Let f:(0, pi) rightarrow mathbb{R} be a funct | vedclass

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(B) For $f$ to be continuous at $x=\frac{\pi}{2}$,we must have $\lim_{x \rightarrow \frac{\pi}{2}^-} f(x) = f(\frac{\pi}{2}) = \lim_{x \rightarrow \fr

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(B) For $f$ to be continuous at $x=\frac{\pi}{2}$,we must have $\lim_{x ightarrow \frac{\pi}{2}^-} f(x) = f(\frac{\pi}{2}) = \lim_{x ightarrow \frac{\pi}{2}^+} f(x)$.$1$. Left-hand limit $(LHL)$: $\lim_{x ightarrow \frac{\pi}{2}^-} (\frac{8}{7})^{\frac{	an 8x}{	an 7x}}$.As $x ightarrow \frac{\pi}{2}$,$	an 8x ightarrow 	an 4\pi = 0$ and $	an 7x ightarrow 	an \frac{7\pi}{2} = \infty$. Thus,the exponent $\frac{	an 8x}{	an 7x} ightarrow 0$. So,$LHL$ $= (\frac{8}{7})^0 = 1$.$2$. Value at $x=\frac{\pi}{2}$: $f(\frac{\pi}{2}) = a-8$.$3$. Right-hand limit $(RHL)$: $\lim_{x ightarrow \frac{\pi}{2}^+} (1+|\cot x|)^{\frac{b}{a}|	an x|}$.Let $t = |\cot x|$. As $x ightarrow \frac{\pi}{2}^+$,$t ightarrow 0$ and $|	an x| = \frac{1}{t}$.The limit becomes $\lim_{t ightarrow 0} (1+t)^{\frac{b}{a} \cdot \frac{1}{t}} = e^{\lim_{t ightarrow 0} \frac{b}{a} \cdot \frac{1}{t} \cdot t} = e^{\frac{b}{a}}$.Equating the values: $1 = a-8 = e^{\frac{b}{a}}$.From $1 = a-8$,we get $a=9$.From $1 = e^{\frac{b}{a}}$,we get $\frac{b}{a} = 0$,so $b=0$.Therefore,$a^2+b^2 = 9^2 + 0^2 = 81$.

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