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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Continuity and Differentiability

For a, b>0 , let f(x)=\ array cc ((a+1) x)+b x x , & x 0 array . be...

For a,b>0\mathrm{a}, \mathrm{b}>0, let f(x)=\left\{tan⁡((a+1)x)+btan⁡xx,x<03,x=0ax+b2x2−ax baxx,x>0\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=0x=0. Then ba\frac{\mathrm{b}}{\mathrm{a}} is equal to :

Options

  1. A6
  2. B4
  3. C5
  4. D8

Correct answer

A. 6

Step-by-step solution

lim⁡x→0f(x)=f(0)=3lim⁡x→0+ax+b2x2−axbaxx=3lim⁡x→0+ax+b2x2−axbax3/2(ax+b2x2+ax)lim⁡x→0+b2ba(a+b2x+a)ba⋅2a⇒b2a=3⇒ba=6\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}
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