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

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

$g'(x) = \frac{(\ln(x) - 2)(3/x) - (3\ln(x) - 7)(1/(x\ln(x) - 2))}{(\ln(x) - 2)^2}$

We need to show that the numerator is positive:

For the first part, simplify the derivative using the product of the derivatives. We know:

• The term ( \ln(x) - 2 ) is positive for ( x > e^2 ) and negative for ( 0 < x < e^2 ),
• The term ( 3/x ) is always positive for positive x.

Evaluating the entire expression for x in the valid range indicates that it retains positivity leading to ( g'(x) > 0 ).

To find the range of values of a for which ( g(a) > 0 ), we analyze the initial function:

( g(a) = \frac{3\ln(a) - 7}{\ln(a) - 2} > 0 )

This condition implies that both the numerator and denominator must either be both positive or both negative. Setting the numerator greater than zero:

1. ( 3\ln(a) - 7 > 0 ) implies ( \ln(a) > \frac{7}{3} ) leading to ( a > e^{\frac{7}{3}} ).

Setting the denominator greater than zero: 2. ( \ln(a) - 2 > 0 ) implies ( \ln(a) > 2 ) leading to ( a > e^2 ).

Hence, for ( g(a) > 0 ), the range of a must satisfy both inequalities, ultimately resulting in:

$a > e^{\frac{7}{3}}$