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

RHL=limx→0+cos2x−sin2−1x2+1−1 =limx→0+ cos2⁡x−sin2⁡x−1x2+1−1=k⇒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 LHL=limx→0-log1+xa-log1−xbx=K ⇒1a+1b=k=−4∴1a+1b+4k=−4−1=−5

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

RHL=limx→0+cos2x−sin2−1x2+1−1 =limx→0+ cos2⁡x−sin2⁡x−1x2+1−1=k⇒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 LHL=limx→0-log1+xa-log1−xbx=K ⇒1a+1b=k=−4∴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

-4

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Detailed Solution

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