$\text{ Let }f\left(x\right)=\left\{\begin{array}{lll}\frac{x}{2}-1,& 0\le x<1& \\ \frac{1}{2} ,& 1\le x\le 2& \mathrm{and} g\left(x\right)=(2x+1)(x-k)+3,0\le x<\infty \text{ then }g\left(f\right(x\left)\right)\text{ is continuous at x=1 if 19k=}\end{array}\right.$

$\text{ g }f\left(x\right)=\left\{\begin{array}{ll}g\left(\frac{x}{2}-1\right),& 0\le x<1\\ g\left(\frac{1}{2}\right),& 1\le x\le 2\end{array}=\left\{\begin{array}{ll}\frac{(x-1)(x-2-2k)}{2}+3& ,0\le x<1\\ 4-2k& ,1\le x<2\end{array}\right.\right.$

$\underset{x\to 1-}{Lt}gf\left(x\right)=3,g\left(f\left(1\right)\right)=4-2k\text{ and }\underset{x\to 1+}{Lt}g\left(f\left(x\right)\right)=4-2k$

$\begin{array}{l}\text{ For }g\left(f\right(x\left)\right)\text{ to be continuous at }x=1,4-2k=3\text{ or }k=\frac{1}{2}\\ 19\mathrm{k}=9.5\end{array}$