\text{b}=\frac{\text{a}}{\text{a}-1}$

Thus, $\frac{\text{a}}{\text{a}-1}$ is inverse of $\text{a}\in\text{R}-\{1\}.$

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Question 664 Marks

If f : R → (-1, 1) defined by $\text{f(x)}=\frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}+10^{-\text{x}}}$ is invertible, find f^-1.

Answer

Injectivity of f: Let x and y be two elements of domain (R), such that

f(x) = f(y)

$\Rightarrow\ \frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}-10^{-\text{x}}}=\frac{10^{\text{y}}-10^{-\text{y}}}{10^{\text{y}}-10^{-\text{y}}}$

$\Rightarrow\ \frac{10^{-\text{x}}(10^{2\text{x}}-1)}{10^{-\text{x}}(10^{2\text{x}}+1)}=\frac{10^{-\text{y}}(10^{2\text{y}}-1)}{10^{-\text{y}}(10^{2\text{y}}+1)}$

$\Rightarrow\ \frac{(10^{2\text{x}}-1)}{(10^{2\text{x}}+1)}=\frac{(10^{2\text{y}}-1)}{(10^{2\text{y}}+1)}$

⇒ (10^2x - 1)(10^2y + 1) = (10^2x + 1)(10^2y - 1)

⇒ 10^2x+2y + 10^2x - 10^2y - 1 = 10^2x+2y - 10^2x + 10^2y - 1

⇒ 2 × 10^2x = 2 × 10^2y

⇒ 10^2x = 10^2y

⇒ 2x = 2y

⇒ x = y

So, f is one-one.

Surjectivity of f: Let y is in the co domain (R), such that f(x) = y

$\Rightarrow\ \frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}+10^{-\text{x}}}=\text{y}$

$\Rightarrow\ \frac{10^{-\text{x}}(10^{2\text{x}}-1)}{10^{-\text{x}}(10^{2\text{x}}+1)}=\text{y}$

$\Rightarrow\ 10^{2\text{x}}-1=\text{y}\times10^{2\text{x}}+\text{y}$

$\Rightarrow\ 10^{2\text{x}}(1-\text{y})=1+\text{y}$

$\Rightarrow\ 10^{2\text{x}}=\frac{1+\text{y}}{1-\text{y}}$

$\Rightarrow\ 2\text{x}=\log\Big(\frac{1+\text{y}}{1-\text{y}}\Big)$

$\Rightarrow\ \text{x}=\frac{1}{2}\log\Big(\frac{1+\text{y}}{1-\text{y}}\Big)\in\text{R}$ (domain)

⇒ f is onto.

So, f is a bijection and hence, it is invertible.

Finding f^-1: Let f^-1(x) = y .......(1)

⇒ f(y) = x

$\Rightarrow\ \frac{10^{\text{y}}-10^{-\text{y}}}{10^{\text{y}}+10^{-\text{y}}}=\text{x}$

$\Rightarrow\ \frac{10^{-\text{y}}(10^{2\text{y}}-1)}{10^{-\text{y}}(10^{2\text{y}}+1)}=\text{x}$

$\Rightarrow\ 10^{2\text{y}}-1=\text{x}\times10^{2\text{y}}+\text{x}$

$\Rightarrow\ 10^{2\text{y}}(1-\text{x})=1+\text{x}$

$\Rightarrow\ 10^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$

$\Rightarrow\ 2\text{y}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

$\Rightarrow\ \text{y}=\frac{1}{2}\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$

So, $\text{f}^{-1}(\text{x})=\frac{1}{2}\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ [from (1)]

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Maths 4 Marks