Solution :
\(\because\) Point (k-1, k) lies inside the parabola \(y^2\) = 4x.
\(\therefore\) \({y_1}^2 – 4ax_1\) < 0
\(\implies\) \(k^2\) – 4(k-1) < 0
\(\implies\) \(k^2\) – 4k + 4 < 0
\((k-2)^2\) < 0 \(\implies\) k \(\in\) \(\phi\)
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