RD Sharma Class 12 Exercise 2.4 Function Solutions Maths - Download PDF Free Online

Functions Excercise: 2.4

functions exercise 2.4 question 1 (I)

Answer:
No, $f$ is many one.
Given:
$f:\{1,2,3,4\}\to \left\{10\right\}withf=\{(1,0),(2,10),(3,10)(4,10)\}$
Hint:
A function has an inverse if it is one-one and onto.
Solution:
Let us check one-one
$f=\{(1,10)(2,10),(3,10),(4,10)\}$

Since all elements have an image, they don't have a unique image.
$\therefore f$ is not one-one
Since, $f$ is not one-one, it doesn't have an inverse
Functions excercise 2.4 question 1 (II)

Answer :
No, $g$ is many one.
Given :
$g=\{5,6,7,8\}\to \{1,2,3,4\}Withg=\{(5,4),(6,3),(7,4)(8,2)\}$
Hint:
A function has an inverse if it is one-one and onto.
Solution:
Let us check one-one
$g=\{(5,4),(6,3),(7,4)(8,2)\}$

Since $5$ and $7$ have same image$4$, $g$ is not one-one.
Since $g$ is not one-one, it doesn't have an inverse.

functions exercise 2.4 question 1 (iii)

Answer:
Yes, $h$ is a bijection.
Given:
$h:\{2,3,4,5\}\to \{7,9,11,13\}withh=\{(2,7),(3,9),(4,11),(5,13)\}$
Hint:
A function has an inverse if it is one-one and onto.
Solution:
Let us check one-one
$h=\{(2,7),(3,9),(4,11),(5,13)\}$


Since all element has unique image $h$ is one-one.
Check onto,

Since each and every image there is a corresponding element.
$\therefore h$ is onto.
Since the function is both one-one and onto, it will have an inverse.
$h=\{(2,7),(3,9),(4,11),(5,13)\}$
$h^{-1}=\{(7,2),(9,3),(11,4),(13,5)\}$

Functions exercise 2.4 question 2 (ii)

Answer:
$f^{-1}$ doesn't exist.
Given:
$A=\{1,3,5,7,9\},B=\{0,1,9,25,49,81\},f\left(x\right)=x^2$
Solution:
Let us check one-one
$f\left(x\right)=x^2,f\left(1\right)=1,f\left(3\right)=3^2=9,f\left(5\right)=5^2=25,f\left(7\right)=7^2=49,f\left(9\right)=9^2=81$
$\therefore f=\{(1,1),(3,9),(5,25),(7,49),(9,81)\}$
The different elements of the domain have different images in the codomain.
Thus, $f$ is one-one.
This is not onto because the element $0$ in the codomain $\left(B\right)$ has no preimage in the domain $\left(A\right)$.
Therefore, $f^{-1}$ doesn't exist.

Functions exercise 2.4 question 3

Answer:
${\left(gof\right)}^{-1}=f^{-1}o{g}^{-1}$
Given:
$f=\{(1,a),(2,b),(c,3)\},g=\{(a,apple),(b,ball),(c,cat)\}$, clearly $f$ and $g$ are bijections.
Hint:
Here, $f$, $g$ and $fog$ are invertible.
Solution:
Now, $f^{-1}=\{(a,1),(b,2),(3,c)\},g^{-1}=\{(apple,a),(ball,b),(cat,c)\}$
So, $f^{-1}o{g}^{-1}=\{(apple,1),(ball,2),(cat,3)\}$
$f:\{1,2,3\}\to \{a,b,c\}andg:\{a,b,c\}\to \{apple,ball,cat\}$
So, $gof:\{1,2,3\}\to \{apple,ball,cat\}$
$\Rightarrow \left(gof\right)\left(1\right)=g\left[f\left(1\right)\right]=g\left(a\right)=apple$
$\left(gof\right)\left(2\right)=g\left[f\left(2\right)\right]=g\left(b\right)=ball$
$\left(gof\right)\left(3\right)=g\left[f\left(3\right)\right]=g\left(c\right)=cat$
$\therefore$ $\left(gof\right)=\{(1,apple),(2,ball),(3,cat)\}$
Clearly, $gof$ is a bijection.
So, $gof$ is invertible.
${\left(gof\right)}^{-1}=\{(apple,1),(ball,2),(cat,3)\}$
$From\left(i\right),\left(iii\right),$ we get,
${\left(gof\right)}^{-1}=f^{-1}o{g}^{-1}$

Functions exercise 2.4 question 4 (0)

Answer:
${\left(gof\right)}^{-1}=f^{-1}o{g}^{-1}$
Given:
$A=\{1,2,3,4\},B=\{3,5,7,9\},C=\{7,23,47,79\}$
$f:A\to B,g:B\to Cdefinedbyf\left(x\right)=2x+1,g\left(x\right)=x^2-2.$
Hint:
Clearly $f$ and $g$ are bijections.
Solution:
$f\left(x\right)=2x+1$
$f=\{(1,2\left(1\right)+1),(2,2\left(2\right)+1),(3,2\left(3\right)+1),(4,2\left(4\right)+1)\}$
$=\{(1,3),(2,5),(3,7),(4,9)\}$
$g\left(x\right)=x^2-2$
$g=\{(3,3^2-2),(5,5^2-2),(7,7^2-2),(9,9^2-2)\}$
$=\{(3,7),(5,23),(7,47),(9,79)\}$
So, $f^{-1}=\{(3,1),(5,2),(7,3),(9,4)\}$
$g^{-1}=\{(7,3),(23,5),(47,7),(79,9)\}$
Now, $\left(f^{-1}0g^{-1}\right):C\to A$
$\left(f^{-1}o{g}^{-1}\right)=\{(7,1),(23,2),(47,3),(79,4)\}$
Also $f:A\to B,g:B\to C$
$gof:A\to C,{\left(gof\right)}^{-1}:C\to A$
So, $f^{-1}o{g}^{-1}and{\left(gof\right)}^{-1}$ have the same domains.
$\left(gof\right)\left(x\right)=g\left(x\right)=g(2x+1)={(2x+1)}^2-2$
$\left(gof\right)\left(x\right)=g\left[f\left(x\right)\right]$
$\left(gof\right)\left(x\right)=4x^2+4x+1-2$
$\left(gof\right)\left(x\right)=4x^2+4x-1$
Then, $\left(gof\right)\left(1\right)=g\left(f\left(1\right)\right)=4+4-1=7$
$\left(gof\right)\left(2\right)=g\left(f\left(2\right)\right)=16+8-1=23$
$\left(gof\right)\left(3\right)=g\left(f\left(3\right)\right)=36+12-1=47$
$\left(gof\right)\left(4\right)=g\left(f\left(4\right)\right)=64+16-1=79$
So, $\left(gof\right)=\{(1,7),(2,23),(3,47),(4,79)\}$
${\left(gof\right)}^{-1}=\{(7,1),(23,2),(47,3),(79,4)\}$
From $\left(i\right),\left(ii\right)$
$\left(i\right),{\left(ii\left(gof\right)\right)}^{-1}=f^{-1}o{g}^{-1}$
$\{(7,1),(23,2),(47,3),(79,4)\}=\{(7,1),(23,2),(47,3),(79,4)\}$

Function Exercise 2.4 Question 5

Answer:
$f^{-1}\left(x\right)=\frac{x-5}{3}$
Given:
$f:Q\to Q$, defined by $f\left(x\right)=3x+5$
Hint:
Solution:
Injectivity of $f$:
Let $x$ and $y$ be two elements of the domain$\left(Q\right)$, such that,
$f\left(x\right)=f\left(y\right)$
$3x+5=3y+5$
$3x=3y$
$x=y$
So, $f$ is one-one.
Subjectivity of $f$
Let $y$ be n the codomain$\left(Q\right)$, $f\left(x\right)=y$.
$3x+5=y$
$3x=y-5$
$x=\frac{y-5}{3}$
$f$ is onto.
So, $f$ is bijection.
Let $f^{-1}\left(x\right)=y$ …(i)
$x=f\left(y\right)$
$x=3y+5$
$x-5=3y$
$y=\frac{x-5}{3}$
$f^{-1}\left(x\right)=\frac{x-5}{3}$ [From (i)]

Function Exercise 2.4 Question 6

Answer:
$f^{-1}\left(x\right)=\frac{x-3}{4}$
Given:
$f\left(x\right)=4x+3,f:R\to R$.
Hint:
Bijection function should be fulfil the injectivity, surjectivity.
Solution:
Let $x,y$ be two elements of domain$\left(R\right)$, such that
$4x+3=4y+3$
$4x=4y$
$x=y$
So, $f$ is one-one.
Let $y$ be in the co-domain$\left(R\right)$, such that $f\left(x\right)=y$.
$4x+3=y$
$4x=y-3$
$x=\frac{y-3}{4}ϵR$
$f$ is onto.
So, $f$ is a bijection.
$f^{-1}\left(x\right)=y$ … (i)
$x=f\left(y\right)$
$x=4y+3$
$x-3=4y$
$y=\frac{x-3}{4}$
$f^{-1}\left(x\right)=\frac{x-3}{4}$

Function Exercise 2.4 Question 7

Answer:
$f^{-1}\left(x\right)=\sqrt{x-4}$
Given:
$f\left(x\right)=x^2+4,f:R\to R_{+}\to [4,\mathrm{\infty }]$.
Hint:
Bijection function should be fulfil the injectivity and surjectivity.
Solution:
Let $x,y$ be two elements of the domain $\left(a\right)$.
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ {x}^{2+4}=y^2+4 \\ {x}^2=y^2 \\ x=y \end{gathered}$$
So $f$is one-one.
Let $y$ be in the co-domain$\left(a\right)$ such that
$$\begin{gathered} f\left(x\right)=y \\ {x}^2+4=y \\ {x}^2=y-4 \\ x=\sqrt{y-4}ϵR \end{gathered}$$
$f$ is onto, so $f$ is bijection.
$$\begin{gathered} f^{-1}\left(x\right)=y \\ x=f\left(y\right) \\ x=y^2+4 \\ x-4=y^{2}y=\sqrt{x-4} \\ {f}^{-1}\left(x\right)=\sqrt{x-4} \end{gathered}$$ … (i)

Function Exercise 2.4 Question 8

Answer:
$f$ is invertible.
Given:
$f\left(x\right)=\frac{4x+3}{6x-4},x\ne \frac{2}{3}$
Hint:
If the function is invertible, $fof=1$.
Solution:
$f\left(x\right)=\frac{4x+3}{6x-4},x\ne \frac{2}{3}$
$\left(fof\right)\left(x\right)=f\left(f\left(x\right)\right)=f\left(\frac{4x+3}{6x-4}\right)$
$\frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}=\frac{16x+12+18x-12}{24x+18-24x+16}=\frac{34x}{34}=x$
$\therefore fof\left(x\right)=x$ for all $x\ne \frac{2}{3}$
$fof=1$
Hence, the given function $f$ is invertible and inverse of $f$ is $f$ itself

Function Exercise 2.4 Question 9

Answer:
$f$is invertible with the function.
$f^{-1}\left(x\right)=\frac{\sqrt{x-6}-1}{3}$
Given:
$f:R_{+}\to [5,\mathrm{\infty }]$ given by $f\left(x\right)=9x^2+6x-5$
Hint:
Bijection function should fulfil the injection and surjection condition.
Solution:
Let $x,y$ be two elements of domain$\left(R^{+}\right)$, such that$f\left(x\right)=f\left(y\right)$.
$$\begin{gathered} 9x^2+6x-5=9y^2+6y-5 \\ 9x^2+6x=9y^2+6y \\ x=y(as,x,yϵR^{+}) \end{gathered}$$
$f$is one-one.
Subjectivity of $f$:
Let $y$ is in the co-domain$\left(Q\right)$ such that $f\left(x\right)=y$
$$\begin{gathered} 9x^2+6x-5=y \\ 9x^2+6x=y+5 \end{gathered}$$
$9x^2+6x+1=y+6$ [Adding $1$ on both sides]
$$\begin{gathered} {(3x+1)}^2=6+y \\ 3x+1=\sqrt{y+6} \\ 3x=\sqrt{y+6}-1 \\ x=\frac{\sqrt{y+6}1}{3}ϵR^{+}\left(domain\right) \end{gathered}$$
$f$ is onto.
So, $f$ is a bijection and hence, it is invertible.
$f^{-1}\left(x\right)=y$ … (i)
$$\begin{gathered} x=f\left(y\right) \\ x=9y^2+6y-5 \\ x+5=9y^2+6y \end{gathered}$$
$x+6={(3y+1)}^2$ [Adding on both sides]


$$\begin{gathered} 3y+1=\sqrt{x+6} \\ 3y=\sqrt{x+6}-1 \\ y=\frac{\sqrt{x+6}-1}{3} \end{gathered}$$
$f^{-1}\left(x\right)=\frac{\sqrt{x+6}-1}{3}$ [From (i)]

Function Exercise 2.4 Question 10

Answer:
$f^{-1}\left(x\right)=\sqrt[3]{x+3},f^{-1}\left(24\right)=3,f^{-1}\left(5\right)=2$
Given:
If $f:R\to R$ be defined by $f\left(x\right)=x^3-3$.
Hint:
Bijection function should be fulfil the injectivity, surjectivity condition.
Solution:
Let $x,y$ be two elements in domain$\left(R\right)$, such that $x^3-3=y^3-3$.
$$\begin{gathered} x^3=y^3 \\ x=y \end{gathered}$$
So, $f$ is one-one.
Surjectivity of $f$.
Let $y$ be in the co-domain$\left(R\right)$such that $f\left(x\right)=y$
$$\begin{gathered} x^3-3=y \\ x=\sqrt[3]{y+3}ϵR \end{gathered}$$
$f$ is onto.
So, $f$is a bijection and hence it’s convertible.
$$\begin{gathered} Let,f^{-1}\left(x\right)=y \\ x=f\left(y\right) \\ x=y^3-3 \\ x+3=y^3 \\ y=\sqrt[3]{x+3}=f^{-1}\left(x\right) \end{gathered}$$ …(i)
From (i)
$$\begin{gathered} f^{-1}\left(x\right)=\sqrt[3]{x+3} \\ {f}^{-1}(24+3)=\sqrt[3]{27}=\sqrt[3]{3^3}=3 \\ {f}^{-1}\left(5\right)=\sqrt[3]{5+3}=\sqrt[3]{8}=2 \end{gathered}$$

Function Exercise 2.4 Question 11

Answer:
Bijection, $f^{-1}\left(3\right)=-1$.
Given:
$f:R\to R$ is defined as $f\left(x\right)=x^3+4$.
Hint:
Bijection function should be fulfil the injection and surjection condition.
Solution:
Let $x,y$ be two elements of domain$\left(R\right)$.
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ {x}^3+4=y^3+4 \\ {x}^3=y^3 \\ x=y \end{gathered}$$
$f$is one-one.

Let $y$ be in the co-domain$\left(R\right)$, such that $f\left(x\right)=y$


$$\begin{gathered} x^2+4=y \\ {x}^3=y-4 \\ x=\sqrt[3]{y-4}ϵRdomain \end{gathered}$$
$f$ is onto.
So, $f$ is a bijection, hence is invertible.
$f^{-1}\left(x\right)=y$ …(i)
$x=f\left(y\right)$
$$\begin{gathered} x=y^3+4 \\ x-4=y^3 \\ y=\sqrt[3]{x-4} \end{gathered}$$
$$\begin{gathered} So{f}^{-1}\left(x\right)=\sqrt[3]{x-4} \\ {f}^{-1}\left(3\right)=\sqrt[3]{3-4}=\sqrt[3]{-1}=-1 \end{gathered}$$ [from (i

Function Exercise 2.4 Question 12

Answer:
${\left(gof\right)}^{-1}=f^{-1}o{g}^{-1}$
Given:
$f:Q\to Q,g:Q\to Q$ defined by $f\left(x\right)=2x,g\left(x\right)=x+2$.
Hint:
Bijective function should be fulfil the injective and surjective function.
Solution:
$x,y$ be two elements of domain$\left(Q\right)$, such
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ 2x=2y \\ x=y \end{gathered}$$
$f$ is one-one.
Let $y$ be in the co-domain$\left(Q\right)$, such that
$f\left(x\right)=y$
$2x=y$
$x=\frac{y}{2}ϵQ$.
$f$ is onto.
So, $f$ is a bijection and hence, it’s invertible
Let _{$f^{-1}\left(x\right)=y$} …(i)
$$\begin{gathered} x=f\left(y\right) \\ y=\frac{x}{2} \\ {f}^{-1}\left(x\right)=\frac{x}{2} \end{gathered}$$ Injectivity of $g$
Let $x,y$ be two elements of domain$\left(Q\right)$.
$$\begin{gathered} g\left(x\right)=g\left(y\right) \\ x+2=y+2 \\ x=y \end{gathered}$$
So, $g$ is one-one.
Subjectivity of $g$:
Let $y$ be in the co-domain$\left(Q\right)$, such that $g\left(x\right)=y$.
$x+2=y$
$x=2-yϵ$ (domain)
$g$ is onto.
So, $g$ is a bijection, hence it’s invertible.
Let $g^{-1}\left(x\right)=y$
$x=g\left(y\right)$ …(ii)
$x=y+2$
$y=x-2$
So, $g^{-1}\left(x\right)=x-2$ [from (ii)]
${\left(gof\right)}^{-1}=f^{-1}o{g}^{-1}$.
$f\left(x\right)=2x,g\left(x\right)=x+2,f^{-1}\left(x\right)=\frac{x}{2},g^{-1}\left(x\right)=x-2$
$\left(f^{-1}o{g}^{-1}\right)\left(x\right)=f^{-1}(x-2)$ $[\because \left(f^{-1}o{g}^{-1}\right)\left(x\right)=f^{-1}\left(g^{-1}\left(x\right)\right)]$
$\left(f^{-1}o{g}^{-1}\right)\left(x\right)=f^{-1}\left(\frac{x-2}{2}\right)$ …(iii)
$\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)$
$=g\left(2x\right)$
$=2x+1$
Let $\left(gof\right)\left(x\right)=y$ …(vi)
$x=\left(gof\right)\left(y\right)$
$x=2y+2$
$y=\frac{x-2}{2}$
${\left(gof\right)}^{-1}=f^{-1}o{g}^{-1}$.

Function Exercise 2.4 Question 13

Answer:
$f^{-1}\left(x\right)=\frac{3x-2}{x-1}$
Given:
$A=R-\left\{3\right\},B=R-\left\{1\right\},f:A\to B$ defined by $f\left(x\right)=\frac{x-2}{x-3}$.
Hint:
One-one means every domain has a distinct range. One means every image has some preimage in the domain function.
Solution:
We have,$A=R-\left\{3\right\},B=R-\left\{1\right\}$.
The function $f:A\to B$ defined by $f\left(x\right)=\frac{x-2}{x-3}$.
Let $x,yϵA$ such that $f\left(x\right)=f\left(y\right)$ then
$\frac{x-2}{x-3}=\frac{y-2}{y-3}$
$xy-3x-2y+6=xy-2x-3y+6$
$-x=-y$
$x=y$
$f$is one-one.
Let $yϵB$, then $y\ne 1$.
The function $f$is onto if there exists $xϵA$ such that $f\left(x\right)=y$.
$\Rightarrow \frac{x-2}{x-3}=y$
$x-2=xy-3y$
$x-xy=2-3y$
$x=\frac{2-3y}{1-y}ϵA(y\ne 1)$
Thus, for any $yϵB$, there exists $\frac{2-3y}{1-y}ϵA$ such that
$f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2+2y}{2-3y-3+3y}=\frac{-y}{-1}=y$
$f$ is onto.
So, $f$ is one-one and onto function.
$x=\frac{2-3y}{1-y}$
$f^{-1}\left(x\right)=\frac{3x-2}{x-1}$

Function Exercise 2.4 Question 14

Answer:
$f^{-1}\left(y\right)g\left(y\right)=\frac{\sqrt{25y+54}-3}{5}$
Given:
$f:R^{+}\to [-9,\mathrm{\infty }]$ given by $f\left(x\right)=5x^2+6x-9$.
We need to prove that $f^{-1}\left(y\right)=\frac{\sqrt{54+25y}-3}{5}$
Solution:
$f\left(x\right)=5x^2+6x-9$
Let $y=5x^2+6x-9$
Dividing by$5$,
$=x^2+\frac{6}{5}x-\frac{9}{5}$
$=x^2+2x\frac{3}{5}+\frac{9}{25}-\frac{9}{25}-\frac{9}{5}$
$=x^2+(2x\times \frac{3}{5})+{\left(\frac{3}{5}\right)}^2-\frac{9}{25}-\frac{9}{5}$ $[\because {(a+b)}^2=a^2+2ab+b^2]$
$={(x+\frac{3}{5})}^2-\frac{9-45}{25}$
$y={(x+\frac{3}{5})}^2-\frac{54}{25}$
$\sqrt{y+\frac{54}{25}}=(x+\frac{3}{5})$
$\sqrt{\frac{25y+54}{25}}=x+\frac{3}{5}$
$x=\frac{\sqrt{25y+54}}{5}-\frac{3}{5}$
$x=\frac{\sqrt{25y+54}-3}{5}$
Let $g\left(y\right)=\frac{\sqrt{25y+54}-3}{5}$
Now,
$fog\left(y\right)=f\left(g\left(y\right)\right)$
$=f\left(\frac{\sqrt{25y+54}-3}{5}\right)$
$=5{\left(\frac{\sqrt{25y+54}-3}{5}\right)}^2+6\left(\frac{\sqrt{25y+54}-3}{5}\right)-9$
$=5\left(\frac{25y+54+9-6\sqrt{25y+54-3}}{25}\right)+\left(\frac{6\sqrt{25y+54}-18}{5}\right)-9$
$=\frac{25y+63-6\sqrt{25y+54}}{5}+\frac{6\sqrt{25y+54}-18}{5}-9$
$=\frac{25y+63-18-45}{5}$
$=\frac{25y}{5}$
$=5y$
Identity function,
Also, $gof\left(x\right)=g\left(f\left(x\right)\right)$
$=g(5x^2+6x-9)$
$=\frac{\sqrt{5(5x^2+6x-9)+54}-3}{5}$
$=\frac{\sqrt{25x^2+30x-45+54}-3}{5}$
$=\frac{\sqrt{25x^2+30x+9}-3}{5}$
$=\frac{\sqrt{{(5x+3)}^2}-3}{5}$
$=\frac{5x+3-3}{5}$
$=x$
So, $f$ is invertible, an identity function.
$f\left(y\right)=g\left(y\right)=\frac{\sqrt{25y+54}-3}{5}$