Q45 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics
$1 \mathrm{f}\left|\begin{array}{ccc}x^2+3 x & x+1 & x-3 \\ x-1 & 2-x & x+4 \\ x-3 & x-3 & 3 x\end{array}\right|=a_0$ $+a_3 x^3+a_4 x^4$, then $\left(a_1+a_3\right)+2\left(a_0+a_2+a_4\right)=$
- Correct answerA-1
- B0
- C1
- D-29
Solution
We have,
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x^2+3 x & x+1 & x-3 \\
x-1 & 2-x & x+4 \\
x-3 & x-3 & 3 x
\end{array}\right| \\
= & a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4
\end{aligned}
$$
Put $x=1$ both sides, we get
$$
\begin{aligned}
& \left|\begin{array}{ccc}
4 & 2 & -2 \\
0 & 1 & 5 \\
-2 & -2 & 3
\end{array}\right|=a_0+a_2+a_2+a_3+a_4 \\
& 4(3+10)-\not 10+Z=a_0+a_2+a_2+a_3+a_4 \\
& 52-24=a_0+a_1+a_2+a_3+a_4 \\
\Rightarrow & a_0+a_1+a_2+a_3+a_4=28
\end{aligned}
$$
Put $x=-1$ both sides, we get
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-2 & 0 & -4 \\
-2 & 3 & 3 \\
-4 & -4 & -3
\end{array}\right|=a_0-a_2+a_2-a_3+a_4 \\
& -2\left(-9+12-4(8+12)=a_0-a_2+a_2-a_3+a_4\right. \\
& -6-80=a_0-a_2+a_2-a_3+a_4 \\
\Rightarrow & a_0-a_2+a_2-a_3+a_4=-86
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
2\left(a_0+a_2+a_4\right)=-58
$$
On subtracting Eq. (ii) from Eq. (i), we get
$$
\begin{array}{cc}
& \mathcal{2}\left(a_1+a_3\right)=28+86=114 \\
\Rightarrow & a_1+a_3=57 \\
\therefore & \left(a_2+a_3\right)+\mathcal{Z}\left(a_0+a_2+a_4\right)=57-58=-1
\end{array}
$$
Hence, the correct answer is option: 'A'.