JEE Main 2024 (Online) 8th April Evening ShiftQuestion 1
For $a,b>0$, let $f(x)=\begin{cases}\frac{\tan((a+1)x)+b\tan x}{x}, & x<0\\ 3, & x=0\\ \frac{\sqrt{ax+b^{2}x^{2}}-\sqrt{ax}}{b\sqrt{a}\,x\sqrt{x}}, & x>0\end{cases}$ be a continous function at $x=0$. Then $\frac{b}{a}$ is equal to :
Limits & Continuity
Solution
Left: $\frac{\tan((a+1)x)+b\tan x}{x}\to a+1+b=3$. Right: $\sqrt{ax+b^2x^2}-\sqrt{ax}=\sqrt{ax}\left(\sqrt{1+b^2x/a}-1\right)\approx\sqrt{ax}\cdot\frac{b^2x}{2a}$, so the quotient $\to\frac{b^2/(2a)}{b}\cdot\frac{\sqrt a\,x^{3/2}}{\sqrt a\,x^{3/2}}=\frac{b}{2a}=3$. So $b=6a$, hence $b/a=6$ (then $7a=2$, $a=2/7>0$, $b=12/7>0$). Option (A).