Question

In a $△ABC,a$ and $b$ are positive integers such that
$\frac{2}{1!9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{2^a}{b!}$
and the distance of a variable point on the curve $x^2-xy+y^2=10$
from the origin is $r$ and $c$ is the maximum value of $r^2$, then $△ABC$ is necessarily scalene.

• A. True
• B. False

Answer 1

The correct option is B False

$\frac{2}{1!9!}+\frac{2}{3!7!}+\frac{1}{5!5!}=\frac{2^a}{b!}$
$\Rightarrow \frac{1}{1!9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!1!}=\frac{2^a}{b!}$
$\Rightarrow \frac{1}{10!}(\frac{10!}{1!9!}+\frac{10!}{3!7!}+\frac{10!}{5!5!}+\frac{10!}{7!3!}+\frac{10!}{9!1!})=\frac{2^a}{b!}$
$\Rightarrow \frac{1}{10!}({10}_{C_1}+{10}_{C_3}+{10}_{C_5}+{10}_{C_7}+{10}_{C_9})=\frac{2^a}{b!}$
$\Rightarrow \frac{2^9}{10!}=\frac{2^a}{b!}$
$\Rightarrow a=9b=10$
$\because x^2-xy+y^2=10$
$\Rightarrow {(x-\frac{y}{2})}^2+\frac{3y^2}{4}=10$
$\Rightarrow \frac{{(x-\frac{y}{2})}^2}{{\left(\sqrt{10}\right)}^2}+\frac{y^2}{{\left(\sqrt{\frac{40}{3}}\right)}^2}=1$ (ellipse)
Centre $(0,0)$ maximum distance of point on the ellipse from origin is $\sqrt{10}$
$\therefore r^2=10=c$
i.e.,$a=9,b=c=10,$ triangle is isosceles.