Question

The graph of the function $f(x)=x^{10}+9 x^9+7 x^8+\ldots \ldots \ldots . . .+a_1 x+a_0$ intersect the line $y=b$ at the points $B _1, B _2, B _3, \ldots \ldots, B _{10}$ (from left to right) and the line $y = c$ at the points $C _1, C _2, C _3, \ldots \ldots$, $C _{10}$ (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 $\displaystyle\sum_{n=1}^{10} \cot \left(\angle B_n C_n P\right)$.

Answer 1

Let $b_1, b_2, \ldots \ldots . . b_{10}$ be the roots of $f(x)=b (b=5)$
$\therefore b_1+b_2+\ldots \ldots+b_{10}=-9 $ (Theory of equation)
and $c _1, c _2, \ldots \ldots . c _{10}$ be the roots of $f ( x )= c ( c =3)$
$\therefore c_1+c_2+\ldots \ldots . .+c_{10}=-9$ 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 $\displaystyle\sum_{n=1}^{10} \cot \left(B_n C_n P\right)=\frac{\left(b_1-c_1\right)+\left(b_2-c_2\right)+\ldots \ldots .+\left(b_{10}-c_{10}\right)}{b-c}$
$=\frac{\left( b _1+ b _2+\ldots \ldots \ldots+ b _{10}\right)+\left( c _1+ c _2+\ldots \ldots \ldots+ c _2\right)}{ b - c }=\frac{(-9)-(-9)}{5-3}=0$