Question

Let $a-2b+c=1$, If $f\left(x\right)=\left|\begin{array}{ccc}x+a& x+2& x+1\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right|,$ then:

• A. $f(-50)=501$
• B. $f(-50)=-1$
• C. $f\left(50\right)=1$
• D. $f\left(50\right)=-501$

Answer 1

The correct option is C $f\left(50\right)=1$

Given $f\left(x\right)=\left|\begin{array}{ccc}x+a& x+2& x+1\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right|,$
$a-2b+c=1$
Applying $R_1\to R_1-2R_2+R_3$
$f\left(x\right)=\left|\begin{array}{ccc}a-2b+c& 0& 0\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right|$
Using $a-2b+c=1$
$\because f\left(x\right)=(x+3{)}^2-(x+2\left)\right(x+4)$
$\Rightarrow f\left(x\right)=(x+3{)}^2-(x+2\left)\right(x+4)$
$\Rightarrow f\left(x\right)=1$
$\Rightarrow f\left(50\right)=1$
$\Rightarrow f(-50)=1$

Alternative:
Take $a=1,b=0=c$ and proceed