Q41 / 90 · JEE Main 2024 (08 Apr Shift 2) · Mathematics

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

  • Correct answer
    A
    6
  • B
    4
  • C
    5
  • D
    8

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} f(x)=f(0)=3 \\ & \lim _{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3 \\ & \lim _{x \rightarrow 0^{+}} \frac{a x+b^2 x^2-a x}{b \sqrt{a} x^{3 / 2}\left(\sqrt{a x+b^2 x^2}+\sqrt{a x}\right)} \\ & \lim _{x \rightarrow 0^{+}} \frac{b^2}{b \sqrt{a}\left(\sqrt{a+b^2 x}+\sqrt{a}\right)} \\ & \frac{b}{\sqrt{a} \cdot 2 \sqrt{a}} \Rightarrow \frac{b}{2 a}=3 \Rightarrow \frac{b}{a}=6\end{aligned}$

Hence, the correct answer is option: 'A'.

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