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(A) For $g(f(x))$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^-} g(f(x)) = \lim_{x 	o 1^+} g(f(x)) = g(f(1))$.First,evaluate the limits o

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(A) For $g(f(x))$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^-} g(f(x)) = \lim_{x 	o 1^+} g(f(x)) = g(f(1))$.First,evaluate the limits of $f(x)$ at $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} \frac{x - 1}{2} = 0$.$\lim_{x 	o 1^+} f(x) = 1/2$.$f(1) = 1/2$.Now,apply the condition for continuity of $g(f(x))$ at $x = 1$:$g(\lim_{x 	o 1^-} f(x)) = g(\lim_{x 	o 1^+} f(x)) = g(f(1))$.$g(0) = g(1/2) = g(1/2)$.Calculate $g(0)$ and $g(1/2)$ using $g(x) = (2x + 1)(x - k) + 3$:$g(0) = (2(0) + 1)(0 - k) + 3 = -k + 3$.$g(1/2) = (2(1/2) + 1)(1/2 - k) + 3 = (1 + 1)(1/2 - k) + 3 = 2(1/2 - k) + 3 = 1 - 2k + 3 = 4 - 2k$.Equating $g(0) = g(1/2)$:$-k + 3 = 4 - 2k$.$2k - k = 4 - 3$.$k = 1$.

If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to:

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