Question:
Let $a-2 b+c=1$
If $f(x)=\left|\begin{array}{lll}x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3\end{array}\right|$, then :
Correct Option: , 3
Solution:
$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}$
$f(x)=\left|\begin{array}{ccc}a+c-2 b & 0 & 0 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3\end{array}\right|$
$=(a+c-2 b)\left((x+3)^{2}-(x+2)(x+4)\right)$
$=x^{2}+6 x+9-x^{2}-6 x-8=1$
$\Rightarrow f(x)=1 \Rightarrow f(50)=1$