QBManager

Sets, Relation and Function

117238 If $f (x) = 3 x + 6, g (x) = 4 x + k$ and fog $(x) =$ gof
(x) then $k =$

1 -9

2 18

3

\frac{1}{9}

4 9

Explanation:

D Given,
$f (x) = 3 x + 6, g (x) = 4 x + k$
$f \circ g (x) = f (g (x)) = f (4 x + k) = 3 (4 x + k) + 6 = 12 x + 3 k + 6$
$g \circ f (x) = g (f (x)) = g (3 x + 6) = 4 (3 x + 6) + k = 12 x + 24 + k$
$∵ f o g (x) = gof (x)$
$12 x + 3 k + 6 = 12 x + 24 + k$
$2 k = 18$
$k = 9$

MHT-CET 2020

Sets, Relation and Function

117239 If $f : R \to R$ is defined by $f (x) = 2 x + 3$ , then $f^{- 1}$ (x)

1 does not exist because '

f

' is not surjective

2 is given by

\frac{x - 3}{2}

3 is given by

\frac{1}{2 x + 3}

4 does not exist because '

f

' is not injective

Explanation:

B Given, $f : R \to R$ is defined by
$f (x) = 2 x + 3$
So, let
$f (x) = y$
$y = 2 x + 3$
$2 x = y - 3$
$x = \frac{y - 3}{2}$
Hence,
$f^{- 1} (x) = \frac{x - 3}{2}$

Karnataka CET 2012 BITSAT - 2006

Sets, Relation and Function

117240 $f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
for the given functions $(g^{- 1}) (10) =$

1 21

2 29

3 7

4

\frac{8}{3}

Explanation:

A Given,
$f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
and find $(g \circ f^{- 1}) (10) =$ ?
Let $f (x) = y$
$y = 3 x + 2$
$3 x = y - 2 \Rightarrow x = \frac{y - 2}{3}$
$f^{- 1} (x) = \frac{x - 2}{3}$
Then, $(g \circ f^{- 1}) (10) = g {f^{- 1} (10)} {∵ x = 10}$
$= g {\frac{10 - 2}{3}}$
$= g {\frac{8}{3}}$
Now,
$g {\frac{8}{3}} = 6 \times \frac{8}{3} + 5$
$= 2 \times 8 + 5$
$= 16 + 5$
$(go f^{- 1}) (10) = 21 .$

GUJCET-2017

Sets, Relation and Function

117241 If $f (x) = \frac{2 x - 3}{3 x + 4}$ , then $f^{- 1} (\frac{- 4}{3}) =$

1 zero

2

\frac{3}{4}

3

- \frac{2}{3}

4 None of these

Explanation:

D Given,
$f (x) = \frac{2 x - 3}{3 x + 4}$
Let, $f (x) = y = \frac{2 x - 3}{3 x + 4}$
On cross multiplication, we get
$3 x y + 4 y = 2 x - 3$
$3 x y - 2 x = - 3 - 4 y$
$x (3 y - 2) = - 3 - 4 y$
$x = f^{- 1} (y) = \frac{- 3 - 4 y}{3 y - 2}$
Put, $y = - \frac{4}{3}$ , we get -
$f^{- 1} (- \frac{4}{3}) = \frac{- 3 - 4 \times (- \frac{4}{3})}{3 (- \frac{4}{3}) - 2}$
$= \frac{- 3 + \frac{16}{3}}{- 4 - 2} = \frac{7}{3 \times (- 6)} = - \frac{7}{18}$

COMEDK 2014

Sets, Relation and Function

117242 $f : R \to R$ and $g : R \to R$ are two function such that $f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$ , then $x^{4} (f \circ g) (x)$ is equal to

1 0

2 1

3

x^{4}

4

x^{2}

Explanation:

B Given that,
$f (x) = x^{2}$
$g (x) = \frac{1}{x^{2}}$
Then,
$(f \circ g) x = f (g (x))$
$f (g (x)) = {(\frac{1}{x^{2}})}^{2} = \frac{1}{x^{4}}$
$x^{4} (f \circ g) (x) = x^{4} \times \frac{1}{x^{4}} = 1$

Shift-II

Sets, Relation and Function

117238 If $f (x) = 3 x + 6, g (x) = 4 x + k$ and fog $(x) =$ gof
(x) then $k =$

1 -9

2 18

3

\frac{1}{9}

4 9

Explanation:

D Given,
$f (x) = 3 x + 6, g (x) = 4 x + k$
$f \circ g (x) = f (g (x)) = f (4 x + k) = 3 (4 x + k) + 6 = 12 x + 3 k + 6$
$g \circ f (x) = g (f (x)) = g (3 x + 6) = 4 (3 x + 6) + k = 12 x + 24 + k$
$∵ f o g (x) = gof (x)$
$12 x + 3 k + 6 = 12 x + 24 + k$
$2 k = 18$
$k = 9$

MHT-CET 2020

Sets, Relation and Function

117239 If $f : R \to R$ is defined by $f (x) = 2 x + 3$ , then $f^{- 1}$ (x)

1 does not exist because '

f

' is not surjective

2 is given by

\frac{x - 3}{2}

3 is given by

\frac{1}{2 x + 3}

4 does not exist because '

f

' is not injective

Explanation:

B Given, $f : R \to R$ is defined by
$f (x) = 2 x + 3$
So, let
$f (x) = y$
$y = 2 x + 3$
$2 x = y - 3$
$x = \frac{y - 3}{2}$
Hence,
$f^{- 1} (x) = \frac{x - 3}{2}$

Karnataka CET 2012 BITSAT - 2006

Sets, Relation and Function

117240 $f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
for the given functions $(g^{- 1}) (10) =$

1 21

2 29

3 7

4

\frac{8}{3}

Explanation:

A Given,
$f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
and find $(g \circ f^{- 1}) (10) =$ ?
Let $f (x) = y$
$y = 3 x + 2$
$3 x = y - 2 \Rightarrow x = \frac{y - 2}{3}$
$f^{- 1} (x) = \frac{x - 2}{3}$
Then, $(g \circ f^{- 1}) (10) = g {f^{- 1} (10)} {∵ x = 10}$
$= g {\frac{10 - 2}{3}}$
$= g {\frac{8}{3}}$
Now,
$g {\frac{8}{3}} = 6 \times \frac{8}{3} + 5$
$= 2 \times 8 + 5$
$= 16 + 5$
$(go f^{- 1}) (10) = 21 .$

GUJCET-2017

Sets, Relation and Function

117241 If $f (x) = \frac{2 x - 3}{3 x + 4}$ , then $f^{- 1} (\frac{- 4}{3}) =$

1 zero

2

\frac{3}{4}

3

- \frac{2}{3}

4 None of these

Explanation:

D Given,
$f (x) = \frac{2 x - 3}{3 x + 4}$
Let, $f (x) = y = \frac{2 x - 3}{3 x + 4}$
On cross multiplication, we get
$3 x y + 4 y = 2 x - 3$
$3 x y - 2 x = - 3 - 4 y$
$x (3 y - 2) = - 3 - 4 y$
$x = f^{- 1} (y) = \frac{- 3 - 4 y}{3 y - 2}$
Put, $y = - \frac{4}{3}$ , we get -
$f^{- 1} (- \frac{4}{3}) = \frac{- 3 - 4 \times (- \frac{4}{3})}{3 (- \frac{4}{3}) - 2}$
$= \frac{- 3 + \frac{16}{3}}{- 4 - 2} = \frac{7}{3 \times (- 6)} = - \frac{7}{18}$

COMEDK 2014

Sets, Relation and Function

117242 $f : R \to R$ and $g : R \to R$ are two function such that $f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$ , then $x^{4} (f \circ g) (x)$ is equal to

1 0

2 1

3

x^{4}

4

x^{2}

Explanation:

B Given that,
$f (x) = x^{2}$
$g (x) = \frac{1}{x^{2}}$
Then,
$(f \circ g) x = f (g (x))$
$f (g (x)) = {(\frac{1}{x^{2}})}^{2} = \frac{1}{x^{4}}$
$x^{4} (f \circ g) (x) = x^{4} \times \frac{1}{x^{4}} = 1$

Shift-II

Sets, Relation and Function

117238 If $f (x) = 3 x + 6, g (x) = 4 x + k$ and fog $(x) =$ gof
(x) then $k =$

1 -9

2 18

3

\frac{1}{9}

4 9

Explanation:

D Given,
$f (x) = 3 x + 6, g (x) = 4 x + k$
$f \circ g (x) = f (g (x)) = f (4 x + k) = 3 (4 x + k) + 6 = 12 x + 3 k + 6$
$g \circ f (x) = g (f (x)) = g (3 x + 6) = 4 (3 x + 6) + k = 12 x + 24 + k$
$∵ f o g (x) = gof (x)$
$12 x + 3 k + 6 = 12 x + 24 + k$
$2 k = 18$
$k = 9$

MHT-CET 2020

Sets, Relation and Function

117239 If $f : R \to R$ is defined by $f (x) = 2 x + 3$ , then $f^{- 1}$ (x)

1 does not exist because '

f

' is not surjective

2 is given by

\frac{x - 3}{2}

3 is given by

\frac{1}{2 x + 3}

4 does not exist because '

f

' is not injective

Explanation:

B Given, $f : R \to R$ is defined by
$f (x) = 2 x + 3$
So, let
$f (x) = y$
$y = 2 x + 3$
$2 x = y - 3$
$x = \frac{y - 3}{2}$
Hence,
$f^{- 1} (x) = \frac{x - 3}{2}$

Karnataka CET 2012 BITSAT - 2006

Sets, Relation and Function

117240 $f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
for the given functions $(g^{- 1}) (10) =$

1 21

2 29

3 7

4

\frac{8}{3}

Explanation:

A Given,
$f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
and find $(g \circ f^{- 1}) (10) =$ ?
Let $f (x) = y$
$y = 3 x + 2$
$3 x = y - 2 \Rightarrow x = \frac{y - 2}{3}$
$f^{- 1} (x) = \frac{x - 2}{3}$
Then, $(g \circ f^{- 1}) (10) = g {f^{- 1} (10)} {∵ x = 10}$
$= g {\frac{10 - 2}{3}}$
$= g {\frac{8}{3}}$
Now,
$g {\frac{8}{3}} = 6 \times \frac{8}{3} + 5$
$= 2 \times 8 + 5$
$= 16 + 5$
$(go f^{- 1}) (10) = 21 .$

GUJCET-2017

Sets, Relation and Function

117241 If $f (x) = \frac{2 x - 3}{3 x + 4}$ , then $f^{- 1} (\frac{- 4}{3}) =$

1 zero

2

\frac{3}{4}

3

- \frac{2}{3}

4 None of these

Explanation:

D Given,
$f (x) = \frac{2 x - 3}{3 x + 4}$
Let, $f (x) = y = \frac{2 x - 3}{3 x + 4}$
On cross multiplication, we get
$3 x y + 4 y = 2 x - 3$
$3 x y - 2 x = - 3 - 4 y$
$x (3 y - 2) = - 3 - 4 y$
$x = f^{- 1} (y) = \frac{- 3 - 4 y}{3 y - 2}$
Put, $y = - \frac{4}{3}$ , we get -
$f^{- 1} (- \frac{4}{3}) = \frac{- 3 - 4 \times (- \frac{4}{3})}{3 (- \frac{4}{3}) - 2}$
$= \frac{- 3 + \frac{16}{3}}{- 4 - 2} = \frac{7}{3 \times (- 6)} = - \frac{7}{18}$

COMEDK 2014

Sets, Relation and Function

117242 $f : R \to R$ and $g : R \to R$ are two function such that $f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$ , then $x^{4} (f \circ g) (x)$ is equal to

1 0

2 1

3

x^{4}

4

x^{2}

Explanation:

B Given that,
$f (x) = x^{2}$
$g (x) = \frac{1}{x^{2}}$
Then,
$(f \circ g) x = f (g (x))$
$f (g (x)) = {(\frac{1}{x^{2}})}^{2} = \frac{1}{x^{4}}$
$x^{4} (f \circ g) (x) = x^{4} \times \frac{1}{x^{4}} = 1$

Shift-II

Sets, Relation and Function

117238 If $f (x) = 3 x + 6, g (x) = 4 x + k$ and fog $(x) =$ gof
(x) then $k =$

1 -9

2 18

3

\frac{1}{9}

4 9

Explanation:

D Given,
$f (x) = 3 x + 6, g (x) = 4 x + k$
$f \circ g (x) = f (g (x)) = f (4 x + k) = 3 (4 x + k) + 6 = 12 x + 3 k + 6$
$g \circ f (x) = g (f (x)) = g (3 x + 6) = 4 (3 x + 6) + k = 12 x + 24 + k$
$∵ f o g (x) = gof (x)$
$12 x + 3 k + 6 = 12 x + 24 + k$
$2 k = 18$
$k = 9$

MHT-CET 2020

Sets, Relation and Function

117239 If $f : R \to R$ is defined by $f (x) = 2 x + 3$ , then $f^{- 1}$ (x)

1 does not exist because '

f

' is not surjective

2 is given by

\frac{x - 3}{2}

3 is given by

\frac{1}{2 x + 3}

4 does not exist because '

f

' is not injective

Explanation:

B Given, $f : R \to R$ is defined by
$f (x) = 2 x + 3$
So, let
$f (x) = y$
$y = 2 x + 3$
$2 x = y - 3$
$x = \frac{y - 3}{2}$
Hence,
$f^{- 1} (x) = \frac{x - 3}{2}$

Karnataka CET 2012 BITSAT - 2006

Sets, Relation and Function

117240 $f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
for the given functions $(g^{- 1}) (10) =$

1 21

2 29

3 7

4

\frac{8}{3}

Explanation:

A Given,
$f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
and find $(g \circ f^{- 1}) (10) =$ ?
Let $f (x) = y$
$y = 3 x + 2$
$3 x = y - 2 \Rightarrow x = \frac{y - 2}{3}$
$f^{- 1} (x) = \frac{x - 2}{3}$
Then, $(g \circ f^{- 1}) (10) = g {f^{- 1} (10)} {∵ x = 10}$
$= g {\frac{10 - 2}{3}}$
$= g {\frac{8}{3}}$
Now,
$g {\frac{8}{3}} = 6 \times \frac{8}{3} + 5$
$= 2 \times 8 + 5$
$= 16 + 5$
$(go f^{- 1}) (10) = 21 .$

GUJCET-2017

Sets, Relation and Function

117241 If $f (x) = \frac{2 x - 3}{3 x + 4}$ , then $f^{- 1} (\frac{- 4}{3}) =$

1 zero

2

\frac{3}{4}

3

- \frac{2}{3}

4 None of these

Explanation:

D Given,
$f (x) = \frac{2 x - 3}{3 x + 4}$
Let, $f (x) = y = \frac{2 x - 3}{3 x + 4}$
On cross multiplication, we get
$3 x y + 4 y = 2 x - 3$
$3 x y - 2 x = - 3 - 4 y$
$x (3 y - 2) = - 3 - 4 y$
$x = f^{- 1} (y) = \frac{- 3 - 4 y}{3 y - 2}$
Put, $y = - \frac{4}{3}$ , we get -
$f^{- 1} (- \frac{4}{3}) = \frac{- 3 - 4 \times (- \frac{4}{3})}{3 (- \frac{4}{3}) - 2}$
$= \frac{- 3 + \frac{16}{3}}{- 4 - 2} = \frac{7}{3 \times (- 6)} = - \frac{7}{18}$

COMEDK 2014

Sets, Relation and Function

117242 $f : R \to R$ and $g : R \to R$ are two function such that $f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$ , then $x^{4} (f \circ g) (x)$ is equal to

1 0

2 1

3

x^{4}

4

x^{2}

Explanation:

B Given that,
$f (x) = x^{2}$
$g (x) = \frac{1}{x^{2}}$
Then,
$(f \circ g) x = f (g (x))$
$f (g (x)) = {(\frac{1}{x^{2}})}^{2} = \frac{1}{x^{4}}$
$x^{4} (f \circ g) (x) = x^{4} \times \frac{1}{x^{4}} = 1$

Shift-II

Sets, Relation and Function

117238 If $f (x) = 3 x + 6, g (x) = 4 x + k$ and fog $(x) =$ gof
(x) then $k =$

1 -9

2 18

3

\frac{1}{9}

4 9

Explanation:

D Given,
$f (x) = 3 x + 6, g (x) = 4 x + k$
$f \circ g (x) = f (g (x)) = f (4 x + k) = 3 (4 x + k) + 6 = 12 x + 3 k + 6$
$g \circ f (x) = g (f (x)) = g (3 x + 6) = 4 (3 x + 6) + k = 12 x + 24 + k$
$∵ f o g (x) = gof (x)$
$12 x + 3 k + 6 = 12 x + 24 + k$
$2 k = 18$
$k = 9$

MHT-CET 2020

Sets, Relation and Function

117239 If $f : R \to R$ is defined by $f (x) = 2 x + 3$ , then $f^{- 1}$ (x)

1 does not exist because '

f

' is not surjective

2 is given by

\frac{x - 3}{2}

3 is given by

\frac{1}{2 x + 3}

4 does not exist because '

f

' is not injective

Explanation:

B Given, $f : R \to R$ is defined by
$f (x) = 2 x + 3$
So, let
$f (x) = y$
$y = 2 x + 3$
$2 x = y - 3$
$x = \frac{y - 3}{2}$
Hence,
$f^{- 1} (x) = \frac{x - 3}{2}$

Karnataka CET 2012 BITSAT - 2006

Sets, Relation and Function

117240 $f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
for the given functions $(g^{- 1}) (10) =$

1 21

2 29

3 7

4

\frac{8}{3}

Explanation:

A Given,
$f : R \to R, f (x) = 3 x + 2$
$g : R \to R, g (x) = 6 x + 5$
and find $(g \circ f^{- 1}) (10) =$ ?
Let $f (x) = y$
$y = 3 x + 2$
$3 x = y - 2 \Rightarrow x = \frac{y - 2}{3}$
$f^{- 1} (x) = \frac{x - 2}{3}$
Then, $(g \circ f^{- 1}) (10) = g {f^{- 1} (10)} {∵ x = 10}$
$= g {\frac{10 - 2}{3}}$
$= g {\frac{8}{3}}$
Now,
$g {\frac{8}{3}} = 6 \times \frac{8}{3} + 5$
$= 2 \times 8 + 5$
$= 16 + 5$
$(go f^{- 1}) (10) = 21 .$

GUJCET-2017

Sets, Relation and Function

117241 If $f (x) = \frac{2 x - 3}{3 x + 4}$ , then $f^{- 1} (\frac{- 4}{3}) =$

1 zero

2

\frac{3}{4}

3

- \frac{2}{3}

4 None of these

Explanation:

D Given,
$f (x) = \frac{2 x - 3}{3 x + 4}$
Let, $f (x) = y = \frac{2 x - 3}{3 x + 4}$
On cross multiplication, we get
$3 x y + 4 y = 2 x - 3$
$3 x y - 2 x = - 3 - 4 y$
$x (3 y - 2) = - 3 - 4 y$
$x = f^{- 1} (y) = \frac{- 3 - 4 y}{3 y - 2}$
Put, $y = - \frac{4}{3}$ , we get -
$f^{- 1} (- \frac{4}{3}) = \frac{- 3 - 4 \times (- \frac{4}{3})}{3 (- \frac{4}{3}) - 2}$
$= \frac{- 3 + \frac{16}{3}}{- 4 - 2} = \frac{7}{3 \times (- 6)} = - \frac{7}{18}$

COMEDK 2014

Sets, Relation and Function

117242 $f : R \to R$ and $g : R \to R$ are two function such that $f (x) = x^{2}$ and $g (x) = \frac{1}{x^{2}}$ , then $x^{4} (f \circ g) (x)$ is equal to

1 0

2 1

3

x^{4}

4

x^{2}

Explanation:

B Given that,
$f (x) = x^{2}$
$g (x) = \frac{1}{x^{2}}$
Then,
$(f \circ g) x = f (g (x))$
$f (g (x)) = {(\frac{1}{x^{2}})}^{2} = \frac{1}{x^{4}}$
$x^{4} (f \circ g) (x) = x^{4} \times \frac{1}{x^{4}} = 1$

Shift-II