Discussion
Let \(f(x) = x^{2} - 10x - 19, x \in \mathbb{R}\) .Then the inverse image of 5, \(f^{-1}(5) =\)
Answer: B
\(f^{-1}(5) = \) \(\{-2, 12\}\).
The inverse image of 5 is the set of all real \(x\) with \(f(x) = 5\):
\[ \begin{aligned} x^{2} - 10x - 19 &= 5 \\ x^{2} - 10x - 24 &= 0 \\ (x - 12)(x + 2) &= 0 \end{aligned} \]
So \(x = 12\) or \(x = -2\). The other sets have a sign wrong on one root (a common slip when factorising \(-24\)), and the empty set is wrong because the quadratic has real roots.
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