Answer:
We have, \[f\,(x)=\left\{ \begin{matrix} 5x-4 & 0<x\le 1 \\ 4{{x}^{2}}+3ax, & 1<x<2 \\ \end{matrix} \right.\] Here, \[f\,(1)=5\,(1)-4=1\] \[\therefore \] \[=\int_{0}^{1}{{{e}^{x-{{[x]}_{dx}}}}}+\int_{1}^{2}{{{e}^{(x-1)}}dx}\] \[=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,4\,{{(1+h)}^{2}}+3a\,(1+h)=4+3a\] and \[\underset{x\,\,\to \,\,{{1}^{-}}}{\mathop{\lim }}\,f\,(x)=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,f\,(1-h)\] \[=\underset{h\,\,\to \,\,0}{\mathop{\lim }}\,5\,(1-h)-4=1\] \[\therefore \]\[f\,(x)\]is continuous at x = 1. \[\therefore \] \[\underset{x\,\,\to \,\,{{1}^{+}}}{\mathop{\lim }}\,f\,(x)=\underset{x\,\,\to \,\,{{1}^{-}}}{\mathop{\lim }}\,f\,(x)=f\,(1)\] \[\Rightarrow \] \[4+3a=1\] \[\Rightarrow \] \[a=\frac{1-4}{3}=-\,1\] \[\therefore \]For \[a=-\,1\]the given function is continuous at x = 1.
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