Question

If $f \left(x\right)=\begin{vmatrix}\left(1+x\right)^{a}&\left(1+2x\right)^{b}&1\\ 1&\left(1+x\right)^{a}&\left(1+2x\right)^{b}\\ \left(1+2x\right)^{b}&1&\left(1+x\right)^{a}\end{vmatrix} $ $a, b$ being positive integers, then

• A. Constant term or $f(x)$ is
• B. Coefficient of $x$ in $f(x)$ is 0
• C. Constant term in $f(x)$ is $a-b$
• D. Coefficient of $x$ in $f(x)$ is -1

Answer 1

Answer: (B)

let $\begin{vmatrix}\left(1+x\right)^{a}&\left(1+2x\right)^{b}&1\\ 1&\left(1+x\right)^{a}&\left(1+2x\right)^{b}\\ \left(1+2x\right)^{b}&1&\left(1+x\right)^{a}\end{vmatrix} =a_{0}+a_{1} x+a_{2} x^{2}\ldots$
Putting x = 0, we get
$a_{0}=\begin{vmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{vmatrix}=0 $
Now differentiating both sides with respect to x and puttingx = 0, we get $a_{1}=\begin{vmatrix}a&2b&0\\ 1&1&1\\ 1&1&1\end{vmatrix}+\begin{vmatrix}1&1&1\\ 0&a&2b\\ 1&1&1\end{vmatrix}+\begin{vmatrix}1&1&1\\ 1&1&1\\ 2b&0&a\end{vmatrix}=0$
Hence, coefficient of $x$ is $0$