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

Correct Answer: 6

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