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Sets, Relation and Function

117074 Let $a, b, c$ be real numbers, each greater than 1, such that $\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$ .
If the value of $b$ is 9 , then the value of ' $a$ ' must be

1

\sqrt[3]{81}

2

\frac{27}{2}

3 18

4 27

Explanation:

D Given,
$b = 9$
$\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$
$a, b, c$ are real
$\frac{2}{3} \log_{9} a + \frac{3}{5} \log_{c} 9 + \frac{5}{2} \log_{a} c = 3$
$\Rightarrow \log_{9} a^{2 / 3} + \log_{c} 9^{3 / 5} + \log_{a} c^{5 / 3} = 3$
As there are two unknown variable a \& c, considering every term is equal to 1 .
$\log_{9} a^{2 / 3} = 1, \log_{c} 9^{3 / 5} = 1, \log_{a} c^{5 / 3} = 1$
$∴ \frac{\log a^{2 / 3}}{\log 9} = 1 (∵ \log_{x} y = \frac{\log y}{\log x})$
$\Rightarrow \log a^{2 / 3} = \log 9$
$\Rightarrow a^{2 / 3} = 9$
$\Rightarrow a = 9^{3 / 2}$
$\Rightarrow a = (3)^{2 \times 3 / 2}$
$= a = (3)^{3} = 27 .$ Hence option (D) is correct.

WB JEE-2021

Sets, Relation and Function

117075 $f : X \to R, X = {x ∣ 0 < x < 1}$ is defined as $f (x) =$ $\frac{2 x}{1 - | 2 x - 1 |}$ . Then

1

f

is only injective

2

f

is only surjective

3

f

is bijective

4

f

is neither injective nor surjective

Explanation:

A Given,
$f (x) = \frac{2 x - 1}{1 - | 2 x - 1 |}$
$f (x) = {\begin{cases} \frac{2 x - 1}{1 + 2 x - 1}, 0 < x < \frac{1}{2} \\ \frac{2 x}{1 - 2 x + 1}, \frac{1}{2} \leq x < 1 \end{cases}$
$f (x) = {\begin{cases} \frac{2 x - 1}{2 x}, 0 < x < \frac{1}{2} \\ \frac{2 x - 1}{2 - 2 x}, \frac{1}{2} \leq x < 1 \end{cases}$
Now, let us draw the graph of above function.
original image
From the above graph it is clear that $f (x)$ is one-one and since the range of $f (x)$ is real number, which implies that codomain is equal to range.
$∴ f (x)$ is one-one and onto both and hence $f (x)$ is bijective.

WB JEE-2022

Sets, Relation and Function

117077 For the mapping f: $R - {1} \to R - {2}$ , given by $f (x) = \frac{2 x}{x - 1}$ . Which of the following is correct ?

1

f

is one- one but not onto

2

f

is onto but not one-one

3

f

is neither one-one nor onto

4

f

is both one-one and onto

Explanation:

D Given,
$f (x) = \frac{2 x}{x - 1}$
$y = \frac{2 x}{x - 1} = \frac{2 x - 2 + 2}{x - 1} = \frac{2 (x - 1) + 2}{(x - 1)}$
$= \frac{2 (x - 1)}{(x - 1)} + \frac{2}{(x - 1)}$
$∴ (y - 2) (x - 1) = 2$
$onto \to co-domain = range$
original image
Thus, $f$ is both one-one and onto.

WB JEE-2022

Sets, Relation and Function

117078 The function $f : X \to Y$ defined by $f (x) = \sin x$ is one-one but not onto, if $X$ and $Y$ are respectively equal to

1

R

and

R

2

[0, π]

and

[0, 1]

3

[0, \frac{π}{2}]

and

[- 1, 1]

4

[\frac{- π}{2}, \frac{π}{2}]

and

[- 1, 1]

Explanation:

C One-one or injective function- Basically denotes the mapping of two sets.
Since, $f : X \to Y$
Then $f (x) = \sin x$
Now check option $C$
Domain $= [0, \frac{π}{2}]$
Range $= [- 1, 1]$
For every value of $x$ , we get unique value of $y$ . But the value of $y$ in $[- 1, 0)$ does not have any pre-image $∴$ Function is one-one but not onto. Hence, option (c) is correct.

Jamia Millia Islamia-2008

Sets, Relation and Function

117074 Let $a, b, c$ be real numbers, each greater than 1, such that $\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$ .
If the value of $b$ is 9 , then the value of ' $a$ ' must be

1

\sqrt[3]{81}

2

\frac{27}{2}

3 18

4 27

Explanation:

D Given,
$b = 9$
$\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$
$a, b, c$ are real
$\frac{2}{3} \log_{9} a + \frac{3}{5} \log_{c} 9 + \frac{5}{2} \log_{a} c = 3$
$\Rightarrow \log_{9} a^{2 / 3} + \log_{c} 9^{3 / 5} + \log_{a} c^{5 / 3} = 3$
As there are two unknown variable a \& c, considering every term is equal to 1 .
$\log_{9} a^{2 / 3} = 1, \log_{c} 9^{3 / 5} = 1, \log_{a} c^{5 / 3} = 1$
$∴ \frac{\log a^{2 / 3}}{\log 9} = 1 (∵ \log_{x} y = \frac{\log y}{\log x})$
$\Rightarrow \log a^{2 / 3} = \log 9$
$\Rightarrow a^{2 / 3} = 9$
$\Rightarrow a = 9^{3 / 2}$
$\Rightarrow a = (3)^{2 \times 3 / 2}$
$= a = (3)^{3} = 27 .$ Hence option (D) is correct.

WB JEE-2021

Sets, Relation and Function

117075 $f : X \to R, X = {x ∣ 0 < x < 1}$ is defined as $f (x) =$ $\frac{2 x}{1 - | 2 x - 1 |}$ . Then

1

f

is only injective

2

f

is only surjective

3

f

is bijective

4

f

is neither injective nor surjective

Explanation:

A Given,
$f (x) = \frac{2 x - 1}{1 - | 2 x - 1 |}$
$f (x) = {\begin{cases} \frac{2 x - 1}{1 + 2 x - 1}, 0 < x < \frac{1}{2} \\ \frac{2 x}{1 - 2 x + 1}, \frac{1}{2} \leq x < 1 \end{cases}$
$f (x) = {\begin{cases} \frac{2 x - 1}{2 x}, 0 < x < \frac{1}{2} \\ \frac{2 x - 1}{2 - 2 x}, \frac{1}{2} \leq x < 1 \end{cases}$
Now, let us draw the graph of above function.
original image
From the above graph it is clear that $f (x)$ is one-one and since the range of $f (x)$ is real number, which implies that codomain is equal to range.
$∴ f (x)$ is one-one and onto both and hence $f (x)$ is bijective.

WB JEE-2022

Sets, Relation and Function

117077 For the mapping f: $R - {1} \to R - {2}$ , given by $f (x) = \frac{2 x}{x - 1}$ . Which of the following is correct ?

1

f

is one- one but not onto

2

f

is onto but not one-one

3

f

is neither one-one nor onto

4

f

is both one-one and onto

Explanation:

D Given,
$f (x) = \frac{2 x}{x - 1}$
$y = \frac{2 x}{x - 1} = \frac{2 x - 2 + 2}{x - 1} = \frac{2 (x - 1) + 2}{(x - 1)}$
$= \frac{2 (x - 1)}{(x - 1)} + \frac{2}{(x - 1)}$
$∴ (y - 2) (x - 1) = 2$
$onto \to co-domain = range$
original image
Thus, $f$ is both one-one and onto.

WB JEE-2022

Sets, Relation and Function

117078 The function $f : X \to Y$ defined by $f (x) = \sin x$ is one-one but not onto, if $X$ and $Y$ are respectively equal to

1

R

and

R

2

[0, π]

and

[0, 1]

3

[0, \frac{π}{2}]

and

[- 1, 1]

4

[\frac{- π}{2}, \frac{π}{2}]

and

[- 1, 1]

Explanation:

C One-one or injective function- Basically denotes the mapping of two sets.
Since, $f : X \to Y$
Then $f (x) = \sin x$
Now check option $C$
Domain $= [0, \frac{π}{2}]$
Range $= [- 1, 1]$
For every value of $x$ , we get unique value of $y$ . But the value of $y$ in $[- 1, 0)$ does not have any pre-image $∴$ Function is one-one but not onto. Hence, option (c) is correct.

Jamia Millia Islamia-2008

Sets, Relation and Function

117074 Let $a, b, c$ be real numbers, each greater than 1, such that $\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$ .
If the value of $b$ is 9 , then the value of ' $a$ ' must be

1

\sqrt[3]{81}

2

\frac{27}{2}

3 18

4 27

Explanation:

D Given,
$b = 9$
$\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$
$a, b, c$ are real
$\frac{2}{3} \log_{9} a + \frac{3}{5} \log_{c} 9 + \frac{5}{2} \log_{a} c = 3$
$\Rightarrow \log_{9} a^{2 / 3} + \log_{c} 9^{3 / 5} + \log_{a} c^{5 / 3} = 3$
As there are two unknown variable a \& c, considering every term is equal to 1 .
$\log_{9} a^{2 / 3} = 1, \log_{c} 9^{3 / 5} = 1, \log_{a} c^{5 / 3} = 1$
$∴ \frac{\log a^{2 / 3}}{\log 9} = 1 (∵ \log_{x} y = \frac{\log y}{\log x})$
$\Rightarrow \log a^{2 / 3} = \log 9$
$\Rightarrow a^{2 / 3} = 9$
$\Rightarrow a = 9^{3 / 2}$
$\Rightarrow a = (3)^{2 \times 3 / 2}$
$= a = (3)^{3} = 27 .$ Hence option (D) is correct.

WB JEE-2021

Sets, Relation and Function

117075 $f : X \to R, X = {x ∣ 0 < x < 1}$ is defined as $f (x) =$ $\frac{2 x}{1 - | 2 x - 1 |}$ . Then

1

f

is only injective

2

f

is only surjective

3

f

is bijective

4

f

is neither injective nor surjective

Explanation:

A Given,
$f (x) = \frac{2 x - 1}{1 - | 2 x - 1 |}$
$f (x) = {\begin{cases} \frac{2 x - 1}{1 + 2 x - 1}, 0 < x < \frac{1}{2} \\ \frac{2 x}{1 - 2 x + 1}, \frac{1}{2} \leq x < 1 \end{cases}$
$f (x) = {\begin{cases} \frac{2 x - 1}{2 x}, 0 < x < \frac{1}{2} \\ \frac{2 x - 1}{2 - 2 x}, \frac{1}{2} \leq x < 1 \end{cases}$
Now, let us draw the graph of above function.
original image
From the above graph it is clear that $f (x)$ is one-one and since the range of $f (x)$ is real number, which implies that codomain is equal to range.
$∴ f (x)$ is one-one and onto both and hence $f (x)$ is bijective.

WB JEE-2022

Sets, Relation and Function

117077 For the mapping f: $R - {1} \to R - {2}$ , given by $f (x) = \frac{2 x}{x - 1}$ . Which of the following is correct ?

1

f

is one- one but not onto

2

f

is onto but not one-one

3

f

is neither one-one nor onto

4

f

is both one-one and onto

Explanation:

D Given,
$f (x) = \frac{2 x}{x - 1}$
$y = \frac{2 x}{x - 1} = \frac{2 x - 2 + 2}{x - 1} = \frac{2 (x - 1) + 2}{(x - 1)}$
$= \frac{2 (x - 1)}{(x - 1)} + \frac{2}{(x - 1)}$
$∴ (y - 2) (x - 1) = 2$
$onto \to co-domain = range$
original image
Thus, $f$ is both one-one and onto.

WB JEE-2022

Sets, Relation and Function

117078 The function $f : X \to Y$ defined by $f (x) = \sin x$ is one-one but not onto, if $X$ and $Y$ are respectively equal to

1

R

and

R

2

[0, π]

and

[0, 1]

3

[0, \frac{π}{2}]

and

[- 1, 1]

4

[\frac{- π}{2}, \frac{π}{2}]

and

[- 1, 1]

Explanation:

C One-one or injective function- Basically denotes the mapping of two sets.
Since, $f : X \to Y$
Then $f (x) = \sin x$
Now check option $C$
Domain $= [0, \frac{π}{2}]$
Range $= [- 1, 1]$
For every value of $x$ , we get unique value of $y$ . But the value of $y$ in $[- 1, 0)$ does not have any pre-image $∴$ Function is one-one but not onto. Hence, option (c) is correct.

Jamia Millia Islamia-2008

Sets, Relation and Function

117074 Let $a, b, c$ be real numbers, each greater than 1, such that $\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$ .
If the value of $b$ is 9 , then the value of ' $a$ ' must be

1

\sqrt[3]{81}

2

\frac{27}{2}

3 18

4 27

Explanation:

D Given,
$b = 9$
$\frac{2}{3} \log_{b} a + \frac{3}{5} \log_{c} b + \frac{5}{2} \log_{a} c = 3$
$a, b, c$ are real
$\frac{2}{3} \log_{9} a + \frac{3}{5} \log_{c} 9 + \frac{5}{2} \log_{a} c = 3$
$\Rightarrow \log_{9} a^{2 / 3} + \log_{c} 9^{3 / 5} + \log_{a} c^{5 / 3} = 3$
As there are two unknown variable a \& c, considering every term is equal to 1 .
$\log_{9} a^{2 / 3} = 1, \log_{c} 9^{3 / 5} = 1, \log_{a} c^{5 / 3} = 1$
$∴ \frac{\log a^{2 / 3}}{\log 9} = 1 (∵ \log_{x} y = \frac{\log y}{\log x})$
$\Rightarrow \log a^{2 / 3} = \log 9$
$\Rightarrow a^{2 / 3} = 9$
$\Rightarrow a = 9^{3 / 2}$
$\Rightarrow a = (3)^{2 \times 3 / 2}$
$= a = (3)^{3} = 27 .$ Hence option (D) is correct.

WB JEE-2021

Sets, Relation and Function

117075 $f : X \to R, X = {x ∣ 0 < x < 1}$ is defined as $f (x) =$ $\frac{2 x}{1 - | 2 x - 1 |}$ . Then

1

f

is only injective

2

f

is only surjective

3

f

is bijective

4

f

is neither injective nor surjective

Explanation:

A Given,
$f (x) = \frac{2 x - 1}{1 - | 2 x - 1 |}$
$f (x) = {\begin{cases} \frac{2 x - 1}{1 + 2 x - 1}, 0 < x < \frac{1}{2} \\ \frac{2 x}{1 - 2 x + 1}, \frac{1}{2} \leq x < 1 \end{cases}$
$f (x) = {\begin{cases} \frac{2 x - 1}{2 x}, 0 < x < \frac{1}{2} \\ \frac{2 x - 1}{2 - 2 x}, \frac{1}{2} \leq x < 1 \end{cases}$
Now, let us draw the graph of above function.
original image
From the above graph it is clear that $f (x)$ is one-one and since the range of $f (x)$ is real number, which implies that codomain is equal to range.
$∴ f (x)$ is one-one and onto both and hence $f (x)$ is bijective.

WB JEE-2022

Sets, Relation and Function

117077 For the mapping f: $R - {1} \to R - {2}$ , given by $f (x) = \frac{2 x}{x - 1}$ . Which of the following is correct ?

1

f

is one- one but not onto

2

f

is onto but not one-one

3

f

is neither one-one nor onto

4

f

is both one-one and onto

Explanation:

D Given,
$f (x) = \frac{2 x}{x - 1}$
$y = \frac{2 x}{x - 1} = \frac{2 x - 2 + 2}{x - 1} = \frac{2 (x - 1) + 2}{(x - 1)}$
$= \frac{2 (x - 1)}{(x - 1)} + \frac{2}{(x - 1)}$
$∴ (y - 2) (x - 1) = 2$
$onto \to co-domain = range$
original image
Thus, $f$ is both one-one and onto.

WB JEE-2022

Sets, Relation and Function

117078 The function $f : X \to Y$ defined by $f (x) = \sin x$ is one-one but not onto, if $X$ and $Y$ are respectively equal to

1

R

and

R

2

[0, π]

and

[0, 1]

3

[0, \frac{π}{2}]

and

[- 1, 1]

4

[\frac{- π}{2}, \frac{π}{2}]

and

[- 1, 1]

Explanation:

C One-one or injective function- Basically denotes the mapping of two sets.
Since, $f : X \to Y$
Then $f (x) = \sin x$
Now check option $C$
Domain $= [0, \frac{π}{2}]$
Range $= [- 1, 1]$
For every value of $x$ , we get unique value of $y$ . But the value of $y$ in $[- 1, 0)$ does not have any pre-image $∴$ Function is one-one but not onto. Hence, option (c) is correct.

Jamia Millia Islamia-2008