The values of parameter $a$ such that the line $[\log_3(1 + 5a - a^2)]x - 5y - (a^2 - 5) = 0$ is a normal to the curve $xy = 1$, may lie in the interval $(c, d)$ then $c - d$ equals to:
Step-by-Step Solution
Key Concept: For a line to be normal to xy = 1, its slope must equal x² (the normal slope at point (x, 1/x)), and the logarithmic base and argument constraints must be simultaneously satisfied: log₃(1 + 5a - a²) > 0 requires 1 + 5a - a² > 1, giving a(5 - a) > 0.
Given curve $xy = 1$, we find $y = \frac{1}{x}$ and $\frac{dy}{dx} = -\frac{1}{x^2}$. The slope of the normal is $x^2 > 0$. For the normal to be tangent to the given line with slope $\frac{\log_2(1+5a-a^2)}{5} > 0$, we require $\log_2(1+5a-a^2) > 0$, which gives $a^2 - 5a < 0$. Therefore $a \in (0,5)$. Since the tangent line $y=1$ has slope zero, $\tan\theta = 2$ gives $\tan[0] = 2$.
Correct Answer: 5