For\ a,b 0,\ let f(x)= begin cases tan\!big((a+1)xbig)+btan x/x, & x<0, \ 10pt] 3, & x=0, \ 10pt] dfrac sqrt ax+b 2 x 2 -√(ax) b√(a)\,x√(x) , & x 0. end cases be a continuous function at x=0 . Then b/a is equal to

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 8 April, Shift 2 — Question 17

For\ a,b>0,\ \text{let}

f(x)= \begin{cases} \dfrac{\tan\!\big((a+1)x\big)+b\tan x}{x}, & x<0, \\[10pt] 3, & x=0, \\[10pt] \dfrac{\sqrt{ax+b^{2}x^{2}}-\sqrt{ax}}{b\sqrt{a}\,x\sqrt{x}}, & x>0. \end{cases}

be a continuous function at $x=0$ . Then $\frac{b}{a}$ is equal to

$\lim _{x \rightarrow 0} f(x)=f(0)=3$

$\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^{2} x^{2}}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3$

$\lim _{x \rightarrow 0^{+}} \frac{a x+b^{2} x^{2}-a x}{b \sqrt{a} x^{3 / 2}\left(\sqrt{a x+b^{2} x^{2}}+\sqrt{a x}\right)}$

$\lim _{x \rightarrow 0^{+}} \frac{b^{2}}{b \sqrt{a}\left(\sqrt{a+b^{2} x}+\sqrt{a}\right)}$

$\frac{\mathrm{b}}{\sqrt{\mathrm{a}} \cdot 2 \sqrt{\mathrm{a}}}$

$\Rightarrow \frac{\mathrm{b}}{2 \mathrm{a}}=3$

$\Rightarrow \frac{\mathrm{~b}}{\mathrm{a}}=6$

Answer key and solution verified before publishing.

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