AP EAMCET · Maths · Determinants
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
- A \(f(1)\)
- B \(f(3)\)
- C \(f(1)+f(3)\)
- D \(f(1)+f(5)\)
Answer & Solution
Correct Answer
(B) \(f(3)\)
Step-by-step Solution
Detailed explanation
\(f(x)=(x-3)(x-5)\left|\begin{array}{ccc}1 & 2(x+3) & 3\left(x^2+9+3 x\right) \\ 1 & 2(x+5) & 4\left(x^2+25+5 x\right) \\ 1 & 2 & 3\end{array}\right|\) \(\Rightarrow \quad f(3)=f(5)=0\) So, \(f(1) f(3)+f(3) \cdot f(5)+f(5) f(1)=0=f(3)\)
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