Question

If $f=\left\{\right(a,0),(b,-2),(c,3\left)\right\}$, $g=\left\{\right(a,-2),(b,0),(c,1\left)\right\}$ then $\frac{f}{g}=$

• A. $\left\{\right(a,-1),(b,-2),(c,4\left)\right\}$
• B. $\left\{\right(a,3),(b,-2),(c,2\left)\right\}$
• C. $\left\{\right(a,0),(c,3\left)\right\}$
• D. does not exist

Answer 1

The correct option is C $\left\{\right(a,0),(c,3\left)\right\}$

$f=\left\{\right(a,0),(b,-2),(c,3\left)\right\}$ and $g=\left\{\right(a,-2),(b,0),(c,1\left)\right\}$

Implies

$f\left(a\right)=0$

$f\left(b\right)=-2$

$f\left(c\right)=3$

Similarly

$g\left(a\right)=-2$

$g\left(b\right)=0$

$g\left(c\right)=1$

Consider a function

$h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

Then

$h\left(a\right)=\frac{f\left(a\right)}{g\left(a\right)}=\frac{0}{-2}=0$ ...(i)
$h\left(b\right)=\frac{f\left(b\right)}{g\left(b\right)}=\frac{-2}{0}=\infty$...(ii)
$g\left(c\right)=\frac{f\left(c\right)}{g\left(c\right)}=\frac{3}{1}=3$...(iii)

From i ii and iii

$h=\left\{\right(a,0),(c,3\left)\right\}$

Or

$\frac{f}{g}=\left\{\right(a,0),(c,3\left)\right\}$