QJEEM24Mab33a: JEE Main 2024 Mathematics - Class 12 JEEM24Mab33a Solution: For , let be a continuous function at . Then is equal to :

$f(x)=\left\{\begin{array}{cc}\frac{\tan ((a+1) x)+b \tan x}{x}, & x<0 \\3 & x=0 \\\frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, & x>0\end{array}\right.$

$f(x)$ is continuous at $x=0$

$\begin{aligned}& \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\& \lim _{x \rightarrow 0^{-}} f(x)=3 \\& \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \\& \Rightarrow a+1+b=3 \\& \Rightarrow a+b=2 \quad \text{..... (1)}\end{aligned}$

also, $$\underset{x\to 0^{+}}{lim}\frac{\sqrt{ax+b^2{x}^2}-\sqrt{ax}}{b\sqrt{a}x\sqrt{x}}=3$$

$\begin{aligned}& =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \\& =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \\& =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \\& \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \\& \Rightarrow \frac{b}{2 a}=3 \\& \Rightarrow \frac{b}{a}=6 \quad \text{..... (2)}\end{aligned}$