Exercise 1.3 Solutions

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Exercise 1.3
1. Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
It is given that f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}
So, gof : {1, 3, 4} → {1, 3}
The function f is given as
f(1) = 2
f(3) = 5
f(4) = 1
This can be diagramatically represented as follows:
f : {1, 3, 4} → {1, 2, 5} and f = {(1, 2), (3, 5), (4, 1)} {1, 3, 4} {1, 2, 5} 1 3 4 1 2 5 f
The function g is given as
g(1) = 3
g(2) = 3
g(5) = 1
This can be diagramatically represented as follows:
g : {1, 2, 5} → {1, 3} and g = {(1, 3), (2, 3), (5, 1)} {1, 2, 5} {1, 2, 3} 1 2 5 1 2 3 g
gof(1) = g(f(1)) = g(2) = 3
gof(3) = g(f(3)) = g(5) = 1
gof(4) = g(f(4)) = g(1) = 3
This can be diagramatically represented as follows:
gof = {(1,3), (3, 1), (4, 3)} {1, 3, 4} {1, 2, 5} {1, 2, 3} 1 3 4 1 2 5 1 2 3 f g
gof = {(1,3), (3, 1), (4, 3)}

2. Let f, g and h be functions from R to R. Show that
(f + g)oh = foh + goh
(f.g)oh = (foh).(goh)
We have
{((f + g)oh)(x)}
{= (f + g)(h(x))}
{= f(h(x)) + g(h(x))}
{= foh(x) + goh(x)}
{∴ (f + g)oh = foh + goh}
Similarly,
((f.g)oh)(x)
{= (f.g)(h(x))}
{= f(h(x)).g(h(x))}
{= foh(x).goh(x)}
{∴ (f.g)oh = (foh).(goh)}

3. Find gof and fog, if
i.
{f(x) = |x|} and {g(x) = |5x - 2|}
ii.
{f(x) = 8x^3} and {g(x) = x^{\frac{1}{3}}}
To find gof and fog when {f(x) = |x|} and {g(x) = |5x - 2|}
We have,
gof(x)
{= g(f(x))}
{= g(|x|)} {(∵ f(x) = |x|)}
{= |5|x| - 2|} {(∵ g(x) = |5x - 2|)}
{∴ gof(x) = |5|x| - 2|}
Similarly,
fog(x)
{= f(g(x))}
{= f(|5x - 2|)} {(∵ g(x) = |5x - 2|)}
{= ||5x - 2||} {(∵ f(x) = |x|)}
{= |5x - 2|}
To find gof and fog when {f(x) = 8x^3} and {g(x) = x^{\frac{1}{3}}}
We have,
gof(x)
{= g(f(x))}
{= g(8x^3)} {(∵ f(x) = 8x^3)}
{= \left(8x^3\right)^\frac{1}{3}} {(∵ g(x) = x^\frac{1}{3})}
{= \left(\left(2x\right)^3\right)^\frac{1}{3}}
{= \left(2x\right)^{3\times\frac{1}{3}}} {(∵ \left(a^m\right)^n = a^{mn}})
{= 2x}
{∴ gof(x) = 2x}
Similarly,
fog(x)
{= f(g(x))}
{= f(x^\frac{1}{3})} {(∵ g(x) = x^\frac{1}{3})}
{= 8\left(x^\frac{1}{3}\right)^3} {(∵ f(x) = 8x^3)}
{= 8x^{\frac{1}{3}\times3}} {(∵ \left(a^m\right)^n = a^{mn})}
{= 8x}
{∴ fog(x) = 8x}

4. If {f(x) = \dfrac{(4x + 3)}{(6x - 4)}}, x\dfrac{2}{3}, show that {fof = x}, for all x\dfrac{2}{3}. What is the inverse of f?
We have,
fof(x)
{= f(f(x))}
{= f\left(\dfrac{4x - 3}{6x - 4}\right)} {\left(∵ f(x) = \dfrac{4x - 3}{6x - 4}\right)}
{= \dfrac{4\left(\dfrac{4x - 3}{6x - 4}\right) - 3}{6\left(\dfrac{4x - 3}{6x - 4}\right) - 4}} {\left(∵ f(x) = \dfrac{4x - 3}{6x - 4}\right)}
{= \dfrac{\dfrac{4(4x - 3) - 3(6x - 4)}{6x - 4}}{\dfrac{6(4x - 3) - 4(6x - 4)}{6x - 4}}}
{= \dfrac{16x - 12 - 18x + 12}{24x - 18 - 24x + 16}} (After cancelling/cutting {6x - 4} in both numerator and denominator)
{= \dfrac{-2x}{-2}}
= x
{∴ fof(x) = x}
{∵ fof(x) = x}, the inverse of f is f itself.
∴ We can say that {f^{-1} = f}

5. State with reason whether following functions have inverse
(i)
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii)
g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii)
h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
As we know, a function f will have an inverse only if the function is both one-one and onto.
Let’s now check one by one whether the given functions f, g and h are both one-one and onto.
To check whether f is one-one
In the function f, we have
{f(1) = 10}
{f(2) = 10}
{f(3) = 10} and
{f(4) = 10}
This can be diagramatically represented as follows:
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)} {1, 2, 3, 4} {10} 1 2 3 4 10 f
As different elements in the domain have the same image in the co-domain, f is not one-one
f does not have an inverse (as f is not one-one, the check for f being an onto is not required)
To check whether g is one-one
In the function g, we have
{g(5) = 4}
{g(6) = 3}
{g(7) = 4}
{g(8) = 2}
This can be diagramatically represented as follows:
g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)} {5, 6, 7, 8} {1, 2, 3, 4} 5 6 7 8 1 2 3 4 g
As two different elements 5 and 7 in the domain have the same image 4 in the co-domain, g is not one-one.
g does not have an inverse (as g is not one-one, the check for g being an onto is not required)
To check whether h is one-one
In the function h, we have
{h(2) = 7}
{h(3) = 9}
{h(4) = 11}
{h(5) = 13}
This can be diagramatically represented as follows:
h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)} {2, 3, 4, 5} {7, 9, 11, 13} 2 3 4 5 7 9 11 13 h
Clearly different elements in the domain have the different images in the co-domain.
h is one-one
To check whether h is onto
As every element in the co-domain {7, 9, 11, 13} has an inverse-image in the domain, h is also onto.
h is both one-one and onto.
h will have an inverse and the inverse of h will be
h^{-1} = {(7, 2), (9, 3), (11, 4), (13, 5)}

6. Show that f : [-1, 1] → R, given by {f(x) = \dfrac{x}{(x + 2)}} is one-one. Find the inverse of the function f : [-1, 1] → Range f.
(Hint: For y ∈ Range f, {y = f(x) = \dfrac{x}{x + 2}}, for some x in [-1, 1], i.e., {x = \dfrac{2}{2 - y}})
The function f is given by
{f(x) = \dfrac{x}{2 + x}}
This can also be written as
{f(x) = \dfrac{x + 2 - 2}{x + 2} = 1 - \dfrac{2}{x + 2}}
To Check whether f is one-one
Consider the two elements in the domain x_1, x_2 ∈ [-1, 1] such that
{f(x_1) = f(x_2)}
{⇒ 1 - \dfrac{2}{x_1 + 2} = 1 - \dfrac{2}{x_2 + 2}}
{⇒ -\dfrac{2}{x_1 + 2} = -\dfrac{2}{x_2 + 2}} (After cutting/cancelling 1 on both sides)
{⇒ \dfrac{1}{x_1 + 2} = \dfrac{1}{x_2 + 2}} (After cutting/cancelling 2 on both sides)
{⇒ x_1 + 2 = x_2 + 2} (After inversing both sides)
{⇒ x_1 = x_2} (After cutting/cancelling 2 on both sides)
f is one-one
To Check whether f is onto
We have {y = f(x) = 1 - \dfrac{2}{x + 2}}
{⇒ \dfrac{2}{x + 2} = 1 - y}
{⇒ \dfrac{x + 2}{2} = \dfrac{1}{1 - y}}
{⇒ x + 2 = \dfrac{2}{1 - y}}
{⇒ x = \dfrac{2}{y - 1} - 2}
{⇒ x = \dfrac{2 - 2(1 - y)}{1 - y}}
{⇒ x = \dfrac{2y}{1 - y}}
As the above equation is valid for all values of y except 1 (∵ when y = 1, 1 – y = 0 and division by 0 is not defined.
Also, as y is defined as {y = \dfrac{x}{x + 2}}, y can never be 1 {(∵ x}{x + 2)}
⇒ Range of f never includes 1.
Thus for every y ∈ Range of f, there exists an inverse image in the domain.
f is onto.
As f is both one-one and onto f is invertible and is given as {f^{-1}(y) = \dfrac{2y}{1 - y}}
7. Consider f : R → R given by {f(x) = 4x + 3}. Show that f is invertible. Find the inverse of f.
Method 1: To show that f is invertible by showing that f is both one-one and onto.
For a function f to be invertible, it should be both one-one and onto.
To check whether f is one-one
The function f is given as {f(x) = 4x + 3}
Consider two elements x_1, x_2 ∈ R such that
{f(x_1) = f(x_2)}
{⇒ 4x_1 + 3 = 4x_2 + 3}
{⇒ 4x_1 = 4x_2} (After cutting/cancelling 3 on both sides)
{⇒ x_1 = x_2} (After cutting/cancelling 4 on both sides)
f is one-one
To check whether f is onto
The function f is given as {y = f(x) = 4x + 3}
{⇒ y = 4x + 3}
{⇒ y - 3 = 4x}
{⇒ 4x = y - 3}
{⇒ x = \dfrac{y - 3}{4}}
The above expression is valid for all real values of y.
In otherwords, every element yR will be an image of an element x in the domain R
f is onto
As f is both one-one and onto, f is invertible and is given as
{f^{-1}(x) = \dfrac{y - 3}{4}}
Method 2: To show that f is invertible by showing that fog = gof = IR by defining the inverse function g
Let y be an arbitrary element in the range of f.
⇒ For some x ∈ R, {y = 4x + 3}
{⇒ y - 3 = 4x}
{⇒ 4x = y - 3}
{⇒ x = \dfrac{y - 3}{4}}
Let’s now define g : R → R given as {g(y) = \dfrac{y - 3}{4}}
Now,
{gof(x) = g(f(x))}
{= g(4x + 3)}
{= \dfrac{(4x + 3) - 3}{4}}
{= \dfrac{4x}{4}}
{= x}
Also,
{fog(y) = f(g(y))}
{= f\left(\dfrac{y - 3}{4}\right)}
{= 4\times\left(\dfrac{y - 3}{4}\right) + 3}
{= y - 3 + 3} (After cancelling/cutting 4 in the numerator and denominator)
{= y}
Hence gof = IR and also fog = IR
f is invertible and is given as {f^{-1} = g}

8. Consider f : R+ → [4, ∞) given by {f(x) = x² + 4}. Show that f is invertible with the inverse {f^{-1}} of f given by {f^{-1}(y) = \sqrt{y - 4}}, where R+ is the set of all non-negative real numbers.
Method 1: To show that f is invertible by showing that f is both one-one and onto.
For a function f to be invertible, it should be both one-one and onto.
To check whether f is one-one
The function f is given as {f(x) = x^2 + 4}
Consider two elements x_1, x_2R+ in the domain such that
{f(x_1) = f(x_2)}
{⇒ x_1^2 + 4 = x_2^2 + 4}
{⇒ x_1^2 = x_2^2} (After cutting/cancelling 4 on both sides)
{⇒ x_1 = \pm x_2}
As it is given that the domain R+ is the set of non-nagative real numbers, we can ignore the negative values of x
f is one-one
To check whether f is onto
The function f is given as {y = f(x) = x^2 + 4}
{⇒ x^2 + 4 = y}
{⇒ x^2 = y - 4}
{⇒ x = \sqrt{y - 4}}
The above equation is valid for all {y - 4 ≥ 0} (∵ square root of negative numbers is not defined in real numbers)
or {y \geq 4}
It is given that the co-domain of f is [4, ∞)
Every y ∈ [4, ∞) in the co-domain will be an image of an element xR+ in the domain.
f is onto.
As f is both one-one and onto, f is invertible and the inverse of f is given as
{f^{-1}(y) = \sqrt{y - 4}}
Method 2: To show that f is invertible by showing that {fof^{-1} = f^{-1}of} = IR+ by using the inverse function f^{-1}
Let y be an arbitrary element in the range of f.
For some xR+, {y = x^2 + 4}
{⇒ y - 4 = x^2}
{⇒ x^2 = y - 4}
{⇒ x = \sqrt{y - 4}}
Let’s now define g : [4, ∞) → R+ given as {g(y) = \sqrt{y - 4}}
Now
{gof(x) = g(f(x))}
{= g(x² + 4)}
{= \sqrt{x^2 + 4 - 4}} {(∵ g(y) = \sqrt{y - 4})}
{= \sqrt{x^2}}
{= x}
Also,
{fog(y) = f(g(y))}
{= f(\sqrt{y - 4})}
{= \left({\sqrt{y - 4}}\right)^2 + 4} {(∵ f(x) = x²)}
{= y - 4 + 4}
{= y}
Hence gof = IR+ and also fog = IR+
f is invertible and is given as {f^{-1} = g}

9. Consider f : R+ → [-5, ∞) given by {f(x) = 9x^2 + 6x - 5}. Show that f is invertible with {f^{-1}(y) = \left(\dfrac{\left(\sqrt{y + 6}\right) - 1}{3}\right)}.
Method 1: To show that f is invertible by showing that f is both one-one and onto.
For a function f to be invertible, it should be both one-one and onto.
To check whether f is one-one
The function f is given as {f(x) = 9x^2 + 6x - 5}
This function can be written as {y = f(x) = 9x² + 6x + -5}
Consider two elements x_1, x_2R+, in the domain R+ such that
{f(x_1) = f(x_2)}
{⇒ 9x_1^2 + 6x_1 - 5 = 9x_2^2 + 6x_2 - 5}
{⇒ 9x_1^2 + 6x_1 = 9x_2^2 + 6x_2} (After cutting/cancelling -5 on both sides)
{⇒ 9x_1^2 - 9x_2^2 + 6x_1 - 6x_2 = 0}
{⇒ 9(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0}
{⇒ 9(x_1 + x_2)(x_1 - x_2) + 6(x_1 - x_2) = 0} {(∵ a^2 - b^2 = (a + b)(a - b))}
{⇒ 3(x_1 - x_2)[3(x_1 + x_2 + 2)] = 0}
{x_1 - x_2 = 0} (∵ xR+ and hence {3x_1 + 3x_2 + 2} being sum of 2 positive real numbers and 2 can never be 0)
{⇒ x_1 = x_2}
f is one-one
To check whether f is onto:
The function f is given as {y = f(x) = 9x^2 + 6x - 5}
{⇒ 9x^2 + 6x - 5 = y}
{⇒ 9x^2 + 6x + 1 - 6 = y} (so that we can use the identity {a^2 + 2ab + b^2 = (a + b)^2})
{⇒ (3x)^2 + 2×(3x)×1 + 1^2 - 6 = y}
{⇒ (3x + 1)² - 6 = y} {(∵ a^2 + 2ab + b^2 = (a + b)^2)}
{⇒ (3x + 1)^2 = y + 6}
{⇒ (3x + 1) = \sqrt{y + 6}}
{⇒ 3x = \left(\sqrt{y + 6}\right) - 1}
⇒ x = \dfrac{\left(\sqrt{y + 6}\right) - 1}{3}
The above equation is valid when {y + 6 \geq 0}
{⇒ y ≥ -6}
As the co-domain of f is given as [-5, ∞), every element in y ∈ [-5, ∞) will have an inverse image in the domain.
f is onto.
As f is both one-one and onto, f is invertible and the inverse of f is given as
{f^{-1}(y)\left(\dfrac{\left(\sqrt{y + 6}\right) - 1}{3}\right)}
Method 2: To show that f is invertible by using the inverse function {f{-1}}
Let y be an arbitrary element in the range of f.
For some xR+, {y = 9x^2 + 6x - 5}
{⇒ y = (3x)^2 + 2×(3x)×1 + 1^2 - 6}
{⇒ y + 6 = (3x + 1)^2}
{⇒ (3x + 1)^2 = y + 6}
{⇒ 3x + 1 = \sqrt{y + 6}}
{⇒ 3x = \sqrt{y + 6} - 1}
{⇒ x = \dfrac{\sqrt{y + 6} - 1}{3}}
Let’s now define g : [-6, ∞) → R+, given as
{g(y) = \dfrac{\sqrt{y + 6} - 1}{3}}
Now
{gof(x) = g(f(x))}
{= g(9x^2 + 6x - 5)} {(∵ f(x) = 9x² + 6x - 5)}
{= \dfrac{\sqrt{9x² + 6x - 5 + 6} - 1}{3}} {(∵ g(y) = \dfrac{\sqrt{y + 6} - 1}{3})}
{= \dfrac{\sqrt{9x² + 6x + 1} - 1}{3}}
{= \dfrac{\sqrt{(3x)² + 2 × 3x × 1 + 1²} - 1}{3}}
{= \dfrac{\sqrt{(3x + 1)²} - 1}{3}}
{= \dfrac{3x + 1 - 1}{3}}
{= \dfrac{3x}{3}}
{= x}
Also,
{fog(y) = f(g(y))}
{= f\left(\dfrac{\sqrt{y + 6} - 1}{3}\right)} {(∵ g(y) = \dfrac{\sqrt{y + 6} - 1}{3}}
{= 9\left(\dfrac{\sqrt{y + 6} - 1}{3}\right)^2 + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5} {(∵ f(x) = 9x^2 + 6x - 5)}
{= {9\left(\dfrac{\left(\sqrt{y + 6}\right)^2 + 2\times\sqrt{y + 6}\times1 + 1^2}{3^2}\right)} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5}
{= {9\left(\dfrac{y + 6 + 2\sqrt{y + 6} + 1}{9}\right)} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5}
{= y + 7 - 2\sqrt{y + 6} + 6\left(\dfrac{\sqrt{y + 6} - 1}{3}\right) - 5} (After cutting/cancelling 9 in the numerator and denominator)
{= \dfrac{3y + 21 - 6\sqrt{y + 6} + 6\sqrt{y + 6} - 6 - 15}{3}}
{= \dfrac{3y}{3}}
{= y}
Hence gof = IR+ and fog = I[-5, ∞)
f is convertible and is given as {f^{-1} = g}

10. Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g_1 and g_2 are two inverses of f. Then for all {y ∈ Y}, fog_1(y) = IY{(y) = fog_2(y)}. Use one-one ness of f).
As f is defined as f : X → Y and also as it is given that f is invertible.
f is both one-one and onto.
As f is one-one, if there exists any two elements x_1, x_2X such that
{f(x_1) = f(x_2)} then {x_1 = x_2}
Let’s now assume that the invertible function f has two different inverses g_1 and g_2 defined as g_1 : Y → X and g_2 : Y → X
{⇒ fog_1(y) = I_Y} and {fog_2(y) = I_Y}
{⇒ fog_1(y) = fog_2(y)}
{⇒ f(g_1(y)) = f(g_2(y))}
{⇒ g_1(y) = g_2(y)} {(∵ f(x_1) = f(x_2) ⇒ x_1 = x_2)}
{⇒ g_1 = g_2}
f has unique inverse.

11. Consider f : {1, 2, 3} → {{a, b, c}} given by {f(1) = a}, {f(2) = b} and {f(3) = c}. Find f^{-1} and show that {(f^{-1})^{-1} = f}
It is given that
{f(1) = a} {⇒ f^{-1}(a) = 1}
{f(2) = b} {⇒ f^{-1}(b) = 2}
{f(3) = c} {⇒ f^{-1}(c) = 3}
This can be diagramatically represented as follows:
f : {1, 2, 3} → {{a, b, c}} given by f(1) = a, f(2) = b and f(3) = c {1, 2, 3} {a, b, c} 1 2 3 a b c f
f^{-1} is defined as {f^{-1} : \{a, b, c\}} → {1, 2, 3} and {f^{-1} = \{(a, 1), (b, 2), (c, 3)\}}
As {f^{-1}(a) = 1 ⇒ (f^{-1})^{-1}(1) = a}
As {f^{-1}(b) = 2 ⇒ (f^{-1})^{-1}(2) = b}
As {f^{-1}(c) = 3 ⇒ (f^{-1})^{-1}(3) = c}
And this can be diagramatically represented as follows:
f^-1 : {a, b, c} → {1, 2, 3} and f^-1 = {(a, 1), (b, 2), (c, 3)} {a, b, c} {1, 2, 3} a b c 1 2 3 f -1
{(f^{-1})^{-1}} is defined as {(f^{-1})^{-1} = \{1, 2, 3\} → \{a, b, c\}} and {(f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\}}
∴ From the above, it is clear that {(f^{-1})^{-1} = f}

12. Let f : X → Y be an invertible function. Show that the inverse of f^{-1} is f, i.e., {(f^{-1})^{-1} = f}.
The inverse of f is f^{-1}
It is given that f is invertible
f is both one-one and onto.
Consider an arbitrary elemnet xX, such that {f(x) = y}
{⇒ f^{-1}(y) = x}
{⇒ (f^{-1})^{-1}(x) = y}
But we know that {f(x) = y}
So {(f^{-1})^{-1}(x) = f(x)}
As x is an arbitrary element in X, the above statement is true for all xX
{∴ (f^{-1})^{-1} = f}
13. If f : R → R be given by {f(x) = \left(3 - x^3\right)^{\frac{1}{3}}}, then fof(x) is
A.
x^\frac{1}{3}
B.
x^3
C.
x
D.
{(3 - x^3)}
{fof(x) = f(f(x))}
{= f\left(\left(3 - x^3\right)^{\frac{1}{3}}\right)}
{= \left(3 - \left(\left(3 - x^3\right)^\frac{1}{3}\right)^3\right)^\frac{1}{3}}
{= \left(3 - \left(3 - x^3\right)^{3\times\frac{1}{3}}\right)^\frac{1}{3}}
{= \left(3 - \left(3 - x^3\right)^1\right)^\frac{1}{3}}
{= \left(3 - \left(3 - x^3\right)\right)^\frac{1}{3}}
{= \left(3 - 3 + x^3\right)^\frac{1}{3}}
{= \left(x^3\right)^\frac{1}{3}}
= x
C is the correct answer.
14. Let f : R – \Bigl\{\dfrac{4}{3}\Bigr\} → R be a function defined as {f(x) = \dfrac{4x}{3x + 4}}. The inverse of f is the map g : Range f → R {- \Bigl\{\dfrac{4}{3}\Bigr\}} given by
A.
{g(y) = \dfrac{3y}{3 - 4y}}
B.
{g(y) = \dfrac{4y}{4 - 3y}}
C.
{g(y) = \dfrac{4y}{3 - 4y}}
D.
{g(y) = \dfrac{3y}{4 - 3y}}

The given function is {y = f(x)}
{⇒ f^{-1}(y) = x}
{⇒ g(y) = x} (∵ The inverse of f is given as g)
Also, it is given that {y = f(x) = \dfrac{4x}{3x + 4}}
{⇒ \dfrac{4x}{3x + 4} = y}
{⇒ 4x = (3x + 4)y}
{⇒ 4x = 3xy + 4y}
{⇒ 4x - 3xy = 4y}
{⇒ x(4 - 3y) = 4y}
{⇒ x = \dfrac{4y}{4 - 3y}}
{⇒ g(y) = \dfrac{4y}{4 - 3y}}
B is the correct answer