Question

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

• A. f (1)
• B. f (3)
• C. f (1) + f (3)
• D. f (1) + f (5)

Answer 1

The correct option is B f (3)

$f\left(x\right)=2(x-3)(x-5);\left|\begin{array}{ccc}1& x+3& 3(x^2+3x+9)\\ 1& x+5& 4(x^2+5x+25)\\ 1& 1& 3\end{array}\right|$
(Taking out (x-3),(x-5) and 2 from $I^{st}$ row, $I{I}^{nd}$ row and $II{I}^{rd}$ column respectively)
$f\left(x\right)=2(x-3)(x-5)\left|\begin{array}{ccc}0& (x+2)& 3(x^2+3x+8)\\ 0& 2& x^2+11x+73\\ 1& 1& 3\end{array}\right|,(R_1\to R_1-R_{3}and{R}_2\to R_2-R_1)=2(x-3)(x-5)\left[1\right(x+2\left)\right(x^2+11x+73)-6(x^2+3x+8\left)\right]=2(x^2-8x+15)(x^3+13x^2+95x+146-6x^2-18x-48)=2(x^2-8x+15)(x^3+7x^2+77x+98)=2(x^5-x^4+36x^3-413x^2+371x+1470)f\left(1\right)=2928,f\left(3\right)=0,f\left(5\right)=0\therefore f\left(1\right).f\left(3\right)+f\left(3\right).f\left(5\right)+f\left(5\right).f\left(1\right)=0+0+0=0=f\left(3\right).$