If a<c<b and if 1−k1<ln⁡ba<k2−1 then k1,k2 is

f(x)=ln⁡x,f1(x)=1x,f(b)−f(a)b−a=1c⇒ln⁡ba=b−ac⇒a<c<b⇒1b<1c<1a⇒b−ab<b−ac<b−aa⇒1−ab<ln⁡ba<ba−1∴K1=ab,K2=ba

If a<c<b and if 1−k1<ln⁡ba<k2−1 then k1,k2 is

f(x)=ln⁡x,f1(x)=1x,f(b)−f(a)b−a=1c⇒ln⁡ba=b−ac⇒a<c<b⇒1b<1c<1a⇒b−ab<b−ac<b−aa⇒1−ab<ln⁡ba<ba−1∴K1=ab,K2=ba

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If $a < c < b$ and if $1 - k_{1} < \ln (\frac{b}{a}) < k_{2} - 1$ then $(k_{1}, k_{2})$ is

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$(\frac{a}{b}, \frac{b}{a})$

$(\frac{b}{a}, \frac{a}{b})$

$(2 a, 2 b)$

$(a, b)$

answer is A.

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Detailed Solution

$f (x) = \ln x, f^{1} (x) = \frac{1}{x}, \frac{f (b) - f (a)}{b - a} = \frac{1}{c} \Rightarrow \ln \frac{b}{a} = \frac{b - a}{c}$

$\Rightarrow a < c < b \Rightarrow \frac{1}{b} < \frac{1}{c} < \frac{1}{a} \Rightarrow \frac{b - a}{b} < \frac{b - a}{c} < \frac{b - a}{a}$

$\Rightarrow 1 - \frac{a}{b} < \ln \frac{b}{a} < \frac{b}{a} - 1 ∴ K_{1} = \frac{a}{b}, K_{2} = \frac{b}{a}$

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If $a < c < b$ and if $1 - k_{1} < \ln (\frac{b}{a}) < k_{2} - 1$ then $(k_{1}, k_{2})$ is