JEE Main 2021 — Mathematics Calculus

$$\underset{x\rightarrow 0}{\mathrm{lim}}\frac{ax-({e}^{4x}-1)}{ax({e}^{4x}-1)}(\frac{0}{0})$$x→0lim​ax(e4x−1)ax−(e4x−1)​(00​)

$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{ax-({e}^{4x}-1)}{ax\cdot 4x}$=x→0lim​ax⋅4xax−(e4x−1)​ Use $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{e}^{4x}-1}{4x}=1$x→0lim​4xe4x−1​=1

Apply $L$L'Hospital Rule

$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{a-4{e}^{4x}}{8ax}$=x→0lim​8axa−4e4x​ ( $\frac{a-4}{0}$0a−4​ form )

limit exists only when $a-4=0\Rightarrow a=4$a−4=0⇒a=4

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{4-4{e}^{4x}}{32x}$$=x→0lim​32x4−4e4x​

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{1-{e}^{4x}}{8x}(\frac{0}{0})$$=x→0lim​8x1−e4x​(00​)

$$=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{-{e}^{4x}\cdot 4}{8}=-\frac{1}{2}\Rightarrow b=-\frac{1}{2}$$=x→0lim​8−e4x⋅4​=−21​⇒b=−21​

$$a-2b=4-2(-\frac{1}{2})$$a−2b=4−2(−21​)

$$=5$$=5