$f\left(-\frac{2}{3}\right)$ এর মান নির্ণয় কর।  

এখানে, $f\left(x\right) = \frac{3x + 2}{3x - 2}$

              $\therefore f\left(\frac{1}{x}\right) = \frac{3 {\displaystyle \frac{1}{x }}+ 2}{3 {\displaystyle \frac{1}{x}} - 2} =\frac{ {\displaystyle \frac{3}{x}}+ 2}{({\displaystyle \frac{3}{x }}- 2}$

বা, $\frac{f\left({\displaystyle \frac{1}{x}}\right)+ 1 }{f\left({\displaystyle \frac{1}{x}}\right) - 1}= \frac{({\displaystyle \frac{3}{x}}+ 2 + {\displaystyle \frac{3}{x}}- 2)}{{\displaystyle \frac{3}{x }}+ 2 - {\displaystyle \frac{3}{x}}+ 2)}$

নির্ণেয় মান $\frac{3}{2x}$

এখানে, $g\left(x\right)=x^4+9x^2+11kx+6k$

$\therefore g(-1)=(-1{)}^4+9*(-1{)}^2+11k(-1)+6k$

                        $= 1 + 9 - 11k + 6k = 10 - 5k$

এখন, $g(- 1) = 0$

$\therefore 10 - 5k = 0$

বা, $k = \frac{10}{5}= 2$

অর্থাৎ $k = 2$ হলে $g(- 1) = 0$ হবে।