Question

If $f\left(x\right)=\left|\begin{array}{ccc}{(1+x)}^a& {(1+2x)}^b& 1\\ 1& {(1+x)}^a& {(1+2x)}^b\\ {(1+bx)}^b& 1& {(1+x)}^a\end{array}\right|$, then find
(i) constant term
(ii) coefficient of $x$

Answer 1

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

for constant term put $x=0⟹f\left(0\right)=\left|\begin{array}{ccc}(1+0{)}^a& (1+2(0){)}^b& 1\\ 1& (1+0{)}^a& (1+2(0){)}^b\\ (1+b(0){)}^b& 1& (1+0{)}^a\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|=0$

for coefficient of $x$ differentiate $f\left(x\right)$ and put $x=0$$f^1\left(x\right)=\left|\begin{array}{ccc}a(1+x{)}^{a-1}& 2b(1+2x{)}^{b-1}& 0\\ 1& (1+x{)}^a& (1+2x{)}^b\\ (1+bx{)}^b& 1& (1+x{)}^a\end{array}\right|+\left|\begin{array}{ccc}(1+x{)}^a& (1+2x{)}^b& 1\\ 0& a(1+x{)}^{a-1}& 2b(1+2x{)}^{b-1}\\ (1+bx{)}^b& 1& (1+x{)}^a\end{array}\right|+\left|\begin{array}{ccc}(1+x{)}^a& (1+2x{)}^b& 1\\ 1& (1+x{)}^a& (1+2x{)}^b\\ b^2(1+bx{)}^{b-1}& 0& a(1+x{)}^{a-1}\end{array}\right|$

$f^1\left(0\right)=\left|\begin{array}{ccc}a& 2b& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ 0& a& 2b\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ b^2& 0& a\end{array}\right|=0$

constant term $=0$

coefficient of $x=0$