Question

The values of parameter ${}^{\prime }{a}^{\prime }$ such that the line $\left({\log }_2(1+5a-a^2)\right)x-5y-(a^2-5)=0$ is a normal to the curve $xy=1$, may lie in the interval

• A. $(-\infty ,0)$
• B. $(0,5)$
• C. $(5,10)$
• D. $(10,\infty )$

Answer 1

The correct option is B $(0,5)$

$xy=1$

$\Rightarrow y=\frac{1}{x}$

$\Rightarrow \frac{dy}{dx}=-\frac{1}{x^2}$

$\Rightarrow \frac{-dx}{dy}=x^2\ge 0$ (slope of normal)

L:-$\left({\log }_2(1+5a-a^2)\right)x-5y-(a^2-5)=0$

$slope=-\frac{{\log }_2(1+5a+a^2)}{-5}$

$\therefore \frac{{\log }_2(1+5a+a^2)}{5}\ge 0$

$\Rightarrow {\log }_2(1+5a+a^2)\ge 0$

$\Rightarrow 1+5a+a^2\ge 2^0$

$\Rightarrow 5a-a^2\ge 0$

$\Rightarrow a^2-5a\le 0$

$a(a-5)\le 0$

$aϵ(0,5)$