KCET · Maths · Determinants
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
- A \(-2183328\)
- B \(2183328\)
- C \(-3183328\)
- D \(3183328\)
Answer & Solution
Correct Answer
(A) \(-2183328\)
Step-by-step Solution
Detailed explanation
\(f(x)=2(x-5)\left|\begin{array}{ccc}x-3 & x^2-9 & 2 x^3-81 \\ 1 & x+5 & 4\left(x^2+5 x+25\right) \\ 1 & 1 & 3\end{array}\right|\)
\(\Rightarrow f(5)=0\)
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, \(f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)\)
\(=2888 \times(-756)+0+0=-2183328\)
\(\Rightarrow f(5)=0\)
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, \(f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)\)
\(=2888 \times(-756)+0+0=-2183328\)
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