If \(\lim _{x\to 0}\frac{{e}^{(a−1)x}+2\cos bx+(c−2){e}^{−x}}{x\cos x−{\log }_{e}(1+x)}=2,\) then \({a}^{2}+{b}^{2}+{c}^…

Given: \(\lim _{x\to 0}\frac{{e}^{(a−1)x}+2\cos bx+(c−2){e}^{−x}}{x\cos x−{\log }_{e}(1+x)}=2,\)

\({\text{⇒lim}}_{x\to 0}\frac{\left(1+\left(a−1\right)x+\frac{{(a−1)}^{2}{x}^{2}}{2!}\right)+2\left(1−\frac{{b}^{2}{x}^{2}}{2!}\right)+\left(c−2\right)\left(1−x+\frac{{x}^{2}}{2!}\right)}{x\left(1−\frac{{x}^{2}}{2!}\right)−\left(x−\frac{{x}^{2}}{2}\ldots \right)}=2\)

\({\text{⇒lim}}_{x\to 0}\frac{\left(1+2+c−2\right)+x\left(a−1−c+2\right)+{x}^{2}\left(\frac{{(a−1)}^{2}}{2}−{b}^{2}+\left(\frac{c−2}{2}\right)\right)}{\frac{{x}^{2}}{2}−\frac{{x}^{3}}{2!}+\ldots }=2\)

For which

\(∵c+1=0\Rightarrow c=−1\)

\(∵a−c=−1\Rightarrow a=−2\)

\(∵\frac{{(\text{a}−1)}^{2}}{2}−{\text{b}}^{2}+\left(\frac{\text{c}−2}{2}\right)=1\)

So, \(\frac{9}{2}−{\text{b}}^{2}−\frac{3}{2}=1\Rightarrow {\text{ b}}^{2}=2\)

Hence, \({a}^{2}+{b}^{2}+{c}^{2}=4+2+1=7\)