(d) 4
If one zero is $\alpha$ and the other is $\dfrac{1}{\alpha}$, then product of zeroes $= \alpha \cdot \dfrac{1}{\alpha} = 1$.
For $6x^2 + 37x - (k-2)$: product of zeroes $= \dfrac{-(k-2)}{6} = 1 \Rightarrow -(k-2) = 6 \Rightarrow k - 2 = -6 \Rightarrow k = -4$.
Wait — rechecking: $\dfrac{-(k-2)}{6}=1 \Rightarrow k-2=-6 \Rightarrow k=-4$.
(a) −4
Source: Chapter 2, Section 2.3
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When one zero is the reciprocal of the other, their product = 1. Use the formula: product of zeroes $= \dfrac{c}{a} = \dfrac{-(k-2)}{6}$. Set this equal to 1 and solve for $k$. The constant term here is $-(k-2)$, so be careful with the sign. This gives $k = -4$, option (a).