If a^2-4a-1=0, a ≠ 0, then the value of a^2+3a+1/a^2-3/a is (a) 24 (b) 26 (c) 28 (d) 30

Given

\[
\begin{aligned}
& a^2-4 a-1=0 \\
& a^2-4 a=1 \\
& a(a-4)=1 \\
& a-4=\frac{1}{a} \\
& a-\frac{1}{a}=4
\end{aligned}
\]

We have \(\mathrm{a}^2+3 \mathrm{a}+\frac{1}{\mathrm{a}^2}-\frac{3}{\mathrm{a}}\)
\[
\begin{aligned}
& \left(a^2+\frac{1}{a^2}\right)+3\left(a-\frac{1}{a}\right) \\
& \left(a-\frac{1}{a}\right)^2+3\left(a-\frac{1}{a}\right)+2 \\
& 4^2+3 \times 4+2=30
\end{aligned}
\]

Therefore, Correct option is (d).