QBManager

Sets, Relation and Function

117148 A binary operation $o$ is defined on the set of integers $I$ by poq $= 3 p^{2} + 2 q^{2} - 5 p q$ If aol $= 1$ , then $a$ is equal to

1 -1

2 1

3 -2

4 none of these.

Explanation:

Exp: (D) : Given,
And,
$poq = 3 p^{2} + 2 q^{2} - 5 pq$
Then,
aol $= 1$
aol $= 3 a^{2} + 2 (1)^{2} - 5 a (1)$
$1 = 3 a^{2} + 2 - 5 a$
$3 a^{2} - 5 a + 1 = 0$
From
$x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}$
$a = \frac{- (- 5) \pm \sqrt{25 - 4 \times 3 \times 1}}{2 \times 3}$
$a = \frac{5 \pm \sqrt{13}}{6}$

[AMU-2008]

Sets, Relation and Function

117149 If $f (x) = \frac{2 x - 1}{x + 5} (x \neq - 5)$ , then $f^{- 1} (x)$ is equal to:

1

\frac{x + 5}{2 x - 1}, x \neq \frac{1}{2}

2

\frac{5 x + 1}{2 - x}, x \neq 2

3

\frac{x - 5}{2 x + 1}, x \neq \frac{1}{2}

4

\frac{5 x - 1}{2 - x}, x \neq 2

Explanation:

Exp: (B) : Given,
$f (x) = \frac{2 x - 1}{x + 5}, (x \neq - 5)$
$f^{- 1} (x) = ?$
Let,
$f (x) = y$
$y = \frac{2 x - 1}{x + 5}$
$x y + 5 y = 2 x - 1$
$x y - 2 x = - 1 - 5 y$
$x (y - 2) = - (1 + 5 y)$
$x (2 - y) = (1 + 5 y)$
$x = \frac{1 + 5 y}{2 - y}$
So, $f^{- 1} (x) = \frac{1 + 5 x}{2 - x}, x \neq 2$
Also written as $f^{- 1} (x) = \frac{5 x + 1}{2 - x}, x \neq 2$

CGPET - 2007]

Sets, Relation and Function

117152 If $f (x) = (4 - (x - 7)^{3})$ , then $f^{- 1} (x) =$

1

x

2

4^{1 / 3} - (x - 7)

3

7 + (4 - x)^{1 / 3}

4

7 - (4 - x)^{1 / 3}

Explanation:

Exp: (C): Given that,
$f (x) = (4 - (x - 7)^{3})$
$y = (4 - (x - 7)^{3})$
$(x - 7)^{3} = 4 - y$
$x - 7 = (4 - y)^{1 / 3}$
$x = 7 + (4 - y)^{1 / 3}$
$f^{- 1} (x) = 7 + (4 - x)^{1 / 3}$

[JCECE-2019]

Sets, Relation and Function

117153 Let * be a binary operation on the set $Q^{+}$ of all +ve rational numbers defined by $a * b = \frac{a b}{100}$ for all $a, b \in Q^{+}$ . The inverse of 0.1 under operation * is

1

10^{5}

2

10^{6}

3

10^{4}

4 None of these

Explanation:

Exp: (A) : We have $a * b = \frac{ab}{100}$
Let e be the identify element.
Then, $e * a = a * e = a \forall a \in Q^{+}$
$a = \frac{ea}{100} \Rightarrow e = 100$
Now, let inverse of 0.1 be $x$ , then $0.1 * x = e$
$\Rightarrow \frac{0.1 x}{100} = 100 x \in Q^{+}$
$x = \frac{10^{4}}{0.1} = 10^{5}$

[AMU-2015]

Sets, Relation and Function

117148 A binary operation $o$ is defined on the set of integers $I$ by poq $= 3 p^{2} + 2 q^{2} - 5 p q$ If aol $= 1$ , then $a$ is equal to

1 -1

2 1

3 -2

4 none of these.

Explanation:

Exp: (D) : Given,
And,
$poq = 3 p^{2} + 2 q^{2} - 5 pq$
Then,
aol $= 1$
aol $= 3 a^{2} + 2 (1)^{2} - 5 a (1)$
$1 = 3 a^{2} + 2 - 5 a$
$3 a^{2} - 5 a + 1 = 0$
From
$x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}$
$a = \frac{- (- 5) \pm \sqrt{25 - 4 \times 3 \times 1}}{2 \times 3}$
$a = \frac{5 \pm \sqrt{13}}{6}$

[AMU-2008]

Sets, Relation and Function

117149 If $f (x) = \frac{2 x - 1}{x + 5} (x \neq - 5)$ , then $f^{- 1} (x)$ is equal to:

1

\frac{x + 5}{2 x - 1}, x \neq \frac{1}{2}

2

\frac{5 x + 1}{2 - x}, x \neq 2

3

\frac{x - 5}{2 x + 1}, x \neq \frac{1}{2}

4

\frac{5 x - 1}{2 - x}, x \neq 2

Explanation:

Exp: (B) : Given,
$f (x) = \frac{2 x - 1}{x + 5}, (x \neq - 5)$
$f^{- 1} (x) = ?$
Let,
$f (x) = y$
$y = \frac{2 x - 1}{x + 5}$
$x y + 5 y = 2 x - 1$
$x y - 2 x = - 1 - 5 y$
$x (y - 2) = - (1 + 5 y)$
$x (2 - y) = (1 + 5 y)$
$x = \frac{1 + 5 y}{2 - y}$
So, $f^{- 1} (x) = \frac{1 + 5 x}{2 - x}, x \neq 2$
Also written as $f^{- 1} (x) = \frac{5 x + 1}{2 - x}, x \neq 2$

CGPET - 2007]

Sets, Relation and Function

117152 If $f (x) = (4 - (x - 7)^{3})$ , then $f^{- 1} (x) =$

1

x

2

4^{1 / 3} - (x - 7)

3

7 + (4 - x)^{1 / 3}

4

7 - (4 - x)^{1 / 3}

Explanation:

Exp: (C): Given that,
$f (x) = (4 - (x - 7)^{3})$
$y = (4 - (x - 7)^{3})$
$(x - 7)^{3} = 4 - y$
$x - 7 = (4 - y)^{1 / 3}$
$x = 7 + (4 - y)^{1 / 3}$
$f^{- 1} (x) = 7 + (4 - x)^{1 / 3}$

[JCECE-2019]

Sets, Relation and Function

117153 Let * be a binary operation on the set $Q^{+}$ of all +ve rational numbers defined by $a * b = \frac{a b}{100}$ for all $a, b \in Q^{+}$ . The inverse of 0.1 under operation * is

1

10^{5}

2

10^{6}

3

10^{4}

4 None of these

Explanation:

Exp: (A) : We have $a * b = \frac{ab}{100}$
Let e be the identify element.
Then, $e * a = a * e = a \forall a \in Q^{+}$
$a = \frac{ea}{100} \Rightarrow e = 100$
Now, let inverse of 0.1 be $x$ , then $0.1 * x = e$
$\Rightarrow \frac{0.1 x}{100} = 100 x \in Q^{+}$
$x = \frac{10^{4}}{0.1} = 10^{5}$

[AMU-2015]

Sets, Relation and Function

117148 A binary operation $o$ is defined on the set of integers $I$ by poq $= 3 p^{2} + 2 q^{2} - 5 p q$ If aol $= 1$ , then $a$ is equal to

1 -1

2 1

3 -2

4 none of these.

Explanation:

Exp: (D) : Given,
And,
$poq = 3 p^{2} + 2 q^{2} - 5 pq$
Then,
aol $= 1$
aol $= 3 a^{2} + 2 (1)^{2} - 5 a (1)$
$1 = 3 a^{2} + 2 - 5 a$
$3 a^{2} - 5 a + 1 = 0$
From
$x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}$
$a = \frac{- (- 5) \pm \sqrt{25 - 4 \times 3 \times 1}}{2 \times 3}$
$a = \frac{5 \pm \sqrt{13}}{6}$

[AMU-2008]

Sets, Relation and Function

117149 If $f (x) = \frac{2 x - 1}{x + 5} (x \neq - 5)$ , then $f^{- 1} (x)$ is equal to:

1

\frac{x + 5}{2 x - 1}, x \neq \frac{1}{2}

2

\frac{5 x + 1}{2 - x}, x \neq 2

3

\frac{x - 5}{2 x + 1}, x \neq \frac{1}{2}

4

\frac{5 x - 1}{2 - x}, x \neq 2

Explanation:

Exp: (B) : Given,
$f (x) = \frac{2 x - 1}{x + 5}, (x \neq - 5)$
$f^{- 1} (x) = ?$
Let,
$f (x) = y$
$y = \frac{2 x - 1}{x + 5}$
$x y + 5 y = 2 x - 1$
$x y - 2 x = - 1 - 5 y$
$x (y - 2) = - (1 + 5 y)$
$x (2 - y) = (1 + 5 y)$
$x = \frac{1 + 5 y}{2 - y}$
So, $f^{- 1} (x) = \frac{1 + 5 x}{2 - x}, x \neq 2$
Also written as $f^{- 1} (x) = \frac{5 x + 1}{2 - x}, x \neq 2$

CGPET - 2007]

Sets, Relation and Function

117152 If $f (x) = (4 - (x - 7)^{3})$ , then $f^{- 1} (x) =$

1

x

2

4^{1 / 3} - (x - 7)

3

7 + (4 - x)^{1 / 3}

4

7 - (4 - x)^{1 / 3}

Explanation:

Exp: (C): Given that,
$f (x) = (4 - (x - 7)^{3})$
$y = (4 - (x - 7)^{3})$
$(x - 7)^{3} = 4 - y$
$x - 7 = (4 - y)^{1 / 3}$
$x = 7 + (4 - y)^{1 / 3}$
$f^{- 1} (x) = 7 + (4 - x)^{1 / 3}$

[JCECE-2019]

Sets, Relation and Function

117153 Let * be a binary operation on the set $Q^{+}$ of all +ve rational numbers defined by $a * b = \frac{a b}{100}$ for all $a, b \in Q^{+}$ . The inverse of 0.1 under operation * is

1

10^{5}

2

10^{6}

3

10^{4}

4 None of these

Explanation:

Exp: (A) : We have $a * b = \frac{ab}{100}$
Let e be the identify element.
Then, $e * a = a * e = a \forall a \in Q^{+}$
$a = \frac{ea}{100} \Rightarrow e = 100$
Now, let inverse of 0.1 be $x$ , then $0.1 * x = e$
$\Rightarrow \frac{0.1 x}{100} = 100 x \in Q^{+}$
$x = \frac{10^{4}}{0.1} = 10^{5}$

[AMU-2015]

Sets, Relation and Function

117148 A binary operation $o$ is defined on the set of integers $I$ by poq $= 3 p^{2} + 2 q^{2} - 5 p q$ If aol $= 1$ , then $a$ is equal to

1 -1

2 1

3 -2

4 none of these.

Explanation:

Exp: (D) : Given,
And,
$poq = 3 p^{2} + 2 q^{2} - 5 pq$
Then,
aol $= 1$
aol $= 3 a^{2} + 2 (1)^{2} - 5 a (1)$
$1 = 3 a^{2} + 2 - 5 a$
$3 a^{2} - 5 a + 1 = 0$
From
$x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}$
$a = \frac{- (- 5) \pm \sqrt{25 - 4 \times 3 \times 1}}{2 \times 3}$
$a = \frac{5 \pm \sqrt{13}}{6}$

[AMU-2008]

Sets, Relation and Function

117149 If $f (x) = \frac{2 x - 1}{x + 5} (x \neq - 5)$ , then $f^{- 1} (x)$ is equal to:

1

\frac{x + 5}{2 x - 1}, x \neq \frac{1}{2}

2

\frac{5 x + 1}{2 - x}, x \neq 2

3

\frac{x - 5}{2 x + 1}, x \neq \frac{1}{2}

4

\frac{5 x - 1}{2 - x}, x \neq 2

Explanation:

Exp: (B) : Given,
$f (x) = \frac{2 x - 1}{x + 5}, (x \neq - 5)$
$f^{- 1} (x) = ?$
Let,
$f (x) = y$
$y = \frac{2 x - 1}{x + 5}$
$x y + 5 y = 2 x - 1$
$x y - 2 x = - 1 - 5 y$
$x (y - 2) = - (1 + 5 y)$
$x (2 - y) = (1 + 5 y)$
$x = \frac{1 + 5 y}{2 - y}$
So, $f^{- 1} (x) = \frac{1 + 5 x}{2 - x}, x \neq 2$
Also written as $f^{- 1} (x) = \frac{5 x + 1}{2 - x}, x \neq 2$

CGPET - 2007]

Sets, Relation and Function

117152 If $f (x) = (4 - (x - 7)^{3})$ , then $f^{- 1} (x) =$

1

x

2

4^{1 / 3} - (x - 7)

3

7 + (4 - x)^{1 / 3}

4

7 - (4 - x)^{1 / 3}

Explanation:

Exp: (C): Given that,
$f (x) = (4 - (x - 7)^{3})$
$y = (4 - (x - 7)^{3})$
$(x - 7)^{3} = 4 - y$
$x - 7 = (4 - y)^{1 / 3}$
$x = 7 + (4 - y)^{1 / 3}$
$f^{- 1} (x) = 7 + (4 - x)^{1 / 3}$

[JCECE-2019]

Sets, Relation and Function

117153 Let * be a binary operation on the set $Q^{+}$ of all +ve rational numbers defined by $a * b = \frac{a b}{100}$ for all $a, b \in Q^{+}$ . The inverse of 0.1 under operation * is

1

10^{5}

2

10^{6}

3

10^{4}

4 None of these

Explanation:

Exp: (A) : We have $a * b = \frac{ab}{100}$
Let e be the identify element.
Then, $e * a = a * e = a \forall a \in Q^{+}$
$a = \frac{ea}{100} \Rightarrow e = 100$
Now, let inverse of 0.1 be $x$ , then $0.1 * x = e$
$\Rightarrow \frac{0.1 x}{100} = 100 x \in Q^{+}$
$x = \frac{10^{4}}{0.1} = 10^{5}$

[AMU-2015]

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