JEE Main 2021 — Mathematics Calculus

If $f(x)$f(x) is continuous at point $x=a,$x=a, then $\underset{x\rightarrow 0}{\mathrm{lim}}f(x)=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}f(x)$x→0lim​f(x)=x→0+lim​f(x)=x→0−lim​f(x)

From the question, we can say that $\underset{x\rightarrow 0}{\mathrm{lim}}f(x)=b$x→0lim​f(x)=b

And $\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}{e}^{\frac{\mathrm{cot}4x}{\mathrm{cot}2x}}=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}{e}^{\frac{\mathrm{tan}2x}{\mathrm{tan}4x}}=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}{e}^{\frac{\frac{\mathrm{tan}2x}{2\times 2x}}{\frac{\mathrm{tan}4x}{4x}}}={e}^{\frac{1}{2}}=b$x→0+lim​ecot2xcot4x​=x→0+lim​etan4xtan2x​=x→0+lim​e4xtan4x​2×2xtan2x​​=e21​=b

And $\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}{(1+|\mathrm{sin}x|)}^{\frac{3a}{|\mathrm{sin}x|}}={e}^{\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}[(|\mathrm{sin}x|)(\frac{3a}{|\mathrm{sin}x|})]}={e}^{3a}$x→0−lim​(1+∣sinx∣)∣sinx∣3a​=ex→0−lim​[(∣sinx∣)(∣sinx∣3a​)]=e3a

From the above we can say that

$${e}^{3a}={e}^{\frac{1}{2}}$$e3a=e21​

$$\Rightarrow a=\frac{1}{6}\Rightarrow 6a=1$$⇒a=61​⇒6a=1

So, $(6a+{b}^{2})=(1+e)$(6a+b2)=(1+e)