Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Here,
$f\left(x\right)=\left|\begin{array}{ccc}(1+x{)}^a& (1+2x{)}^b& 1\\ 1& (1+x{)}^2& (1+2x{)}^b\\ (1+2x{)}^b& 1& (1+x{)}^a\end{array}\right|=A+Bx+C{x}^2+\dots$ {i}
Putting $x=0$, we get
$f\left(0\right)=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|=A+B\left(0\right)+C(0{)}^2+....$
$\Rightarrow A=0$
Differentiating Eq. (i} w.r.t. $x$, we get
$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}a(1+x{)}^{a-1}& 2b(1+2x{)}^{b-1}& 0\\ 1& (1+x{)}^a& (1+2x{)}^b\\ (1+2x{)}^b& 1& (1+x{)}^a\end{array}\right|+\left|\begin{array}{ccc}(1+x{)}^a& (1+2x{)}^b& 1\\ 0& a(1+x{)}^{a-1}& 2b(1+2x{)}^{b-1}\\ (1+2x{)}^b& 1& (1+x{)}^a\end{array}\right|+\left|\begin{array}{ccc}(1+x{)}^a& (1+2x{)}^b& 1\\ 1& (1+x{)}^a& (1+2x{)}^b\\ 2b(1+2x{)}^{b-1}& 0& a(1+x{)}^{a-1}\end{array}\right|=B+2Cx+.....$
$f^{\prime }\left(0\right)=\left|\begin{array}{ccc}a& 2b& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ 0& a& 2b\\ 1& 1& 1\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 2b& 0& a\end{array}\right|=B$

$\Rightarrow B=0$ .....(ii)
$\therefore$ Constant term $+$ coefficient of $x$ $=0$.