MCQ
If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
- Af(a) = 0
- Bf(b) = 0
- ✓f(0) = 0
- Df(1) = 0
✓
Answer
Correct option: C.
f(0) = 0
Let $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
Now, $\text{f(a)}=\begin{vmatrix}0&\text{a}-\text{a}&\text{a}-\text{b}\\\text{a}+\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&0&\text{a}-\text{b}\\2\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=(\text{a}-\text{b})(2\text{a}^2+2\text{ac})\neq0$
$\text{f(b)}=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}-\text{b}\\\text{b}+\text{a}&0&\text{b}-\text{c}\\\text{b}+\text{b}&\text{b}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&\text{b}-\text{a}&0\\\text{b}+\text{a}&0&\text{b}-\text{c}\\2\text{a}&\text{b}+\text{c}&0\end{vmatrix}$
$=(\text{b}-\text{a})(2\text{ab}-2\text{ac})\neq0$
$\text{f(0)}=\begin{vmatrix}0&\text{0}-\text{a}&\text{0}-\text{b}\\\text{0}+\text{a}&0&\text{0}-\text{c}\\\text{0}+\text{b}&\text{0}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&\text{c}\\\text{b}&\text{c}&0\end{vmatrix}$
$=\text{a}(\text{bc})-\text{b}(\text{ac})=0$
Hence, the correct option is (c)
Now, $\text{f(a)}=\begin{vmatrix}0&\text{a}-\text{a}&\text{a}-\text{b}\\\text{a}+\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&0&\text{a}-\text{b}\\2\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$
$=(\text{a}-\text{b})(2\text{a}^2+2\text{ac})\neq0$
$\text{f(b)}=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}-\text{b}\\\text{b}+\text{a}&0&\text{b}-\text{c}\\\text{b}+\text{b}&\text{b}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&\text{b}-\text{a}&0\\\text{b}+\text{a}&0&\text{b}-\text{c}\\2\text{a}&\text{b}+\text{c}&0\end{vmatrix}$
$=(\text{b}-\text{a})(2\text{ab}-2\text{ac})\neq0$
$\text{f(0)}=\begin{vmatrix}0&\text{0}-\text{a}&\text{0}-\text{b}\\\text{0}+\text{a}&0&\text{0}-\text{c}\\\text{0}+\text{b}&\text{0}+\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&\text{c}\\\text{b}&\text{c}&0\end{vmatrix}$
$=\text{a}(\text{bc})-\text{b}(\text{ac})=0$
Hence, the correct option is (c)
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