If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^2-18 & 3 x^3-81 \\ x-5 & 2 x^2-50 & 4 x^3-500 \\ 1 & 2 &…

If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^2-18 & 3 x^3-81 \\ x-5 & 2 x^2-50 & 4 x^3-500 \\ 1 & 2 & 3\end{array}\right|\), then \(f(\mathrm{l}) f(3)+f(3) f(5)+f(5) f(\mathrm{l})\) is equal to
  1. \(f(1)\)
  2. \(f(3)\)
  3. \(f(1)+f(3)\)
  4. \(f(1)+f(5)\)

Solution

\(f(x)=(x-3)(x-5)\left|\begin{array}{ccc}1 & 2(x+3) & 3\left(x^2+9+3 x\right) \\ 1 & 2(x+5) & 4\left(x^2+25+5 x\right) \\ 1 & 2 & 3\end{array}\right|\) \(\Rightarrow \quad f(3)=f(5)=0\) So, \(f(1) f(3)+f(3) \cdot f(5)+f(5) f(1)=0=f(3)\)

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

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