Given \(f(x) = x^2 - 1\) and \(g(x)= 1 - 2x\)

Now
\(g(k) = 1 - 2k\)
\(f(g(k)) = (1 - 2k)^2 - 1\)

But \(f(g(k)) = 3\)

so,
\(3 = (1 - 2k)^2 - 1\)

or \(4k^2 - 4k - 3 = 0\)

or \(4k^2 -6k + 2k - 3 = 0\)

or \(k^2 - \frac{3k}{2} +\frac{k}{2} - \frac{3}{4} = 0\)

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


we get \(k = -\frac{1}{2} or \frac{3}{2}\)

Only Option D is correct