Question

If $\alpha = \lim_{x \to 0^+} \left( \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}} \right)$ $\beta = \lim_{x \to 0} (1 + \sin x)^{\frac{1}{2\cot x}}$ are the roots of the quadratic equation $ax^2 + bx - \sqrt{e} = 0$, then $12 \log_e (a + b)$ is equal to _________.

Answer 1

Given: The given expression for $\alpha$ is:

$\alpha = \lim_{x \to 0^+} \frac{\sqrt{x} \left( e^{\sqrt{\tan x} - \sqrt{x}} - 1 \right)}{\sqrt{\tan x} - \sqrt{x}}.$

As $x \to 0^+$, we observe that $\sqrt{\tan x} \to \sqrt{x}$. The expression simplifies as the numerator and denominator converge to zero:

$\alpha = 1.$

The given expression for $\beta$ is:

$\beta = \lim_{x \to 0} \left( 1 + \sin x \right)^{\frac{1}{2} \cot x}.$

Using the approximation $\sin x \approx x$ and $\cot x \approx \frac{1}{x}$ as $x \to 0$, we rewrite the expression:

$\beta = \left( 1 + x \right)^{\frac{1}{2x}}.$

Using the standard limit $\lim_{x \to 0} \left( 1 + x \right)^{\frac{1}{x}} = e$, we get:

$\beta = e^{\frac{1}{2}}.$

The quadratic equation is:

$x^2 - \left( 1 + \sqrt{e} \right) x + \sqrt{e} = 0.$

Compare with the general quadratic form:

$ax^2 + bx + c = 0.$

From the given equation:

$a = 1, \quad b = -(1 + \sqrt{e}), \quad c = \sqrt{e}.$

Substitute the values of $a$ and $b$:

$a + b = 1 - (1 + \sqrt{e}) = -\sqrt{e}.$

$\ln(a + b) = \ln(-\sqrt{e}) = \ln(\sqrt{e}) + \ln(-1) = \frac{1}{2} \ln(e).$

Finally, calculate:

$12 \ln(a + b) = 12 \times \frac{1}{2} = 6.$

Answer: 6