Q60 / 160 · WBJEE 2015 · Mathematics
$$
\text { If } f(x)=\begin{array}{|ccc|}
1 & x & x+1\\
2 x & x(x-1) & (x+1)x\\
3 x(x-1) & x(x-1)(x-2) & (x+1)x(x-1)
\end{array}
$$ Then, $f(100)$ is equal to
- Correct answerA0
- B1
- C100
- D10
Solution
We have, $f(x)= \begin{array}{|ccc|}1 & x &x+1 \\ 2 x & x(x-1) & (x+1)x\\ 3 x(x-1) & x(x-1)(x-2) &(x+1)x(x-1)\end{array}$
Taking common $x,(x-1)$ from $R_{2}$ and $R_{3}$ respectively, we ge $f(x)=x(x-1)\left|\begin{array}{ccc}1 & x & x+1 \\ 2 & x-1 & x+1 \\ 3 x & x(x-2) & (x+1) x\end{array}\right|$
Taking commen $(x+1)$ from $C_{2}$, we get
$f(x)=x(x-1)(x+1)\left|\begin{array}{ccc}1 & x & 1 \\ 2 & x-1 & 1 \\ 3 x & x(x-2) & x\end{array}\right|$
Taking common $x$ from $R_{3}$, we get $f(x)=x^{2}(x-1)(x+1)\left|\begin{array}{ccc}1 & x & 1 \\ 2 & x-1 & 1 \\ 3 & x-2 & 1\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-R_{2}, R_{2}$ we get
$f(x)=x^2(x-1)(x+1)\left|\begin{array}{ccc}-1 & 1 & 0 \\ -1 & 1 & 0 \\ 3 & (x-2) & 1\end{array}\right|$
$\Rightarrow f(100)=0$
Hence, the correct answer is option: 'A'.