If `f(x)=|0x-a x-b x+a0x-c x+b x+c0|,t h e n` `f(x)=0` (b) `f(b)=0` (c) `f(0)=0` `f(1)=0`

Explanation

Correct answer is a `f(x)=0`

$$f(x)=\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|$$
$$\Rightarrow f(a)=\left|\begin{array}{ccc}0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0\end{array}\right|$$
$$[(\mathrm{a}-\mathrm{b})\{2 \mathrm{a} .(\mathrm{a}+\mathrm{v})\}] \equiv 0$$
$$\therefore f(b)\left|\begin{array}{ccc}0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 b & b+c & 0\end{array}\right|$$
$$=-(\mathrm{b}-\mathrm{a})[2 \mathrm{~b}(\mathrm{~b}-\mathrm{c})]$$
$$=-2 b(b-a)(b-c) \equiv 0$$
$$\therefore f(0)=\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|$$
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