ln ⁡ b − ln ⁡ a b − a \frac{\ln b-\ln a}{b-a}

Calculus Level 2

For real numbers a a and b b ( a < b ) a<b) in the interval [ 4 , 11 ] , [4,11], the range of the real number k k such that ln ⁡ b − ln ⁡ a b − a = k \frac{\ln b-\ln a}{b-a}=k can be expressed as α < k < β . \alpha < k < \beta. What is 1 α β ? \frac{1}{\alpha\beta}?

44 55 33 22

Ajay Dutta by Ajay Dutta , 24, India

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2 solutions

Using Lagrange mean value theorem,

f ′ ( c ) = f ( b ) − f ( a ) b − a f'(c) = \dfrac {f(b) - f(a)}{b -a}

in the interval [ a , b ] = [ 4 , 11 ] [a, b]=[4,11] such that a < c < b a<c<b

Here, f ( x ) = ln ⁡ x ⇒ f ′ ( x ) = 1 x = k f(x) = \ln x \Rightarrow f'(x) = \dfrac 1x = k

Now, 1 4 < k < 1 11 \dfrac 14 < k < \dfrac {1}{11}

Therefore, 1 α β = 1 1 4 × 1 11 = 44 \dfrac {1}{\alpha \beta } = \dfrac {1}{\frac 14 \times \frac {1}{11} } = \boxed {44}

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Rohit Shah
Mar 8, 2014

By Lagrange's mean value theorem

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