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

Answer 1

The correct option is C

$f\left(50\right)=1$

Explanation for Correct answer:

Finding the value:

Given,

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

Applying row operation $R_1\to R_1-2R_2+R_3$

$$\begin{gathered} f\left(x\right)=\left|\begin{array}{ccc}\left(x+a\right)-2\left(x+b\right)+\left(x+c\right)& \left(x+2\right)-2\left(x+3\right)+\left(x+4\right)& \left(x+1\right)-2\left(x+2\right)+\left(x+3\right)\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right| \\ =\left|\begin{array}{ccc}a-2b+c& 0& 0\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right| \\ =\left|\begin{array}{ccc}1& 0& 0\\ x+b& x+3& x+2\\ x+c& x+4& x+3\end{array}\right|\left[\because \mathrm{Given}a-2b+c=1\right] \\ f\left(x\right)=1\left[\left(x+3\right)\left(x+3\right)-\left(x+4\right)\left(x+2\right)\right] \\ =1\left[x^2+6x+9-x^2+6x-8\right] \\ =1 \end{gathered}$$

$f\left(x\right)=1$

Therefore $f\left(50\right)=1$

Hence, option (C) is the correct answer.