JEE Main 20248 Apr, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002For a,b>0\mathrm{a}, \mathrm{b}>0a,b>0, let f(x)={tan((a+1)x)+btanxx,x<03,x=0ax+b2x2−ax baxx,x>0f(x)= \begin{cases}\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{cases}f(x)=⎩⎨⎧xtan((a+1)x)+btanx,3, baxxax+b2x2−ax,x<0x=0x>0 be a continuous function at x=0x=0x=0. Then ba\frac{\mathrm{b}}{\mathrm{a}}ab is equal to :AOption A. 4BOption B. 5COption C. 8DOption D. 6CheckSkip · show solution
