Let $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\mathrm{x}}{2}-1,0\le \mathrm{x}<1\\ \frac{1}{2},1\le \mathrm{x}\le 2\end{array}\right.$ and g(x) = (2x + 1) (x - k) +3, 0 ≤ x ∞. Then g(f(x)) is continuous at x - 1 if k is equal to _____ .

$\begin{array}{l}\mathrm{g}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\left\{\begin{array}{c}\mathrm{g}\left(\frac{\mathrm{x}}{2}-1\right),0\le \mathrm{x}<1\\ \mathrm{g}\left(\frac{1}{2}\right),1\le \mathrm{x}\le 2\end{array}\right.\\ =\left\{\begin{array}{c}\frac{(\mathrm{x}-1)(\mathrm{x}-2-2\mathrm{k})}{2}+3,0\le \mathrm{x}<1\\ 4-2\mathrm{k},1\le \mathrm{x}<2\end{array}\right.\end{array}$
$\underset{\mathrm{x}\to 1^{-}}{lim} \mathrm{g}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=3,\mathrm{g}\left(\mathrm{f}\right(1\left)\right)=4-2\mathrm{k}$
and $\underset{\mathrm{x}\to 1^{+}}{lim} \mathrm{g}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=4-2\mathrm{k}$
For g(f(x)) to be continuous at x = 1,4 -2k = 3 or $\mathrm{k}=\frac{1}{2}$.