Relations and Functions 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}
gof:{1, 3, 4} → {1, 3}
The function f is given as
f(1) = 2
f(3) = 5
f(4) = 1
The function g is given as
g(1) = 3
f(2) = 3
f(5) = 1
gof(1) = g(f(1)]) = g(2) = 3
gof(3) = g(f(3)]) = g(5) = 1
gof(4) = g(f(4)]) = g(1) = 3
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

  1. f(x) = |x| and g(x) = |5x – 2|
  2. f(x) = 8x³ 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) = 8x³ and g(x) = x^{\frac{1}{3}}
We have
gof(x)
= g(f(x))
= g(8x³) (∵ f(x) = 8x³)
= \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) = 8x³)
= 8x^{\frac{1}{3}\times3} (∵ {\left(a^m\right)^n = a^{mn}})
= 8x
fog(x) = 8x

4. If f(x) = \frac{(4x + 3)}{(6x - 4)}, x ≠ \frac{2}{3}, show that fof = x, for all x ≠ \frac{2}{3}. What is the inverse of f ?
We have,
fof(x)
= f(f(x))
= f\left(\dfrac{4x - 3}{6x - 4}\right) (∵ f(x) = \dfrac{4x - 3}{6x - 4})
= \dfrac{4\left(\dfrac{4x - 3}{6x - 4}\right) - 3}{6\left(\dfrac{4x - 3}{6x - 4}\right) - 4} (∵ f(x) = \dfrac{4x - 3}{6x - 4})
= {\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

  1. f:{1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
  2. g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
  3. 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
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
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
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 x1, x2 ∈ [-1, 1] such that
f(x1) = f(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)
x1 + 2 = x2 + 2 (After inversing both sides)
x1 = x2 (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
\frac{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 x1, x2 ∈ R such that
f(x1) = f(x2)
4x1 + 3 = 4x2 + 3
4x1 = 4x2 (After cutting/cancelling 3 on both sides)
x1 = x2 (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 y ∈ R will have an inverse image 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 = \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) = x² + 4
Consider two elements x1, x2 ∈ R+ in the domain such that
f(x1) = f(x2)
x1² + 4 = x2² + 4
x1² = x2² (After cutting/cancelling 4 on both sides)
x1 = ±x2
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) = x² + 4
x² + 4 = y
x² = 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 ≥ 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 x ∈ R+ 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-1of = IR+ by using the inverse function f-1
Let y be an arbitrary element in the range of f.
For some x ∈ R+, y = x² + 4
y – 4 = x²
x² = 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) = 9x² + 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 if is given as f(x) = 9x² + 6x – 5
This function can be written as y = f(x) = 9x² + 6x + -5
Consider two elements x1, x2 ∈ R+, in the domain R+ such that
f(x1) = f(x2)
9x1² + 6x1 – 5 = 9x2² + 6x2 – 5
9x1² + 6x1 = 9x2² + 6x2 (After cutting/cancelling -5 on both sides)
9x1² – 9x2² + 6x1 – 62 = 0
9(x1² – x2²) + 6(x1 – x2) = 0
9(x1 + x2)(x1 – x2) + 6(x1 – x2) = 0(∵ a² – b² = (a + b)(a – b))
3(x1 – x2)[3(x1 + x2) + 2] = 0
x1 – x2 = 0(∵ x ∈ R+ and hence 3x1 + 3x2 + 2 being sum of 2 positive real numbers and 2 can never be 0)
x1 = x2
∴ f is one-one
To check whether f is onto:
The function f is given as y = f(x) = 9x² + 6x – 5
9x² + 6x – 5 = y
9x² + 6x + 1 – 6 = y (so that we can use the identity a² + 2ab + b² = (a + b)²)
(3x)² + 2×(3x)×1 + 1² – 6 = y
(3x + 1)² – 6 = y (∵ a² + 2ab + b² = (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 ≥ 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 hte inverse function f-1
Let y be an arbitrary element in the range of f.
For some x ∈ R+, y = 9x² + 6x – 5
y = (3x)² + 2×(3x)×1 + 1² – 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:[-5, ∞) → R+, given as
g(y) = \dfrac{\sqrt{y + 6} - 1}{3}
Now
gof(x) = g(f(x))
= g(9x² + 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\times3x\times1 + 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) = 9x² + 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 g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(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 x1, x2 ∈ X such that
f(x1) = f(x2) then x1 = x2
Let’s now assume that the invertible function f has two different inverses g1 and g2 defined as g1:Y → X and g2:Y → X
fog1(y) = IY and fog2(y) = IY
fog1(y) = fog2(y)
f(g1(y)) = f(g2(y))
g1(y) = g2(y)(∵ f(x1) = f(x2) ⇒ x1 = x2)
g1 = g2
∴ 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
∴ 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
∴ (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 x ∈ X, 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 = f(x)
As x is an arbitrary element in X, the above statement is true for all x ∈ X
(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

  1. x^\frac{1}{3}
  2. x
  3. (3 – x³)
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 – \{\dfrac{4}{3}\} → R be a function defined as f(x) = \dfrac{4x}{3x + 4}. The inverse of f is the map g: Range f → R – \{\dfrac{4}{3}\} given by

  1. g(y) = \dfrac{3y}{3 - 4y}
  2. g(y) = \dfrac{4y}{4 - 3y}
  3. g(y) = \dfrac{4y}{3 - 4y}
  4. 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}4y
g(y) = \dfrac{4y}{4 - 3y}4y
∴ B is the correct answer