Exercise 1.5 - Chapter 1 Relations & Functions 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.5$
Question 1
Using the functions $f$ and $g$ given below, find fog and gof. Check whether $f \circ g=g o f$.
(i) $f(x)=x-6, g(x)=x^{2}$
(ii) $f(x)=\frac{2}{x}, g(x)=2 x^{2}-1$
(iii) $f(x)=\frac{x+6}{3} g(x)=3-x$
(iv) $f(x)=3+x, g(x)=x-4$
(v) $f(x)=4 x^{2}-1, g(x)=1+x$
Solution:
(i) $f(x)=x-6, g(x)=x^{2}$
$f \circ g(x)=f(g(x))=f\left(x^{2}\right)=x^{2}-6$
$g \circ f(x)=g(f(x))=g(x-6)=(x-6)^{2}$
$=x^{2}+36-12 x=x^{2}-12 x+36$
(1) $\neq(2)$
$\therefore$ fog $(x) \neq \operatorname{gof}(x)$

(ii) $\begin{aligned}
f(x) &=\frac{2}{x}, g(x)=2 x^{2}-1 \\
f o g(x) &=f(g(x))=f\left(2 x^{2}-1\right)=\frac{2}{2 x^{2}-1} \\
g o f(x) &=g(f(x))=g\left(\frac{2}{x}\right)=2\left(\frac{2}{x}\right)^{2}-1 \\
&=2\left(\frac{4}{x^{2}}\right)-1=\frac{8}{x^{2}}-1
\end{aligned}$
(iii) $f(x)=\frac{x+6}{3} g(x)=3-x$
$(1) \neq(2)$
$\begin{aligned}
f \circ g(x) &=f(g(x)) \\
&=\frac{9-x}{3}
\end{aligned}$
$\begin{aligned}
g \circ f(x) &=g(f(x))=g\left(\frac{x+6}{3}\right)=3-\frac{x+6}{3} \\
&=\frac{9-x-6}{3}=\frac{3-x}{3} \\
\neq(2) & f \circ g(x) \neq g \circ f(x)
\end{aligned}$
(iv) $f(x)=3+x, g(x)=x-4$
$f \circ g(x)=f(g(x))=f(x-4)=3+x-4$
$=x-1 \ldots . . . . . .(1)$
$g \circ f(x)=g(f(x))=g(3+x)=3+x-4$
$=x-1 \ldots . . \ldots . . \ldots . .(2)$
Here $f \circ g(x)=g \circ f(x)$
(v) $f(x)=4 x^{2}-1, g(x)=1+x$
$f \circ g(x)=f(g(x))=f(1+x)=4(1+x)^{2}-1$
$=4\left(1+x^{2}+2 x\right)-1=4+4 x^{2}+8 x-1$
$=4 x^{2}+8 x+3$ (1)

$\begin{aligned}
&g \circ f(x)=g(f(x))=g\left(4 x^{2}-1\right) \\
&=1+4 x^{2}-1=4 x^{2} \ldots \ldots . \\
&(1) \neq(2) \\
&\therefore f \circ g(x) \neq g \circ f(x)
\end{aligned}$

Question 2.
Find the value of $k$, such that $f \circ g=g \circ f$
(i) $f(x)=3 x+2, g(x)=6 x-k$
Answer:
$f(x)=3 x+2 ; g(x)=6 x-k$ $f o g=f[g(x)]$ $=f(6 x-k)$ $=3(6 x-k)+2$ $=18 x-3 K+2$ $g 0 f=g[f(x)]$ $=g(3 x+2)$ $=6(3 x+2)-k$ $=18 x+12-k$ But given fog $=g o f$ $18 x-3 x+2=18 x+12-k$ $-3 k+2=12-k$ $-3 k+k=12-2$ $-2 k=10$ $k=\frac{-10}{2}=-5$
The value of $k=-5$
(ii) $f(x)=2 x-k, g(x)=4 x+5$
Answer:
$\begin{aligned}
&f(x)=2 x-k ; g(x)=4 x+5 \\
&f \circ g=f[g(x)] \\
&=f(4 x+5) \\
&=2(4 x+5)-k \\
&=8 x+10-k \\
&g o f=g[f(x)] \\
&=g(2 x-k) \\
&=4(2 x-k)+5 \\
&=8 x-4 k+5 \\
&\text { But fog }=g o f \\
&8 x+10-k=8 x-4 k+5 \\
&-k+4 k=5-10 \\
&3 k=-5
\end{aligned}$

$k=\frac{-5}{3}$
The value of $k=\frac{-5}{3}$

Question $3 .$
if $f(x)=2 x-1, g(x)=\frac{x+1}{2}$, show that fog $=g \circ f=x$
Solution:
$f(x)=2 x-1, g(x)=\frac{x+1}{2}, f \circ g=g \circ f=x$
$f(x)=2 x-1, g(x)=\frac{x+1}{2}, f o g=g o f=x$
$f \circ g(x)=f(g(x))=f\left(\frac{x+1}{2}\right)$
$=2\left(\frac{x+1}{2}\right)-1=x$
$\begin{aligned} g o f(x) &=g(f(x))=\\ &=\frac{2 x}{2}=x \end{aligned}$
(1) $=(2)$
$f o g=g o f=x$
Hence proved.

Question $4 .$
(i) If $f(x)=x^{2}-1, g(x)=x-2$ find $a$, if $g \circ f(a)=1$.
(a) Find $k$, if $f(k)=2 k-1$ and
fof $(k)=5$.
Answer:
(i) $f(x)=x^{2}-1 ; g(x)=x-2$.
$\begin{aligned}
&g \circ f=g[f(x)] \\
&=g\left(x^{2}-1\right) \\
&=x^{2}-1-2 \\
&=x^{2}-3
\end{aligned}$

given gof $(a)=1$
$\begin{aligned}
&a^{2}-3=1 \text { [But } g \circ f(x)=x^{2}-3 \text { ] } \\
&a^{2}=4 \\
&a=\sqrt{4}=\pm 2
\end{aligned}$

The value of $a=\pm 2$
$\begin{aligned}
&\text { (ii) } f(\mathrm{k})=2 \mathrm{k}-1 ; f \mathrm{fof}(\mathrm{k})=5 \\
&\text { fof }=\mathrm{f}[\mathrm{f}(\mathrm{k})] \\
&=\mathrm{f}(2 \mathrm{k}-1) \\
&=2(2 \mathrm{k}-1)-1 \\
&=4 \mathrm{k}-2-1 \\
&=4 \mathrm{k}-3 \\
&\mathrm{fof}(\mathrm{k})=5 \\
&4 \mathrm{k}-3=5 \\
&4 \mathrm{k}=5+3 \\
&4 \mathrm{k}=8 \\
&\mathrm{k}=\frac{8}{4}=2
\end{aligned}$
The value of $k=2$

Question $5 .$
Let $\mathrm{A}, \mathrm{B}, \mathrm{C} \subset \mathrm{N}$ and a function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ be defined $\mathrm{by} \mathrm{f}(\mathrm{x})=2 \mathrm{x}+1$ and $\mathrm{g}: \mathrm{B} \rightarrow \mathrm{C}$ be defined by $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}$. Find the range of fog and gof
Solution:

$f(x)=2 x+1$
$g(x)=x^{2}$
$f o g(x)=f g(x))=f\left(x^{2}\right)=2 x^{2}+1$
$g o f(x)=g(f(x))=g(2 x+1)=(2 x+1)^{2}$
$=4 x^{2}+4 x+1$
Range of fog is
$\left\{y / y=2 x^{2}+1, x \in N\right\}$
Range of gof is
$\left\{y / y=(2 x+1)^{2}, x \in N\right\}$

Question $6 .$
Let $f(x)=x^{2}-1$. Find (i) fof (ii) fofof
Answer:
$f(x)=x^{2}-1$
(i) $f \circ f=f[f\{x)]$
$=f\left(x^{2}-1\right)$
$=\left(x^{2}-1\right)^{2}-1$
$=x^{4}-2 x^{2}+1-1$
$=x^{4}-2 x^{2}$
(ii) fofof $=$ fof $[f(x)]$
$=f$ of $\left(x^{2}-1\right)$
$\begin{aligned}
&=f\left(x^{2}-1\right)^{2}-1 \\
&=f\left(x^{4}-2 x^{2}+1-1\right) \\
&=f\left(x^{4}-2 x^{2}\right) \\
&\text { fofof }=\left(x^{4}-2 x^{2}\right)^{2}-1
\end{aligned}$

Question 7.
If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ are defined $b \mathrm{f} f(\mathrm{x})=\mathrm{x}^{5}$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{4}$ then check if $\mathrm{f}, \mathrm{g}$ are one-one and fog is one-one?
Solution:
$\begin{aligned}
&f(x)=x^{5} \\
&g(x)=x^{4} \\
&f o g=f o g(x)=f(g(x))=f\left(x^{4}\right) \\
&=\left(x^{4}\right)^{5}=x^{20}
\end{aligned}$
$f$ is one-one, $g$ is not one-one.
$\because \mathrm{g}(1)=1^{4}=1$

$g(-1)=(-1)^{4}=1$
Different elements have same images
fog is not one-one. $[\because$ fog $(1)=$ fog $(-1)=1]$
 

Question 8.
Consider the functions $f(x), g(x), h(x)$ as given below. Show that (f $\circ g) \circ h=f \circ(g \circ h)$ in each case.
(i) $f(x)=x-1, g(x)=3 x+1$ and $h(x)=x^{2}$
(ii) $f(x)=x^{2}, g(x)=2 x$ and $h(x)=x+4$
(iii) $f(x)=x-4, g(x)=x^{2}$ and $h(x)=3 x-5$
Answer:
(i) $f(x)=x-1, g(x)=3 x+1, h(x)=x^{2}$
$f \circ g(x)=f[g(x)]$
$=f(3 x+1)$
$=3 x+1-1$
$f o g=3 x$
(fog) $\circ h(x)=f o g[h(x)]$,
$=f o g\left(x^{2}\right)$
$=3\left(\mathrm{x}^{2}\right)$
(fog) $o h=3 x^{2} \ldots$ (1)
$g \circ h(x)=g[h(x)]$
$=g\left(x^{2}\right)$
$=3\left(x^{2}\right)+1$
$=3 x^{2}+1$
$f o(g o h) x=f[g o h(x)]$
$=\mathrm{f}\left[3 \mathrm{x}^{2}+1\right]$
$=3 x^{2}+1-1$
$=3 x^{2} \ldots(2)$
From (1) and (2) we get
(fog) $\mathrm{oh}=\mathrm{fo}$ (goh)
Hence it is verified

(ii) $f(x)=x^{2} ; g(x)=2 x$ and $h(x)=x+4$
$(f \circ g) x=f[g(x)]$
$=f(2 x)$
$=(2 x)^{2}$
$=4 x^{2}$
(fog) $o h(x)=f o g[h(x)]$
$=f o g(x+4)$
$=4(x+4)^{2}$
$=4\left[x^{2}+8 x+16\right]$
$=4 x^{2}+32 x+64 \ldots . .$ (1)
$g \circ h(x)=g[h(x)]$
$=g(x+4)$
$=2(x+4)$
$=2 x+8$
fo $(g o h) x=$ fo $[g o h(x)]$
$=\mathrm{f}[2 \mathrm{x}+8]$
$=(2 x+8)^{2}$
$=4 \times 2+32 x+64 \ldots .$ (2)
From (1) and (2) we get
$(f \circ g) o h=f o(g \circ h)$
(iii) $f(x)=x-4 ; g(x)=x^{2} ; h(x)=3 x-5$
$f \circ g(x)=f[g(x)]$
$=f\left(x^{2}\right)$
$=x^{2}-4$
(fog) oh $(x)=f o g[h(x)]$
$=f o g(3 x-5)$
$=(3 x-5)^{2}-4$
$=9 x^{2}-30 x+25-4$
$=9 x^{2}-30 x+21 \ldots$ (1)
$g \circ h(x)=g[h(x)]$

$=g(3 x-5)$ $=(3 x-5)^{2}$ $=9 x^{2}+25-30 x$ fo $(g o h) x=f[g o h(x)]$ $=f\left[9 x^{2}-30 x+25\right]$ $=9 x^{2}-30 x+25-4$ $=9 x^{2}-30 x+21 \ldots(2)$ From $(1)$ and $(2)$ we get (fog) oh $=$ fo(goh)
$(f \circ g) o h=f o(g o h)$
 

Question $9 .$
Let $f=\{(-1,3),(0,-1),(2,-9)\}$ be a linear function from $Z$ into $Z$. Find $f(x)$.
Solution:
$f=\{(-1,3),(0,-1), 2,-9)$ $f(x)=(a x)+b \ldots \ldots \ldots .(1)$ is the equation of all linear functions. $\therefore f(-1)=3$ $f(0)=-1$ $f(2)=-9$ $f(x)=a x+b$ $f(-1)=-a+b=3 \ldots \ldots \ldots \ldots . .(2)$ $f(0)=b=-1$ $-a-1=3[\because$ substituting $b=-1$ in $(2)]$ $-a=4$ $a=-4$
The linear function is $-4 x-1$. [From (1)]

Question 10.
In electrical circuit theory, a circuit $\mathrm{C}(\mathrm{t})$ is called a linear circuit if it satisfies the superposition principle given by $\mathrm{C}\left(\mathrm{at}_{1}+\mathrm{bt}_{2}\right)=\mathrm{aC}\left(\mathrm{t}_{1}\right)+\mathrm{bC}\left(\mathrm{t}_{2}\right)$, where $\mathrm{a}, \mathrm{b}$ are constants. Show that the circuit $\mathrm{C}(\mathrm{t})=31$ is linear.
Answer:
Given $C(t)=3 t$
$\left.\mathrm{C}\left(\mathrm{at} \mathrm{t}_{1}\right)=3 \mathrm{at}_{1} \ldots . \mathrm{l}\right)$
$\mathrm{C}\left(\mathrm{bt}_{2}\right)=3 \mathrm{bt}_{2} \ldots$ (2)
Add (1) and (2)
$\mathrm{C}\left(\mathrm{at} \mathrm{t}_{1}\right)+\mathrm{C}\left(\mathrm{bt_{2 }}\right)=3 \mathrm{at}_{1}+3 \mathrm{bt}_{2}$
$\mathrm{C}\left(\mathrm{at} \mathrm{t}_{1}+\mathrm{bt}_{2}\right)=3 \mathrm{at}_{1}+3 \mathrm{bt}_{2}$
$=\mathrm{Cat}_{1}+\mathrm{Cbt}_{2}[$ from (1) and (2) $]$
$\therefore \mathrm{C}\left(\mathrm{at}_{1}+\mathrm{bt}_{2}\right)=\mathrm{C}\left(\mathrm{at_{1 }}+\mathrm{bt}_{2}\right)$

Superposition principle is satisfied. $\therefore \mathrm{C}(\mathrm{t})=3 \mathrm{t}$ is a linear function.