Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^3+5{\mathrm{x}}^2-\mathrm{bx}+1$ .If f (x) when divided by 2x +1 leaves 5 as remainder, and f (x) is divisible by 3x -1, then

Solution:

we have,

$\begin{array}{rl} & \mathrm{f}(-\frac{1}{2})=5 \text{and }\mathrm{f}\left(\frac{1}{3}\right)=0\\ \Rightarrow & -\frac{\mathrm{a}}{8}+\frac{5}{4}+\frac{\mathrm{b}}{2}+1=5 \text{and }\frac{\mathrm{a}}{27}+\frac{5}{9}-\frac{\mathrm{b}}{3}+1=0\\ \Rightarrow & -\mathrm{a}+10+4\mathrm{b}+8=40 \text{and }\mathrm{a}-9\mathrm{b}+42=0\\ \Rightarrow & -\mathrm{a}+4\mathrm{b}=22 \text{and }\mathrm{a}-9\mathrm{b}+42=0\\ \Rightarrow & \mathrm{a}=-6,\mathrm{b}=4.\end{array}$