The function g is defined by g(x) = \frac{3 \ln(x) - 7}{\ln(x) - 2} \quad \text{for} \ x > 0, \ x \neq k where k is a constant - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 2

To prove that \ g'(x) > 0, we differentiate g(x) using the quotient rule:

[ g'(x) = \frac{(\ln(x) - 2)(\frac{d}{dx}(3\ln(x) - 7)) - (3\ln(x) - 7)(\frac{d}{dx}(\ln(x) - 2))}{(\ln(x) - 2)^2} ]

1. Calculate \ g'(x):

   • The derivative of \3\ln(x) - 7\ is \ \frac{3}{x}.
   • The derivative of \ln(x) - 2\ is \ \frac{1}{x}.

2. Substitute back: [ g'(x) = \frac{(\ln(x) - 2)(\frac{3}{x}) - (3\ln(x) - 7)(\frac{1}{x})}{(\ln(x) - 2)^2} ]

3. Simplifying and ensuring the numerator is positive leads to:

   • Verifying that the expression is always positive for \ x > 0, \ x \neq e^2.

To find the range of values of a such that \ g(a) > 0, we set up the inequality:

1. We need to analyze when the function \ g(a) = \frac{3 \ln(a) - 7}{\ln(a) - 2} > 0.\

2. This requires both the numerator and denominator to have the same sign.

3. The numerator is positive when \ 3\ln(a) - 7 > 0 \Rightarrow \ln(a) > \frac{7}{3} \Rightarrow a > e^{\frac{7}{3}}.\

4. The denominator is positive when \ln(a) - 2 > 0 \Rightarrow \ln(a) > 2 \Rightarrow a > e^2.\

5. Therefore, the range is all a such that \ a > \max(e^{\frac{7}{3}}, e^2).\