Question

The graph of the function $f(x)=x^{10}+9x^9+7x^8+\dots \dots \dots ...+a_{1}x+a_0$ intersect the line $y=b$ at the points $B_1,B_2,B_3,\dots \dots ,B_{10}$ (from left to right) and the line $y=c$ at the points $C_1,C_2,C_3,\dots \dots$, $C_{10}$ (from left to right). Let $P$ be a point on the line $y=c$ to the right of the point $C_{10}$. If $b=5$ and $c=3$, then find the sum $\sum _{n=1}^{10}\cot \left(\angle B_n{C}_{n}P\right)$.

Answer 1

Answer: 0

Let $b_1,b_2,\dots \dots ..b_{10}$ be the roots of $f(x)=b(b=5)$
$\therefore b_1+b_2+\dots \dots +b_{10}=-9$ (Theory of equation)
and $c_1,c_2,\dots \dots .c_{10}$ be the roots of $f(x)=c(c=3)$
$\therefore c_1+c_2+\dots \dots ..+c_{10}=-9$ then $\cot \left(B_1{C}_{1}P\right)=\frac{\left(b_1-c_1\right)}{(b-c)};\cot \left(B_2{C}_{2}P\right)=\frac{\left(b_2-c_2\right)}{(b-c)}$, & so on
Hence $\sum _{n=1}^{10}\cot \left(B_n{C}_{n}P\right)=\frac{\left(b_1-c_1\right)+\left(b_2-c_2\right)+\dots \dots .+\left(b_{10}-c_{10}\right)}{b-c}$
$=\frac{\left(b_1+b_2+\dots \dots \dots +b_{10}\right)+\left(c_1+c_2+\dots \dots \dots +c_2\right)}{b-c}=\frac{(-9)-(-9)}{5-3}=0$