Mathematics - Functions Question with Solution | TestHub

Let $\mathrm{f}(\mathrm{x})={\mathrm{e}}^{\mathrm{mx}}$ be a trial solution. Then $f(x+2)=e^{m(x+2)}=e^{mx}\cdot e^{2m},f(x+1)=e^{mx}\cdot e^m$.

Therefore, $\mathrm{f}(\mathrm{x}+2)-5\mathrm{f}(\mathrm{x}+1)+6\mathrm{f}(\mathrm{x})=0$

$$\begin{array}{rlr}& \Rightarrow e^{mx}\cdot e^{2m}-5e^{mx}\cdot e^m+6e^{mx}=0\Rightarrow e^{mx}\left[e^{2m}-5e^m+6\right]=0& \end{array}$$

$$\Rightarrow e^{2m}-5e^m+6=0;e^{mx}\ne 0$$

Let ${\mathrm{e}}^{\mathrm{m}}=\mathrm{u}$. Then from the above equation,

$$\begin{array}{rlrl}& u^2-5u+6=0\Rightarrow (u-2)(u-3)=0u=2\text{ or }u=3.& & \end{array}$$

$$\therefore e^m=2or{e}^m=3\Rightarrow e^{mx}=2^{x}or{e}^{mx}=3^x$$

Hence $f(x)=A\cdot 2^x+B\cdot 3^x$.