Question

If $f\left(x\right)=\left|\begin{array}{ccc}0& x-a& x-b\\ x+a& 0& x-c\\ x+b& x+c& 0\end{array}\right|$, then

(a) f(a) = 0
(b) f(b) = 0
(c) f(0) = 0
(d) f(1) = 0

Answer 1

Let $f\left(x\right)=\left|\begin{array}{ccc}0& x-a& x-b\\ x+a& 0& x-c\\ x+b& x+c& 0\end{array}\right|$.

Now,
$$\begin{gathered} f\left(a\right)=\left|\begin{array}{ccc}0& a-a& a-b\\ a+a& 0& a-c\\ a+b& a+c& 0\end{array}\right| \\ =\left|\begin{array}{ccc}0& 0& a-b\\ 2a& 0& a-c\\ a+b& a+c& 0\end{array}\right| \\ =\left(a-b\right)\left(2a^2+2ac\right)\ne 0 \\ f\left(b\right)=\left|\begin{array}{ccc}0& b-a& b-b\\ b+a& 0& b-c\\ b+b& b+c& 0\end{array}\right| \\ =\left|\begin{array}{ccc}0& b-a& 0\\ b+a& 0& b-c\\ 2a& b+c& 0\end{array}\right| \\ =\left(b-a\right)\left(2ab-2ac\right)\ne 0 \\ f\left(0\right)=\left|\begin{array}{ccc}0& 0-a& 0-b\\ 0+a& 0& 0-c\\ 0+b& 0+c& 0\end{array}\right| \\ =\left|\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right| \\ =a\left(bc\right)-b\left(ac\right)=0 \end{gathered}$$

Hence, the correct option is (c).