Q62 / 180 · KCET 2010 · Mathematics
The constant term of the polynomial $\left|\begin{array}{ccc}x+3 & x & x+2 \\ x & x+1 & x-1 \\ x+2 & 2 x & 3 x+1\end{array}\right|$ is
- A0
- B2
- Correct answerC$-1$
- D1
Solution
$$
\left|\begin{array}{ccc}
x+3 & x & x+2 \\
x & x+1 & x-1 \\
x+2 & 2 x & 3 x+1
\end{array}\right|=f(x)
$$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$
$$
f(x)=\left|\begin{array}{ccc}
x+3 & x & x+2 \\
-3 & 1 & -3 \\
-1 & x & 2 x-1
\end{array}\right|
$$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}$ and $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$
$$
f(x)=\left|\begin{array}{ccc}
1 & -2 & x+2 \\
0 & 4 & -3 \\
-2 x & 1-x & 2 x-1
\end{array}\right|
$$
Expand w.r.t. ' $\mathrm{C}_{1}$,
$$
f(x)=[4(2 x-1)+3(1-x)]
$$
$$
+(-2 \mathrm{x})[6-4(\mathrm{x}+2)]
$$
$f(x)=[8 x-4+3-3 x]+[-2 x][-4 x-2]$
$f(x)=(5 x-1)+\left(8 x^{2}+4 x\right)$
$$
f(x)=8 x^{2}+9 x-1
$$
Hence, the constant term of quadratic equation is $-1$.
Hence, the correct answer is option: 'C'.