Question

$f\left(x\right)=\left|\begin{array}{ccc}x+c_1& x+a& x+a\\ x+b& x+c_2& x+a\\ x+b& x+b& x+c_3\end{array}\right|$ and $g\left(x\right)=(c_1-x)(c_2-x)(c_3-x)$

Which of the following is not a constant term in $f\left(x\right)$?

• A. $\frac{bg\left(a\right)-ag\left(b\right)}{b-a}$
• B. $\frac{bg\left(a\right)-af(-b)}{b-a}$
• C. $\frac{bf(-a)-ag\left(b\right)}{b-a}$
• D. none of these

Answer 1

The correct option is A $\frac{bg\left(a\right)-ag\left(b\right)}{b-a}$

Given,$g\left(x\right)=(c_1-x)(c_2-x)(c_3-x)$.
$g\left(a\right)=(c_1-a)(c_2-a)(c_3-a)$ ....(i)
$g\left(b\right)=(c_1-b)(c_2-b)(c_3-b)$ ....(ii)
$f\left(x\right)=\left|\begin{array}{ccc}x+c_1& x+a& x+a\\ x+b& x+c_2& x+a\\ x+b& x+b& x+c_3\end{array}\right|$
Clearly,$f\left(x\right)$ is linear in $x$.
Let $f\left(x\right)=\alpha x+\beta$
$f(-a)=-a\alpha +\beta$
and $f(-b)=-b\alpha +\beta$
$\Rightarrow \beta =\frac{bf(-a)-af(-b)}{b-a}$ ....(iii)
Now, $f(-a)=\left|\begin{array}{ccc}{c}_1-a& 0& 0\\ b-a& c_2-a& 0\\ b-a& b-a& c_3-a\end{array}\right|$
$\Rightarrow f(-a)=(c_1-a)(c_2-a)(c_3-a)$
$\Rightarrow f(-a)=g\left(a\right)$ (using (i)) .....(iv)
$f(-b)=\left|\begin{array}{ccc}{c}_1-b& a-b& a-b\\ 0& c_2-b& a-b\\ 0& 0& c_3-a\end{array}\right|$
$\Rightarrow f(-b)=(c_1-b)(c_2-b)(c_3-b)$
$\Rightarrow f(-b)=g\left(b\right)$ .....(v)
Put this value in (iii), we get
$\beta =\frac{bg\left(a\right)-ag\left(b\right)}{b-a}$