Question

If the polynomial $f\left(x\right)=\left|\begin{array}{ccc}(1+x{)}^a& (2+x{)}^b& 1\\ 1& (1+x{)}^a& (2+x{)}^b\\ (2+x{)}^b& 1& (1+x{)}^a\end{array}\right|,$ then the constant term of $f\left(x\right)$ is

• A. $2-{3.2}^b+2^{3b}$
• B. $2+{3.2}^b+2^{3b}$
• C. $2+{3.2}^b-2^{3b}$
• D. $2-{3.2}^b–2^{3b}$

Answer 1

The correct option is A $2-{3.2}^b+2^{3b}$

$f\left(x\right)=\left|\begin{array}{ccc}(1+x{)}^a& (2+x{)}^b& 1\\ 1& (1+x{)}^a& (2+x{)}^b\\ (2+x{)}^b& 1& (1+x{)}^a\end{array}\right|$

For constant term put, $x=0$
$f\left(0\right)=\left|\begin{array}{ccc}1& 2^b& 1\\ 1& 1& 2^b\\ 2^b& 1& 1\end{array}\right|$
$C_1\to C_1-C_2$
$=\left|\begin{array}{ccc}1-2^b& 2^b& 1\\ 0& 1& 2^b\\ 2^b-1& 1& 1\end{array}\right|$
$=(1-2^b)(1-2^b)+(2^b-1)(2^{2b}-1)$
$=2-3.2^b+2^{3b}$