Question

Let \( f: (0, \pi) \to \mathbb{R} \) be a function given by
\[ f(x) = \begin{cases} \left(\frac{8}{7}\right)^{\tan 8x / \tan 7x}, & 0 < x < \frac{\pi}{2} \\ a - 8, & x = \frac{\pi}{2} \\ \left(1 + |\cot x|\right)^{b^{\lfloor \tan x \rfloor}}, & \frac{\pi}{2} < x < \pi \end{cases} \]
Where \( a, b \in \mathbb{Z} \). If \( f \) is continuous at \( x = \frac{\pi}{2} \), then \( a^2 + b^2 \) is equal to __________.

Answer 1

Correct Answer: 81

Left-Hand Limit (LHL) at \( x = \frac{\pi}{2} \)

The left-hand limit is:

\[ \lim_{x \to \frac{\pi}{2}^-} \left( \frac{8}{7} \right)^{\tan \frac{8x}{\tan 7x}}. \]

As \( x \to \frac{\pi}{2}^- \), both \( \tan 8x \to \infty \) and \( \tan 7x \to \infty \), so:

\[ \frac{\tan 8x}{\tan 7x} \to \frac{8}{7}. \]

Thus:

\[ \lim_{x \to \frac{\pi}{2}^-} \left( \frac{8}{7} \right)^{\frac{\tan 8x}{\tan 7x}} = \left( \frac{8}{7} \right)^0 = 1. \]

Right-Hand Limit (RHL) at \( x = \frac{\pi}{2} \)

The right-hand limit is:

\[ \lim_{x \to \frac{\pi}{2}^+} \left( 1 + \lvert \cot x \rvert \right)^{\frac{b \tan \lvert x \rvert}{a}}. \]

As \( x \to \frac{\pi}{2}^+ \), \( \cot x \to 0 \) and \( \tan \lvert x \rvert \to \infty \). This simplifies to:

\[ \lim_{x \to \frac{\pi}{2}^+} \left( 1 + \lvert \cot x \rvert \right)^{\frac{b \tan \lvert x \rvert}{a}} = \frac{b}{e^a}. \]

Continuity Condition For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \):

\[ LHL = f\left( \frac{\pi}{2} \right) = RHL. \]

Substitute:

\[ 1 = a - 8 = \frac{b}{e^a}. \]

From \( a - 8 = 1 \):

\[ a = 9. \]

Substitute \( a = 9 \) into \( \frac{b}{e^a} = e^{\frac{b}{a}} \), giving:

\[ e^{\frac{b}{9}} = 1 \implies \frac{b}{9} = 0 \implies b = 0. \]

Final Calculation

\[ a^2 + b^2 = 9^2 + 0^2 = 81. \]

Final Answer is : 81.