1

Question

Classify the following functions as injection, surjection or bijection :

(i) f : N → N given by f(x) = x^{2}

(ii) f : Z → Z given by f(x) = x^{2}

(iii) f : N → N given by f(x) = x^{3}

(iv) f : Z → Z given by f(x) = x^{3}

(v) f : R → R, defined by f(x) = |x|

(vi) f : Z → Z, defined by f(x) = x^{2} + 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) = x^{3} + 1

(x) f : R → R, defined by f(x) = x^{3} − x

(xi) f : R → R, defined by f(x) = sin^{2}x + cos^{2}x

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

(xiii) f : Q → Q, defined by f(x) = x^{3} + 1

(xiv) f : R → R, defined by f(x) = 5x^{3} + 4

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

(xvi) f : R → R, defined by f(x) = 1 + x^{2}

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

(i) f : N → N given by f(x) = x

(ii) f : Z → Z given by f(x) = x

(iii) f : N → N given by f(x) = x

(iv) f : Z → Z given by f(x) = x

(v) f : R → R, defined by f(x) = |x|

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

(x) f : R → R, defined by f(x) = x

(xi) f : R → R, defined by f(x) = sin

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

(xiii) f : Q → Q, defined by f(x) = x

(xiv) f : R → R, defined by f(x) = 5x

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

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

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

Open in App

Solution

(i) f : **N** → **N**, given by f(x) = x^{2}

Injection test:

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

f(x)=f(y)

${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=y\text{(We do not get \xb1 because}x\text{and}y\text{are in}\mathbf{N}\text{)}$

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

${x}^{2}=y\phantom{\rule{0ex}{0ex}}x=\sqrt{y},\text{which may not be in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt{3}\text{is not in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

So, f is not a surjection.

So, f is not a bijection.

(ii) f :**Z** → **Z**, given by f(x) = x^{2}

Injection test:

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

f(x) = f(y)

${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=\pm y$

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

${x}^{2}=y\phantom{\rule{0ex}{0ex}}x=\pm \sqrt{y}\text{which may not be in}\mathbf{Z}.\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\pm \sqrt{3}\text{is not in}\mathbf{Z}\text{.}$

So, f is not a surjection.

So, f is not a bijection.

(iii) f :**N** → **N**, given by f(x) = x^{3}

Injection test:

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

f(x) = f(y)

${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

${x}^{3}=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y}\text{which may not be in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{3}\text{is not in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

(iv) f :**Z** → **Z**, given by f(x) = x^{3}

Injection test:

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

f(x) = f(y)

${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

${x}^{3}=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y}\text{which may not be in}\mathbf{Z}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{3}\text{is not in}\mathbf{Z}\text{.}$

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)

$\left|x\right|=\left|y\right|\phantom{\rule{0ex}{0ex}}x=\pm y$

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

$\left|x\right|=y\phantom{\rule{0ex}{0ex}}x=\pm y\in \mathbf{Z}$

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

(vi) f :**Z** → **Z**, defined by f(x) = x^{2} + 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)

${x}^{2}+x={y}^{2}+y\phantom{\rule{0ex}{0ex}}\text{Here, we cannot say that}x\text{=}y\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, x = 2 and y = - 3}\phantom{\rule{0ex}{0ex}}\text{Then},\phantom{\rule{0ex}{0ex}}{x}^{2}+x={2}^{2}+2=6\phantom{\rule{0ex}{0ex}}{y}^{2}+y={\left(-3\right)}^{2}-3=6\phantom{\rule{0ex}{0ex}}\text{So, we have two numbers 2 and -3 in the domain}\mathbf{Z}\text{whose image is same as 6.}$

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

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

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)

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

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) = x^{3} + 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)

${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1}\in \mathbf{R}$

So, f is a surjection.

So, f is a bijection.

(x) f :**R** → **R**, defined by f(x) = x^{3} − 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)

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

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

${x}^{3}-x=y\phantom{\rule{0ex}{0ex}}\text{By observation we can say that there exist some}x\text{in}\mathbf{R},\text{such that}{x}^{3}\text{-x=y.}$

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

(xi) f :**R** → **R**, defined by f(x) = sin^{2}x + cos^{2}x

f(x) = sin^{2}x + cos^{2}x = 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)

$\frac{2x+3}{x-3}=\frac{2y+3}{y-3}\phantom{\rule{0ex}{0ex}}\left(2x+3\right)\left(y-3\right)=\left(2y+3\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}2xy-6x+3y-9=2xy-6y+3x-9\phantom{\rule{0ex}{0ex}}9x=9y\phantom{\rule{0ex}{0ex}}x=y$

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

$\frac{2x+3}{x-3}=y\phantom{\rule{0ex}{0ex}}2x+3=xy-3y\phantom{\rule{0ex}{0ex}}2x-xy=-3y-3\phantom{\rule{0ex}{0ex}}x\left(2-y\right)=-3\left(y+1\right)\phantom{\rule{0ex}{0ex}}x=\frac{3\left(y+1\right)}{y-2},\text{which is not defined at}y\text{=2.}\phantom{\rule{0ex}{0ex}}$

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

(xiii) f :**Q** → **Q**, defined by f(x) = x^{3} + 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)

${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1,}\text{which may not be in}\mathbf{Q}.\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{= 8,}\phantom{\rule{0ex}{0ex}}{x}^{3}+1=8\phantom{\rule{0ex}{0ex}}{x}^{3}=7\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{7},\text{which is not in}\mathbf{\text{Q}}\text{.}$

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) = 5x^{3} + 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)

$5{x}^{3}+4=5{y}^{3}+4\phantom{\rule{0ex}{0ex}}5{x}^{3}=5{y}^{3}\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

$5{x}^{3}+4=y\phantom{\rule{0ex}{0ex}}5{x}^{3}=y-4\phantom{\rule{0ex}{0ex}}{x}^{3}=\frac{y-4}{5}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{\frac{y-4}{5}}\in \mathbf{R}$

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)

$3-4x=3-4y\phantom{\rule{0ex}{0ex}}-4x=-4y\phantom{\rule{0ex}{0ex}}x=y$

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

$3-4x=y\phantom{\rule{0ex}{0ex}}4x=3-y\phantom{\rule{0ex}{0ex}}x=\frac{3-y}{4}\in \mathbf{R}$

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

(xvi) f :**R** → **R**, defined by f(x) = 1 + x^{2}

Injection test:

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

f(x) = f(y)

$1+{x}^{2}=1+{y}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=\pm y$

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

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

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)

$\frac{x}{{x}^{2}+1}=\frac{y}{{y}^{2}+1}\phantom{\rule{0ex}{0ex}}x{y}^{2}+x={x}^{2}y+y\phantom{\rule{0ex}{0ex}}x{y}^{2}-{x}^{2}y+x-y=0\phantom{\rule{0ex}{0ex}}-xy\left(-y+x\right)+1\left(x-y\right)=0\phantom{\rule{0ex}{0ex}}\left(x-y\right)\left(1-xy\right)=0\phantom{\rule{0ex}{0ex}}x=y\mathrm{or}x=\frac{1}{y}$

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

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

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

Injection test:

Let x and y be any two elements in the domain (

f(x)=f(y)

${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=y\text{(We do not get \xb1 because}x\text{and}y\text{are in}\mathbf{N}\text{)}$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{2}=y\phantom{\rule{0ex}{0ex}}x=\sqrt{y},\text{which may not be in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt{3}\text{is not in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

So, f is not a surjection.

So, f is not a bijection.

(ii) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=\pm y$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{2}=y\phantom{\rule{0ex}{0ex}}x=\pm \sqrt{y}\text{which may not be in}\mathbf{Z}.\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\pm \sqrt{3}\text{is not in}\mathbf{Z}\text{.}$

So, f is not a surjection.

So, f is not a bijection.

(iii) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{3}=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y}\text{which may not be in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{3}\text{is not in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

(iv) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{3}=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y}\text{which may not be in}\mathbf{Z}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{3}\text{is not in}\mathbf{Z}\text{.}$

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

(v) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

$\left|x\right|=\left|y\right|\phantom{\rule{0ex}{0ex}}x=\pm y$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

$\left|x\right|=y\phantom{\rule{0ex}{0ex}}x=\pm y\in \mathbf{Z}$

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

(vi) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

${x}^{2}+x={y}^{2}+y\phantom{\rule{0ex}{0ex}}\text{Here, we cannot say that}x\text{=}y\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, x = 2 and y = - 3}\phantom{\rule{0ex}{0ex}}\text{Then},\phantom{\rule{0ex}{0ex}}{x}^{2}+x={2}^{2}+2=6\phantom{\rule{0ex}{0ex}}{y}^{2}+y={\left(-3\right)}^{2}-3=6\phantom{\rule{0ex}{0ex}}\text{So, we have two numbers 2 and -3 in the domain}\mathbf{Z}\text{whose image is same as 6.}$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

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

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

(vii) f :

Injection test:

Let x and y be any two elements in the domain (

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 (

f(x) = y

x $-$ 5 = y

x = y + 5, which is in

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

(viii) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

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

So, f is not an injection .

Surjection test:

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

Co-domain of f =

Both are not same.

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

(ix) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1}\in \mathbf{R}$

So, f is a surjection.

So, f is a bijection.

(x) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

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

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{3}-x=y\phantom{\rule{0ex}{0ex}}\text{By observation we can say that there exist some}x\text{in}\mathbf{R},\text{such that}{x}^{3}\text{-x=y.}$

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

(xi) f :

f(x) = sin

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

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 =

Both are not same.

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

(xii) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

$\frac{2x+3}{x-3}=\frac{2y+3}{y-3}\phantom{\rule{0ex}{0ex}}\left(2x+3\right)\left(y-3\right)=\left(2y+3\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}2xy-6x+3y-9=2xy-6y+3x-9\phantom{\rule{0ex}{0ex}}9x=9y\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (

f(x) = y

$\frac{2x+3}{x-3}=y\phantom{\rule{0ex}{0ex}}2x+3=xy-3y\phantom{\rule{0ex}{0ex}}2x-xy=-3y-3\phantom{\rule{0ex}{0ex}}x\left(2-y\right)=-3\left(y+1\right)\phantom{\rule{0ex}{0ex}}x=\frac{3\left(y+1\right)}{y-2},\text{which is not defined at}y\text{=2.}\phantom{\rule{0ex}{0ex}}$

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

(xiii) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1,}\text{which may not be in}\mathbf{Q}.\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{= 8,}\phantom{\rule{0ex}{0ex}}{x}^{3}+1=8\phantom{\rule{0ex}{0ex}}{x}^{3}=7\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{7},\text{which is not in}\mathbf{\text{Q}}\text{.}$

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

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

(xiv) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

$5{x}^{3}+4=5{y}^{3}+4\phantom{\rule{0ex}{0ex}}5{x}^{3}=5{y}^{3}\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

$5{x}^{3}+4=y\phantom{\rule{0ex}{0ex}}5{x}^{3}=y-4\phantom{\rule{0ex}{0ex}}{x}^{3}=\frac{y-4}{5}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{\frac{y-4}{5}}\in \mathbf{R}$

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

(xv) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

$3-4x=3-4y\phantom{\rule{0ex}{0ex}}-4x=-4y\phantom{\rule{0ex}{0ex}}x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (

f(x) = y

$3-4x=y\phantom{\rule{0ex}{0ex}}4x=3-y\phantom{\rule{0ex}{0ex}}x=\frac{3-y}{4}\in \mathbf{R}$

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

(xvi) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

$1+{x}^{2}=1+{y}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=\pm y$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (

f(x) = y

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

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

(xvii) f :

Injection test:

Let x and y be any two elements in the domain (

f(x) = f(y)

$\frac{x}{{x}^{2}+1}=\frac{y}{{y}^{2}+1}\phantom{\rule{0ex}{0ex}}x{y}^{2}+x={x}^{2}y+y\phantom{\rule{0ex}{0ex}}x{y}^{2}-{x}^{2}y+x-y=0\phantom{\rule{0ex}{0ex}}-xy\left(-y+x\right)+1\left(x-y\right)=0\phantom{\rule{0ex}{0ex}}\left(x-y\right)\left(1-xy\right)=0\phantom{\rule{0ex}{0ex}}x=y\mathrm{or}x=\frac{1}{y}$

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (

f(x) = y

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

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

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