Q59 / 90 · JEE Main 2024 (09 Apr Shift 1) · Mathematics
Let $f:(0, \pi) \rightarrow \mathbf{R}$ be a function given by $f(x)=$
\(\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+|\cot x|)^{\frac{\mathrm{b}}{}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.\)
where $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to
where $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to
Correct answer: 81
Solution
LHL at $\mathrm{x}=\frac{\pi}{2}$
$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}=\left(\frac{8}{7}\right)^0=1$
RHL at $\mathrm{x}=\frac{\pi}{2}$ $\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}}(1+|\cot x|)^{\frac{b}{\operatorname{lan}} \tan x} \\ & =e^{\lim _{x-\frac{\pi}{2}}|\cot x| \frac{b}{a}|\tan x|}=e^{\frac{b}{a}} \end{aligned}$ $\begin{aligned} & \Rightarrow 1=a-8=e^{\frac{b}{a}} \\ & \Rightarrow a=9, b=0 \\ & \Rightarrow a^2+b^2=81\end{aligned}$
RHL at $\mathrm{x}=\frac{\pi}{2}$ $\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}}(1+|\cot x|)^{\frac{b}{\operatorname{lan}} \tan x} \\ & =e^{\lim _{x-\frac{\pi}{2}}|\cot x| \frac{b}{a}|\tan x|}=e^{\frac{b}{a}} \end{aligned}$ $\begin{aligned} & \Rightarrow 1=a-8=e^{\frac{b}{a}} \\ & \Rightarrow a=9, b=0 \\ & \Rightarrow a^2+b^2=81\end{aligned}$
Hence, the correct answer is 81.