QBManager

Sets, Relation and Function

117164 Let $\boldsymbol{f}: \mathrm{N} \times \mathrm{N} \rightarrow \mathrm{N}$ be a function such that $\mathrm{f}(1$, 1) $=2$ and $f((\mathrm{~m}+1, \mathrm{n}))=f((\mathrm{~m}, \mathrm{n}))+2(\mathrm{~m}+\mathrm{n})$ and $f((\mathrm{~m}, \mathrm{n}+1))=f((\mathrm{~m}, \mathrm{n}))+2(\mathrm{~m}+\mathrm{n}-1), \forall$ $\mathrm{m}, \mathrm{n} \in \mathrm{N}$, then find $\boldsymbol f (2, 2)$

1 8

2 7

3 9

4 10

Explanation:

Exp: (D) : Given,
$f : N \times N \to N$
And, $f (1, 1) = 2$
And also given,
$f (m + 1, n) = f (m, n) + 2 (m + n)$
$∴$
$f (2, 1) = f (1, 1) + 2 \times (1 + 1)$
$f (2, 1) = 2 + 2 \times 2$
$f (2, 1) = 6$
Also,
$f (m, n + 1) = f (m, n) + 2 (m + n - 1)$
$∴$
$f (1, 2) = f (1, 1 + 1) = f (1, 1) + 2 \times 1$
$= 2 + 2 = 4$
So, $f (2, 2) = f (1 + 1, 2) = f (1, 2) + 2 (1 + 2)$
$= 4 + 6 = 10$

[AP EAMCET-22.09.2020

Sets, Relation and Function

117165 In the set $Q^{+}$ of all positive rational numbers, the operation * is defined by the formula $a * b = \frac{a b}{6}$ . Then, the inverse of 9 with respect to * is

1 4

2 3

3

\frac{1}{9}

4

\frac{1}{3}

Explanation:

Exp: (A) : Given, * is a binary operation on $Q^{+}$ which is set of all positive rational number such that, $a * b = \frac{a b}{2}$ Let e be the identity,
$∴ a * e = a$
$\Rightarrow \frac{ae}{6} = a$
$\Rightarrow e = 6$
$∵ a^{*} a^{- 1} = e$
$∴ a^{*} a^{- 1} = 6$
$\Rightarrow \frac{{aa}^{- 1}}{6} = 6$
$\Rightarrow a^{- 1} = \frac{6 \times 6}{a}$
So, $9^{- 1} = \frac{6 \times 6}{9} = 4$

[Manipal UGET-2012

Sets, Relation and Function

117166 Which of the following functions is inverse itself?

1

f (t) = \frac{(1 - t)}{(1 + t)}

2

f (t) = \frac{(1 - t^{2})}{(1 + t^{2})}

3

f (t) = 4^{\log t}

4

f (t) = 2^{t}

Explanation:

Exp: (A) : Let, $f (t) = y$
$y = \frac{1 - t}{1 + t}$
$\frac{1}{y} = \frac{1 + t}{1 - t}$
$\frac{1 - y}{1 + y} = \frac{2 t}{2}$
$t = \frac{1 - y}{1 + y}$
$f^{- 1} (y) = \frac{1 - y}{1 + y}$
$f^{- 1} (t) = \frac{1 - t}{1 + t}$ So, $f (t) = \frac{1 - t}{1 + t}$ is inverse itself.

[J and K CET-2017

Sets, Relation and Function

117168 Let $A = {1, 2, 3, 4}$ and $R$ be the relation on $A$ defined by ${(a, b) : a, b \in A, a \times b$ is an even number $}$ , then find the range of $R$ .

1

{1, 2, 3, 4}

2

{2, 4}

3

{2, 3, 4}

4

{1, 2, 4}

Explanation:

Exp: (A) : Given,
$A = {1, 2, 3, 4}$
$R = {a \times b even number}$
$= {(1, 2), (1, 4), (2, 1), (2, 3), (2, 4), (3, 2), (3, 4), (4, 1)$ , $(4, 2), (4, 3)}$
So, the range of $R = {1, 2, 3, 4}$ .

[J and K CET-2014]

Sets, Relation and Function

117164 Let $\boldsymbol f : N \times N \to N$ be a function such that $f (1$ , 1) $= 2$ and $f ((m + 1, n)) = f ((m, n)) + 2 (m + n)$ and $f ((m, n + 1)) = f ((m, n)) + 2 (m + n - 1), \forall$ $m, n \in N$ , then find $\boldsymbol f (2, 2)$

1 8

2 7

3 9

4 10

Explanation:

Exp: (D) : Given,
$f : N \times N \to N$
And, $f (1, 1) = 2$
And also given,
$f (m + 1, n) = f (m, n) + 2 (m + n)$
$∴$
$f (2, 1) = f (1, 1) + 2 \times (1 + 1)$
$f (2, 1) = 2 + 2 \times 2$
$f (2, 1) = 6$
Also,
$f (m, n + 1) = f (m, n) + 2 (m + n - 1)$
$∴$
$f (1, 2) = f (1, 1 + 1) = f (1, 1) + 2 \times 1$
$= 2 + 2 = 4$
So, $f (2, 2) = f (1 + 1, 2) = f (1, 2) + 2 (1 + 2)$
$= 4 + 6 = 10$

[AP EAMCET-22.09.2020

Sets, Relation and Function

117165 In the set $Q^{+}$ of all positive rational numbers, the operation * is defined by the formula $a * b = \frac{a b}{6}$ . Then, the inverse of 9 with respect to * is

1 4

2 3

3

\frac{1}{9}

4

\frac{1}{3}

Explanation:

Exp: (A) : Given, * is a binary operation on $Q^{+}$ which is set of all positive rational number such that, $a * b = \frac{a b}{2}$ Let e be the identity,
$∴ a * e = a$
$\Rightarrow \frac{ae}{6} = a$
$\Rightarrow e = 6$
$∵ a^{*} a^{- 1} = e$
$∴ a^{*} a^{- 1} = 6$
$\Rightarrow \frac{{aa}^{- 1}}{6} = 6$
$\Rightarrow a^{- 1} = \frac{6 \times 6}{a}$
So, $9^{- 1} = \frac{6 \times 6}{9} = 4$

[Manipal UGET-2012

Sets, Relation and Function

117166 Which of the following functions is inverse itself?

1

f (t) = \frac{(1 - t)}{(1 + t)}

2

f (t) = \frac{(1 - t^{2})}{(1 + t^{2})}

3

f (t) = 4^{\log t}

4

f (t) = 2^{t}

Explanation:

Exp: (A) : Let, $f (t) = y$
$y = \frac{1 - t}{1 + t}$
$\frac{1}{y} = \frac{1 + t}{1 - t}$
$\frac{1 - y}{1 + y} = \frac{2 t}{2}$
$t = \frac{1 - y}{1 + y}$
$f^{- 1} (y) = \frac{1 - y}{1 + y}$
$f^{- 1} (t) = \frac{1 - t}{1 + t}$ So, $f (t) = \frac{1 - t}{1 + t}$ is inverse itself.

[J and K CET-2017

Sets, Relation and Function

117168 Let $A = {1, 2, 3, 4}$ and $R$ be the relation on $A$ defined by ${(a, b) : a, b \in A, a \times b$ is an even number $}$ , then find the range of $R$ .

1

{1, 2, 3, 4}

2

{2, 4}

3

{2, 3, 4}

4

{1, 2, 4}

Explanation:

Exp: (A) : Given,
$A = {1, 2, 3, 4}$
$R = {a \times b even number}$
$= {(1, 2), (1, 4), (2, 1), (2, 3), (2, 4), (3, 2), (3, 4), (4, 1)$ , $(4, 2), (4, 3)}$
So, the range of $R = {1, 2, 3, 4}$ .

[J and K CET-2014]

Sets, Relation and Function

117164 Let $\boldsymbol f : N \times N \to N$ be a function such that $f (1$ , 1) $= 2$ and $f ((m + 1, n)) = f ((m, n)) + 2 (m + n)$ and $f ((m, n + 1)) = f ((m, n)) + 2 (m + n - 1), \forall$ $m, n \in N$ , then find $\boldsymbol f (2, 2)$

1 8

2 7

3 9

4 10

Explanation:

Exp: (D) : Given,
$f : N \times N \to N$
And, $f (1, 1) = 2$
And also given,
$f (m + 1, n) = f (m, n) + 2 (m + n)$
$∴$
$f (2, 1) = f (1, 1) + 2 \times (1 + 1)$
$f (2, 1) = 2 + 2 \times 2$
$f (2, 1) = 6$
Also,
$f (m, n + 1) = f (m, n) + 2 (m + n - 1)$
$∴$
$f (1, 2) = f (1, 1 + 1) = f (1, 1) + 2 \times 1$
$= 2 + 2 = 4$
So, $f (2, 2) = f (1 + 1, 2) = f (1, 2) + 2 (1 + 2)$
$= 4 + 6 = 10$

[AP EAMCET-22.09.2020

Sets, Relation and Function

117165 In the set $Q^{+}$ of all positive rational numbers, the operation * is defined by the formula $a * b = \frac{a b}{6}$ . Then, the inverse of 9 with respect to * is

1 4

2 3

3

\frac{1}{9}

4

\frac{1}{3}

Explanation:

Exp: (A) : Given, * is a binary operation on $Q^{+}$ which is set of all positive rational number such that, $a * b = \frac{a b}{2}$ Let e be the identity,
$∴ a * e = a$
$\Rightarrow \frac{ae}{6} = a$
$\Rightarrow e = 6$
$∵ a^{*} a^{- 1} = e$
$∴ a^{*} a^{- 1} = 6$
$\Rightarrow \frac{{aa}^{- 1}}{6} = 6$
$\Rightarrow a^{- 1} = \frac{6 \times 6}{a}$
So, $9^{- 1} = \frac{6 \times 6}{9} = 4$

[Manipal UGET-2012

Sets, Relation and Function

117166 Which of the following functions is inverse itself?

1

f (t) = \frac{(1 - t)}{(1 + t)}

2

f (t) = \frac{(1 - t^{2})}{(1 + t^{2})}

3

f (t) = 4^{\log t}

4

f (t) = 2^{t}

Explanation:

Exp: (A) : Let, $f (t) = y$
$y = \frac{1 - t}{1 + t}$
$\frac{1}{y} = \frac{1 + t}{1 - t}$
$\frac{1 - y}{1 + y} = \frac{2 t}{2}$
$t = \frac{1 - y}{1 + y}$
$f^{- 1} (y) = \frac{1 - y}{1 + y}$
$f^{- 1} (t) = \frac{1 - t}{1 + t}$ So, $f (t) = \frac{1 - t}{1 + t}$ is inverse itself.

[J and K CET-2017

Sets, Relation and Function

117168 Let $A = {1, 2, 3, 4}$ and $R$ be the relation on $A$ defined by ${(a, b) : a, b \in A, a \times b$ is an even number $}$ , then find the range of $R$ .

1

{1, 2, 3, 4}

2

{2, 4}

3

{2, 3, 4}

4

{1, 2, 4}

Explanation:

Exp: (A) : Given,
$A = {1, 2, 3, 4}$
$R = {a \times b even number}$
$= {(1, 2), (1, 4), (2, 1), (2, 3), (2, 4), (3, 2), (3, 4), (4, 1)$ , $(4, 2), (4, 3)}$
So, the range of $R = {1, 2, 3, 4}$ .

[J and K CET-2014]

Sets, Relation and Function

117164 Let $\boldsymbol f : N \times N \to N$ be a function such that $f (1$ , 1) $= 2$ and $f ((m + 1, n)) = f ((m, n)) + 2 (m + n)$ and $f ((m, n + 1)) = f ((m, n)) + 2 (m + n - 1), \forall$ $m, n \in N$ , then find $\boldsymbol f (2, 2)$

1 8

2 7

3 9

4 10

Explanation:

Exp: (D) : Given,
$f : N \times N \to N$
And, $f (1, 1) = 2$
And also given,
$f (m + 1, n) = f (m, n) + 2 (m + n)$
$∴$
$f (2, 1) = f (1, 1) + 2 \times (1 + 1)$
$f (2, 1) = 2 + 2 \times 2$
$f (2, 1) = 6$
Also,
$f (m, n + 1) = f (m, n) + 2 (m + n - 1)$
$∴$
$f (1, 2) = f (1, 1 + 1) = f (1, 1) + 2 \times 1$
$= 2 + 2 = 4$
So, $f (2, 2) = f (1 + 1, 2) = f (1, 2) + 2 (1 + 2)$
$= 4 + 6 = 10$

[AP EAMCET-22.09.2020

Sets, Relation and Function

117165 In the set $Q^{+}$ of all positive rational numbers, the operation * is defined by the formula $a * b = \frac{a b}{6}$ . Then, the inverse of 9 with respect to * is

1 4

2 3

3

\frac{1}{9}

4

\frac{1}{3}

Explanation:

Exp: (A) : Given, * is a binary operation on $Q^{+}$ which is set of all positive rational number such that, $a * b = \frac{a b}{2}$ Let e be the identity,
$∴ a * e = a$
$\Rightarrow \frac{ae}{6} = a$
$\Rightarrow e = 6$
$∵ a^{*} a^{- 1} = e$
$∴ a^{*} a^{- 1} = 6$
$\Rightarrow \frac{{aa}^{- 1}}{6} = 6$
$\Rightarrow a^{- 1} = \frac{6 \times 6}{a}$
So, $9^{- 1} = \frac{6 \times 6}{9} = 4$

[Manipal UGET-2012

Sets, Relation and Function

117166 Which of the following functions is inverse itself?

1

f (t) = \frac{(1 - t)}{(1 + t)}

2

f (t) = \frac{(1 - t^{2})}{(1 + t^{2})}

3

f (t) = 4^{\log t}

4

f (t) = 2^{t}

Explanation:

Exp: (A) : Let, $f (t) = y$
$y = \frac{1 - t}{1 + t}$
$\frac{1}{y} = \frac{1 + t}{1 - t}$
$\frac{1 - y}{1 + y} = \frac{2 t}{2}$
$t = \frac{1 - y}{1 + y}$
$f^{- 1} (y) = \frac{1 - y}{1 + y}$
$f^{- 1} (t) = \frac{1 - t}{1 + t}$ So, $f (t) = \frac{1 - t}{1 + t}$ is inverse itself.

[J and K CET-2017

Sets, Relation and Function

117168 Let $A = {1, 2, 3, 4}$ and $R$ be the relation on $A$ defined by ${(a, b) : a, b \in A, a \times b$ is an even number $}$ , then find the range of $R$ .

1

{1, 2, 3, 4}

2

{2, 4}

3

{2, 3, 4}

4

{1, 2, 4}

Explanation:

Exp: (A) : Given,
$A = {1, 2, 3, 4}$
$R = {a \times b even number}$
$= {(1, 2), (1, 4), (2, 1), (2, 3), (2, 4), (3, 2), (3, 4), (4, 1)$ , $(4, 2), (4, 3)}$
So, the range of $R = {1, 2, 3, 4}$ $R = {1, 2, 3, 4}$ .

[J and K CET-2014]

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: