Aspire Faculty ID #14480 · Topic: JEE Main 2024 (8 April Evening Shift) · Just now
JEE Main 2024 (8 April Evening Shift)

$\mathrm{a}, \mathrm{b}>0$, let $f(x)= \begin{cases}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x< 0 \\ 3, & x=0 \\ \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{~b} \sqrt{\mathrm{a}} x \sqrt{x}}, & x> 0\end{cases}$ be a continuous function at $x=0$. Then $\frac{\mathrm{b}}{\mathrm{a}}$ is equal to :

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