Let be a function given by f ( x ) = { ( 8 7 ) tan 8 x tan 7 x , 0 < x < 2 a - 8 , x = 2 ( 1 + | c o t x | ) b a | tan x | , 2 < x < where a , b ? Z . If f is continuous at x = 2 , then a 2 + b 2 is equal to _ _ _ _ _ . [2024]

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Solution

Q.
Let $f : (0, π) \to R$ be a function given by

$f (x)$ $= {\begin{matrix} {(\frac{8}{7})}^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{π}{2} \\ a - 8, & x = \frac{π}{2} \\ {(1 + | c o t x |)}^{\frac{b}{a} | \tan x |}, & \frac{π}{2} < x < π \end{matrix}$

$where a, b \in Z . If f is continuous at x = \frac{π}{2}, then a^{2} + b^{2} is equal to_____.$ [2024]

Ans.
(81)

$Since, f is continuous at x = \frac{π}{2}$

$∴ \lim_{x \to \frac{π^{-}}{2}} f (x) = f (\frac{π}{2}) = \lim_{x \to \frac{π^{+}}{2}} f (x)$

$Now, \lim_{x \to \frac{π^{-}}{2}} {(\frac{8}{7})}^{(\frac{\tan 8 x}{\tan 7 x})}$

$= \lim_{h \to 0} {(\frac{8}{7})}^{\frac{\tan (4 π - 8 h)}{\tan (3 π + \frac{π}{2} - 7 h)}} = \lim_{h \to 0} {(\frac{8}{7})}^{\frac{\tan (- 8 h)}{\cot (7 h)}} = {(\frac{8}{7})}^{0} = 1$

$\Rightarrow a - 8 = 1 \Rightarrow a = 9$

$Also, \lim_{π \to {\frac{π}{2}}^{+}} f (x) = \lim_{π \to \frac{π}{2}} {(1 + | \cot x |)}^{\frac{b}{a} | \tan x |}$

$= \lim_{h \to 0} {(1 - \tan h)}^{- \frac{b}{9} \cot h}$

$= \lim_{h \to 0} {(1 - \tan h)}^{(\frac{1}{\tan h}) \cdot (\tan h) \cdot (\frac{- b}{9} \cot h)} = e^{\frac{b}{9}} = 1 \Rightarrow b = 0$

Hence, $a^{2} + b^{2} = 81 + 0 = 81$

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