Let $f(x)=\left\{\begin{matrix}\frac{x}{2}-1&;0≤x≤1\\\frac{1}{2}&;1<x≤2\end{matrix}\right.$ & $g(x) = (2x + 1) (x – k) + 3 ; 0 ≤ x < ∞$ then g(f(x)) is continuous at x = 1, if k equals

$\frac{1}{2}$

$g (f(x)) = (2f(x) + 1) (f(x) – k) + 3$

$\left\{\begin{matrix}(2(\frac{x}{2}-1)+1)(\frac{x}{2}-1-k)+3;&0≤x≤1\\(2×\frac{1}{2}+1)(\frac{1}{2}-k)+3;&1<x≤2\end{matrix}\right.$

∴ g(f(x)) is continuous at x = 1 ∴ f(1) = R.H.L.

$0 + 3 = 2(\frac{1}{2}-k)+3⇒k=\frac{1}{2}$