Question

If $f=\left\{\right(-2,4),(0,6),(2,8\left)\right\}\text{and}g=\left\{\right(-2,-1),(0,3),(2,5\left)\right\}$ then $(\frac{2f}{3g}+\frac{3g}{2f})\left(0\right)$ is

• A. $\frac{1}{12}$
• B. $\frac{25}{12}$
• C. $\frac{5}{12}$
• D. $\frac{13}{12}$

Answer 1

Answer: (B)

$\frac{25}{12}$

We have,

$\begin{array}{c}\left(\frac{2f}{3g}+\frac{3g}{2f}\right)\left(0\right)\\ =\frac{2f\left(0\right)}{3g\left(0\right)}+\frac{3g\left(0\right)}{2f\left(0\right)}\\ =\frac{2.6}{3.3}+\frac{3.3}{2.6}\left(f\left(0\right)=6andg\left(0\right)=3\right)\\ =\frac{4}{3}+\frac{3}{4}\\ =\frac{16+9}{12}\\ =\frac{25}{12}\end{array}$

Hence, the option $B$ is the correct answer.