3 Marks

Question 13 Marks

A function $f: R \rightarrow R$ given by $f(x)=2 x+3$. Prove that $f$ is invertible.

Answer

$f: R \rightarrow R$ and $f(x)=2 x+3$
suppose $x_1, x_2 \in R$ such that$
\begin{array}{l}
f\left(x_1\right)=f\left(x_2\right) \\
\Rightarrow \quad 2 x_1+3=2 x_2+3 \\
\Rightarrow \quad 2 x_1=2 x_2 \\
\Rightarrow \quad x_1=x_2 \\
\therefore f \text { is one-one. } \\
\text { Again suppose } y=2 x+3 \\
\therefore \quad x=\frac{y-3}{2} \\
\text { now } \quad f\left(\frac{y-3}{2}\right)=2\left(\frac{y-3}{2}\right)+3=y-3+3=y \\
\Rightarrow \quad f(x)=y
\end{array}
$
hence function is onto.
f is one-one onto hence $f$ is invertible so, $f^{-1}$ will exist.$
\begin{array}{l}
f^{-1}(y)=x=\frac{y-3}{2} \\
f^{-1}(x)=\frac{x-3}{2}
\end{array}
$
Hence Proved.

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Question 23 Marks

Prove that in set $Z$ relation $R$ defined as $a R b$ $\Leftrightarrow a-b$ is divisible by 3 , is equivalence relation.

Answer

R is reflexive, because every $a \in Z , 3$ divides $(a-a)$. hence $(a, a) \in R$
again if $(a, b) \in R$, then 3 divides $a-b$. so 3 also divides $b-a$. hence $(b, a) \in R$. It is prove that R is symmetric thus if $(a, b) \in R$ and $(b, c) \in R$ then $a$ $-b$ and $b-c$ dividied by 3 now $a-c=a-b+$ $b-c$ is even. Sum of even number is even and even of odd number is also even hence $(a-c)$ also divided by 3 . this proves that R is transitive. Hence given relation is reflexive, symmetric and transitive. So, in set $Z , R$ is equivalence relation.

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Question 33 Marks

Prove that $f: R \rightarrow R , f(x)=x^3+x$ is one-one onto function.

Answer

One-one : Suppose $x, y \in R$ such that $f(x)=f(y)$$
\begin{aligned}
\because f(x)=f(y) & \Rightarrow x^3+x=y^3+y \\
& \Rightarrow x^3-y^3+x-y=0 \\
& \Rightarrow(x-y)\left(x^2+x y+y^2\right)-(x-y)=0 \\
& \Rightarrow(x-y)\left(x^2+x y+y^2+1\right)=0 \\
& \Rightarrow x-y=0 \\
& \quad\left[\begin{array}{l}
\because x^2+x y+y^2 \geq 0 \forall x, y \in R \\
\therefore x^2+x y+y^2 \geq 1 \forall x, y \in R
\end{array}\right]
\end{aligned}
$thus $f(x)=f(y) \Rightarrow x=y \forall x, y \in R$
$\therefore f$, is one-one function.
Onto : Suppose $y$ is a arbitrary constant in R then$
f(x)=y
$
$
\begin{array}{l}
\Rightarrow x^3+x=y \\
\Rightarrow x^3+x-y=0
\end{array}
$
We know that is every equation of odd power has a real root then for every real value of $y$, equation $x^3+$ $x-y=0$ has a real root.Suppose this root is $\alpha$ then$
\begin{array}{ll}
\alpha^3+\alpha-y=0 & \Rightarrow \alpha^3+\alpha=y \\
& \Rightarrow f(x)=y
\end{array}
$
thus for every $y \in R$, there exist $\alpha \in R$ then $f(x)=$ $y$. Hence $f$ is onto. Thus $f: R \rightarrow R$ is one-one onto function.

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Question 43 Marks

Suppose $A=R-\{2\}$ and $B=R-\{1\}$, if a function $f: A \rightarrow B$ is defined such that $f(x)=\frac{x-1}{x-2}$, then prove that $f$ is one-one onto.

Answer

One-one : Suppose $x, y \in A$ such that $f(x)=f(y)$$
\begin{array}{rlrl}
\because & & f(x)=f(y) \\
\Rightarrow & & \frac{x-1}{x-2} & =\frac{y-1}{x-2} \\
\Rightarrow & & (x-1)(y-2) & =(x-2)(y-1) \\
\Rightarrow & x y-y-2 x+2 & =x y-x-2 y+2 \\
\Rightarrow & & x & =y
\end{array}
$thus, $f(x)=f(y) \Rightarrow x=y \forall x, y \in A$
hence $f$ is one-one.
Onto : Suppose $y$ is a arbitrary element of B. then$
\begin{aligned}
f(x) & =y \Rightarrow \frac{x-1}{x-2}=y \Rightarrow(x-1)=y(x-2) \\
\Rightarrow \quad x & =\frac{1-2 y}{1-y}
\end{aligned}$
cleary, $x=\frac{1-2 y}{1-y}, y \neq 1$ is real number and $\frac{1-2 y}{1-y}$
$\neq 2$.
Hence for every element of $y$ is B has exist one $x$ in A. So, $f$ is onto function. Thus $f$ is one-one onto function.

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Question 53 Marks

If $f, g: R \rightarrow R$ function is defined such that $f(x)=x^2+1, g(x)=2 x-3$ then find $f o g(x)$, gof $(x)$ and $\operatorname{gog}(3)$.

Answer

Given that :$
f(x)=x^2+1, g(x)=2 x-3
$
To find: $: f o g(x), \operatorname{gof}(x)$ and $\operatorname{gog}(3)$$
\begin{aligned}
f o g(x) & =f(g(x))=f(2 x-3)=(2 x-3)^2+1 \\
& =4 x^2-12 x+9+1=4 x^2-12 x+10
\end{aligned}
$

$
\begin{array}{l}
\operatorname{gof}(x)=g(f(x))=g\left(x^2+1\right)=2\left(x^2+1\right)-3 \\
\operatorname{gof}(x)=2 x^2+2-3=2 x^2-1 \quad \text { } \\
\operatorname{gog}(3)=g(g(3))=g(2 \times 3-3)=g(3)
\end{array}
$$\therefore \quad \operatorname{gog}(3)=2 \times 3-3=6-3=3 \quad$

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Question 63 Marks

If $g(x)=x^2+x-2$ and $(g o f)(x)=4 x^2-10 x$ +4 , find $f(x)$.

Answer

given :$
\begin{array}{l}
g(x)=x^2+x-2 \\
(g o f)(x)=g[f(x)] \\
4 x^2-10 x+4=g[f(x)] \\
\Rightarrow \quad 4 x^2-10 x+4=g(y) \\
\Rightarrow \quad 4 x^2-10 x+4=y^2+y-2 \\
\Rightarrow y^2+y+\left(10 x-4 x^2-6\right)=0 \\
\text { or } \\
y=\frac{-1 \pm \sqrt{1-4\left(10 x-4 x^2-6\right)}}{2.1}
\end{array}
$
$\begin{array}{ll}\text { or } & f(x)=\frac{-1 \pm \sqrt{16 x^2-40 x+25}}{2} \\ \text { or } & f(x)=\frac{-1 \pm(4 x-5)}{2} \\ \text { or } & f(x)=2 x-3 \\ \text { or } & f(x)=2-2 x\end{array}$

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Question 73 Marks

If $f : A \rightarrow B$ is a one-one onto function, then prove that :$
\left(f^{-1}\right)^{-1}=f
$

Answer

$\because f: A \rightarrow B$, is a one-one onto function.
$\therefore$ there is inverse function of this funciton $f^{-1}: B$
$\rightarrow A$ and $f^{-1}$ will be also one-one onto function.
suppose $f^{-1}=g$
then $\quad\left(f^{-1}\right)^{-1}=g^{-1}$
Again suppose that $g^{-1}(x)=y$
then $x=g(y) \Rightarrow x=f^{-1}(y) \Rightarrow f(x)=y$$
\begin{array}{lll}
\Rightarrow & f(x)=g^{-1}(x) & {[\text { from equation (2)] }} \\
\Rightarrow & & f=g^{-1} \Rightarrow f=\left(f^{-1}\right)^{-1} \\
& & {[\text { from equation (1)] }}
\end{array}
$

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Question 83 Marks

If a relation $R$ is defined in set $A$, prove that $R$ will be symmetric if and only if $R=R^{-1}$.

Answer

Here we prove that :
(i) If R is symmetric, then $R = R ^{-1}$ and
(ii) If $R = R ^{-1}$ then R will be symmetric.
(i) Suppose $(x, y) \in R$, then$
\begin{array}{l}
(x, y) \in R \Rightarrow(y, x) \in R \\
\Rightarrow \quad(x, y) \in R^{-1} \\
\{\because R \text { is symmetric }\} \\
\therefore \\
R \subset R^{-1} \\
\text { (by definition) } \\
\text { and } \\
(y, x) \in R^{-1} \Rightarrow(x, y) \in R \text { (by definition) } \\
\Rightarrow \quad(y, x) \in R \quad \because R \text { is symmetric } \\
\therefore \\
R^{-1} \subset R
\end{array}
$from equation (1) and (2)$
R=R^{-1}
$(ii) Suppose $(x, y) \in R$ then$
\begin{aligned}
(x, y) \in R \Rightarrow(y, x) \in & R^{-1} \\
& (\text { by definition }) \\
\Rightarrow(y, x) \in & R \\
& {\left[\because R^{-1}=R\right] }
\end{aligned}
$$\therefore R$ is symmetric relation.

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Maths 3 Marks

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