$\underset{\mathrm{x}\to 0}{lim} \frac{{\mathrm{e}}^{\mathrm{ax}}-\cos \left(\mathrm{bx}\right)-\frac{{\mathrm{cxe}}^{-\mathrm{cx}}}{2}}{1-\cos \left(2\mathrm{x}\right)}=17$, then 5a² + b² is equal to

$\begin{array}{l}\begin{array}{rl} & \underset{\mathrm{x}\to 0}{lim} \left(\frac{{\mathrm{e}}^{\mathrm{ax}}-\cos \mathrm{bx}-\frac{\mathrm{cx}}{2}{\mathrm{e}}^{-\mathrm{cx}}}{{\mathrm{x}}^2\left(\frac{1-\cos 2\mathrm{x}}{{\mathrm{x}}^2}\right)}\right)\\ & =\frac{1}{2}\underset{\mathrm{x}\to 0}{lim} \left(\frac{{\mathrm{e}}^{\mathrm{ax}}-\cos \mathrm{bx}-\frac{\mathrm{cx}}{2}{\mathrm{e}}^{-\mathrm{cx}}}{{\mathrm{x}}^2}\right) \left(\frac{0}{0}\right)\\ & =\frac{1}{2}\underset{\mathrm{x}\to 0}{lim} \left(\frac{{\mathrm{ae}}^{\mathrm{ax}}+\mathrm{bsin}\mathrm{bx}-\frac{\mathrm{c}}{2}\left(-{\mathrm{xce}}^{\mathrm{cx}}+{\mathrm{e}}^{-\mathrm{cx}}\right)}{2\mathrm{x}}\right)\left(\frac{0}{0}\right)\end{array}\\ \text{ As }\mathrm{Dr}\to 0\text{ as }\mathrm{x}=0,\mathrm{Nr}\to 0\\ \mathrm{a}+0-\frac{\mathrm{c}}{2}=0\Rightarrow 2\mathrm{a}=\mathrm{c}\end{array}$
Again use L’ Hospitals rule and put x = 0
$\begin{array}{l}\frac{1}{2}\left({\mathrm{a}}^2+{\mathrm{b}}^2+\frac{{\mathrm{c}}^2}{2}(1-0+1)\right)=34\\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=68\\ \Rightarrow 5{\mathrm{a}}^2+{\mathrm{b}}^2=68\end{array}$