For $a , b 0$, let $f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, & x0\end{cases}$ be a continous function at $x=0$. Then $\frac{ b }{ a }$ is equal to: [JEE Main 2024, 08 Apr (Shift 2)]

Question

For $a , b >0$, let $f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, & x<0 \ 3, & x=0 \ \frac{\sqrt{ a x+ b ^2 x^2}-\sqrt{ a x}}{ b \sqrt{ a } x \sqrt{x}}, & x>0\end{cases}$ be a continous function at $x=0$. Then $\frac{ b }{ a }$ is equal to: [JEE Main 2024, 08 Apr (Shift 2)]

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For \(a , b >0\), let \(f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, & x<0 \\ 3, & x=0 \\ \fra…

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For \(a , b &gt;0\), let \(f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, &amp; x&lt;0 \\ 3, &amp; x=0 \\ \fra…

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For \(a , b >0\), let \(f(x)= \begin{cases}\frac{\tan (( a +1) x)+ b \tan x}{x}, & x<0 \\ 3, & x=0 \\ \fra…