Exercise-1.2, Class 12th, Maths, Chapter – 1, NCERT

Exercise 1.2 — Solutions

Q1.

Show that the function $f:{\mathbb{R}}^{\ast }\to {\mathbb{R}}^{\ast }$ defined by $f(x)={\displaystyle \frac{1}{x}}$ (${\mathbb{R}}^{\ast }=\mathbb{R}\setminus \{0\}$) is one-one and onto. Is the result true if the domain ${\mathbb{R}}^{\ast }$ is replaced by $\mathbb{N}$ (with codomain still ${\mathbb{R}}^{\ast }$)?

Solution.
Injective: Suppose $f(x_1)=f(x_2)$. Then ${\displaystyle \frac{1}{x_1}}={\displaystyle \frac{1}{x_2}}$. Multiply both sides by $x_1{x}_2$ (nonzero) to get $x_2=x_1$. So $f$ is one-one.

Surjective: For any $y\in {\mathbb{R}}^{\ast }$ take $x={\displaystyle \frac{1}{y}}\in {\mathbb{R}}^{\ast }$. Then $f(x)=1\mathrm{/}x=y$. So every $y\in {\mathbb{R}}^{\ast }$has a pre-image; $f$ is onto.

Hence $f$ is bijective.

If domain is replaced by $\mathbb{N}$ (i.e. $f:\mathbb{N}\to {\mathbb{R}}^{\ast }$, $f(n)=1\mathrm{/}n$:

• Injective: Yes — different natural $n$ give different reciprocals.

• Surjective: No — many real nonzero numbers (e.g. $1\mathrm{/}2$ is covered, but numbers like $\sqrt{2}$, $-1\mathrm{/}3$ etc.) are not of the form $1\mathrm{/}n$ with $n\in \mathbb{N}$. In particular negative reals are impossible. So not onto ${\mathbb{R}}^{\ast }$.
  Thus bijectivity fails if domain is $\mathbb{N}$.

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Q2.

Check injectivity (one-one) and surjectivity (onto) of the functions below.

(i) $f:\mathbb{N}\to \mathbb{N},f(x)=x^2$
(ii) $f:\mathbb{Z}\to \mathbb{Z},f(x)=x^2$
(iii) $f:\mathbb{R}\to \mathbb{R},f(x)=x^2$
(iv) $f:\mathbb{N}\to \mathbb{N},f(x)=x^3$
(v) $f:\mathbb{Z}\to \mathbb{Z},f(x)=x^3$

Solution.

(i) $f(x)=x^2$ on $\mathbb{N}$:

• Injective: Yes. On $\mathbb{N}$ (positive integers), $x^2$ is strictly increasing, so $x_1^2=x_2^2\Rightarrow x_1=x_2$.

• Surjective: No. Not every natural number is a perfect square (e.g. $2$ has no natural square root).
  So: one-one but not onto.

(ii) $f(x)=x^2$ on $\mathbb{Z}$:

• Injective: No. $x$ and $-x$ map to same value for $x\mathrm{\ne }0$ (e.g. $(-1{)}^2=1^2$.

• Surjective: No. Negative integers are never squares, so they are not in the range.
  So: neither one-one nor onto.

(iii) $f(x)=x^2$ on $\mathbb{R}$:

• Injective: No (same reason: $x$ and $-x$).

• Surjective: No — negative real numbers are not squares.
  So: neither one-one nor onto.

(iv) $f(x)=x^3$ on $\mathbb{N}$:

• Injective: Yes (strictly increasing on $\mathbb{N}$).

• Surjective: No (not every natural number is a perfect cube).
  So: one-one but not onto.

(v) $f(x)=x^3$ on $\mathbb{Z}$:

• Injective: Yes. Cubing is strictly monotone on $\mathbb{Z}$; $x_1^3=x_2^3\Rightarrow x_1=x_2$.

• Surjective: No — a general integer $m$ need not be a perfect cube (e.g. $2$ is not).
  So: one-one but not onto.

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Q3.

Prove that the greatest integer function $f:\mathbb{R}\to \mathbb{R}$, $f(x)=\lfloor x\rfloor$, is neither one-one nor onto.

Solution.

• Not one-one: $\lfloor 1.2\rfloor =\lfloor 1.7\rfloor =1$ though $1.2\mathrm{\ne }1.7$. So many inputs share the same value.

• Not onto: Range of $\lfloor x\rfloor$ is the set of integers $\mathbb{Z}$. Non-integer real numbers (e.g. $0.5$) are not attained. Hence not onto $\mathbb{R}$.

Therefore neither injective nor surjective.

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Q4.

Show that the modulus function $f:\mathbb{R}\to \mathbb{R},f(x)=\mathrm{\mid }x\mathrm{\mid }$ is neither one-one nor onto.

Solution.

• Not one-one: $\mathrm{\mid }1\mathrm{\mid }=1=\mathrm{\mid }-1\mathrm{\mid }$ with $1\mathrm{\ne }-1$

• Not onto: Negative reals are not attained (no $x$ with $\mathrm{\mid }x\mathrm{\mid }=-1$). So not onto $\mathbb{R}$.

Hence neither injective nor surjective.

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Q5.

Show that the signum function $\mathrm{sgn}:\mathbb{R}\to \mathbb{R}$ defined by

$$\mathrm{sgn}(x)=\{\begin{array}{ll}1,& x>0,\\ 0,& x=0,\\ -1,& x<0,\end{array}$$

is neither one-one nor onto.

Solution.

• Not one-one: All positive numbers map to $1$, so many inputs share the same image.

• Not onto: The range is $\{-1,0,1\}$, a proper subset of $\mathbb{R}$; reals like $2$ are not attained. So not onto.

Therefore neither injective nor surjective.

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Q6.

Let $A=\{1,2,3\},B=\{4,5,6,7\}$ and $f=\{(1,4),(2,5),(3,6)\}$ as a function $A\to B$. Show that $f$ is one-one.

Solution.
The images are $4,5,6$ which are distinct, so distinct domain elements have distinct images. Thus $f$ is injective. $f$ is not onto $B$ because $7$ is not an image.

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Q7.

Decide whether the given functions are one-one, onto, or bijective:

(i) $f:\mathbb{R}\to \mathbb{R},f(x)=3-4x$
(ii) $f:\mathbb{R}\to \mathbb{R},f(x)=1+x^2$

Solution.

(i) $f(x)=3-4x$: linear with slope $-4\mathrm{\ne }0$.

• Injective: Yes. If $3-4x_1=3-4x_2$ then $x_1=x_2$.

• Surjective: Yes. Given any $y$, solve $y=3-4x$ ⇒ $x=(3-y)\mathrm{/}4\in \mathbb{R}$. So every real $y$ has a pre-image.
  Thus bijective (one-one and onto).

(ii) $f(x)=1+x^2$:

• Injective: No because $x$ and $-x$ give same value for $x\mathrm{\ne }0$.

• Surjective: No because range is $[1,\mathrm{\infty })$, so values $<1$ are not attained.
  Thus neither one-one nor onto.

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Q8.

Let $A,B$ be sets. Show $f:A\times B\to B\times A$ defined by $f(a,b)=(b,a)$ is bijective.

Solution.
Define $g:B\times A\to A\times B$ by $g(b,a)=(a,b)$. Then $g\circ f(a,b)=(a,b)$ and $f\circ g(b,a)=(b,a)$, so $g=f^{-1}$. Hence $f$ has an inverse and is bijective (both one-one and onto).

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Q9.

Let $f:\mathbb{N}\to \mathbb{N}$ be defined by

$$f(n)=\{\begin{array}{ll}{\displaystyle \frac{n+1}{2}},& \text{if }n\text{ is odd},\\ {\displaystyle \frac{n}{2}},& \text{if }n\text{ is even},\end{array}\text{for all }n\in \mathbb{N}\mathrm{.}$$

State whether $f$ is bijective. Justify your answer.

Solution.
For a fixed $m\in \mathbb{N}$ both $2m-1$ (odd) and $2m$ (even) map to $m$:

$$f(2m-1)=\frac{(2m-1)+1}{2}=m,f(2m)=\frac{2m}{2}=m\mathrm{.}$$

So every $m$ has at least one preimage (in fact two), hence $f$ is onto.

But $f(1)=1$ and $f(2)=1$ show $f$ is not one-one. Therefore onto but not one-one; hence not bijective.

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Q10.

(From the PDF — formula hard to render.) Let $A=\mathbb{R}\setminus \{3\}$, $B=\mathbb{R}\setminus \{1\}$ and (interpreting the PDF) consider

$$f:A\to B,f(x)=\frac{2x-3}{x-3}\mathrm{.}$$

Is $f$ one-one and onto?

Solution – Under the interpretation $f(x)={\displaystyle \frac{2x-3}{x-3}}$

1. Injectivity: Suppose $f(x_1)=f(x_2)$. Then

$$\frac{2x_1-3}{x_1-3}=\frac{2x_2-3}{x_2-3}\mathrm{}$$

Cross-multiply and simplify:

$$(2x_1-3)(x_2-3)=(2x_2-3)(x_1-3)\mathrm{}$$

Expanding and cancelling leads to $(x_1-x_2)\cdot (2-\frac{3}{x_1-3}-\cdots )=0$. Working the algebra carefully yields $x_1=x_2$ (there is no other solution in $A$). Thus $f$ is injective. (One can solve for $x$ in terms of $y$ below to make injectivity explicit.)

2. Find range (surjectivity): Solve $y={\displaystyle \frac{2x-3}{x-3}}$ for $x$:

$$y(x-3)=2x-3\Rightarrow yx-3y=2x-3\Rightarrow x(y-2)=3(y-1)\mathrm{}$$

So if $y\mathrm{\ne }2$ then $x={\displaystyle \frac{3(y-1)}{y-2}}$ and this $x\in A$ (provided $x\mathrm{\ne }3$). The algebra shows every $y\mathrm{\ne }2$ is attained by some $x\in A$. But $y=2$ yields no solution (division by zero). Thus the range = $\mathbb{R}\setminus \{2\}$.

Hence with our interpreted formula:

• $f$ is one-one.

• The image is $\mathbb{R}\setminus \{2\}$, so $f$ is onto $\mathbb{R}\setminus \{2\}$ but not onto the given codomain $B=\mathbb{R}\setminus \{1\}$(because $1\in \mathbb{R}\setminus \{2\}$ is in the image). Therefore $f$ is not onto the stated $B$ (but would be onto $\mathbb{R}\setminus \{2\}$).

Important: If the rational expression in your book is different from $(2x-3)\mathrm{/}(x-3)$, tell me the exact formula shown in the PDF and I will re-evaluate Q10 precisely.

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Q11.

Let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=x^4$. Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
$x^4=(-x{)}^4$, so $f$ is many-one (not injective). Range is $[0,\mathrm{\infty })$, so negative reals are not attained ⇒ not onto $\mathbb{R}$. So answer is (D): neither one-one nor onto.

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Q12.

Let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=3x$. Choose the correct option:
(A) one-one onto, (B) many-one onto, (C) one-one but not onto, (D) neither one-one nor onto.

Solution.
Linear map with nonzero slope: injection holds and for any $y$ we have $x=y\mathrm{/}3$ so surjection holds. Thus one-one and onto, option (A).