$f(x)=x^{2}-2 x$ $\therefore f^{\prime}(x)=2 x-2$ $f(x)$ is strictly decreasing, when $f^{\prime}(x) < 0$ $\quad f^{\prime}(x) < 0$ $\Rightarrow \quad 2(x-1) < 0$ $\Rightarrow \quad x < 1$ Hence, $f(x)$ is strictly decreasing in the interval $(-\infty, 1) .$ $(-\infty, 1)$.