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

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.