If the function fx=1xloge1+xa1−xb , x 0 Is continuous at x = 0, then 1a+1b+4k is equal to

limx→0−log1+xa−log1−xbx=klimx→0−1xlog1+xa-1xlog1-xb=1a+1b⇒k=1a+1b=limx→0+cos2x−sin2x−1x2+1−1 ⇒limx→0+ −2sin2⁡xx2+1−1=k⇒limx→0+ −2sin2⁡xx2+12−12x2+1+1=k⇒limx→0+ −2sin2⁡xx2x2+1+1=k⇒k=−4⇒1a+1b=k=−4. use L.H. rule∴ 1a+1b+4k=−4−1=−5.

If the function fx=1xloge1+xa1−xb , x 0 Is continuous at x = 0, then 1a+1b+4k is equal to

limx→0−log1+xa−log1−xbx=klimx→0−1xlog1+xa-1xlog1-xb=1a+1b⇒k=1a+1b=limx→0+cos2x−sin2x−1x2+1−1 ⇒limx→0+ −2sin2⁡xx2+1−1=k⇒limx→0+ −2sin2⁡xx2+12−12x2+1+1=k⇒limx→0+ −2sin2⁡xx2x2+1+1=k⇒k=−4⇒1a+1b=k=−4. use L.H. rule∴ 1a+1b+4k=−4−1=−5.

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If the function $f (x) = \{\begin{cases} \frac{1}{x} \log_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), x < 0 \\ k, x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2} + 1} - 1}, x > 0 \end{cases}$ Is continuous at x = 0, then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to

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-5

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Detailed Solution

$\begin{array}{l} \lim_{x \to 0^{-}} \frac{\log (1 + \frac{x}{a}) - \log (1 - \frac{x}{b})}{x} = k \\ \lim_{x \to 0^{-}} \frac{1}{x} \log (1 + \frac{x}{a}) - \frac{1}{x} \log (1 - \frac{x}{b}) = \frac{1}{a} + \frac{1}{b} \Rightarrow k = \frac{1}{a} + \frac{1}{b} \\ = \lim_{x \to 0^{+}} \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2} + 1} - 1} \\ \begin{array}{l} \Rightarrow lim_{x \to 0^{+}} \frac{- 2 \sin^{2} x}{\sqrt{x^{2} + 1} - 1} = k \\ \Rightarrow lim_{x \to 0^{+}} (\frac{- 2 \sin^{2} x}{{(\sqrt{x^{2} + 1})}^{2} - 1^{2}}) (\sqrt{x^{2} + 1} + 1) = k \\ \Rightarrow lim_{x \to 0^{+}} (\frac{- 2 \sin^{2} x}{x^{2}}) (\sqrt{x^{2} + 1} + 1) = k \\ \Rightarrow k = - 4 \end{array} \\ \Rightarrow \frac{1}{a} + \frac{1}{b} = k = - 4 . (use L . H . rule) \\ ∴ \frac{1}{a} + \frac{1}{b} + \frac{4}{k} = - 4 - 1 = - 5 . \end{array}$