Ans: B
$\begin{array}{cl}
& f(1 + \beta) – f(1 – \beta) \\
= & [3(1 + \beta)^2 – (1+\beta) – 2] – [3(1 – \beta)^2 – (1-\beta) – 2] \\
= & (3 + 6\beta + 3\beta^2 – 1 – \beta – 2) – (3 – 6\beta + 3\beta^2 – 1 + \beta – 2) \\
= & 10 \beta
\end{array}$
$\begin{array}{cl}
& f(1 + \beta) – f(1 – \beta) \\
= & [3(1 + \beta)^2 – (1+\beta) – 2] – [3(1 – \beta)^2 – (1-\beta) – 2] \\
= & (3 + 6\beta + 3\beta^2 – 1 – \beta – 2) – (3 – 6\beta + 3\beta^2 – 1 + \beta – 2) \\
= & 10 \beta
\end{array}$
