Question

If the function $f(x)=\{\begin{array}{ll}\frac{1}{x}{\log }_e\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right)& ,x<0\\ k& ,x=0\\ \frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1}& ,x>0\end{array}$
is continuous at $x=0$, then $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ is equal to:

• A. -5
• B. 5
• C. -4
• D. 4

Answer 1

Answer: (A)

If $f(x)$ is continuous at $x=0$, $RHL=LHL=f(0)$
$\underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1}\cdot \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$ (Rationalisation)
$\underset{x\to 0^{+}}{\lim }-\frac{2{\sin }^{2}x}{x^2}\cdot \left(\sqrt{x^2+1}+1\right)=-4$
$\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }\frac{1}{x}\ell n\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right)$
$\underset{x\to 0^{-}}{\lim }\frac{\ell n\left(1+\frac{x}{a}\right)}{\left(\frac{x}{a}\right)\cdot a}+\frac{\ell n\left(1-\frac{x}{b}\right)}{\left(-\frac{x}{b}\right)\cdot b}$
$=\frac{1}{a}+\frac{1}{b}$
So $\frac{1}{a}+\frac{1}{b}=-4=k$
$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{4}{k}$
$=-4-1=-5$