Question

Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : N → N given by f(x) = x³
(iv) f : Z → Z given by f(x) = x³
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x² + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x³ + 1
(x) f : R → R, defined by f(x) = x³ − x
(xi) f : R → R, defined by f(x) = sin²x + cos²x
(xii) f : Q − {3} → Q, defined by $f\left(x\right)=\frac{2x+3}{x-3}$
(xiii) f : Q → Q, defined by f(x) = x³ + 1
(xiv) f : R → R, defined by f(x) = 5x³ + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x²
(xvii) f : R → R, defined by f(x) = $\frac{x}{x^2+1}$ [NCERT EXEMPLAR]

Answer 1

(i) f : N → N, given by f(x) = x²

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x)=f(y)

$$\begin{gathered} x^2=y^2 \\ x=y\text{(We do not get ± because}x\text{and}y\text{are in}\mathbf{N}\text{)} \end{gathered}$$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

$$\begin{gathered} x^2=y \\ x=\sqrt{y},\text{which may not be in}\mathbf{N}\text{.} \\ \text{For example, if}y\text{=3,} \\ x=\sqrt{3}\text{is not in}\mathbf{N}\text{.} \end{gathered}$$

So, f is not a surjection.

So, f is not a bijection.

(ii) f : Z → Z, given by f(x) = x²

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} x^2=y^2 \\ x=\pm y \end{gathered}$$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

$$\begin{gathered} x^2=y \\ x=\pm \sqrt{y}\text{which may not be in}\mathbf{Z}. \\ \text{For example, if}y\text{=3,} \\ x=\pm \sqrt{3}\text{is not in}\mathbf{Z}\text{.} \end{gathered}$$

So, f is not a surjection.

So, f is not a bijection.

(iii) f : N → N, given by f(x) = x³

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} x^3=y^3 \\ x=y \end{gathered}$$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

$$\begin{gathered} x^3=y \\ x=\sqrt[3]{y}\text{which may not be in}\mathbf{N}\text{.} \\ \text{For example, if}y\text{=3,} \\ x=\sqrt[3]{3}\text{is not in}\mathbf{N}\text{.} \end{gathered}$$

So, f is not a surjection and f is not a bijection.

(iv) f : Z → Z, given by f(x) = x³

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x) = f(y)

$$\begin{gathered} x^3=y^3 \\ x=y \end{gathered}$$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

$$\begin{gathered} x^3=y \\ x=\sqrt[3]{y}\text{which may not be in}\mathbf{Z}\text{.} \\ \text{For example, if}y\text{=3,} \\ x=\sqrt[3]{3}\text{is not in}\mathbf{Z}\text{.} \end{gathered}$$

So, f is not a surjection and f is not a bijection.

(v) f : R → R, defined by f(x) = |x|

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

f(x) = f(y)

$$\begin{gathered} \left|x\right|=\left|y\right| \\ x=\pm y \end{gathered}$$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} \left|x\right|=y \\ x=\pm y\in \mathbf{Z} \end{gathered}$$

So, f is a surjection and f is not a bijection.

(vi) f : Z → Z, defined by f(x) = x² + x

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} x^2+x=y^2+y \\ \text{Here, we cannot say that}x\text{=}y\text{.} \\ \text{For example, x = 2 and y = - 3} \\ \text{Then}, \\ {x}^2+x=2^2+2=6 \\ {y}^2+y={\left(-3\right)}^2-3=6 \\ \text{So, we have two numbers 2 and -3 in the domain}\mathbf{Z}\text{whose image is same as 6.} \end{gathered}$$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

$$\begin{gathered} x^2+x=y \\ \text{Here, we cannot say}x\in \mathbf{Z}. \\ \text{For example,}y\text{=-4.} \\ {x}^2+x=-4 \\ {x}^2+x+4=0 \\ x=\frac{-1\pm \sqrt{-15}}{2}=\frac{-1\pm i\sqrt{15}}{2}\text{which is not in}\mathbf{Z}. \end{gathered}$$

So, f is not a surjection and f is not a bijection.

(vii) f : Z → Z, defined by f(x) = x − 5

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x $-$ 5 = y $-$ 5

x = y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x $-$ 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

(viii) f : R → R, defined by f(x) = sinx

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} \sin x=\sin y \\ \text{Here,}\mathit{\text{x}}\text{may not be equal to}\mathit{\text{y}}\text{because}\sin 0=\mathrm{sin\pi }. \\ \text{So, 0 and π have the same image 0.} \end{gathered}$$

So, f is not an injection .

Surjection test:

Range of f = [$-$1, 1]

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) f : R → R, defined by f(x) = x³ + 1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} x^3+1=y^3+1 \\ {x}^3=y^3 \\ x=y \end{gathered}$$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} x^3+1=y \\ x=\sqrt[3]{y-1}\in \mathbf{R} \end{gathered}$$

So, f is a surjection.

So, f is a bijection.

(x) f : R → R, defined by f(x) = x³ − x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} x^3-x=y^3-y \\ \text{Here, we cannot say}x=y. \\ \text{For example,}x\text{=1 and}\mathit{\text{y}}\text{=-1} \\ {x}^3-x=1-1=0 \\ {y}^3-y={\left(-1\right)}^3-\left(-1\right)-1+1=0 \\ \text{So, 1 and -1 have the same image 0.} \end{gathered}$$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} x^3-x=y \\ \text{By observation we can say that there exist some}x\text{in}\mathbf{R},\text{such that}{x}^3\text{-x=y.} \end{gathered}$$

So, f is a surjection and f is not a bijection.

(xi) f : R → R, defined by f(x) = sin²x + cos²x

f(x) = sin²x + cos²x = 1

So, f(x) = 1 for every x in R.

So, for all elements in the domain, the image is 1.

So, f is not an injection.

Range of f = {1}

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(xii) f : Q − {3} → Q, defined by $f\left(x\right)=\frac{2x+3}{x-3}$

Injection test:

Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} \frac{2x+3}{x-3}=\frac{2y+3}{y-3} \\ \left(2x+3\right)\left(y-3\right)=\left(2y+3\right)\left(x-3\right) \\ 2xy-6x+3y-9=2xy-6y+3x-9 \\ 9x=9y \\ x=y \end{gathered}$$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

$$\begin{gathered} \frac{2x+3}{x-3}=y \\ 2x+3=xy-3y \\ 2x-xy=-3y-3 \\ x\left(2-y\right)=-3\left(y+1\right) \\ x=\frac{3\left(y+1\right)}{y-2},\text{which is not defined at}y\text{=2.} \end{gathered}$$

So, f is not a surjection and f is not a bijection.

(xiii) f : Q → Q, defined by f(x) = x³ + 1

Injection test:

Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} x^3+1=y^3+1 \\ {x}^3=y^3 \\ x=y \end{gathered}$$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).

f(x) = y

$$\begin{gathered} x^3+1=y \\ x=\sqrt[3]{y-1,}\text{which may not be in}\mathbf{Q}. \\ \text{For example, if}y\text{= 8,} \\ {x}^3+1=8 \\ {x}^3=7 \\ x=\sqrt[3]{7},\text{which is not in}\mathbf{\text{Q}}\text{.} \end{gathered}$$

So, f is not a surjection and f is not a bijection.

So, f is a surjection and f is a bijection.

(xiv) f : R → R, defined by f(x) = 5x³ + 4

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} 5x^3+4=5y^3+4 \\ 5x^3=5y^3 \\ {x}^3=y^3 \\ x=y \end{gathered}$$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} 5x^3+4=y \\ 5x^3=y-4 \\ {x}^3=\frac{y-4}{5} \\ x=\sqrt[3]{\frac{y-4}{5}}\in \mathbf{R} \end{gathered}$$

So, f is a surjection and f is a bijection.

(xv) f : R → R, defined by f(x) = 3 − 4x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} 3-4x=3-4y \\ -4x=-4y \\ x=y \end{gathered}$$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} 3-4x=y \\ 4x=3-y \\ x=\frac{3-y}{4}\in \mathbf{R} \end{gathered}$$

So, f is a surjection and f is a bijection.

(xvi) f : R → R, defined by f(x) = 1 + x²

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} 1+x^2=1+y^2 \\ {x}^2=y^2 \\ x=\pm y \end{gathered}$$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} 1+x^2=y \\ {x}^2=y-1 \\ x=\pm \sqrt{y-1}\text{which may not be in}\mathbf{R} \\ \text{For example, if}y\text{=0,} \\ x=\pm \sqrt{-1}=\pm i\text{is not in}\mathbf{R}. \end{gathered}$$

So, f is not a surjection and f is not a bijection.

(xvii) f : R → R, defined by f(x) = $\frac{x}{x^2+1}$

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

$$\begin{gathered} \frac{x}{x^2+1}=\frac{y}{y^2+1} \\ x{y}^2+x=x^{2}y+y \\ x{y}^2-x^{2}y+x-y=0 \\ -xy\left(-y+x\right)+1\left(x-y\right)=0 \\ \left(x-y\right)\left(1-xy\right)=0 \\ x=y\mathrm{or}x=\frac{1}{y} \end{gathered}$$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

$$\begin{gathered} \frac{x}{x^2+1}=y \\ y{x}^2-x+y=0 \\ x=\frac{-\left(-1\right)\pm \sqrt{1-4y^2}}{2y},\mathrm{if}y\ne 0 \\ =\frac{1\pm \sqrt{1-4y^2}}{2y},\mathrm{which}\mathrm{may}\mathrm{not}\mathrm{be}\mathrm{in}\mathbf{R} \\ \mathrm{For}\mathrm{example},\mathrm{if}y=1,\mathrm{then} \\ x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm \mathrm{i}\sqrt{3}}{2},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{in}\mathbf{R} \\ \mathrm{So},f\mathrm{is}\mathrm{not}\mathrm{surjection}\mathrm{and}f\mathrm{is}\mathrm{not}\mathrm{bijection}. \end{gathered}$$

So, f is not a surjection and f is not a bijection.