Question

If the function $f ( x )=\begin{cases}\frac{1}{ x } \log _{ e }\left(\frac{1+\frac{ x }{ a }}{1-\frac{ x }{ b }}\right) & , \quad x <0 \\ k & , \quad x =0 \\ \frac{\cos ^{2} x -\sin ^{2} x -1}{\sqrt{ x ^{2}+1}-1} & , \quad x >0\end{cases}$
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 $x=0$, $RHL =L H L=f(0)$
$\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} \cdot \frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}+1}$ (Rationalisation)
$\displaystyle\lim _{x \rightarrow 0^{+}}-\frac{2 \sin ^{2} x}{x^{2}} \cdot\left(\sqrt{x^{2}+1}+1\right)=-4$
$\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{x \rightarrow 0^{-}} \frac{1}{x} \ell n\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right)$
$\displaystyle\lim _{x \rightarrow 0^{-}} \frac{\ell n\left(1+\frac{x}{a}\right)}{\left(\frac{x}{a}\right) \cdot a}+\frac{\ell n\left(1-\frac{x}{b}\right)}{\left(-\frac{x}{b}\right) \cdot b}$
$=\frac{1}{a}+\frac{1}{b}$
So $\frac{1}{a}+\frac{1}{b}=-4=k$
$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{4}{k}$
$=-4-1=-5$