Q84 / 180 · KCET 2024 · Mathematics
If $f(x)=\left|\begin{array}{ccc}x-3 & 2 x^2-18 & 2 x^3-81 \\ x-5 & 2 x^2-50 & 4 x^3-500 \\ 1 & 2 & 3\end{array}\right|$ then
$f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)$ is
- Correct answerA$-2183328$
- B$2183328$
- C$-3183328$
- D$3183328$
Solution
$f(x)=2(x-5)\left|\begin{array}{ccc}x-3 & x^2-9 & 2 x^3-81 \\ 1 & x+5 & 4\left(x^2+5 x+25\right) \\ 1 & 1 & 3\end{array}\right|$
$\Rightarrow f(5)=0$
and $f(1)=2(-4)\left[\begin{array}{ccc}-2 & -8 & -79 \\ 1 & 6 & 124 \\ 1 & 1 & 3\end{array}\right]$
$=-8[212-968+395]=2888$
$f(3)=2(-2)\left[\begin{array}{ccc}0 & 0 & -27 \\ 1 & 8 & 196 \\ 1 & 1 & 3\end{array}\right]$
$=-4[0+0-27(1-8)]=-756$
So, $f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)$
$=2888 \times(-756)+0+0=-2183328$
Hence, the correct answer is option: 'A'.