Relations and Functions

One-One:

Let $f\left(x_1\right)=f\left(x_2\right)$.

$4x_1+3=4x_2+3$

$x_1=x_2$

Onto:

$y=4x+3$

$x=\frac{y-3}{4}$

$x\in [0,\infty )\Rightarrow y\ge 3$

Range $=[3,\infty )\ne R$

Given, $\mathit{\text{R}}=\left\{\left(x,y\right):4y=5x-3,x,y\in \text{M}\right\}$ and $\mathit{\text{M}}=\left\{1,2,3,\dots ,16\right\}$

$R=\left\{\left(3,3\right),\left(7,8\right),\left(11,13\right)\right\}$

To make it symmetric $(8, 7), (13, 11)$ must be added.

$f\left(x\right)=x^2-4x+5$

$$\begin{gathered} f\left(x_1\right)=f\left(x_2\right) \\ {x}_1^2-4x_1+5=x_1^2-4x_2+5 \\ {x}_1^2-x_2^2=4\left(x_1-x_2\right) \\ \left(x_1-x_2\right)\left(x_1+x_2\right)=4\left(x_1-x_2\right) \\ {x}_1+x_2=4 \\ {x}_1=4-x_2 \end{gathered}$$

$x_1\ne x_2$, $f\left(x\right)$ is not an injective mapping.

Now check for surjective.

$$\begin{gathered} y=x^2-4x+5 \\ y=(x-2{)}^2+1 \\ y-1=(x-2{)}^2 \\ \mathrm{x}=\sqrt{(y-1)}+2 \end{gathered}$$

For any value $y<1$, $x\notin R$.

Thus, $f\left(x\right)$ is not surjective as well.

\( f(x)=\left|2 x^2+3 x-2\right|+\sin x \cos x \)
\( =|(2 x-1)(x+2)|+\sin x \cos x \)
\( f(x)=\left\{\begin{array}{l}
-2 x^2-3 x+2+\sin x \cos x, \quad 0\leq x<\frac{1}{2} \\
2 x^2+3 x-2+\sin x \cos x, \quad \frac{1}{2} \leq x<1
\end{array}\right. \)
\( f^{\prime}(x)=\left\{\begin{array}{l}
-4 x-3+\cos 2 x, \quad 0\leq x<\frac{1}{2} \\
4 x+3+\cos 2 x, \quad \frac{1}{2} \leq x<1
\end{array}\right. \)

For \(0 \leq x<\frac{1}{2} \Rightarrow f^{\prime}(x)<0\)

\(\Rightarrow f(x)\) is decreasing

For \(\frac{1}{2}<x \leq 1 \Rightarrow f^{\prime}(x)>0\)

\(\Rightarrow f(x)\) is increasing

\(f(0)=2\)
\( f(1)=3+\sin 1 \cos 1 \text { and } f\left(\frac{1}{2}\right)=\sin \frac{1}{2} \cos \frac{1}{2}\)

Hence required sum

\( f(1)+f\left(\frac{1}{2}\right)=3+\frac{\sin 2}{2}+\frac{\sin 1}{2}=3+\frac{1}{2}(\sin 1+\sin 2) \)
\( =3+\frac{\sin 1}{2}(1+2 \cos 1)\)

Given, $f\left(x\right)=9x^2+6x-5={\left(3x+1\right)}^2-6$

$\Rightarrow$one-one

Now, vertex at

$x=-\frac{b}{2a}=-\frac{6}{18}=-\frac{1}{3}<0$

$\Rightarrow$ min on D at $x = 0$

$f_{\min }=f\left(0\right)=-5,$

$\underset{x\to \infty }{\lim }f\left(x\right)=\infty$

$\Rightarrow$ Range of $f=\left[5,\infty \right)$

Range = Codomain $=\left[5,\infty \right)$

$\Rightarrow$ onto

$\Rightarrow f$ is bijective

\(f(x)=9 x^2+6 x-5, f^{\prime}(x)=18 x+6\)

for all \( x \in[0, \infty), f^{\prime}(x) \geq 6>0 \Rightarrow f\) strictly increasing

⇒ one-one

Vertex at \(x=-\frac{b}{2 a}=-\frac{6}{18}=-\frac{1}{3}<0\)

\(\Rightarrow \min\) on \(D\) at \(x=0\)

\(f_{\min }=f(0)=-5, \lim _{x \rightarrow \infty} f(x)=\infty\)

⇒ Range \((f)=[-5, \infty)\)

Range \(=\) Codomain \(=[-5, \infty) \Rightarrow\) onto

\(\Rightarrow f\) is bijective

Given: $f\left(x\right)=x^2-4x+5$

Check for injective:

Let, $f\left(x_1\right)=f\left(x_2\right)$

$$\begin{gathered} \Rightarrow x_1^2-4x_1+5=x_1^2-4x_2+5 \\ \Rightarrow x_1^2-x_2^2=4\left(x_1-x_2\right) \\ \Rightarrow \left(x_1-x_2\right)\left(x_1+x_2\right)=4\left(x_1-x_2\right) \\ \Rightarrow x_1+x_2=4 or x_1-x_2=0 \\ \Rightarrow x_1=4-x_2 or x_1=x_2 \end{gathered}$$

Hence, $f\left(x\right)$ is not an injective mapping.

Now check for surjective:

Let, $y=x^2-4x+5$

$$\begin{gathered} \Rightarrow y=(x-2{)}^2+1 \\ \Rightarrow y-1=(x-2{)}^2 \\ \Rightarrow \mathrm{x}=\pm \sqrt{(y-1)}+2 \end{gathered}$$

For any value $y<1$, $x\notin R$.

Thus, $f\left(x\right)$ is not surjective as well.

\(\begin{aligned}
& n=1 \Rightarrow m \in \phi \\
& n=2 \Rightarrow m=1 \Rightarrow \frac{m}{n} \text { can be } \frac{1}{2} = 1 \\
& n=3 \Rightarrow m=1,2 \Rightarrow \frac{m}{n} \text { can be } \frac{1}{3}, \frac{2}{3} = 2 \\
& n=4 \Rightarrow m=1,3 \Rightarrow \frac{m}{n} \text { can be } \frac{1}{4}, \frac{3}{4} = 2 \\
& n=5 \Rightarrow m=1,2,3,4 \Rightarrow \frac{m}{n}=\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} = 4 \\
& n=6 \Rightarrow m=1,5 \Rightarrow \frac{m}{n}=\frac{1}{6}, \frac{5}{6} \ldots 2 \\
& n=7 \Rightarrow m=1,2,3,4,5,6 \Rightarrow \frac{m}{n}=\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} \\
& = 6 \\
& n=8 \Rightarrow m=1,3,5,7 \Rightarrow \frac{m}{n}=\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8} = 4 \\
& n=9 \Rightarrow m=1,2,4,5,7,8 \Rightarrow \frac{m}{n}=\frac{1}{9}, \frac{2}{9}, \frac{3}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9} \\
& = 6 \\
& n=10 \Rightarrow m=1,3,7,9 \Rightarrow \frac{m}{n}=\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10} = 4
\end{aligned}\)

no. of elements in \(\mathrm{B}=1+2+2+4+2+6+4+6+4=31\)

Check One-One:

Let $f\left(x_1\right)=f\left(x_2\right)$.

$\Rightarrow 4x_1+3=4x_2+3$

$\Rightarrow x_1=x_2$

\(\Rightarrow\) one-one

Check Onto:

Let $y=4x+3$

$\Rightarrow x=\frac{y-3}{4}$

So, $x\in [0,\infty )\Rightarrow y\ge 3$

Range $=[3,\infty )\ne R$

\(\Rightarrow\) Not onto