JEE Main 2021 — Mathematics Calculus

Given that

f(x)={\begin{matrix}\frac{1}{x}{\mathrm{log}}_{e}(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}),x<0 \\ k,x=0 \\ \frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1},x>0\end{matrix}

$f(x)$f(x) is continuous at $x=0$x=0

$$L.H.L=R.H.L=f(0)$$L.H.L=R.H.L=f(0)

$$\mathrm{LHL}=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\mathrm{log}(\frac{1+\frac{x}{a}}{1-\frac{x}{b}})}{x}$$LHL=x→0−lim​xlog(1−bx​1+ax​​)​

$$=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\mathrm{log}(1+\frac{x}{a})-\mathrm{log}(1-\frac{x}{b})}{x}$$=x→0−lim​xlog(1+ax​)−log(1−bx​)​

$$=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\mathrm{log}(1+\frac{x}{a})}{\frac{x}{a}}\times \frac{1}{a}-\frac{\mathrm{log}(1-\frac{x}{b})}{-\frac{x}{b}}\times -\frac{1}{b}=\frac{1}{b}+\frac{1}{a}$$=x→0−lim​ax​log(1+ax​)​×a1​−−bx​log(1−bx​)​×−b1​=b1​+a1​

$$\mathrm{RHL}=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1}\times \frac{\sqrt{{x}^{2}+1}+1}{\sqrt{{x}^{2}+1}+1}$$RHL=x→0+lim​x2+1​−1cos2x−sin2x−1​×x2+1​+1x2+1​+1​

$$=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}-\frac{2{\mathrm{sin}}^{2}x}{{x}^{2}}(\sqrt{{x}^{2}+1}+1)=-4$$=x→0+lim​−x22sin2x​(x2+1​+1)=−4

So, $\frac{1}{b}+\frac{1}{a}=k=-4.$b1​+a1​=k=−4..

$$k=-4\Rightarrow \frac{4}{k}=-1.$$k=−4⇒k4​=−1.

$$\frac{1}{a}+\frac{1}{b}=-4.$$a1​+b1​=−4.

$$\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=-5.$$∴a1​+b1​+k4​=−5.